% Main source. Build with pdfLaTeX, BibTeX, pdfLaTeX, pdfLaTeX. % Keep references.bib beside this file. No figures or shell escape are required. % Human contributor metadata and distribution licence are supplied at deposit. \documentclass[11pt,a4paper]{article} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage[margin=20mm]{geometry} \usepackage{mathpazo} \renewcommand{\sfdefault}{phv} \renewcommand{\ttdefault}{pcr} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{booktabs,array,longtable,enumitem} \usepackage{microtype} \usepackage{xcolor} \usepackage[colorlinks=true,linkcolor=blue!45!black,citecolor=blue!45!black,urlcolor=blue!45!black,unicode]{hyperref} \usepackage{bookmark} \usepackage{xurl} \urlstyle{same} \usepackage{needspace} \usepackage{alphalph} \setlength{\parskip}{3pt} \setlength{\parindent}{1em} \setlength{\emergencystretch}{3em} \setlist{itemsep=1pt,topsep=3pt} \allowdisplaybreaks[2] \newcommand{\Q}{\mathbb Q} \newcommand{\Z}{\mathbb Z} \newcommand{\F}{\mathbb F} \newcommand{\R}{\mathbb R} \newcommand{\Cl}{\operatorname{Cl}} \newcommand{\Pic}{\operatorname{Pic}} \newcommand{\Jac}{\operatorname{Jac}} \newcommand{\divv}{\operatorname{div}} \newcommand{\Sha}{\mathop{\mathrm{Sha}}} \newcommand{\code}[1]{\textsf{certificate #1}} \newcommand{\res}[1]{\textsf{result #1}} \theoremstyle{plain} \newtheorem{thmL}{Proposition} \renewcommand{\thethmL}{\AlphAlph{\value{thmL}}} \newtheorem{lemL}[thmL]{Lemma} \newtheorem{lem}{Lemma}[subsection] \newtheorem{prop}[lem]{Proposition} \newtheorem{thm}[lem]{Theorem} \newtheorem{introthm}{Theorem} \newtheorem{cor}[lem]{Corollary} \theoremstyle{definition} \newtheorem{defn}[lem]{Definition} \theoremstyle{remark} \newtheorem*{rem}{Remark} \setcounter{tocdepth}{1} \title{Long solutions to small equations} \author{Bogdan Grechuk, Eugene Go and Avraham Eisenberg} \date{23 October 2026} \hypersetup{pdftitle={Long solutions to small equations},pdfauthor={Bogdan Grechuk, Eugene Go and Avraham Eisenberg},pdfsubject={Diophantine equations; explicit descent; Chabauty methods; reciprocity; modular methods}} \begin{document} \maketitle \vspace{-2em} \begin{abstract} We give complete solutions of thirty-eight Diophantine equations left open in Grechuk's 2024 monograph: fifteen affine equations, five generalized Fermat equations, eight plane quartics, and ten further homogeneous equations. All proofs were generated by AI and require substantial computer algebra computations. \end{abstract} \noindent\textbf{Mathematics Subject Classification (2020).} 11D25, 11D41, 11G30, 11Y50. \noindent\textbf{Keywords.} Diophantine equations; explicit descent; Chabauty method; Brauer--Manin obstruction. %\tableofcontents \section{Introduction}\label{sec:introduction} The theory of Diophantine equations contains many problems that are easy to state but exceptionally difficult to solve. Prominent examples include Fermat's Last Theorem, which asserts that $x^n+y^n=z^n$ has no solution in positive integers for any integer $n\geq3$~\cite{Wiles95,TW95}, and Mih\u ailescu's theorem, formerly Catalan's conjecture, which states that $8$ and $9$ are the only consecutive perfect powers greater than~$1$~\cite{Mihailescu04}. Grechuk's monograph~\cite{Grechuk24} develops a systematic approach to finding such problems among polynomial Diophantine equations \begin{equation}\label{eq:diofgen} P(x_1,\ldots,x_n)=0, \end{equation} where $P$ has integer coefficients. Write $P$ as a sum of distinct monomials with nonzero coefficients $a_1,\ldots,a_k$ and total degrees $d_1,\ldots,d_k$. Its size is defined by \[ H(P)=\sum_{i=1}^k |a_i|2^{d_i}. \] After identifying equations related by renaming variables, changing the signs of variables, or multiplying the equation by $-1$, and omitting variables that do not occur, there are only finitely many equations of any given size. This provides an ordering in which to study them systematically. The same programme can be pursued within particular classes, such as equations in two variables, homogeneous equations, or generalized Fermat equations. The meaning of ``solving'' an equation depends on the class. An absolutely irreducible affine plane curve of positive genus has only finitely many integer points by Siegel's theorem~\cite[Theorem~3.59]{Grechuk24}; the task is to determine all of them. For a generalized Fermat equation \[ ax^n+by^m=cz^r, \] with fixed nonzero integer coefficients and positive integer exponents satisfying $1/n+1/m+1/r<1$, Darmon and Granville~\cite{DG95} prove that the set of primitive integer solutions is finite. Here a solution is primitive if $\gcd(x,y,z)=1$. A nonzero solution generates infinitely many solutions by weighted scaling, so the primitive solutions are the natural objects to classify. For a homogeneous equation, the zero vector is always a solution; one may instead ask whether a solution exists with at least one, or with every, coordinate nonzero. These two conditions are distinguished throughout. In each class, the monograph resolves the relevant problem below a specified size and records the smallest outstanding cases. Subsequent progress is described in~\cite{Wilcox24,RG25,grechuk2026shortest,grechuk2026cubic,grechuk2026polynomial,GrechukOpen}. In this paper we present AI-assisted proofs for thirty-eight equations explicitly left open in the book: fifteen affine equations, five generalized Fermat equations, eight plane quartics, and ten further homogeneous equations. For two-variable equations of size $H\leq31$, Table~3.25 of~\cite{Grechuk24} contains thirty-one cases not resolved in that chapter. One is subsequently settled in Proposition~7.7 of the same book, leaving thirty open cases. Six were completed after publication of the book and before this paper~\cite{grechuk2026certifying,Wilcox24}; we prove fourteen more, leaving ten open. Tables~\ref{tab:H31Solved} and~\ref{tab:H31Open} list the twenty completed cases and the ten remaining ones. In all tables, signs within a tuple vary independently; a sign before an entire tuple changes all its coordinates simultaneously. \Needspace{5\baselineskip} \begingroup\small \setlength{\tabcolsep}{3pt} \renewcommand{\arraystretch}{1.25} \setlength{\LTcapwidth}{\textwidth} \setlength{\LTpre}{6pt} \setlength{\LTpost}{8pt} \begin{longtable}{|c|l|l|l|} \caption{Completed two-variable equations of size $H\leq31$ left open in~\cite{Grechuk24}.}\label{tab:H31Solved}\\ \hline $H$ & Equation & Solutions, $(x,y)$ & Reference \\ \hline\hline \endfirsthead \caption[]{Completed two-variable equations of size $H\leq31$ left open in the book (continued).}\\ \hline $H$ & Equation & Solutions, $(x,y)$ & Reference \\ \hline\hline \endhead \hline\multicolumn{4}{r}{\footnotesize Continued on the next page.}\\ \endfoot \endlastfoot % Book-open entry: A01; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $28$ & $x^4-y^3-x+y=0$ & $(0,\pm1),(0,0),(1,0),(1,\pm1)$ & Thm.~\ref{thm:1.1.1} \\ \hline % Book-open entry: A02; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $28$ & $x^4-y^3+x-y=0$ & $(-1,0),(0,0),(1,1)$ & Thm.~\ref{thm:1.1.2} \\ \hline % Book-open entry: A03; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $29$ & $x^4+y^3+xy+1=0$ & $(0,-1),(1,-1)$ & Thm.~\ref{thm:1.1.3} \\ \hline % Book-open entry: A05; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $30$ & $x^4-y^3+xy+x=0$ & $(-1,0),(0,0)$ & \cite{Wilcox24} \\ \hline % Book-open entry: A06; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $30$ & $x^4+y^3-y^2+x=0$ & $(-4,-6),(-1,0),(-1,1),(0,0),(0,1),(1,-1)$ & Thm.~\ref{thm:1.1.7} \\ \hline % Book-open entry: A07; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $30$ & $x^4+y^3+x-2y=0$ & $(-1,0),(0,0)$ & \cite{grechuk2026certifying} \\ \hline % Book-open entry: A12; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $30$ & $x^4+y^3+2x-y=0$ & $(-8,-16),(0,0),(0,\pm1)$ & Thm.~\ref{thm:1.1.4} \\ \hline % Book-open entry: A13; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $30$ & $x^4+y^3+2x+y=0$ & $(0,0),(8,-16)$ & Thm.~\ref{thm:1.1.5} \\ \hline % Book-open entry: A14; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $30$ & $x^4+y^3+y^2+x=0$ & $(-1,-1),(-1,0),(0,-1),(0,0),(2,-3)$ & Thm.~\ref{thm:1.1.6} \\ \hline % Book-open entry: A15; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $30$ & $x^4+y^3+xy-y=0$ & $(0,\pm1),(0,0),(1,-1),(-1,1)$ & \cite{Wilcox24} \\ \hline % Book-open entry: A16; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $30$ & $x^4+y^3+xy+y=0$ & $(-1,-1),(0,0)$ & \cite{Wilcox24} \\ \hline % Book-open entry: A17; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $30$ & $x^4+y^3+xy+x=0$ & $(-1,\pm1),(-1,0),(0,0),(1,-1)$ & \cite{Wilcox24} \\ \hline % Book-open entry: A19; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $31$ & $x^4-y^3+xy+x-1=0$ & $(0,-1)$ & Thm.~\ref{thm:1.1.12} \\ \hline % Book-open entry: A20; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $31$ & $x^4+y^3+x-2y+1=0$ & $(-1,1),(0,1)$ & \cite{grechuk2026certifying} \\ \hline % Book-open entry: A23; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $31$ & $x^4+y^3+2x-y+1=0$ & $(-1,\pm1),(-1,0)$ & Thm.~\ref{thm:1.1.8} \\ \hline % Book-open entry: A24; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $31$ & $x^4+y^3+2x+y+1=0$ & $(-1,0)$ & Thm.~\ref{thm:1.1.9} \\ \hline % Book-open entry: A25; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $31$ & $x^4+y^3+2x+y-1=0$ & $(-1,1),(1,-1)$ & Thm.~\ref{thm:1.1.10} \\ \hline % Book-open entry: A26; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $31$ & $x^4+y^3+xy-y-1=0$ & $(\pm1,0)$ & Thm.~\ref{thm:1.1.11} \\ \hline % Book-open entry: A29; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $31$ & $x^4+y^3+xy+y-1=0$ & $(\pm1,0),(3,-4)$ & Thm.~\ref{thm:1.1.13} \\ \hline % Book-open entry: A30; Section 3.5.1, Table 3.25 and Exercise 3.78; Table 8.8, p. 640 (excluding the case solved in Proposition 7.7) $31$ & $x^4+y^3+xy+x+1=0$ & $(0,-1)$ & Thm.~\ref{thm:1.1.14} \\* \cline{1-4} \end{longtable} \endgroup \Needspace{10\baselineskip} \begingroup\small \setlength{\tabcolsep}{3pt} \renewcommand{\arraystretch}{1.25} \setlength{\LTcapwidth}{\textwidth} \begin{longtable}{|c|l|l|l|} \caption{Remaining open two-variable equations of size $H\leq31$.}\label{tab:H31Open}\\\hline $H$ & Equation & $H$ & Equation \\ \hline\hline % Book-open entries: A04, A18 $29$ & $x^4+y^3+xy-1=0$ & $31$ & $x^4-y^3+xy+x+1=0$ \\ \hline % Book-open entries: A11, A28 $30$ & $x^4+y^3+x+2y=0$ & $31$ & $x^4+y^3+xy-3=0$ \\ \hline % Book-open entries: A08, A21 $30$ & $x^4+y^3+x-y-2=0$ & $31$ & $x^4+y^3+x-2y-1=0$ \\ \hline % Book-open entries: A09, A31 $30$ & $x^4+y^3+x+y+2=0$ & $31$ & $x^4+y^3+xy+x-1=0$ \\ \hline % Book-open entries: A10, A22 $30$ & $x^4+y^3+x+y-2=0$ & $31$ & $x^4+y^3+x+2y+1=0$ \\* \cline{1-4} \end{longtable} \endgroup \begin{introthm}\label{th:h31} For each equation in the second column of Table~\ref{tab:H31Solved}, the integer solutions are exactly the pairs $(x,y)$ in the third column. \end{introthm} \begin{proof} See the references in the fourth column of Table~\ref{tab:H31Solved}. \end{proof} All equations in Tables~\ref{tab:H31Solved} and~\ref{tab:H31Open} have smooth projective models of genus~$3$. The smallest genus-two equation left open in the book is~\cite[Equation~(3.237)]{Grechuk24} \begin{equation}\label{eq:h32gen2} x^4-x^2+xy+y^3=0, \end{equation} of size $H=32$. Its plane model is singular, and its normalization has genus~$2$. We determine its integer solutions as well. \begin{introthm}\label{th:h32genus2} The only integer solutions to equation~\eqref{eq:h32gen2} are \[ (x,y)=(0,0),\ (1,0),\ (-1,0),\ (-1,1),\ (-1,-1),\ (2,-2),\ (4,-6). \] \end{introthm} \begin{proof} See Theorem~\ref{thm:1.1.15}. \end{proof} Open Question~4.52 of~\cite{Grechuk24} asks for all primitive solutions of the sixteen generalized Fermat equations in its Table~4.16. Table~\ref{tab:FermatSolved} gives their complete solution sets. Eight entries were resolved after publication of the book and before this paper: three in the appendix added to Wilcox~\cite[Propositions~A.2--A.4]{Wilcox24}, two through Grechuk's reductions~\cite{GrechukMO26} to results of Bruin~\cite{Bruin03} and Brown~\cite{Brown12}, and three through the two separate papers~\cite{hexagon:2609.00132v1,hexagon:2609.00130v1}. The latter papers determine the primitive solutions of $x^4+3y^3+2z^3=0$ and $x^4+4y^3+z^3=0$, respectively. The second also settles $2x^4+2y^3+z^3=0$: parity forces $z=2u$, and division by~$2$ gives $x^4+4u^3+y^3=0$. This correspondence preserves primitivity, so the two presentations count as two table entries but require only one result. Three further entries follow from results published before the book, by Zhang--Bai~\cite{ZhangBai14}, Bennett--Vatsal--Yazdani~\cite{BVY04}, and Bennett--Ellenberg--Ng~\cite{BEN10}; their relevance is recorded in~\cite{Wilcox24,RG25}. The five proofs in Section~\ref{sec:1.2} settle the five remaining entries. No entry of the book's Table~4.16 remains open. \Needspace{5\baselineskip} \begingroup\small \setlength{\tabcolsep}{3pt} \renewcommand{\arraystretch}{1.25} \setlength{\LTcapwidth}{\textwidth} \setlength{\LTpre}{6pt} \setlength{\LTpost}{8pt} \begin{longtable}{|c|l|l|l|} \caption{Completed generalized Fermat equations from~\cite[Open Question~4.52]{Grechuk24}. All listed solutions are primitive.}\label{tab:FermatSolved}\\ \hline $H$ & Equation & Solutions, $(x,y,z)$ & Reference \\ \hline\hline \endfirsthead \caption[]{Completed generalized Fermat equations from~\cite[Open Question~4.52]{Grechuk24}. All listed solutions are primitive (continued).}\\ \hline $H$ & Equation & Solutions, $(x,y,z)$ & Reference \\ \hline\hline \endhead \hline\multicolumn{4}{r}{\footnotesize Continued on the next page.}\\ \endfoot \endlastfoot % Book-open entry: F01; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $40$ & $x^4+2y^3+z^3=0$ & $(\pm1,0,-1),(\pm1,-1,1),(\pm43,-203,237)$ & \cite{GrechukMO26} \\ \hline % Book-open entry: F02; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $48$ & $x^4+2y^3+2z^3=0$ & $\pm(0,1,-1)$ & \cite[Prop.~A.4]{Wilcox24} \\ \hline % Book-open entry: F03; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $48$ & $x^4+3y^3+z^3=0$ & $(\pm1,0,-1),(\pm895,-6202,4199)$ & Thm.~\ref{thm:1.2.1} \\ \hline % Book-open entry: F04; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $48$ & $2x^4+y^3+z^3=0$ & $\pm(0,1,-1),(\pm1,-1,-1)$ & \cite{ZhangBai14} \\ \hline % Book-open entry: F05; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $48$ & $x^4-y^4+2z^3=0$ & $(\pm1,\pm1,0)$ & \cite[Prop.~A.2]{Wilcox24} \\ \hline % Book-open entry: F06; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $48$ & $x^4+y^4+2z^3=0$ & $(\pm1,\pm1,-1)$ & \cite[Thm.~8.1]{BVY04} \\ \hline % Book-open entry: F07; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $56$ & $x^4+3y^3+2z^3=0$ & $(\pm1,-1,1)$ & \cite{hexagon:2609.00132v1} \\ \hline % Book-open entry: F08; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $56$ & $x^4+4y^3+z^3=0$ & $(\pm1,0,-1)$ & \cite{hexagon:2609.00130v1} \\ \hline % Book-open entry: F09; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $56$ & $2x^4+2y^3+z^3=0$ & $(\pm1,-1,0)$ & \cite{hexagon:2609.00130v1} \\ \hline % Book-open entry: F10; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $56$ & $x^4-y^4+3z^3=0$ & $(\pm1,\pm1,0)$ & \cite[Prop.~A.3]{Wilcox24} \\ \hline % Book-open entry: F11; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $56$ & $2x^4-y^4+z^3=0$ & $(0,\pm1,1),(\pm1,\pm1,-1)$ & Thm.~\ref{thm:1.2.3} \\ \hline % Book-open entry: F12; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $56$ & $x^5+2y^3+z^3=0$ & $\pm(1,0,-1),\pm(1,-1,1)$ & \cite{GrechukMO26} \\ \hline % Book-open entry: F13; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $56$ & $x^5+y^4-2z^2=0$ & $(-1,\pm1,0),(1,\pm1,\pm1)$ & Thm.~\ref{thm:1.2.4} \\ \hline % Book-open entry: F14; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $56$ & $x^5+y^4+2z^2=0$ & $(-1,\pm1,0),(-3,\pm1,\pm11)$ & \cite[Thm.~1]{BEN10} \\ \hline % Book-open entry: F15; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $60$ & $x^5+y^4-3z^2=0$ & $(-1,\pm1,0),(11,\pm29,\pm538)$ & Thm.~\ref{thm:1.2.6} \\ \hline % Book-open entry: F16; Open Question 4.52, p. 345; Table 8.8, pp. 641–642 $60$ & $x^5+y^4+3z^2=0$ & $(-1,\pm1,0)$ & Thm.~\ref{thm:1.2.7} \\* \cline{1-4} \end{longtable} \endgroup \begin{introthm}\label{th:Fermat} For each equation in the second column of Table~\ref{tab:FermatSolved}, the primitive integer solutions are exactly the triples $(x,y,z)$ in the third column. \end{introthm} \begin{proof} See the references in the fourth column. The row for $2x^4+2y^3+z^3=0$ follows from~\cite{hexagon:2609.00130v1} by the change of variables described above. \end{proof} For homogeneous equations, Open Question~6.8 of~\cite{Grechuk24} asks whether the equation \begin{equation}\label{eq:homogeneousOpen} x^4+x^3y-y^4+y^3z+z^4=0 \end{equation} of size $H=80$ has an integer solution other than $(0,0,0)$. This question remains open. Open Question~6.9 of~\cite{Grechuk24} asks whether any of the three next-smallest equations in this category has a nonzero integer solution. We settle all three, as summarized in Table~\ref{tab:CubicSolved}. \Needspace{10\baselineskip} \begingroup\small \setlength{\tabcolsep}{3pt} \renewcommand{\arraystretch}{1.25} \setlength{\LTcapwidth}{\textwidth} \setlength{\LTpre}{6pt} \setlength{\LTpost}{8pt} \begin{longtable}{|c|l|l|} \caption{Equations left open in~\cite[Open Question~6.9]{Grechuk24}. Each has only the integer solution $(x,y,z,t)=(0,0,0,0)$.}\label{tab:CubicSolved}\\ \hline $H$ & Equation & Reference \\ \hline\hline \endfirsthead \caption[]{Equations left open in~\cite[Open Question~6.9]{Grechuk24}. Each has only the integer solution $(x,y,z,t)=(0,0,0,0)$ (continued).}\\ \hline $H$ & Equation & Reference \\ \hline\hline \endhead \hline\multicolumn{3}{r}{\footnotesize Continued on the next page.}\\ \endfoot \endlastfoot % Book-open entry: S09; Open Questions 6.8–6.9, pp. 485–486; Table 8.8, pp. 642–643 $96$ & $x^3+2x^2y+y^3+y^2z+xz^2+z^3+t^2x-t^2y-t^2z+tyz+t^3=0$ & Thm.~\ref{thm:1.4.1} \\ \hline % Book-open entry: S10; Open Questions 6.8–6.9, pp. 485–486; Table 8.8, pp. 642–643 $96$ & $2x^3+x^2y+xy^2+y^3+y^2z-2xz^2-z^3+t^2y+tz^2-t^3=0$ & Thm.~\ref{thm:1.4.2} \\ \hline % Book-open entry: S11; Open Questions 6.8–6.9, pp. 485–486; Table 8.8, pp. 642–643 $96$ & $2x^3+2x^2y+2y^3+y^2z-2xz^2-z^3+tz^2-t^3=0$ & Thm.~\ref{thm:1.4.3} \\* \cline{1-3} \end{longtable} \endgroup \begin{introthm}\label{th:Cubic} Each equation in Table~\ref{tab:CubicSolved} has only the integer solution $(x,y,z,t)=(0,0,0,0)$. \end{introthm} \begin{proof} See Theorems~\ref{thm:1.4.1}--\ref{thm:1.4.3}. \end{proof} Open Question~6.20 of~\cite{Grechuk24} concerns the five diagonal quintic equations in its Table~6.6: does any have a nonzero integer solution? Table~\ref{tab:QuinticSolved} records our negative answer in each case, so none of these five questions remains open. \Needspace{10\baselineskip} \begingroup\small \setlength{\tabcolsep}{3pt} \renewcommand{\arraystretch}{1.25} \setlength{\LTcapwidth}{\textwidth} \setlength{\LTpre}{6pt} \setlength{\LTpost}{8pt} \begin{longtable}{|c|l|l|} \caption{Diagonal quintic equations left open in~\cite[Open Question~6.20]{Grechuk24}. Each has only the integer solution $(x,y,z)=(0,0,0)$.}\label{tab:QuinticSolved}\\ \hline $H$ & Equation & Reference \\ \hline\hline \endfirsthead \caption[]{Diagonal quintic equations left open in~\cite[Open Question~6.20]{Grechuk24}. Each has only the integer solution $(x,y,z)=(0,0,0)$ (continued).}\\ \hline $H$ & Equation & Reference \\ \hline\hline \endhead \hline\multicolumn{3}{r}{\footnotesize Continued on the next page.}\\ \endfoot \endlastfoot % Book-open entry: D01; Open Question 6.20, p. 508; Table 8.8, p. 643 $704$ & $14x^5+5y^5+3z^5=0$ & Thm.~\ref{thm:1.4.5} \\ \hline % Book-open entry: D02; Open Question 6.20, p. 508; Table 8.8, p. 643 $832$ & $17x^5+6y^5+3z^5=0$ & Thm.~\ref{thm:1.4.7} \\ \hline % Book-open entry: D03; Open Question 6.20, p. 508; Table 8.8, p. 643 $928$ & $14x^5+13y^5+2z^5=0$ & Thm.~\ref{thm:1.4.9} \\ \hline % Book-open entry: D04; Open Question 6.20, p. 508; Table 8.8, p. 643 $960$ & $21x^5+5y^5+4z^5=0$ & Thm.~\ref{thm:1.4.6} \\ \hline % Book-open entry: D05; Open Question 6.20, p. 508; Table 8.8, p. 643 $992$ & $17x^5+9y^5+5z^5=0$ & Thm.~\ref{thm:1.4.8} \\* \cline{1-3} \end{longtable} \endgroup \Needspace{5\baselineskip} \begin{introthm}\label{th:Quintic} Each equation in Table~\ref{tab:QuinticSolved} has only the integer solution $(x,y,z)=(0,0,0)$. \end{introthm} \begin{proof} See Theorems~\ref{thm:1.4.5}--\ref{thm:1.4.9}. \end{proof} Open Question~6.21 of~\cite{Grechuk24} asks the analogous existence question for six diagonal quartic equations in four variables, all of size $H=288$. Three were subsequently resolved by explicit nonzero solutions attributed to Eugene Go in~\cite[Sections~8.9--8.10]{GrechukOpen}. We prove that a fourth has only the zero solution. Table~\ref{tab:QuarticSolved} combines these four resolutions: a nonzero tuple is an existence witness, whereas the zero tuple is the complete solution set for its row. The two equations in Table~\ref{tab:QuarticOpen} remain open; for each, the question is whether any nonzero integer solution exists. \Needspace{11\baselineskip} \begingroup\small \setlength{\tabcolsep}{3pt} \renewcommand{\arraystretch}{1.25} \setlength{\LTcapwidth}{\textwidth} \setlength{\LTpre}{6pt} \setlength{\LTpost}{8pt} \begin{longtable}{|c|l|l|l|} \caption{Resolved existence questions for diagonal quartics~\cite[Open Question~6.21]{Grechuk24}. Nonzero rows give one witness each; the zero row gives all solutions.}\label{tab:QuarticSolved}\\ \hline $H$ & Equation & Solutions, $(x,y,z,t)$ & Reference \\ \hline\hline \endfirsthead \caption[]{Resolved existence questions for diagonal quartics~\cite[Open Question~6.21]{Grechuk24}. Nonzero rows give one witness each; the zero row gives all solutions (continued).}\\ \hline $H$ & Equation & Solutions, $(x,y,z,t)$ & Reference \\ \hline\hline \endhead \hline\multicolumn{4}{r}{\footnotesize Continued on the next page.}\\ \endfoot \endlastfoot % Book-open entry: Q01; Open Question 6.21, p. 510; Table 8.8, p. 643 $288$ & $11x^4+4y^4+2z^4-t^4=0$ & $(32519425,30486957,21147587,63400357)$ & Go, \cite[\S8.10]{GrechukOpen} \\ \hline % Book-open entry: Q02; Open Question 6.21, p. 510; Table 8.8, p. 643 $288$ & $8x^4+7y^4+2z^4-t^4=0$ & $(0,0,0,0)$ & Thm.~\ref{thm:1.4.4} \\ \hline % Book-open entry: Q05; Open Question 6.21, p. 510; Table 8.8, p. 643 $288$ & $7x^4+5y^4+4z^4-2t^4=0$ & $(7337397,6619385,8035045,12352344)$ & Go, \cite[\S8.9]{GrechukOpen} \\ \hline % Book-open entry: Q06; Open Question 6.21, p. 510; Table 8.8, p. 643 $288$ & $8x^4+7y^4+z^4-2t^4=0$ & $(60428235,11074181,51780065,86883262)$ & Go, \cite[\S8.10]{GrechukOpen} \\* \cline{1-4} \end{longtable} \endgroup \Needspace{8\baselineskip} \begingroup\small \setlength{\tabcolsep}{3pt} \renewcommand{\arraystretch}{1.25} \setlength{\LTcapwidth}{\textwidth} \setlength{\LTpre}{6pt} \setlength{\LTpost}{8pt} \begin{longtable}{|c|l|} \caption{Remaining open diagonal quartic equations from~\cite[Open Question~6.21]{Grechuk24}. The question is the existence of a nonzero integer solution.}\label{tab:QuarticOpen}\\ \hline $H$ & Equation \\ \hline\hline \endfirsthead \caption[]{Remaining open diagonal quartic equations from~\cite[Open Question~6.21]{Grechuk24}. The question is the existence of a nonzero integer solution (continued).}\\ \hline $H$ & Equation \\ \hline\hline \endhead \hline\multicolumn{2}{r}{\footnotesize Continued on the next page.}\\ \endfoot \endlastfoot % Book-open entry: Q03; Open Question 6.21, p. 510; Table 8.8, p. 643 $288$ & $7x^4+6y^4+4z^4-t^4=0$ \\ \hline % Book-open entry: Q04; Open Question 6.21, p. 510; Table 8.8, p. 643 $288$ & $7x^4+6y^4+3z^4-2t^4=0$ \\* \cline{1-2} \end{longtable} \endgroup \begin{introthm}\label{th:Quartic} The equation $8x^4+7y^4+2z^4-t^4=0$ in Table~\ref{tab:QuarticSolved} has only the integer solution $(x,y,z,t)=(0,0,0,0)$. Each of the other three equations in that table has a nonzero integer solution. \end{introthm} \begin{proof} The nonexistence assertion is Theorem~\ref{thm:1.4.4}; the three displayed witnesses are verified by substitution. \end{proof} Open Question~6.28 of~\cite{Grechuk24} asks whether any of the eight plane quartics in its Table~6.10 has an integer solution with $xyz\ne0$. Table~\ref{tab:PlaneSolved} records a negative answer in every case. The cited theorems determine all integer solutions, including those with a zero coordinate, and no case in that table remains open. \Needspace{14\baselineskip} \begingroup\small \setlength{\tabcolsep}{3pt} \renewcommand{\arraystretch}{1.25} \setlength{\LTcapwidth}{\textwidth} \setlength{\LTpre}{6pt} \setlength{\LTpost}{8pt} \begin{longtable}{|c|l|l|} \caption{Plane quartics listed in~\cite[Open Question~6.28]{Grechuk24}. None has an integer solution with $xyz\ne0$.}\label{tab:PlaneSolved}\\ \hline $H$ & Equation & Reference \\ \hline\hline \endfirsthead \caption[]{Plane quartics listed in~\cite[Open Question~6.28]{Grechuk24}. None has an integer solution with $xyz\ne0$ (continued).}\\ \hline $H$ & Equation & Reference \\ \hline\hline \endhead \hline\multicolumn{3}{r}{\footnotesize Continued on the next page.}\\ \endfoot \endlastfoot % Book-open entry: P01; Open Question 6.28, p. 520; Table 8.8, p. 642 $64$ & $x^4+x^3y+xy^3-z^4=0$ & Thm.~\ref{thm:1.3.1} \\ \hline % Book-open entry: P02; Open Question 6.28, p. 520; Table 8.8, p. 642 $64$ & $x^4+x^2y^2+y^3z-yz^3=0$ & Thm.~\ref{thm:1.3.3} \\ \hline % Book-open entry: P03; Open Question 6.28, p. 520; Table 8.8, p. 642 $64$ & $x^4-x^2y^2+y^3z+yz^3=0$ & Thm.~\ref{thm:1.3.5} \\ \hline % Book-open entry: P04; Open Question 6.28, p. 520; Table 8.8, p. 642 $64$ & $x^4+x^2y^2+y^3z+z^4=0$ & Thm.~\ref{thm:1.3.6} \\ \hline % Book-open entry: P05; Open Question 6.28, p. 520; Table 8.8, p. 642 $64$ & $x^4-y^4+x^2yz+yz^3=0$ & Thm.~\ref{thm:1.3.2} \\ \hline % Book-open entry: P06; Open Question 6.28, p. 520; Table 8.8, p. 642 $64$ & $x^4+y^4+x^2yz-yz^3=0$ & Thm.~\ref{thm:1.3.4} \\ \hline % Book-open entry: P07; Open Question 6.28, p. 520; Table 8.8, p. 642 $64$ & $x^3z+y^3x+y^2z^2-z^3y=0$ & Thm.~\ref{thm:1.3.7} \\ \hline % Book-open entry: P08; Open Question 6.28, p. 520; Table 8.8, p. 642 $64$ & $x^4-y^3z+xyz^2+xz^3=0$ & Thm.~\ref{thm:1.3.8} \\* \cline{1-3} \end{longtable} \endgroup \Needspace{5\baselineskip} \begin{introthm}\label{th:Plane} None of the equations in Table~\ref{tab:PlaneSolved} has an integer solution with $xyz\ne0$. \end{introthm} \begin{proof} See Theorems~\ref{thm:1.3.1}--\ref{thm:1.3.8}. \end{proof} Two related questions in four variables remain open. Open Question~6.32 of~\cite{Grechuk24} asks whether \begin{equation}\label{eq:allNonzeroOpen1} x^4+xy^3+z^4+t^4=0 \end{equation} has an integer solution with $xyzt\ne0$, and Open Question~6.33 asks the same question for \begin{equation}\label{eq:allNonzeroOpen2} x^4+y^4+z^3t-zt^3=0. \end{equation} Both equations have size $H=64$ and remain open. \Needspace{11\baselineskip} Finally, the book~\cite{Grechuk24} leaves open the existence of an integer solution with $xyz\ne0$ to its equation~(6.138), namely \begin{equation}\label{eq:quintic608} 4x^5+4y^5+11z^5=0, \end{equation} of size $H=608$. We determine its complete solution set. \begin{introthm}\label{th:quintic608} The integer solutions to equation~\eqref{eq:quintic608} are exactly $(x,y,z)=(n,-n,0)$, where $n\in\Z$. \end{introthm} \begin{proof} See Theorem~\ref{thm:1.4.10}. \end{proof} The preceding discussion covers seventy-three entries explicitly left open in the book. Thirty-eight are proved here, seventeen are classified as completed after publication of the book and before this paper, three follow from pre-book literature, and fifteen remain open. The seventeen earlier entries include the two equivalent presentations settled by~\cite{hexagon:2609.00130v1} in Table~\ref{tab:FermatSolved}. These counts concern the families discussed above, rather than every problem in the book. See the continuously updated arXiv preprint \cite{GrechukOpen} for the full list of the current remaining open equations in this project. The rest of the paper provides proofs for each of the thirty-eight equations. Shared results are collected in Section~\ref{sec:known}; the proofs for affine curves, generalized Fermat equations, plane quartics, and the remaining homogeneous equations appear in Sections~\ref{sec:1.1}--\ref{sec:1.4}, respectively. Established methods are cited where used, and the equation-specific computations have certificate identifiers linked to the accompanying reproducibility material. GPT-6 Astra Pro, GPT-5.6 Sol Ultra and Claude Opus 5.5 were used in developing, reviewing and editing the arguments. The computer-algebra computations used Magma and PARI/GP, together with the accompanying verification programs. \section{Shared tools and certificate lemmas}\label{sec:known} We cite standard theory at the point of use. For clarity, only the specializations needed to interpret the finite certificates are stated below. Proofs are retained for the elementary reductions and for consequences not quoted verbatim from a published theorem. \paragraph{Standard theory.} Explicit descent is used in the forms of Bruin--Poonen--Stoll \cite[Sections 6, 9 and 12]{BPS16}, Poonen--Schaefer \cite{PS97}, Stoll \cite[Sections 4--5]{Stoll01} and Simon \cite{Simon02}. The deficiency correction to a genus-two rank bound is from Poonen--Stoll \cite[Theorem 11 and Corollary 12]{PS99}; two-cover descent is from \cite{BS09}. Chabauty--Coleman integration and the residue-disk bounds are used as in \cite[Sections 4--5]{MP12}, with Strassmann's theorem \cite{Strassmann28} when the prime or disk requires a coefficient calculation. Coleman--Gross heights \cite{CG89}, including tangent-vector normalization \cite{Gross86,BB15}, provide the quadratic Chabauty functions. The Mordell--Weil sieve is used as in \cite{BS10}. For the Prym constructions we use \cite[Theorem 5.1 and Section 6]{Bruin08} and \cite[Theorem 1.1]{RR18}. The determinant defining the genus-two curve uses the three-variable quadratic forms (their $3\times3$ matrices); the tangent calculation on a genus-five cover uses the $5\times5$ matrices of its defining quadrics. These two matrix constructions are distinct. In the ramified cases the rational maps used below are verified explicitly over the function field, including their quadratic twist. Torsion reduction is from \cite[Appendix]{Katz81}, and extension across a smooth model uses the N\'eron mapping property \cite[Section 1.2]{BLR90}. For cyclic algebras we use local and global reciprocity \cite[Chapters V--VI]{Neukirch99}; for quadratic Hilbert symbols we use \cite[Chapter III, Sections 1--2]{Serre73}. The quintic arguments use modularity \cite{BCDT01}, level lowering in the form of \cite[Theorem 3 and Proposition 5.1]{Sik12}, Mazur's torsion theorem \cite{Mazur77}, and the Sturm bound \cite{Sturm87}. The calculations specific to the six Frey families are recorded once in the \hyperref[sec:frey]{common Frey-curve argument} below. \begin{thmL}[2-descent for plane quartics with a rational hyperflex]\label{thm:A} Let $C/\Q$ be a smooth plane quartic (genus $3$), and let the rational point $O$ be a hyperflex; that is, the tangent line at $O$ meets $C$ at $O$ with multiplicity 4. Choose coordinates so that $O=(0:1:0)$ and its tangent line is $z=0$. The other 27 bitangents do not pass through $O$, so they have the form $y=mx+b$. Suppose that these 27 slopes are the roots of a single irreducible polynomial $f\in\Q[m]$ of degree 27, and let $K=\Q[m]/(f)$. Then the following hold. \begin{enumerate}[label=(\arabic*)] \item $J(\Q)[2]=0$. \item From the generic bitangent $y=mx+b$ ($b\in K$) form $h=c\,(y-mx-b)\in K(C)$. The constant $c\in\Q^\times$ is chosen so that the leading coefficient at $O$ is $1$ (the square class of the leading coefficient does not depend on the local parameter). Then $\operatorname{div}h=2(P_m+Q_m-2O)$ ($P_m,Q_m$ are the points of tangency). Sending a degree 0 divisor $\sum n_iP_i$ that avoids the zeros and poles of $h$ to $\prod h(P_i)^{n_i}$, and setting the value at $O$ to be $1$, we obtain an injective homomorphism \[ \delta:J(\Q)/2J(\Q)\hookrightarrow K^\times/K^{\times2},\qquad \delta([P-O])=h(P) \] \item Let $S=\{2\}\cup\{p:\ p\text{ is odd and the Tamagawa number }c_p(J)\text{ is even}\}$. The elements of the image of $\delta$ have norm in $\Q^{\times2}$, and the image lies in \[ K(S,2)=\{\alpha\in K^\times/K^{\times2}:\ \text{for every prime ideal }\mathfrak p\text{ not above }S\text{, }v_{\mathfrak p}(\alpha)\text{ is even}\} \] Moreover, at every prime $p\le\infty$, the local map $\delta_p:J(\Q_p)/2J(\Q_p)\to(K\otimes\Q_p)^\times/(K\otimes\Q_p)^{\times2}$ defined by the same formula satisfies $\operatorname{res}_p\circ\delta=\delta_p\circ\operatorname{res}_p$. \item Hence, if we set \[ \operatorname{Sel}=\{\alpha\in K(S,2):\ N_{K/\Q}(\alpha)\in\Q^{\times2},\ \operatorname{res}_p(\alpha)\in\delta_p(J(\Q_p))\ (p\le\infty)\} \] then $\operatorname{rank}J(\Q)\le\dim_{\F_2}\operatorname{Sel}$. Here, if $K$ has signature $(r_1,r_2)$, then $\dim_{\F_2}K(S,2)=r_1+r_2+\#\{\mathfrak p\mid S\}+\dim_{\F_2}\Cl_S(K)[2]$, and the last term is $0$ if $\Cl(K)[2]=0$. \item (Local size) If $A$ is an abelian variety of dimension $g$ and $F$ is a finite extension of $\Q_p$, then \[ \dim_{\F_2}A(F)/2A(F)=\begin{cases}\dim_{\F_2}A(F)[2], & p\ne2,\\ \dim_{\F_2}A(F)[2]+g\,[F:\Q_2], & p=2\end{cases} \] and $\dim_{\F_2}A(\mathbb R)/2A(\mathbb R)=\dim_{\F_2}A(\mathbb R)[2]-g$, $A(\mathbb C)/2A(\mathbb C)=0$. \end{enumerate} \end{thmL} \begin{proof}[Outline of the proof of (1), of the injectivity in (2), and of $S$ in (3)] The 27 bitangents correspond to the 27 nonzero isotropic vectors of the odd quadratic refinement on $V=J[2]\cong\F_2^6$ (determined by the rational theta characteristic $2O$). The Weil pairing embeds $V$ into the permutation module $E=\F_2^{27}$, and the nonzero vectors of the image have weight 12 or 16 (this can be checked by enumerating all 64 vectors). Since $f$ is irreducible, the Galois image $G$ is transitive on the 27 bitangents, and so is its Sylow 3-subgroup $P$. Hence $E^P$ consists only of the constant vectors; the constant vector has weight 27, which is odd, so it is not in $V$, and therefore $V^P=0$. Since $P$ has odd order, taking $P$-invariants is exact, so $(E/V)^G$ is generated by the image of the constant vector and $E^G\to(E/V)^G$ is surjective. By the long exact sequence, $H^1(\Q,V)\to H^1(\Q,E)=K^\times/K^{\times2}$ is injective, and (1) follows from $V^G=0$. $S$ in (3): if $c_p$ is odd at an odd prime $p$, then $J(\Q_p)=J^0(\Q_p)+2J(\Q_p)$. On the connected Néron component $J^0$, $[2]$ is étale and surjective, so the Kummer classes of $J^0(\Q_p)$ are unramified, and their images have even valuation at every prime ideal above $p$. \end{proof} This is the rational-hyperflex specialization of \cite[Sections 6, 9 and 12]{BPS16}. The injectivity argument is included because it is essential to the rank bounds. \begin{thmL}[Explicit formula criterion for the 2-part of the class group]\label{thm:B} Let $K$ be a number field of degree $n$ and signature $(r_1,r_2)$, let $D\ge|d_K|$ (for example $D=|d_K|$, or $D=|\operatorname{disc}R|$ for an order $R\subseteq\mathcal O_K$ of rank $n$, since $\operatorname{disc}R=[\mathcal O_K:R]^2d_K$), and let $\alpha=r_1/n$. Let $T>0$, and let $v\ge0$ be a nonzero step function supported on an interval of length $T$. Let $f$ be the normalized autocorrelation of $v$, \[ f(x)=\frac{\int_{\mathbb R}v(t)v(t+x)\,dt}{\int_{\mathbb R}v(t)^2\,dt} \] For each prime power $q$, let $a(q)$ be the number of prime ideals of norm $q$, and let $k(q)\le a(q)$ be the number of those whose classes have been proved to lie in $2\Cl(K)$ (for example, those proved to be principal). Set $c_m(q)=a(q)$ if $m\ge1$ is even and $c_m(q)=2k(q)-a(q)$ if $m$ is odd, and let \[ C_\infty=\gamma+\log 4\pi+\alpha-\frac{\log D}{n},\qquad W(x)=\frac{1}{\sinh x}+\frac{\alpha}{2\cosh^2(x/2)} \] ($\gamma$ is Euler's constant). Then, if \[ B^-(f)=C_\infty-\int_0^\infty(1-f(x))W(x)\,dx+\frac4n\sum_{q^m0 \] we have $\Cl(K)[2]=0$. The sum runs over all prime powers $q$ and all $m\ge1$ with $q^m0$. \end{proof} The explicit formula is from \cite{Weil52,Poitou77}; compare \cite[Sections 4--5]{Miller14} and \cite[Theorem 5.1]{BDF08}. The character bound and the partial-principality version are consequences proved here, not the statements of those references. \begin{lemL}[Strong induction on the norm]\label{lem:C} Let $K$ be a number field and $B\ge2$. Suppose that for each prime ideal $\mathfrak p$ of norm at most $B$ an element $0\ne g_{\mathfrak p}\in\mathfrak p$ is given such that, in the prime factorization $\prod q^{e_q}$ of $n_{\mathfrak p}=|N(g_{\mathfrak p})|/N\mathfrak p$, every $q^{e_q}v$, a contradiction. \begin{lemL}[Logarithm of the formal group]\label{lem:O} Let $p$ be an odd prime, $\mathcal A/\Z_p$ an abelian scheme of relative dimension $g$ (e.g.\ a Jacobian with good reduction at $p$), and $z=(z_1,\dots,z_g)$ formal parameters at the origin for which the formal group law $\mathcal F(X,Y)$ has coefficients in $\Z_p$. \begin{enumerate}[label=(\arabic*)] \item There is a unique $L\in\Q_p[[X]]^g$ with $L(X)\equiv X$ (modulo terms of degree $\ge2$) and $L(\mathcal F(X,Y))=L(X)+L(Y)$. The coefficients of the homogeneous part of degree $n$ of $L$ lie in $\tfrac1n\Z_p$. The coefficients of degree $n$ of the inverse $E=L^{-1}$ have valuation at least $-v_p(n!)\ge-\tfrac{n-1}{p-1}$. \item If $r\ge1$, then $L$ and $E$ converge on $(p^r\Z_p)^g$ and are mutually inverse bijections of this set onto itself. If $\min_iv_p(X_i)=r$, then $L(X)\equiv X$, $E(X)\equiv X\pmod{p^{r+1}}$. \item If $Q\in\mathcal A(\Z_p)$ reduces to $0$ and $r=\min_iv_p(z_i(Q))\ge1$, then for $m\in\Z\setminus\{0\}$ we have $z([m]Q)=E(m\,L(z(Q)))$ and \[ z([m]Q)\equiv m\,z(Q)\pmod{p^{v_p(m)+r+1}},\qquad \min_iv_p(z_i([m]Q))=v_p(m)+r \] In particular, if $r=1$, then $z([m]Q)\equiv m\,z(Q)\pmod{p^{v_p(m)+2}}$. The same formula $E(mL(z(Q)))$ defines $[m]Q$ for $m\in\Z_p$. \end{enumerate} \end{lemL} For the formal logarithm and these coefficient estimates see \cite{Hazewinkel78}; compare \cite[Chapter IV]{Silverman09}. \begin{thmL}[Size of local images and generation by points over finite fields]\label{thm:S} Let $C/\Q$ be a curve of genus $g$, $J=\operatorname{Jac}(C)$, and $p$ a prime. \begin{enumerate}[label=(\arabic*)] \item $\dim_{\F_2}J(\Q_p)/2J(\Q_p)=\dim_{\F_2}J(\Q_p)[2]+g\cdot[p=2]$ (Proposition~\ref{thm:A}(5)). \item If $p$ is odd and $C$ has a smooth model $\mathcal C/\Z_p$ at $p$, then reduction gives an isomorphism $J(\Q_p)/2J(\Q_p)\cong J(\F_p)/2J(\F_p)$, and every point of $C(\F_p)$ lifts to $C(\Q_p)$. \item In addition to (2), assume that $O\in C(\Q_p)$ and $p+1>(8g-2)\sqrt p$ (for $g=3$: $p\ge487$). Then the classes $[\tilde P-\tilde O]$ for $\tilde P\in C(\F_p)$ generate $J(\F_p)/2J(\F_p)$. Hence the classes $[P-O]$, for lifts $P$ of the points of $C(\F_p)$, generate $J(\Q_p)/2J(\Q_p)$, and the local image $\delta_p(J(\Q_p))$ (Theorems~\ref{thm:A}, \ref{thm:T}) is generated by their values. \end{enumerate} \end{thmL} \begin{proof} (2) The reduction map is surjective, and its kernel is a pro-$p$ group consisting of points of the formal group (Lemma~\ref{lem:O}). Since $p$ is odd, the kernel is uniquely divisible by 2. (3) Suppose they do not generate. Then there is a nonzero $\chi:J(\F_p)\to\Z/2$ that vanishes on all $[\tilde P-\tilde O]$. By class field theory, $\chi$ gives an unramified double cover $D\to C$, and $\tilde P\in C(\F_p)$ splits in $D$ exactly when $\chi([\tilde P-\tilde O])=0$. Since there is an $\F_p$-point above $\tilde O$, $D$ is geometrically irreducible, and by Riemann--Hurwitz its genus is $2g-1$. Since all $\F_p$-points split, $\#D(\F_p)=2\#C(\F_p)$, and by \cite{Weil48}, $2(p+1-2g\sqrt p)\le p+1+(4g-2)\sqrt p$, that is, $p+1\le(8g-2)\sqrt p$. This contradicts the assumption. \end{proof} The local dimension is the standard local Kummer formula \cite[Section 4]{Stoll01}. The finite-field generation criterion is a consequence of function-field class field theory \cite[Chapter VI]{Serre88} and the Weil bound \cite{Weil48}, as shown above. \begin{thmL}[Descent for a general bitangent algebra]\label{thm:T} In the situation of Proposition~\ref{thm:A}, suppose that the slope polynomial $f$ of the 27 bitangents is squarefree but not necessarily irreducible. Let $f=\prod_if_i$ be its factorization into irreducibles and $L=\Q[m]/(f)=\prod_iL_i$ ($L_i=\Q[m]/(f_i)$). Let $G$ be the Galois image on the 27 bitangents, $E=\F_2^{27}$ the permutation module, and $V=J[2]\hookrightarrow E$ the injection as in the proof of Proposition~\ref{thm:A}. Then $H^1(\Q,E)=L^\times/L^{\times2}$. \begin{enumerate}[label=(\arabic*)] \item $J(\Q)[2]=V^G$. The map $\delta:J(\Q)/2J(\Q)\to L^\times/L^{\times2}$ obtained by evaluating the normalized bitangent function $h$ componentwise is the composition of the Kummer map and $H^1(\Q,V)\to H^1(\Q,E)$. The kernel of the latter map is the quotient of $(E/V)^G$ by the image of $E^G$. In particular, if $E^G\to(E/V)^G$ is surjective, then $\delta$ is injective. \item The image of $\delta$ satisfies $N_{L/\Q}(\delta(x))\in\Q^{\times2}$, lies in $L(S,2)=\prod_iL_i(S,2)$ for the $S$ of Proposition~\ref{thm:A}(3), and is compatible with the local images at all $p\le\infty$. Hence, if $\delta$ is injective, then for $\operatorname{Sel}\subset L(S,2)$ defined as in Proposition~\ref{thm:A}(4) we have $\operatorname{rank}J(\Q)+\dim_{\F_2}V^G\le\dim_{\F_2}\operatorname{Sel}$. \item If $G_p$ is the image of the decomposition group at $p$, then $\dim J(\Q_p)[2]=\dim V^{G_p}$, and the dimension of $J(\Q_p)/2J(\Q_p)$, an upper bound for that of the local image, is given by Proposition~\ref{thm:S}(1). \end{enumerate} \end{thmL} This is the reducible-bitangent-algebra form of \cite[Sections 6, 9 and 12]{BPS16}; the injectivity test is the cohomology exact sequence for $0\to V\to E\to E/V\to0$. \begin{thmL}[Explicit formula inequality for a single quadratic character]\label{thm:U} Use the notation of Proposition~\ref{thm:B} ($n$, $r_1$, $D$, $\alpha$, $T$, $f$, $C_\infty$, $W$), and let $\chi:\Cl(K)\to\{\pm1\}$ be a nontrivial character. Then \[ C_\infty-\int_0^\infty(1-f(x))W(x)\,dx+\frac4n\sum_{\mathfrak p}\ \sum_{m\ge1:\ N\mathfrak p^m\frac4{27}$, it is smaller than both critical values $\pm\frac2{3\sqrt3}$ of $h$. So $\{v:h(v)\ge m_0\}$ is a single half-line; on it, the two real roots of $g(u)=h(v)$ meet at the endpoint, and both tend to $O$ as $v\to-\infty$. Hence $C(\mathbb R)$ is a single loop. The rest is \cite{GH81}. \end{proof} \begin{lem}\label{lem:1.1.1:strict} If $\dim\operatorname{Sel}^{(2)}(J/\Q)>3$, then there is a nontrivial character $\chi:\Cl(K)\to\{\pm1\}$ of $\Cl(K)$ such that (i) $\chi=1$ at prime ideals that are principal, and (ii) if $p\nmid\operatorname{disc}(p_K)$ and $r(p)$ is the number of roots of $p_K$ modulo $p$, then $\sum_{\mathfrak p\mid p,\,N\mathfrak p=p}\chi(\mathfrak p)\ge\kappa(r(p))$. \end{lem} \begin{proof} Let $\gamma$ be a Selmer element outside the span of $\delta(E),\delta(B),\delta(D)$. By Lemma~\ref{lem:1.1.1:desc}(3) there are $a,b,d\in\F_2$ such that $\gamma'=\gamma-a\delta(E)-b\delta(B)-d\delta(D)$ becomes trivial over $\Q_2$, and $\gamma'\ne0$. At the real place $\gamma'$ is trivial (Lemma~\ref{lem:1.1.1:real}), and at odd primes it is unramified, since the reduction is good or the component group is trivial. By Lemma~\ref{lem:1.1.1:desc}(1), the image $\epsilon\in K^\times/K^{\times2}$ of $\gamma'$ is not 1, it is a square above 2 and positive at the real embeddings, and $K(\sqrt\epsilon)/K$ is an unramified quadratic extension that splits at the real places. Its character is $\chi$. (ii) At $p$, $\gamma'$ is determined by the Frobenius $g\in\operatorname{GL}(V)$ and a cocycle value $c\in V$. The prime ideals of norm $p$ correspond to the $w\in\Omega$ fixed by $g$, and there $\chi=(-1)^{\langle c,w\rangle}$; so the sum is the sum of Lemma~\ref{lem:1.1.1:code} for $U=\ker(g-1)$, $\lambda=\langle c,\cdot\rangle$. \end{proof} \paragraph{Step 3: rank.} \begin{lem}\label{lem:1.1.1:rank} $\dim\operatorname{Sel}^{(2)}(J/\Q)=\operatorname{rank}J(\Q)=3$. \end{lem} \begin{proof} Suppose not, and take $\chi$ as in Lemma~\ref{lem:1.1.1:strict}. For a nontrivial quadratic character $\chi$ of $\Cl(K)$, the proof of Proposition~\ref{thm:B} gives \[ 0\ \ge\ C_\infty-\int_0^\infty(1-f)W\,dx+\frac4n\sum_{\mathfrak p,\,m\ge1}\chi(\mathfrak p)^m\frac{\log N\mathfrak p}{N\mathfrak p^{m}+1}\,f(m\log N\mathfrak p) \tag{$*$} \] We take $n=27$, $\alpha=1/9$, $D$ as above, $X=10^{10}$, $T=\log X$, and let $f$ be the normalized autocorrelation of a symmetric step function (with positive integer heights) on 128 cells of width $T/128$. With $M=10^7$ and $R=1.1\cdot10^8$ we split the prime ideals into four groups. All the weights are nonnegative. (a) The $664{,}464$ prime ideals with $N\mathfrak p\le M$ are all principal. For $588{,}850$ of them this is proved with witnesses as in Lemma~\ref{lem:C}, and for $75{,}614$ with generators $g$ satisfying $|N(g)|=N\mathfrak p$ (\code{1.1.1d}). The sum over this part is computed from the counts per norm (\code{1.1.1e}). Norms $q\le10^5$ are included exactly, together with all powers $q^mR(1-\frac1{5\log^2R})$ ($c=\frac{36261}{36260}$, \cite[Section 1]{Dusart10}). By partial summation, $\sum_{RX$, the terms with $m\ge2$ occur only in (a). All prime factors of $\operatorname{disc}(p_K)=2^{312}3^{54}157^6\,64679^6\,333337^2\,172659335137^2$ are less than $M$ or greater than $X$, so (ii) can be used in (c) and (d). The values computed with 224-bit interval arithmetic are as follows (\code{1.1.1f}). \[ \begin{array}{ccccc} \mathcal A(f)&P_-&H&E&-\tfrac{20}{27}\mathcal T_R\\\hline -1.0229409565\ldots&1.1435757488\ldots&-0.0000100075\ldots&-0.0340749409\ldots&-0.0598580721\ldots \end{array} \] Their sum is $\mathcal B>0.0266917716>0$. The right-hand side of ($*$) is at least $\mathcal B$, a contradiction. Hence the Selmer dimension is at most 3, and the conclusion follows from Lemma~\ref{lem:1.1.1:desc}(3) and $J(\Q)[2]=0$. \end{proof} \paragraph{Step 4: $p$-adic heights.} Fix $p\in\{3,5\}$ and use the Coleman--Gross height $h=\sum_vh_v$ ($\log_pp=0$) of \cite{CG89,BBM16}. When supports overlap we normalize by tangent vectors: at affine points we use the dual of $\omega=du/(3v^2-1)=-dv/(4u^3+1)$, and at $O$ the dual of $t=-u/v$. \begin{lem}\label{lem:1.1.1:away} For an integral point $P$ and a prime $\ell\ne p$ we have $h_\ell(P-O,P-O)=0$, and for distinct $Q,Q'$ among $E,B,D$ we have $h_\ell(Q-O,Q'-O)=0$. \end{lem} \begin{proof} Since the two nodes have thickness 1, the plane model is regular over $\Z$; the integral sections do not pass through the nodes, and since all fibers are irreducible, the vertical correction is 0. On the smooth affine part one of $F_u$, $F_v$ is a unit, so the tangent vector dual to $\omega$ is primitive and $(P\cdot P)_\omega=0$. Near $O$ we have $\operatorname{div}\omega=4O$, the leading coefficient $-1$ is a unit, and $t$ is a relative parameter, so $(O\cdot O)_t=0$. Integral points do not meet $O$. Any two of $E,B,D$ differ by $\pm1$ in one coordinate, so their sections do not meet. \end{proof} \begin{lem}\label{lem:1.1.1:kernel} Let $\omega_i=(1,u,v)\,\omega$ and $\eta_i=(5u^2v,2uv,u^2)\,\omega$. The $\eta_i$ are differentials of the second kind with poles only at $O$, and the cup product matrix with respect to $\operatorname{Res}_O((\int\alpha)\beta)$ is $\left(\begin{smallmatrix}0&-I_3\\I_3&0\end{smallmatrix}\right)$. For $P=(a,b)$, $\nu_P(Q)=\frac{v^2+bv+b^2-1}{u-a}\,\omega(Q)$ is a differential of the third kind with residue divisor $P-O$. Let $I_i(P)=\int_O^P\omega_i$, $J_i(P)=\int_O^P\eta_i$, $K(P,Q)=\int_O^Q\nu_P$ (normalized by the tangent vectors above). Then the local height at $p$ with respect to $W=\langle\eta_i\rangle$ is $h_p(P-O,Q-O)=H(P,Q):=K(P,Q)-\sum_iJ_i(P)I_i(Q)$, and $\tau(P)=H(P,P)$ satisfies $d\tau=-2\sum_i\omega_iJ_i$ on the affine disks on which $3v^2-1$ is a unit. \end{lem} \begin{proof} Expanding at $O$ with $u=t^{-3}\rho$, $v=-t^{-4}\rho$, $\rho^2(\rho-1)=-t^8-t^9$, the leading terms of $\omega_0,\omega_1,\omega_2,\eta_0,\eta_1,\eta_2$ are $-t^4dt$, $-t\,dt$, $dt$, $5t^{-6}dt$, $2t^{-3}dt$, $-t^{-2}dt$, respectively, and the corrections start in degree 8; this gives the cup product matrix. On the curve, $(b-v)(v^2+bv+b^2-1)=(u-a)(u^3+u^2a+ua^2+a^3+1)$, so the only poles of $\nu_P$ are $P$ and $O$. The bidifferential identity $d_P\nu_P(Q)-d_Q\nu_Q(P)=\sum_i(\eta_i(P)\omega_i(Q)-\omega_i(P)\eta_i(Q))$ holds, since after dividing both sides by $\omega(P)\omega(Q)$ each side becomes $5a^2b-5u^2v+2abu-2auv+a^2v-bu^2$ (\code{1.1.1g}). At $O$, the coefficients of $t^0,t^1,t^4$ in $\nu_P=(-t^{-1}-at^2+bt^3-a^2t^5+\cdots)dt$ are 0, so the de Rham class of $\nu_P-\sum_iJ_i(P)\omega_i$ lies in $W$, and by the definition in \cite{CG89,BBM16}, $h_p=H$. Integrating the identity gives $d_PH(P,Q)=\nu_Q(P)-\sum_i\omega_i(P)J_i(Q)$. On the diagonal, the regular part $2\cdot(-3bb'/\beta)\,da$ ($\beta=3b^2-1$, $b'=db/da$) of $\nu_P(Q),\nu_Q(P)$ cancels against the $d\log\beta$ added by the tangent vector normalization, and we obtain $d\tau=-2\sum_i\omega_iJ_i$. \end{proof} \begin{lem}\label{lem:1.1.1:qcfun} Let $I_B$ be the matrix with rows $I(E),I(B),I(D)$, let $H_B=(H(Q,Q'))_{Q,Q'\in\{E,B,D\}}$, and assume that $I_B$ is invertible. If $M=I_B^{-1}H_B(I_B^{\mathsf T})^{-1}$ and $q(P)=\tau(P)-I(P)\,M\,I(P)^{\mathsf T}$, then $q(P)=0$ at every integral point $P$. \end{lem} \begin{proof} $h$ is a symmetric bilinear form on $J(\Q)$ (\cite{CG89,BBM16}), and since the rank is 3, it is a quadratic form in $\log$. By Lemma~\ref{lem:1.1.1:away}, $h=h_p$ among $E,B,D$, so its matrix is $M$, and at an integral point $P$ we have $h(P-O,P-O)=\tau(P)$. \end{proof} \paragraph{Step 5: $p=3$.} $C$ has good reduction at 3, and the only affine points modulo 3 are the six reductions of $A,\dots,F_1$. We parametrize the disk with center $P_j=(a_j,b_j)$ by $u=a_j+3s$, $s\in\Z_3$, and let $q_j(s)=q(P_j(s))$. \begin{lem}\label{lem:1.1.1:qc3} (1) $I_B/3\equiv\left(\begin{smallmatrix}1&1&0\\0&2&2\\1&1&2\end{smallmatrix}\right)\pmod3$ (the determinant is a unit), and $\min v_3(M_{ij})=-1$. (2) The minimal valuation $e_j$ of the coefficients of $q_j$ and the reductions are as follows, and the roots of $q_j/s$ modulo 3 are all simple. \[ \begin{array}{c|cccccc} P_j&A&B&C_1&D&E&F_1\\\hline e_j&2&1&1&1&1&1\\ 3^{-e_j}q_j\bmod3&2s^2+2s^3&2s+2s^2&2s+s^2&2s+2s^2&s&s^2\\ \text{extra roots}&2&1&1&1&0&1 \end{array} \] (3) The positions $s$ of the six extra roots and their coordinates $\mathbf a_i=I(P)I_B^{-1}$ (basis $E,B,D$) are, modulo $3^8$, as follows. \[ \begin{array}{c|c|r|rrr} i&\text{disk}&s&\multicolumn{3}{c}{\mathbf a_i}\\\hline 0&A&4680&5946&6246&1629\\ 1&A&1274&4511&1997&707\\ 2&B&5882&1542&2265&5452\\ 3&C_1&2620&1475&1164&2808\\ 4&D&1442&1405&250&6058\\ 5&F_1&5667&6218&5349&6041 \end{array} \] \end{lem} \begin{proof} Choose six functions $f_j$ in $L(7O)$ with $\operatorname{div}f_j=4P_j+R_{j,1}+R_{j,2}+R_{j,3}-7O$. The remaining zeros $R_{j,k}$ are simple modulo 3 and lie in three other disks. Writing the $R_{j,k}$ in terms of the disk centers and tiny integrals, the coefficient matrix of the principal divisor relations has determinant $3185\in\Z_3^\times$. The right-hand sides are given exactly by Coleman reciprocity: for $\rho$ of the second kind it is $\operatorname{Res}_O(\log f_j\cdot\rho)$, and for $\nu_{P_i}$ of the third kind it is the $\log_3$ of the ratio of the leading coefficients with respect to the tangent vectors. Solving with precision $3^{20}$ determines $M$ modulo $3^{17}$ (\code{1.1.1h}). On a disk, $\tau(P_j(s))=\tau(P_j)-2\sum_i\int_0^s\omega_i(J_i(P_j)+\int\eta_i)$, and the expansion coefficients lie in $\Z_3$, so $v_3([s^n]q_j)\ge n-1-2\lfloor\log_3n\rfloor$. This is at least 3 for $n\ge6$, so the reductions in the table take the tail into account. We have $q_j(0)=0$, and (2), (3) follow from Hensel's lemma. The eight digits in the table are within the remaining precision. \end{proof} \paragraph{Step 6: $p=5$.} The affine points modulo 5 are the reductions of the six points and $(3,2)$; the latter lifts to $P_6=(3,\beta)$, $\beta^3-\beta+84=0$, $\beta\equiv309087\pmod{5^8}$. We parametrize by $u=a_j+5s$. \begin{lem}\label{lem:1.1.1:qc5} (1) $5^{-1}I_B\equiv\left(\begin{smallmatrix}3&3&2\\2&0&0\\1&3&0\end{smallmatrix}\right)\pmod5$ (the determinant is a unit), and $\min v_5(M_{ij})=-1$. (2) The minimal valuation of the coefficients of every $q_j$ is 1, and \[ \begin{array}{c|ccccccc} P_j&A&B&C_1&D&E&F_1&P_6\\\hline 5^{-1}q_j\bmod5&4s+s^2&3s+2s^2&3s+s^2&3s+3s^2&2s&4s+3s^2&4s+4s^2\\ \text{extra roots}&1&1&1&1&0&1&2 \end{array} \] ($P_6$ is not a rational point, so both of its roots are extra roots). (3) The positions and the coordinates $\mathbf a'_j$ of the seven extra roots are, modulo $5^8$, as follows. \[ \begin{array}{c|c|r|rrr} j&\text{disk}&s&\multicolumn{3}{c}{\mathbf a'_j}\\\hline 0&A&88061&182268&389041&117968\\ 1&B&51781&9880&216347&307132\\ 2&C_1&291387&125051&375010&46930\\ 3&D&332159&759&70457&158427\\ 4&F_1&49757&79377&350530&127046\\ 5&P_6&325535&352126&189681&390420\\ 6&P_6&235554&361850&252022&388570 \end{array} \] \end{lem} \begin{proof} As in Step 5. We use seven functions in $L(NO)$ over $\Q(\beta)$, $N=6,7,7,7,7,7,9$, and the determinant of the relation matrix is $452\in\Z_5^\times$. For the tail, $v_5([s^n]q_j)\ge n-1-2\lfloor\log_5n\rfloor$, which is at least 2 for $n\ge3$ (\code{1.1.1i}). \end{proof} \paragraph{Step 7: torsion, saturation, Mordell--Weil sieve.} \begin{lem}\label{lem:1.1.1:sat} $J(\Q)$ is torsion-free, and the index of $\Gamma=\langle E,B,D\rangle$ is prime to 30. \end{lem} \begin{proof} $\#J(\F_3)=91$ and $\#J(\F_5)=226$ (\code{1.1.1j}) are coprime, so by \cite[Appendix]{Katz81} the torsion is 0. If a prime $\ell$ divides the index, then there is $R$ with $\ell R=a_1E+a_2B+a_3D$ where the $a_k$ are not all multiples of $\ell$. If $\ell=2$, this contradicts Lemma~\ref{lem:1.1.1:desc}(3). If $\ell=3$: since $3\nmid91$, we have $\log J(\Q_3)\subset3\Z_3^3$ (Lemma~\ref{lem:O}), and since the rows of $I_B$ form a basis of $3\Z_3^3$ (Lemma~\ref{lem:1.1.1:qc3}(1)), we get $a_k/3\in\Z_3$, a contradiction. The case $\ell=5$ is the same. \end{proof} So every $R\in J(\Q)$ can be written as $R=a_1E+a_2B+a_3D$ with $a_k\in\Z_{(2)}\cap\Z_{(3)}\cap\Z_{(5)}$, and at an auxiliary prime $q$ the projection of $R$ onto the $\{2,3,5\}$-primary part is determined by the residue classes of the $a_k$. Arithmetic in $J(\F_q)$ is done by Riemann--Roch computations with the basis $u^iv^j$ ($j\le2$, $3i+4j\le m$) of $L(mO)$, and elements are compared exactly by their reduced representatives. Minimality forces $h^0(D)=1$: otherwise a section vanishing at $O$ would give a representative of smaller degree. \begin{lem}\label{lem:1.1.1:sieve} The only integral points of $C$ are $A,B,C_1,D,E,F_1$. \end{lem} \begin{proof} If there is another integral point $P$, then by Lemmas~\ref{lem:1.1.1:qcfun}, \ref{lem:1.1.1:qc3}, \ref{lem:1.1.1:qc5}, $P$ is a 3-adic extra root $i$ and also a 5-adic extra root $j$ (if it were the root at a center, it would be equal in $\Q_p$ to a known point). The coefficient vector $\mathbf a$ of $P-O$ is $\mathbf a_i$ modulo $3^8$ and $\mathbf a'_j$ modulo $5^8$. By \cite{BS10}, at each $q$ the reduction of $P$ is a point of $C(\F_q)$, and its primary projection must be the value determined by $\mathbf a$. The data for the primes used are as follows (\code{1.1.1j}). \[ \begin{array}{r|rrr|r} q&\#C(\F_q)&\#C(\F_{q^2})&\#C(\F_{q^3})&\#J(\F_q)\\\hline 109&109&12243&1296898&1303500\\ 233&259&54447&12648664&14102424\\ 277&300&77454&21247200&23117400 \end{array} \] Imposing the 3-primary and the 5-primary conditions on the same curve point at $q=109,277$, the values of $j$ that remain for each $i$ are as follows. \[ \begin{array}{c|cccccc} i&0&1&2&3&4&5\\\hline q=109&\{2,6\}&\{0,4,5\}&\{0,2\}&\{0,4,5\}&\{0,2\}&\{0,2\}\\ q=277&\{4\}&\varnothing&\{2,6\}&\varnothing&\{4\}&\{4\} \end{array} \] The only remaining pair is $(2,2)$, and its coefficients are $\mathbf a\equiv(3,78,25)\pmod{81}$, $\mathbf a\equiv(51,10,55)\pmod{125}$. At $q=233$ the 2-primary subgroup is $(\Z/2)^3$, and the projections of $E,B,D$ form a basis of it. The only curve points with matching 3-primary coordinates are $(20,173)$ and $(78,144)$, and their 2-primary coordinates are $B$ and 0, so $a_1\equiv a_3\equiv0\pmod2$. At $q=277$ the only curve point satisfying both conditions is $(44,20)$; the 2-primary subgroup is cyclic of order 8, generated by the projection $U$ of $E$, and $B\mapsto4U$, $D\mapsto5U$, $(44,20)-O\mapsto3U$. Hence $a_1+4a_2+5a_3\equiv3\pmod8$, that is, $a_1+a_3\equiv1\pmod2$, a contradiction. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.1.1}] Under $(x,y)=(-u,-v)$ the six points of Lemma~\ref{lem:1.1.1:sieve} become the six solutions in the theorem. $J(\Q)\cong\Z^3$ follows from Lemmas~\ref{lem:1.1.1:rank}, \ref{lem:1.1.1:sat}. \end{proof} \subsection{\texorpdfstring{$x^4-y^3+x-y=0$ \ ($H=28$)}{x^4-y^3+x-y=0 (H=28)}}\label{sec:1.1.2} \begin{thm}\label{thm:1.1.2} The only integer solutions of $x^4-y^3+x-y=0$ are $(x,y)=(-1,0),(0,0),(1,1)$. \end{thm} \paragraph{Setup.} Setting $u=-y$, $z=x$, the equation becomes \[ C:\quad u^3+u+z^4+z=0 \] The projective closure $z^4+u^3w+uw^3+zw^3=0$ is a smooth plane quartic (genus 3). At the point at infinity $\infty=(1:0:0)$ the tangent line $w=0$ meets the curve with multiplicity 4, as $z^4=0$, so $\infty$ is a rational hyperflex. Theorem~\ref{thm:1.1.2} follows immediately from the following proposition. \begin{prop}\label{prop:1.1.2} $C(\Q)=\{\infty,(0,0),(0,-1),(-1,1)\}$. \end{prop} Indeed, the affine points $(u,z)=(0,0),(0,-1),(-1,1)$ give $(x,y)=(z,-u)=(0,0),(-1,0),(1,1)$. The proof has four steps. We determine the bitangent field $K$, show $\Cl(K)[2]=0$ with Proposition~\ref{thm:B} (Lemma~\ref{lem:1.1.2:class}), use this to show by 2-descent that the rank is at most 2 (Lemma~\ref{lem:1.1.2:rank}), and finish with Chabauty--Coleman at $p=3$ (Lemma~\ref{lem:1.1.2:chab}). \paragraph{Step 1: the bitangent field.} The slopes of the 27 bitangents other than the tangent line at $\infty$ are the roots of \[ f(m)=m^{27}+480m^{19}-1344m^{18}-34560m^{11}-70656m^{10}-40704m^{9}-16384m^{3}+12288m-4096 \] $f$ is irreducible (\code{1.1.2a}), and $K=\Q[m]/(f)$ has signature $(3,12)$ and discriminant $D=2^{24}\cdot277^{6}\cdot66943^{6}$. The bad primes of the curve are 277 and 66943. At both of them the reduction has only one node, of thickness 1, so the Tamagawa number is 1, and hence the descent image is unramified at these two primes. Therefore $S=\{2\}$ suffices. There is only one prime ideal above 2 (ramification index 9, residue degree 3). \paragraph{Step 2: the class group.} \begin{lem}\label{lem:1.1.2:class} $\Cl(K)[2]=0$. \end{lem} \begin{proof} We apply Proposition~\ref{thm:B} with $n=27$, $(r_1,r_2)=(3,12)$, $D$ as above, and $T=\log10^8$. The window is a step function on 256 cells of equal width, with positive integer heights (the input of \code{1.1.2d}). Two kinds of data are needed. \begin{enumerate}[label=(\roman*)] \item There are $5{,}763{,}020$ prime ideals of norm less than $10^8$ in total, and all the counts $a(q)$ by norm are computed (\code{1.1.2b}). \item All $349{,}551$ prime ideals of norm at most $5\cdot10^6$ have a witness as in Lemma~\ref{lem:C} (\code{1.1.2c}). So we set $k(q)=a(q)$ if $q\le5\cdot10^6$, and $k(q)=0$ for larger $q$. \end{enumerate} With these data, $B^-(f)>0.000875203229\ldots>0$ (\code{1.1.2d}). The computation uses exact rational coefficients and directed-rounding interval arithmetic (192 bits), so rounding errors cannot change the conclusion. By Proposition~\ref{thm:B}, $\Cl(K)[2]=0$. \end{proof} \paragraph{Step 3: an upper bound for the rank.} \begin{lem}\label{lem:1.1.2:rank} $\operatorname{rank}J(\Q)\le2$. \end{lem} \begin{proof} The hypotheses of Proposition~\ref{thm:A} were checked above. So $J(\Q)[2]=0$ and $\delta$ is injective. Since $\Cl(K)[2]=0$ by Lemma~\ref{lem:1.1.2:class}, \[ \dim_{\F_2}K(S,2)=(r_1+r_2)+\#\{\mathfrak p\mid2\}=15+1=16 \] Indeed, 16 explicit elements supported on $S$ form a basis of this space. Their independence is checked with quadratic characters at the primes $487$, $491$, $499$, $503$, $509$, $521$, $523$, $541$ (\code{1.1.2a}). We impose conditions on this 16-dimensional space one after another. After imposing the condition that the norm is a square, we impose the local conditions at the good primes $487$, $491$, $499$, $503$ in turn, and the dimension of the candidate space becomes $10,6,4,3$. At 2 the bitangent orbit is transitive, of size 27. So, because the weights of the bitangent code are $12,16$, we have $J(\Q_2)[2]=0$ and $\dim J(\Q_2)/2J(\Q_2)=g=3$. The three 2-adic images are independent, so they fill this space, and this condition reduces the candidate dimension from 3 to 2. Since $J(\Q)[2]=0$, $\operatorname{rank}J(\Q)\le2$. \end{proof} \paragraph{Step 4: Chabauty--Coleman at $p=3$.} $C$ has good reduction at 3, because on the affine part $\partial F/\partial u=3u^2+1\equiv1\pmod3$, and in the chart at infinity the derivative is also a unit. An integral basis of the regular differentials is \[ \omega_0=\frac{dz}{3u^2+1},\quad \omega_1=\frac{z\,dz}{3u^2+1},\quad \omega_2=\frac{u\,dz}{3u^2+1} \] $C(\F_3)$ consists of the four points $\infty,(0,0),(0,2),(2,1)$, so there are also four residue disks. The four known rational points lie one in each disk, since $(0,-1)\equiv(0,2)$ and $(-1,1)\equiv(2,1)$. \begin{lem}\label{lem:1.1.2:chab} $\operatorname{rank}J(\Q)=2$. Moreover, a 3-adic lift of the differential $\omega=2\omega_0+\omega_1+\omega_2$ annihilates all logarithms of $J(\Q)$, and it is nonzero modulo 3 on all four disks. \end{lem} \begin{proof} For $P_0=(0,0)$, $P_1=(0,-1)$, explicit functions $h_i$ satisfy \[ \operatorname{div}(h_i)=28P_i+E_i-31\infty \] Here $E_i$ is an irreducible divisor of degree 3. This identity is checked exactly through the pole order 31, the Taylor order 28 at $P_i$, and the norm identity $\operatorname{Res}_u(F,h_i)=c_i(z-z(P_i))^{28}d_i(z)$ (\code{1.1.2e}). $E_i$ is unramified at 3, and its $z$-coordinates reduce to the fibers $z=1$ and $z=0$, respectively. Both fibers are linearly equivalent to $3\infty$. Hence the 3-adic logarithm of \[ 28[P_i-\infty]=-[E_i-3\infty] \] is computed as a sum (trace) of tiny integrals starting from the points of $E_i$, and dividing it by $-28$ gives the logarithm of $[P_i-\infty]$. Computing modulo $3^{16}$ with 18 terms, the logarithms with respect to $(\omega_0,\omega_1,\omega_2)$ are, modulo $3^{10}$, \[ \log[P_0-\infty]\equiv(57309,48831,7839),\qquad \log[P_1-\infty]\equiv(40674,56691,3162) \] (\code{1.1.2e}). The discarded $k$-th term has valuation at least $k-v_3(k)$, which is at least 17 for $k\ge19$, so this precision is guaranteed. The two logarithms are linearly independent, so $\operatorname{rank}J(\Q)\ge2$, and together with Lemma~\ref{lem:1.1.2:rank} we get $\operatorname{rank}J(\Q)=2$. The components of the cross product of the two vectors have common valuation 2, and the primitive annihilator obtained by dividing by $3^2$ is $(2,1,1)$ modulo 3. The annihilator kills the subgroup generated by the two logarithms, and this subgroup has finite index in $J(\Q)$, so $\omega$ kills all logarithms of $J(\Q)$. Finally we evaluate $\omega\bmod3=(2+z+u)\,dz$ at the centers of the disks. It is 2 at $(0,0)$, 1 at $(0,2)$, and 2 at $(2,1)$. In the chart at infinity, with $t=z/u$, $w=1/u$, the basis becomes $(-w,-t,-1)\,dt/(1+3(1+t)w^2)$, and the value at the center is $-1\equiv2$. All four values are nonzero. \end{proof} \begin{proof}[Proof of Proposition~\ref{prop:1.1.2}] Take one disk, and let $t$ be a local parameter that vanishes at the known rational point $P$ in it. At points of the disk $t\in3\Z_3$, so we write $t=3s$, $s\in\Z_3$. By Lemma~\ref{lem:1.1.2:chab}, $\omega=\sum_{k\ge0}a_kt^k\,dt$ with $a_0$ a 3-adic unit. Integrating from $P$ gives \[ \int\omega=\sum_{k\ge1}\frac{a_{k-1}}{k}(3s)^{k} \] and here the coefficient of the linear term in $s$ has valuation exactly 1. The coefficients of the terms with $k\ge2$ have valuation at least $k-v_3(k)\ge2$. The constant term is 0. By Strassmann's theorem this integral has at most one zero in the disk ($s=0$). By \cite{Coleman85,MP12}, every rational point in the disk is a zero of this integral, so each disk contains at most one rational point. Each of the four disks contains one known point, so $C(\Q)$ consists of these four points only. \end{proof} \subsection{\texorpdfstring{$x^4+y^3+xy+1=0$ \ ($H=29$)}{x^4+y^3+xy+1=0 (H=29)}}\label{sec:1.1.3} \begin{thm}\label{thm:1.1.3} The only integer solutions of $x^4+y^3+xy+1=0$ are $(x,y)=(0,-1),(1,-1)$. \end{thm} \paragraph{Setup.} Let $J$ denote the Jacobian of the smooth projective model of the curve. We do not change coordinates. \[ C:\quad F(x,y)=x^4+y^3+xy+1=0,\qquad P_0=(0,-1),\quad P_1=(1,-1). \] The projective closure $X^4+Y^3Z+XYZ^2+Z^4=0$ has a single point at infinity, $O=(0:1:0)$. In the chart $t=x/y$, $w=1/y$ the equation is $t^4+w+tw^2+w^4=0$; at $O$ the derivative with respect to $w$ is 1, and the tangent line $w=0$ meets the curve as $t^4=0$, so $O$ is a smooth rational hyperflex. In characteristic other than 2, 3, 5, an affine singular point satisfies, from $F_x=4x^3+y=0$, $F_y=3y^2+x=0$, $F=0$, the relations $y=-4x^3$, $x^5=-1/48$, $x^4=3/5$, that is, $x=-5/144$. Hence the characteristic divides $3\cdot144^4-5^5=17^2\cdot4463467$ (4463467 is prime). In characteristics 2 and 3 the only point at which $F_x$, $F_y$ vanish is $(0,0)$, but $F(0,0)=1$; in characteristic 5 and at infinity the curve is also smooth (\code{1.1.3a}). At the two bad primes the only singular point is the reduction of $(x_c,y_c)=(-5/144,\,-4x_c^3)$, and it is an ordinary node with unit Hessian. At a lift of the critical point the valuation of $F$ is 2 at 17 and 1 at 4463467, so the thickness of the node is 2 and 1, respectively (\code{1.1.3a}). A plane quartic with only one node is geometrically irreducible (two distinct components meet at least three times, counted with multiplicity). So at 4463467, where the node has thickness 1, the component group of the Jacobian is trivial. At 17 the component group may be nontrivial, so we put 17 into $S$. The proof has four steps. We determine the bitangent field $K$ (Step 1), show $\Cl(K)[2]=0$ (Step 2), obtain $\operatorname{rank}J(\Q)\le2$ by 2-descent (Step 3), and finish with Chabauty--Coleman at $p=3$ (Step 4). \paragraph{Step 1: the bitangent field.} If the line $y=mx+b$ is a bitangent, then the slope $m$ is a root of \[ \begin{aligned} p_K(m)={}&m^{27}-48m^{22}+144m^{17}+17280m^{15}+832m^{12}+64512m^{10}\\ &-2560m^7+27648m^5-110592m^3-1024m^2+4096 \end{aligned} \] $p_K$ is irreducible, and for each root $m$ there is one $b\in\Q(m)$ for which $F(x,mx+b)$ is the square of a monic quadratic (\code{1.1.3a}). So the 27 finite bitangents form a single Galois orbit, and the remaining one is the tangent line at $O$. $K=\Q[m]/(p_K)$ has signature $(3,12)$ and discriminant $D=2^{20}\cdot4463467^6$. The prime ideals above 2 have $(e,f)$ equal to $(1,1),(1,1),(5,1),(5,1),(5,1),(5,2)$, and there are 9 prime ideals above 17. The descent value of an affine point is $-(y-mx-b)$, and its leading coefficient at $O$ is 1. \paragraph{Step 2: the class group.} \begin{lem}\label{lem:1.1.3:class} $\Cl(K)[2]=0$. \end{lem} \begin{proof} We apply Proposition~\ref{thm:B} with $n=27$, $(r_1,r_2)=(3,12)$, $D$ as above, and $T=\log10^6$. The window is a symmetric step function on 64 cells of equal width, with positive integer heights. Two kinds of data are needed. \begin{enumerate}[label=(\roman*)] \item There are $78{,}493$ prime ideals of norm less than $10^6$ in total, and all the counts $a(q)$ by norm are computed (\code{1.1.3b}). \item Each of the $1{,}260$ prime ideals of norm at most $10^4$ contains some $g\in\mathfrak p$ with $|N(g)|=N\mathfrak p$ (\code{1.1.3b}). This is a witness for Lemma~\ref{lem:C} with cofactor 1, so these ideals are principal. So we set $k(q)=a(q)$ if $q\le10^4$, and $k(q)=0$ for larger $q$. \end{enumerate} Computing in 224-bit interval arithmetic gives $B^-(f)>0.0642182715>0$ (\code{1.1.3c}). By Proposition~\ref{thm:B}, $\Cl(K)[2]=0$. \end{proof} \paragraph{Step 3: an upper bound for the rank.} \begin{lem}\label{lem:1.1.3:rank} $\operatorname{rank}J(\Q)\le2$. \end{lem} \begin{proof} By Proposition~\ref{thm:A}, $J(\Q)[2]=0$, and the descent map sends $J(\Q)/2J(\Q)$ injectively into the part of $K(S,2)$, $S=\{2,17\}$, on which the norm is a square. By Lemma~\ref{lem:1.1.3:class}, \[ \dim_{\F_2}K(S,2)=(r_1+r_2)+\#\{\mathfrak p\mid 2\cdot17\}=15+15=30 \] There are 30 explicit elements whose ideals involve only prime ideals above 2 and 17, and the matrix formed by their valuation parities and their quadratic residue symbols at auxiliary primes has rank 30, so they form a basis (\code{1.1.3d}). We impose conditions on this space one after another. (a) The condition that the norm is a square (sign, and parity of the valuations at 2 and 17) brings the dimension to 27. (b) At a good prime $p$ the local image is the image of $J(\Q_p)/2J(\Q_p)\cong J(\F_p)/2J(\F_p)$, and for $p\ge487$ it is generated by the images of the points of $C(\F_p)$. Otherwise there would be a connected étale double cover (of genus 5) in which all $\F_p$-points split, and the Weil bound (\cite{Weil48}) would require $2(p+1-6\sqrt p)\le p+1+10\sqrt p$, so $p<484$. If a coordinate vanishes at a point of tangency with a bitangent, that coordinate is recovered from the condition that the norm is a square. Imposing the conditions at $p=487,491,499,503,509,521$ in turn, the dimension becomes $25,22,16,14,12,7$ (\code{1.1.3d}). (c) At 2. The orbits of the wild inertia group on the bitangents have size a power of 2 that divides the ramification index 1 or 5, so they have size 1. The action on the bitangents is faithful, so the inertia group is tame and cyclic, and its image has order 5. The nontrivial irreducible $\F_2$-representation of the cyclic group of order 5 is 4-dimensional, so the inertia-invariant subspace of $J[2]$ is 2-dimensional. The two bitangents corresponding to the two prime ideals with $(e,f)=(1,1)$ are distinct nonzero vectors fixed by the decomposition group, so they span this 2-dimensional space. Hence $\dim J(\Q_2)[2]=2$ and $\dim J(\Q_2)/2J(\Q_2)=3+2=5$. The images of the five 2-adic points, determined modulo $2^{44}$, \[ \begin{gathered} (2,\ 12109782689869),\quad (4,\ 6831052190827),\quad (3,\ 10504516484706),\\ (477564860141,\ 1),\quad (7164285329697,\ 3). \end{gathered} \] are independent. At each prime ideal $\mathfrak p$ above 2 the square class of a unit is determined modulo $\mathfrak p^{2e+1}$, and the valuation of the approximation error is greater than the valuation of the value plus $2e$, so these square classes are exact. Hence these five images span the whole local image, and this condition brings the dimension from 7 down to 2 (\code{1.1.3d}). Since $J(\Q)[2]=0$, $\operatorname{rank}J(\Q)\le2$. \end{proof} \paragraph{Step 4: Chabauty--Coleman at $p=3$.} $C$ has good reduction at 3. An integral basis of the regular differentials is $(\omega_0,\omega_1,\omega_2)=(1,x,y)\,dx/(3y^2+x)$, or $-(1,x,y)\,dy/(4x^3+y)$ where the denominator vanishes. The only affine points modulo 3 are $(0,2)$ and $(1,2)$, which are the reductions of $P_0$ and $P_1$, respectively. Let $L_i=\log[P_i-O]\in\Q_3^3$. \begin{lem}\label{lem:1.1.3:logs} $\operatorname{rank}J(\Q)=2$. Moreover, $L_0\equiv(0,6,1)\pmod 9$, $v_3((L_0)_0)\ge4$, and there is a vector $T$ with $3L_1+L_0+T=0$, $T\equiv(0,6,0)\pmod9$. The vector $\lambda=(L_0\times T)/3$ lies in $\Z_3^3$, satisfies $\lambda\equiv(1,0,0)\pmod3$, and annihilates all logarithms of $J(\Q)$. \end{lem} \begin{proof} For the line $y=x/3-1$ we have $F(x,x/3-1)=x^3(x+1/27)$, so its divisor is $3P_0+Q-4O$ with $Q=(-1/27,-82/81)$. Since $t(Q)=3/82\in3\Z_3$, $Q$ lies in the disk of $O$, and $3L_0=-\int_O^Q\omega$. In the chart at infinity, $\omega=-(w,t,1)\,dt/(1+2tw+4w^3)$ and $w\in-t^4+t^5\Z_3[[t]]$, so the inverse of the denominator lies in $1+t^5\Z_3[[t]]$. For $t\in3\Z_3$ the integrated term of degree $n$ has valuation at least $n-v_3(n)$. So the first component consists only of terms of degree at least 5, and its valuation is at least 4 even after dividing by $3$; modulo 9, the second component is $t(Q)^2/6\equiv6$ and the third is $t(Q)/3=1/82\equiv1$. The tangent line $y=-(3x+1)/4$ at $P_1$ gives $F(x,-(3x+1)/4)=(x-1)^2(64x^2+101x+63)/64$. The roots of the quadratic factor lie in $\Z_3$, and $a_0\equiv0$, $a_1\equiv7\pmod9$. The corresponding points $A$, $B$ reduce to $P_0$, $P_1$, respectively. From $\operatorname{div}=2P_1+A+B-4O$ we get $3L_1+L_0+T=0$ with $T=\int_{P_0}^A\omega+\int_{P_1}^B\omega$. With the parameter $s=y+1$ at $P_0$ we have $s(A)\equiv3\pmod9$, and the constant term of the differential is $(1,0,-1)\,ds$. With $s=x-1$ at $P_1$ we have $s(B)\equiv6\pmod9$, and the constant term is $(1,1,-1)\,ds/4$. The higher coefficients lie in $\Z_3$, and $n-v_3(n)\ge2$ for $n\ge2$, so $T\equiv(0,6,0)\pmod9$ (\code{1.1.3e}). Since $L_0\times T\equiv(3,0,0)\pmod9$, the vectors $L_0$ and $T$, and hence $L_0$ and $L_1$, are independent. So $\operatorname{rank}J(\Q)\ge2$, and together with Lemma~\ref{lem:1.1.3:rank} the rank is 2. The three components of $L_0\times T$ have valuations 1, 2, 3, so $\lambda\in\Z_3^3$ and $\lambda\equiv(1,0,0)\pmod3$. $\lambda$ annihilates $L_0$ and $T$, and hence $L_1$. Since the rank is 2, $\lambda$ annihilates all logarithms of $J(\Q)$. \end{proof} \begin{lem}\label{lem:1.1.3:disks} The only integral points of $C$ are $P_0$, $P_1$. \end{lem} \begin{proof} An integral point reduces modulo 3 to an affine point, so it lies in the disk of $P_0$ or in the disk of $P_1$. On the disk of $P_0$ we have $F_x\equiv-1$, so $s=y+1$ is an integral parameter; on the disk of $P_1$ we have $F_y\equiv1$, so $s=x-1$ is one. By Lemma~\ref{lem:1.1.3:logs}, $\lambda\cdot\omega=(c_0+c_1s+\cdots)\,ds$ with $c_k\in\Z_3$, and $c_0$ is a unit, since it is $\lambda\cdot(1,0,-1)\equiv1$ at $P_0$ and $\lambda\cdot(1,1,-1)/4\equiv1/4$ at $P_1$. Setting $s=3z$, $z\in\Z_3$, and integrating from the center gives \[ \int\lambda\cdot\omega=3c_0z+\sum_{n\ge2}\frac{3^nc_{n-1}}{n}z^n \] and each term of the sum has valuation at least 2. After division by $3z$ we get a unit on all of $\Z_3$, so the only zero is $z=0$. By \cite{Coleman85,MP12}, the rational points in a disk are zeros of this integral, so the only integral point in each disk is its center. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.1.3}] This follows from Lemma~\ref{lem:1.1.3:disks} and Lemma~\ref{lem:1.1.3:logs}. Substitution shows that the two points are solutions. \end{proof} \subsection{\texorpdfstring{$x^4+y^3+2x-y=0$ \ ($H=30$)}{x^4+y^3+2x-y=0 (H=30)}}\label{sec:1.1.4} \begin{thm}\label{thm:1.1.4} The only integer solutions of $x^4+y^3+2x-y=0$ are $(-8,-16)$, $(0,-1)$, $(0,0)$ and $(0,1)$. \end{thm} \paragraph{Setup.} Let $J$ denote the Jacobian of the smooth projective model of the curve. Setting $u=-y$, $v=x$, the equation becomes \[ C:\quad F(u,v)=v^4-u^3+u+2v=0 \] and the integer solutions correspond bijectively to the integral points of $C$ via $(x,y)=(v,-u)$. The known points are \[ P_0=(0,0),\quad P_1=(1,0),\quad P_{-1}=(-1,0),\quad P_*=(16,-8) \] which in the original coordinates are $(0,0),(0,-1),(0,1),(-8,-16)$, respectively. There is a single point at infinity, $O$. In the chart $t=v/u$, $w=1/u$ the equation is $t^4-w+(1+2t)w^3=0$, $w=t^4+\cdots$, so $O$ is a smooth rational hyperflex. Let $G_i=[P_i-O]$. Since $\operatorname{div}(u+2v)=3P_0+P_*-4O$ and $\operatorname{div}(v)=P_{-1}+P_0+P_1-3O$, we have $G_*=-3G_0$, $G_{-1}=-G_0-G_1$. In odd characteristic other than 3, an affine singular point satisfies $3u^2=1$, $2v^3=-1$, $12F-4uF_u-3vF_v=8u+18v=0$, and hence $243v^2-16=0$. We have $\operatorname{Res}(243v^2-16,\,2v^3+1)=14332523=83\cdot172681$. In characteristic 3, $F_u=1-3u^2\equiv1$, and the point at infinity is smooth in every odd characteristic. At 83 and 172681 the only singular point is the reduction of $(u_c,v_c)=(2187/128,-243/32)$, which is an ordinary node, and since $F(u_c,v_c)=-3482803089/2097152$ has valuation 1, its thickness is 1. So the two fibers are geometrically irreducible and the component groups are trivial (\code{1.1.4a}). Modulo 2 the plane model is singular at $(1,0)$. Therefore in Step 4 we use explicit analytic disks instead of 2-adic residue disks. The proof has six steps. We determine the bitangent field (Step 1), show $\Cl(K)[2]=0$ (Step 2), and show by 2-descent that the rank is at most 2 (Step 3). We fix 2-adic analytic disks covering the integral points (Step 4), obtain from 2-adic logarithms that the rank is 2, together with the logarithm lattice (Step 5), and determine the integral points disk by disk (Step 6). \paragraph{Step 1: the bitangent field.} If the line $u=mv+\eta$ is a bitangent, then $m$ is a root of \[ p_K(m)=m^{27}-480m^{19}+2688m^{18}-34560m^{11}-141312m^{10}-162816m^{9}+16384m^{3}-49152m+32768 \] and for each root there is one $\eta\in\Q(m)$ for which $F(mv+\eta,v)$ is the square of a monic quadratic (\code{1.1.4a}). Since $p_K$ is irreducible, the 27 finite bitangents form a single Galois orbit, and the remaining one is the tangent line at $O$. $K=\Q[m]/(p_K)$ has signature $(3,12)$ and discriminant $D=2^{28}\cdot83^6\cdot172681^6$, and the three prime ideals above 2 have $(e,f)$ equal to $(1,1),(10,1),(16,1)$. The descent value is $u-mv-\eta$ (with leading coefficient 1 at $O$), and at $G_0$, $G_1$ it is $-\eta$, $1-\eta$, respectively. \begin{lem}\label{lem:1.1.4:desc} $J(\Q)[2]=0$, and $\operatorname{Sel}^{(2)}(J/\Q)$ maps injectively into $K(\{2\},2)$. Moreover, $\dim_{\F_2}J(\Q_2)[2]\le1$ and $\dim_{\F_2}J(\Q_2)/2J(\Q_2)\le4$. \end{lem} \begin{proof} The first claim is Proposition~\ref{thm:A}, and $S=\{2\}$ because the component groups at the odd bad primes are trivial. The codeword given by an element of $J(\Q_2)[2]$ is invariant under the decomposition group, so its support is a union of orbits, which have sizes $1,10,16$. Nonzero codewords have weight 12 or 16, so the only possibility is the single orbit of size 16. Hence there is at most one nonzero invariant element, and $\dim J(\Q_2)/2J(\Q_2)=g+\dim J(\Q_2)[2]\le4$. \end{proof} \paragraph{Step 2: the class group.} \begin{lem}\label{lem:1.1.4:class} $\Cl(K)[2]=0$. \end{lem} \begin{proof} We apply Proposition~\ref{thm:B} with $n=27$, $(r_1,r_2)=(3,12)$, $D$ as above, and $T=\log(5\cdot10^8)$. The window is a symmetric step function on 64 cells of equal width, with positive integer heights. Two kinds of data are needed. \begin{enumerate}[label=(\roman*)] \item There are $26{,}362{,}584$ prime ideals of norm less than $5\cdot10^8$ in total, and all the counts $a(q)$ by norm are computed (\code{1.1.4b}). \item All $150{,}219$ prime ideals of norm at most $2.02\cdot10^6$ have a witness as in Lemma~\ref{lem:C} (\code{1.1.4c}). So we set $k(q)=a(q)$ if $q\le2.02\cdot10^6$, and $k(q)=0$ for larger $q$. \end{enumerate} Computing with directed-rounding interval arithmetic gives $B^-(f)>1.0015\cdot10^{-4}>0$ (\code{1.1.4d}). By Proposition~\ref{thm:B}, $\Cl(K)[2]=0$. \end{proof} \paragraph{Step 3: an upper bound for the rank.} \begin{lem}\label{lem:1.1.4:rank} $\operatorname{rank}J(\Q)\le2$. \end{lem} \begin{proof} By Lemma~\ref{lem:1.1.4:class}, $\dim K(\{2\},2)=r_1+r_2+3=18$. There are 18 explicit elements whose ideals involve only prime ideals above 2, and their square classes in the three completions above 2 alone already have rank 18, so they form a basis (\code{1.1.4e}). We impose conditions on this space one after another. (a) Sign of the norm and parity of the 2-adic valuation of the norm: dimension 16. (b) Images of the points of $C(\F_p)$ at the good primes $p=487,491,499,503,509,521,523$: dimensions $12,9,6,4,4,4,4$ (\code{1.1.4e}). For $p\ge487$ the images of these points generate the whole local image. Otherwise there would be a connected étale double cover (of genus 5) in which all $\F_p$-points split, which contradicts the Weil bound (\cite{Weil48}). The one coordinate that is missing at a point of tangency with a bitangent is recovered from the norm condition. (c) At 2: the images of the 2-adic points $(0,0)$, $(1,0)$, the point with $v=2$ and $u$ even ($u\equiv29760316588\pmod{2^{36}}$), and the point with $t=2$ in the chart at infinity ($w\equiv59134464016\pmod{2^{36}}$) are independent in the three completions above 2 (\code{1.1.4e}). In each approximation the valuation of the error is greater than the valuation of the value plus $2e$, so the square classes are exact. By Lemma~\ref{lem:1.1.4:desc} these four span the whole local image, and this condition brings the dimension from 4 down to 2. Since $J(\Q)[2]=0$, $\operatorname{rank}J(\Q)\le2$. \end{proof} \paragraph{Step 4: 2-adic analytic disks.} \begin{lem}\label{lem:1.1.4:charts} An integral point $(u,v)$ of $C$ lies in one of the three disks of (i), (ii) below. (i) and (ii) are analytic parametrizations that converge on a disk larger than the closed unit disk, and (iii) is one that converges on the open unit disk $|t|<1$. \begin{enumerate}[label=(\roman*)] \item Even disk: $v=2t$, $t\in\Z_2$, $u=u_0(v)$. Here $u_0\in\Z_2[[v]]$ is the solution of $u-u^3=-2v-v^4$, $u_0(0)=0$. \item Odd disks ($\varepsilon=\pm1$): $u=\varepsilon+4r$, $v=4t$, $t\in\Z_2$. Here $r=r_\varepsilon(t)\in\Z_2[[t]]$ is the solution of $r+6\varepsilon r^2+8r^3=t+32t^4$, $r(0)=0$, and $v_2([t^j]r)\ge j-1$. \item Disk at infinity: $t=v/u$, $w=1/u$, and $w=w(t)\in t^4+t^5\Z_2[[t]]$ is the solution of $t^4-w+(1+2t)w^3=0$. \end{enumerate} \end{lem} \begin{proof} Modulo 2, $v^4\equiv u^3-u\equiv0$, so $v$ is even. If $u$ is even, the derivative of $u-u^3$ is a unit at $u\equiv0$, so $u=u_0(v)$ by Hensel's lemma. If $u$ is odd, then $8\mid u-u^3$, so $v^4+2v\equiv0\pmod8$ gives $4\mid v$. Dividing the equation by $-8$ gives the equation of (ii), and its derivative with respect to $r$ is a unit, so the solution is unique. The coefficient inequality follows by induction from $r=t+32t^4-6\varepsilon r^2-8r^3$. (iii) holds because the derivative with respect to $w$ is $-1$. \end{proof} \paragraph{Step 5: 2-adic logarithms.} We use $\omega=(\omega_0,\omega_1,\omega_2)=(1,v,u)\,dv/(1-3u^2)$ as a basis of the regular differentials, and let $L_i=\log G_i\in\Q_2^3$. All integrals below are computed as antiderivatives of power series on the analytic disks of Lemma~\ref{lem:1.1.4:charts}, and their values equal the abelian integrals (\cite[Section 3]{KRZB16}). \begin{lem}\label{lem:1.1.4:logs} (1) $v_2(L_0)=(1,3,4)$, $v_2(L_1)=(-1,1,1)$, and $\Delta=L_{0,0}L_{1,2}-L_{1,0}L_{0,2}$ has valuation 2. Hence $\operatorname{rank}J(\Q)=2$. (2) The annihilating vector $a$ of $L_0,L_1$, normalized by $a_1=1$, lies in $\Z_2^3$, and $a\equiv(4,1,2)\pmod8$. (3) $\log J(\Q)\subseteq\Z_2L_0+\Z_2L_1$. \end{lem} \begin{proof} (1) From $\operatorname{div}(u+2v)$ we get $4L_0=-\int_{P_0}^{P_*}\omega$, and $P_*$ lies in the same even disk as $P_0$ ($t=-4$). The function of pole order 10 \[ g_{10}=-13043u^2+(27304v^2+22862v+19496)u-5780v^3-47008v^2-16272v-6453 \] satisfies $\operatorname{div}g_{10}=7P_1+Q_1+Q_2+Q_3-10O$. $Q_1$ lies in the odd disk of $P_1$, and $Q_2,Q_3$ are conjugate over a 2-adic quadratic extension and lie in the disk at infinity ($v_2(t)=3/2$, $v_2(w)=6$). Hence $8L_1=-\bigl(\int_{P_1}^{Q_1}\omega+\operatorname{Tr}\int_O^{Q_2}\omega\bigr)$. Computing with the valuations of all discarded tails bounded, we obtain the valuations in (1) (\code{1.1.4f}). The two products in $\Delta$ have valuations 2 and 3, so $v_2(\Delta)=2$, and $L_0,L_1$ are independent. So the rank is at least 2, and together with Lemma~\ref{lem:1.1.4:rank} it is 2. (2) is the same computation (\code{1.1.4f}). (3) In Step 3 (c) the 2-adic images of $G_0,G_1$ are independent, so the two are independent in $J(\Q)/2J(\Q)$. The torsion of $J(\Q)$ has odd order ($J(\Q)[2]=0$) and the rank is 2, so $\langle G_0,G_1\rangle$ has odd index modulo torsion. The logarithm of a torsion point is 0, so (3) follows. \end{proof} \paragraph{Step 6: integral points.} \begin{lem}\label{lem:1.1.4:even} The only integral points in the even disk are $P_0$ and $P_*$. \end{lem} \begin{proof} Let $E(t)$ be the integral from $t=0$ of the pullback of $a\cdot\omega$. At rational points, $E(t)=a\cdot(\log[P(t)-O]-L_0)=0$. The coefficients are \[ [t]E=2a_0,\quad[t^2]E=2a_1-4a_2,\quad[t^3]E=32a_0,\quad[t^4]E=48a_1-128a_2 \] and by Lemma~\ref{lem:1.1.4:logs}(2) their valuations are $3,1,7,4$. For $m\ge5$, since the coefficients of $u_0$ are integers, $v_2([t^m]E)\ge m-v_2(m)\ge5$. The minimal valuation occurs only in degree 2 (\code{1.1.4g}), so by Strassmann's theorem $E$ has at most two zeros in $\Z_2$. These two are $t=0$ ($P_0$) and $t=-4$ ($P_*$). \end{proof} \begin{lem}\label{lem:1.1.4:odd} The only integral point in the odd disk $\varepsilon$ is $P_\varepsilon$. \end{lem} \begin{proof} Since $1-3u^2=-2D_\varepsilon$ with $D_\varepsilon=1+12\varepsilon r+24r^2$, the pullback of $\omega$ is $-2(1,4t,\varepsilon+4r)\,dt/D_\varepsilon$, and $v_2([t^j]D_\varepsilon^{-1})\ge j+1$ ($j\ge1$). Let $H_\varepsilon$ ($H_\varepsilon(0)=0$) be the antiderivative of $a\cdot\omega$ divided by $-4$. Then $H_\varepsilon'=(C_\varepsilon/2+2a_1t+2a_2r)/D_\varepsilon$ with $C_\varepsilon=a_0+\varepsilon a_2$, and $v_2(C_\varepsilon)=1$. So $[t]H_\varepsilon=C_\varepsilon/2$ and $[t^2]H_\varepsilon=a_1+a_2-3\varepsilon C_\varepsilon$ are odd, and for $m\ge3$, $v_2([t^m]H_\varepsilon)\ge m-v_2(m)\ge2$. That is, $H_\varepsilon\equiv t+t^2\pmod2$ (\code{1.1.4g}). By Strassmann's theorem and Hensel's lemma, the zeros of $H_\varepsilon$ in $\Z_2$ are one with $t$ even ($t=0$) and one with $t$ odd. Suppose that $P(t)$ is a rational point for an odd $t$. For $I=\int_{P_\varepsilon}^{P(t)}\omega$, the coefficient inequalities above give $I_0/(-2)\equiv t$, $I_2/(-2)\equiv\varepsilon t\pmod2$, so $v_2(I_0)=v_2(I_2)=1$. By Lemma~\ref{lem:1.1.4:logs}(3), $I=b_0L_0+b_1L_1$ with $b_k\in\Z_2$. But in Cramer's formula $b_0=(I_0L_{1,2}-L_{1,0}I_2)/\Delta$ the two terms of the numerator have valuations 2 and 0, so $v_2(b_0)=0-2=-2$. This is a contradiction. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.1.4}] By Lemmas~\ref{lem:1.1.4:charts}, \ref{lem:1.1.4:even}, \ref{lem:1.1.4:odd}, the only integral points of $C$ are $P_0,P_1,P_{-1},P_*$, and under $(x,y)=(v,-u)$ they become the four solutions in the theorem. The rank follows from Lemma~\ref{lem:1.1.4:logs}(1). \end{proof} \subsection{\texorpdfstring{$x^4+y^3+2x+y=0$ \ ($H=30$)}{x^4+y^3+2x+y=0 (H=30)}}\label{sec:1.1.5} \begin{thm}\label{thm:1.1.5} The only integer solutions of $x^4+y^3+2x+y=0$ are $(x,y)=(0,0),(8,-16)$. \end{thm} \paragraph{Setup.} Put $u=y$, $z=x$. Then the equation becomes \[ C:\quad F(u,z)=u^3+u+z^4+2z=0 \] The projective completion $z^4+u^3w+uw^3+2zw^3=0$ is a smooth plane quartic over $\Q$ (genus 3). At the point at infinity $O=(1:0:0)$ the line $w=0$ meets the curve with multiplicity 4, since $z^4=0$; so $O$ is a rational hyperflex. Near $O$ put $t=z/u$, $w=1/u$. Then the equation is \[ w+t^4+(1+2t)w^3=0 \] The partial derivative with respect to $w$ is 1 at $(t,w)=(0,0)$, so $t$ is a local parameter, and $\operatorname{ord}_Oz=-3$, $\operatorname{ord}_Ou=-4$. $J$ is the Jacobian of $C$. \begin{prop}\label{prop:1.1.5} The only integral points of $C$ are $P=(0,0)$ and $R=(-16,8)$. \end{prop} Since $(x,y)=(z,u)$, the points $P,R$ are $(x,y)=(0,0),(8,-16)$, and Theorem~\ref{thm:1.1.5} is equivalent to Proposition~\ref{prop:1.1.5}. This section does not determine all of $C(\Q)$. Modulo 2 the point $(u,z)=(1,0)$ is singular, so $C$ has bad reduction at 2. Therefore we argue in a single 2-adic disk that contains all integral points. The proof has five steps. We fix the bitangent field $K$, show $\Cl(K)[2]=0$ (Lemma~\ref{lem:1.1.5-class}), and obtain $\operatorname{rank}J(\Q)\le2$ by 2-descent (Lemma~\ref{lem:1.1.5-rank}). Then we construct an annihilating differential from 2-adic logarithms (Lemma~\ref{lem:1.1.5-log}) and apply Strassmann's theorem on the disk that contains all integral points (Lemma~\ref{lem:1.1.5-disk}). \paragraph{Step 1: the bitangent field and bad primes.} The 27 bitangents other than the tangent line at $O$ have the form $u=mz+b$, and the slope $m$ is a root of \[ \begin{aligned} g(m)={}&m^{27}+480m^{19}-2688m^{18}-34560m^{11}-141312m^{10}\\ &-162816m^{9}-16384m^{3}+49152m-32768 \end{aligned} \] The intercept $b$ is an explicit element of $\Q(m)$, and substituting $u=mz+b$ into $F$ gives the square of a monic quadratic with nonzero discriminant (\code{1.1.5a}). Since $g$ is irreducible, the 27 bitangents form one Galois orbit. The field $K=\Q[m]/(g)$ has signature $(3,12)$ and discriminant $D=2^{28}\cdot14365291^{6}$. There are three prime ideals above 2; their ramification indices are $1,10,16$, and all residue degrees are 1. We find the bad primes of $C$. Because of the identity \[ 12F-4uF_u-3zF_z=8u+18z \] a singular point in odd characteristic satisfies $u=-9z/4$. In characteristic 3 we have $F_u\equiv1$, so there is no singular point. In characteristic different from 2 and 3, $F_u=3u^2+1=0$ and $F_z=4z^3+2=0$ give $z^2=-16/243$, $z^3=-1/2$, and hence $z=243/32$. For this value to satisfy $z^2=-16/243$, the number $243^3+2^{14}=14365291$ must be 0. The number $q=14365291$ is prime. Modulo $q$ the only singular point is the reduction of $(u_c,z_c)=(-2187/128,\,243/32)$, and at this point the $q$-adic valuations of $(F,F_u,F_z,\det\operatorname{Hess}F)$ are $(1,1,1,0)$. Hence the singular fiber has only one singular point, an ordinary double point (node) of thickness 1. This fiber is geometrically irreducible: if it were reducible, Bézout's theorem would force further intersection points between the components. The component group of the Néron model is trivial, so the descent image is unramified at $q$, and $S=\{2\}$ suffices in Proposition~\ref{thm:A}. \paragraph{Step 2: the class group.} \begin{lem}\label{lem:1.1.5-class} $\Cl(K)[2]=0$. \end{lem} \begin{proof} We apply Proposition~\ref{thm:B} with $n=27$, $(r_1,r_2)=(3,12)$, the $D$ above, and $T=\log10^8$. The window is a step function on 64 cells of equal width, and its heights are positive integers, symmetric from left to right (the input of \code{1.1.5d}). Two pieces of data are needed. \begin{enumerate}[label=(\roman*)] \item There are $5{,}764{,}533$ prime ideals of norm less than $10^8$ in total, and we count the number $a(q)$ for every norm (\code{1.1.5b}). \item Every one of the $412{,}122$ prime ideals of norm at most $6\cdot10^6$ has a witness for Lemma~\ref{lem:C} (\code{1.1.5c}). Therefore we set $k(q)=a(q)$ if $q\le6\cdot10^6$, and $k(q)=0$ otherwise. \end{enumerate} With these data the computation gives $B^-(f)>0.000226856>0$ (\code{1.1.5d}). The computation uses exact rational coefficients and directed interval arithmetic, so rounding errors cannot change the conclusion. By Proposition~\ref{thm:B}, $\Cl(K)[2]=0$. \end{proof} \paragraph{Step 3: an upper bound for the rank.} \begin{lem}\label{lem:1.1.5-rank} $\operatorname{rank}J(\Q)\le2$. \end{lem} \begin{proof} The hypotheses of Proposition~\ref{thm:A} were checked in the setup and in Step 1. Hence $J(\Q)[2]=0$, and the bitangent function $h=-(u-mz-b)$, normalized to have leading coefficient 1 at $O$, gives an injective homomorphism $\delta:J(\Q)/2J(\Q)\to K^\times/K^{\times2}$. By Lemma~\ref{lem:1.1.5-class}, \[ \dim_{\F_2}K(S,2)=(r_1+r_2)+\#\{\mathfrak p\mid2\}=15+3=18 \] Eighteen explicit elements supported on $S$ form a basis of this space, because the matrix formed from their 2-adic valuations and their quadratic characters at the primes 7, 11, 13, 17 has rank 18 (\code{1.1.5a}). We impose conditions on this space one at a time. The condition that the norm is a square reduces the dimension to 16. Imposing in turn the local conditions at the good primes 487, 491, 499, 503 reduces the dimension to $11,9,6,4$. At each prime the local image is generated by the values of $h$ computed at all points of $C(\F_p)$ (Proposition~\ref{thm:S}). At 2, the bitangent code maps $J[2]$ Galois-equivariantly and injectively into $\F_2^{27}$, and the nonzero codewords have weights 12 and 16. An element of $J(\Q_2)[2]$ goes to a codeword that is constant on each local orbit, and the local orbits have sizes $1,10,16$. Among the unions of these orbits, the only one of weight 12 or 16 is the single orbit of size 16. Hence $\dim J(\Q_2)[2]\le1$ and $\dim J(\Q_2)/2J(\Q_2)=3+\dim J(\Q_2)[2]\le4$. The images of the following four local classes are independent (\code{1.1.5a}). \begin{enumerate}[label=(\alph*)] \item $[P-O]$; \item $[Q-O]$ for the point $Q$ with $t=2$ in the chart at infinity; \item $[D_1-2O]$, where $D_1$ is the sum of the $\Q_2(\sqrt2)$-point with $z=\sqrt2$ and its conjugate; \item $[D_2-2O]$, where $\theta^2+\theta+1=0$ and $D_2$ is the sum of the $\Q_2(\theta)$-point with $z=\theta$ and its conjugate. \end{enumerate} The coordinates of the points are Hensel lifts modulo $2^{36}$. The relative valuation that this error contributes to the values of $h$ exceeds twice the ramification index at each 2-adic prime ideal, so the local square classes are determined. The local image has dimension at most 4 and contains four independent elements, so the 2-adic local image is the space spanned by these four elements. In the 4-dimensional candidate space above, the subspace whose restriction at 2 lies in this image is 2-dimensional. Since $\delta$ is injective and $J(\Q)[2]=0$, we get $\operatorname{rank}J(\Q)\le2$. \end{proof} \paragraph{Step 4: 2-adic logarithms.} Let $D$ be the rational divisor of degree 2 defined by $z^2-z+1=0$, $u=1-z$. Since $F(1-z,z)=(z^2-z+1)(z^2+2)$, the divisor $D$ lies on $C$. We use the basis of holomorphic differentials \[ \boldsymbol\omega=(\omega_0,\omega_1,\omega_2)=(1,z,u)\frac{dz}{3u^2+1} \] The $P$-disk consists of the points with $|z|_2<1$, $u=u(z)$. Here $u(z)\in\Z_2[[z]]$ is the unique solution with $u(0)=0$ ($F_u(0,0)=1$). On this disk the coefficients of $\boldsymbol\omega$ lie in $\Z_2$. On the $O$-disk we have $|t|_2<1$, $w=w(t)\in\Z_2[[t]]$, and $\boldsymbol\omega=-(w,t,1)\,dt/(1+3(1+2t)w^2)$, so again the coefficients are integral. Both disks are analytic open disks, so Abelian integrals on them are equal to formal antiderivatives (\cite[Section 3]{KRZB16}). We write $\log X\in\Q_2^3$ for the logarithm of $X\in J(\Q_2)$. \begin{lem}\label{lem:1.1.5-log} \begin{enumerate}[label=(\arabic*)] \item $\operatorname{div}(u+2z)=3P+R-4O$. Moreover, there is a rational function $h_D$ with $\operatorname{div}(h_D)=14D+E_D-31O$, where $E_D$ is an effective divisor of degree 3, irreducible over $\Q$. \item Modulo $2^{16}$, $L_P=\log[P-O]$ and $L_D=\log[D-2O]$ are \[ L_P\equiv(510,\,3064,\,32784),\qquad L_D\equiv(12669,\,59033,\,38772) \] \item $\operatorname{rank}J(\Q)=2$. Moreover, there is a primitive vector $\lambda=(a_0,a_1,a_2)\in\Z_2^3$ with $\lambda\equiv(0,4,1)\pmod8$ such that $\lambda\cdot\boldsymbol\omega$ annihilates all logarithms of $J(\Q)$. \end{enumerate} \end{lem} \begin{proof} (1) Substituting $u=-2z$ gives $F(-2z,z)=z^3(z-8)$, and $u+2z$ has a pole of order 4 at $O$. Hence the first equality holds. We find $h_D$ in the basis $z^au^b$ ($b\le2$, $3a+4b\le31$; 29 elements) of $L(31O)$ from the condition that it vanishes to order 14 at the two points of $D$; the solution space is 1-dimensional. The coefficient of $uz^9$, the unique monomial of weight 31, is nonzero, so the pole has order exactly 31. Further, $\operatorname{Res}_u(F,h_D)=c\,(z^2-z+1)^{14}q_D(z)$, where $q_D$ is an irreducible cubic. At the roots of $q_D$ the greatest common divisor of $F$ and $h_D$ is linear in $u$, so the $u$-coordinates of $E_D$ are also determined. Hence the divisor equality accounts for all zeros and poles (\code{1.1.5e}). (2) $R$ is the point of the $P$-disk with $z=8$. From the first equality in (1), $4[P-O]+[R-P]=0$, so \[ L_P=-\tfrac14\int_P^R\boldsymbol\omega \] The three roots of $q_D$ have distinct 2-adic valuations $1,2,-6$, so all of them lie in $\Q_2$. The first two points lie in the $P$-disk, and the third lies in the $O$-disk ($v_2(t)=2$). The class of the base divisor $2P+O$ is $2[P-O]$. Let $T_D$ be the sum of the tiny integrals from $P,P,O$ to the three points of $E_D$. Then the second equality in (1) gives \[ 14L_D=-T_D-2L_P \] The differentials in the charts have integral coefficients and the parameters of the endpoints have valuation at least 1, so the $n$th term of the integral series has valuation at least $n-v_2(n)$. We compute 60 terms; the discarded terms have valuation at least 55, and the values modulo $2^{16}$ above are determined even after division by 4 and by 14 (\code{1.1.5e}). (3) The valuations of the components of $L_P$ are $(1,3,4)$, and those of $L_D$ are $(0,0,2)$. Since $L_P/2\equiv(1,0,0)$ and $L_D\equiv(1,1,0)\pmod2$ are independent, $[P-O]$ and $[D-2O]$ have infinite order and are independent. Together with Lemma~\ref{lem:1.1.5-rank}, this gives $\operatorname{rank}J(\Q)=2$. The cross product of $L_P/2$ and $L_D$ is an integral vector whose third component is a unit. Let $\lambda$ be this vector divided so that its third component is 1; then $\lambda\equiv(0,4,1)\pmod8$ (\code{1.1.5e}). The vector $\lambda$ annihilates the two logarithms, the subgroup generated by the two classes has finite index in $J(\Q)$, and the logarithm of a torsion element is 0. Therefore $\lambda\cdot\boldsymbol\omega$ annihilates all logarithms of $J(\Q)$. \end{proof} \paragraph{Step 5: the disk containing the integral points.} \begin{lem}\label{lem:1.1.5-disk} All integral points of $C$ lie in the $P$-disk. That is, they can be written as $z=2t$, $t\in\Z_2$, $u=u(2t)$. \end{lem} \begin{proof} Since $u^3+u\equiv u(u+1)^2\equiv0\pmod2$, we have $z^4\equiv0$ modulo 2, that is, $z$ is even. If $u$ is odd, then $u^3+u=u(u^2+1)\equiv2\pmod4$, while $z^4+2z\equiv0\pmod4$, a contradiction. Hence both $u$ and $z$ are even. For each even $z$, the equation $F(u,z)=0$ has only one even solution by Hensel's lemma ($F_u\equiv1\pmod2$), and this solution is $u(z)$. \end{proof} \begin{proof}[Proof of Proposition~\ref{prop:1.1.5}] On the $P$-disk let \[ \Phi(t)=\int_P^{(u(2t),\,2t)}\lambda\cdot\boldsymbol\omega=\sum_{n\ge1}c_nt^n \] The constant term is 0. The exact Taylor expansion gives \[ c_1=2a_0,\qquad c_2=2a_1-4a_2,\qquad c_3=-32a_0,\qquad c_4=-48a_1+128a_2 \] (\code{1.1.5f}). Since $\lambda\equiv(0,4,1)\pmod8$, their valuations are, in order, at least 4, exactly 2, at least 8, and exactly 6. Since the coefficients of $\boldsymbol\omega$ in $z$ are integral, for $n\ge5$ the valuation of $c_n$ is at least $n-v_2(n)\ge5$. Hence $c_2$ is the only coefficient of smallest valuation, and by Strassmann's theorem $\Phi$ has at most two zeros in $\Z_2$. If an integral point $Q$ lies at $t=t_Q$, then $[Q-P]\in J(\Q)$, and by \cite[Section 3]{KRZB16}, $\Phi(t_Q)$ is equal to $\lambda\cdot\log[Q-P]$. By Lemma~\ref{lem:1.1.5-log}(3) this value is 0 (the argument of \cite{Coleman85,MP12}). The points $P$ and $R$ give the distinct zeros $t=0$ and $t=4$. By Lemma~\ref{lem:1.1.5-disk} all integral points lie in this disk, so there are no other integral points. \end{proof} \subsection{\texorpdfstring{$x^4+y^3+y^2+x=0$ \ ($H=30$)}{x^4+y^3+y^2+x=0 (H=30)}}\label{sec:1.1.6} \begin{thm}\label{thm:1.1.6} The equation $x^4+y^3+y^2+x=0$ has only the following integer solutions: $(x,y)=(-1,-1),(-1,0),(0,-1),(0,0),(2,-3)$. \end{thm} \paragraph{Setup.} Put $u=y$, $z=x$. Then the equation becomes \[ C:\quad F(u,z)=u^3+u^2+z^4+z=0 \] The projective completion $u^3w+u^2w^2+z^4+zw^3=0$ is a smooth plane quartic (genus 3), and the point at infinity $\infty=(1:0:0)$ is a rational hyperflex. With $t=z/u$, $w=1/u$, the equation near $\infty$ is $w+w^2+tw^3+t^4=0$, and $t$ is a local parameter. $C$ has good reduction at 2, because $F_z=1+4z^3$ is a unit in the affine chart and $F_w=1+2w+3tw^2$ is a unit in the chart at infinity. An integral basis of holomorphic differentials is \[ \boldsymbol\omega=(\omega_0,\omega_1,\omega_2)=(1,z,u)\frac{dz}{3u^2+2u}=-(1,z,u)\frac{du}{1+4z^3}=-(w,t,1)\frac{dt}{1+2w+3tw^2} \] and in the last two expressions the denominators are units on the 2-adic residue disks. \begin{prop}\label{prop:1.1.6} $C(\Q)=\{\infty,P_0,P_1,A,B,R\}$, where $P_0=(0,0)$, $P_1=(0,-1)$, $A=(-1,0)$, $B=(-1,-1)$, $R=(-3,2)$. \end{prop} Since $(x,y)=(z,u)$, the five affine points are $(x,y)=(0,0),(-1,0),(0,-1),(-1,-1),(2,-3)$, and Theorem~\ref{thm:1.1.6} follows directly from Proposition~\ref{prop:1.1.6}. The proof has six steps. We obtain $\operatorname{rank}J(\Q)\le2$ from $\Cl(K)[2]=0$ (Lemma~\ref{lem:1.1.6-class}) and 2-descent (Lemma~\ref{lem:1.1.6-rank}), and we compute 2-adic logarithms using principal divisors to obtain saturation at 2 and an annihilator (Lemmas~\ref{lem:1.1.6-log}, \ref{lem:1.1.6-sat}). 2-adic Chabauty reduces the candidates to 11 (Lemma~\ref{lem:1.1.6-zeros}), and a sieve modulo 23 eliminates the five that remain (Lemma~\ref{lem:1.1.6-sieve}). \paragraph{Step 1: the bitangent field and bad primes.} The 27 bitangents other than the tangent line at $\infty$ have the form $u=mz+\eta(m)$, where $m$ is a root of \[ \begin{aligned} f(M)={}&M^{27}-160M^{19}-1344M^{18}+1280M^{15}-3840M^{11}\\ &+23552M^{10}-40704M^9+4096M^7-8192M^6-4096M-4096 \end{aligned} \] The element $\eta\in\Q(m)$ is given explicitly, and we check exactly the two coefficient conditions which say that $z^4+(mz+\eta)^3+(mz+\eta)^2+z$ is the square of a quadratic (\code{1.1.6a}). Since $f$ is irreducible, the 27 bitangents form one Galois orbit. The field $K=\Q[M]/(f)$ has signature $(3,12)$ and discriminant $D_K=2^{24}\cdot3^6\cdot601^6\cdot857^6$. There is one prime ideal above 2 (ramification index 9, residue degree 3), and there are eight above 3. A singular point satisfies $F_u=u(3u+2)=0$ and $F_z=4z^3+1=0$. In characteristic 2 we have $F_z\equiv1$. In characteristic different from 2 and 3, if $u=0$ then $F=z(z^3+1)=3z/4\ne0$. If $u=-2/3$ then $F=4/27+3z/4$, so $z=-16/81$, and $\operatorname{Res}_Z(81Z+16,4Z^3+1)=515057=601\cdot857$. At 601 and 857 the only singular point is the reduction of $(-2/3,-16/81)$. There the Hessian determinant $-2048/2187$ is a unit, and the value $-2060228/43046721$ of $F$ has valuation 1. Hence there is only one singular point, an ordinary node of thickness 1, the singular fiber is geometrically irreducible, and the component group is trivial. Therefore $S=\{2,3\}$ suffices in Proposition~\ref{thm:A}. \paragraph{Step 2: the class group.} \begin{lem}\label{lem:1.1.6-class} $\Cl(K)[2]=0$. \end{lem} \begin{proof} We apply Proposition~\ref{thm:B} with a window consisting of a single cell. That is, $v$ is the indicator function of an interval of length $T$, and $f(x)=(1-|x|/T)_+$. We take $T=13$ ($e^{13}=442413.4\ldots$). Every one of the $37{,}489$ prime ideals of norm at most $450{,}000$ has an explicit generator $g$ with $(g)=\mathfrak p$ (\code{1.1.6b}; this is the case of Lemma~\ref{lem:C} with cofactor 1). Hence $c_m(q)=a(q)$ in every term of the sum. For this window the archimedean term has a closed form, and \[ B^-(f)=B_0+\frac2T\sum_{j\ \text{odd}}\frac{e^{-jT}}{j^2}+\frac{2\alpha}T\log(1+e^{-T})>B_0, \] where \[ B_0=\gamma+\log4\pi+\alpha-\frac{\pi^2}{4T}-\frac{2\alpha\log2}{T}-\frac{\log D_K}{n}+\frac4n\sum_{q^m0.0264813>0$ (\code{1.1.6c}). By Proposition~\ref{thm:B}, $\Cl(K)[2]=0$. \end{proof} \paragraph{Step 3: an upper bound for the rank.} \begin{lem}\label{lem:1.1.6-rank} $\operatorname{rank}J(\Q)\le2$. \end{lem} \begin{proof} By Proposition~\ref{thm:A}, $J(\Q)[2]=0$, and $h_m=-(u-mz-\eta(m))$, normalized to have leading coefficient 1 at $\infty$, gives an injective homomorphism $\delta:J(\Q)/2J(\Q)\to K^\times/K^{\times2}$. By Lemma~\ref{lem:1.1.6-class}, $\dim K(S,2)=15+9=24$. Twenty-four explicit elements supported on the nine prime ideals above $S$ form a basis. The support is checked by exact ideal equalities, and the independence by the rank 24 of the matrix formed from valuation characters and quadratic characters at good primes (\code{1.1.6a}). The condition that the norm is a square (the sign, and the parity of the valuations at 2 and 3) reduces the dimension to 21. Imposing in turn the local conditions at the good primes $487$, $491$, $499$, $503$, $509$, $521$, $523$ reduces the dimension to $18$, $15$, $12$, $10$, $8$, $6$, $5$ (the local images have dimensions $1,1,1,1,0,1,2$, in this order). Each local image is generated by the values at all points of $C(\F_p)$ (Proposition~\ref{thm:S}). There is only one prime ideal above 2, so the decomposition group acts transitively on the 27 bitangents. An invariant codeword has weight 0 or 27, while the nonzero codewords have weights 12 and 16; hence $J(\Q_2)[2]=0$ and $\dim J(\Q_2)/2J(\Q_2)=3$. Local square classes are determined by the parity of the valuation and by the characters of $(\mathcal O_K/\mathfrak p^{19})^\times$ ($2e+1=19$). The images of the three $\Q_2$-points with $(u,z)\equiv(0,0),(1,398796270),(1,823490285)$ modulo $2^{30}$ are independent. The approximation error contributes valuation at least 274 to the value of $h_m$, and the value itself has valuation $-2$, so the square class is determined. Hence these three images span the whole 2-adic local image, and the 5-dimensional space shrinks to dimension 2. Since $\delta$ is injective, $\operatorname{rank}J(\Q)\le2$. \end{proof} \paragraph{Step 4: principal divisors and 2-adic logarithms.} Let $D_i=[P_i-\infty]$ ($i=0,1$). The equalities $\operatorname{div}(z)=2P_0+A-3\infty$, $\operatorname{div}(z+1)=2P_1+B-3\infty$, $\operatorname{div}(u+z+1)=2A+P_1+R-4\infty$ are checked by substitution (the last one comes from $F(-z-1,z)=z^2(z-2)(z+1)$), and they give \[ [A-\infty]=-2D_0,\qquad[B-\infty]=-2D_1,\qquad[R-\infty]=4D_0-D_1 \] \begin{lem}\label{lem:1.1.6-log} \begin{enumerate}[label=(\arabic*)] \item For $i=0,1$ there is a polynomial $h_i(u,z)$ with $\operatorname{div}(h_i)=35P_i+E_i-38\infty$, where $E_i$ is an effective divisor of degree 3, irreducible over $\Q$. \item Modulo $2^{24}$, \[ \log D_0\equiv(15495542,1371452,14073150),\qquad \log D_1\equiv(13080018,15362130,15080040). \] \end{enumerate} \end{lem} \begin{proof} (1) Among the combinations of the 36 basis elements $z^au^b$ ($b\le2$, $3a+4b\le38$) of $L(38\infty)$, those whose first 35 Taylor coefficients at $P_i$ vanish form a 1-dimensional space, and its element is $h_i$. Exact computation confirms the order 35 at $P_i$, the pole of order 38, and $\operatorname{Res}_u(F,h_i)=c_i(z-z(P_i))^{35}d_i(z)$. Here $d_i$ is an irreducible monic cubic, and in $\Q[z]/(d_i)$ the common factor of $F$ and $h_i$ is linear in $u$. Hence the divisor equality accounts for all zeros and poles, and $[E_i-3\infty]=-35D_i$ (\code{1.1.6d}). (2) The minimal polynomial $p_i$ of the $u$-coordinate of $E_i$ is integral at 2 and satisfies $p_i(U)\equiv U^2(U+1)\pmod2$, and the $z$-coordinate is a polynomial in $u$ with even coefficients. Hence the three points of $E_i$ reduce modulo 2 to $P_0,P_0,A$. The class of the base divisor $2P_0+A=\operatorname{div}(z)+3\infty$ is 0, so $\log D_i$ is the sum of the tiny integrals from $P_0,P_0,A$ to the three points of $E_i$, divided by $-35$. We compute this sum by a trace in the cubic algebra $\mathcal A_i=\Z_2[U]/(p_i)$. Let $e$ be the idempotent lifting $U^2$ from modulo 2, and take the base coordinate $u=-e$ and $\delta=U+e$. Then $\delta^2\in2\mathcal A_i$, so the $k$th term of the integral has valuation at least $\lfloor k/2\rfloor-v_2(k)$. We compute 80 terms modulo $2^{40}$; the discarded terms have valuation at least 34, and the division loses at most 6 digits, so the values modulo $2^{24}$ above are determined (\code{1.1.6d}). \end{proof} \begin{lem}\label{lem:1.1.6-sat} $\operatorname{rank}J(\Q)=2$, and $H=\langle D_0,D_1\rangle$ has odd index in $J(\Q)$. Hence for each $X\in J(\Q)$ there are unique $c_0,c_1\in\Z_2$ with $\log X=c_0\log D_0+c_1\log D_1$, and if $NX=aD_0+bD_1$ with $N$ odd and $a,b$ integers, then $(c_0,c_1)=(a/N,b/N)$. Moreover, for the primitive vector $\mathbf a$ with $\mathbf a\equiv(699041,3772347,3371045)\pmod{2^{22}}$ and $\mathbf a\equiv(1,1,1)\pmod2$, the differential $\omega=\mathbf a\cdot\boldsymbol\omega$ annihilates all logarithms of $J(\Q)$. \end{lem} \begin{proof} Since $\log D_0/2\equiv(1,0,1)$ and $\log D_1/2\equiv(1,1,0)\pmod2$ are independent, $D_0,D_1$ are independent elements of infinite order, and together with Lemma~\ref{lem:1.1.6-rank} the rank is 2. The numbers of points over $\F_2,\F_4,\F_8$ are $5,9,5$, and $\#J(\F_2)=35$ (\code{1.1.6d}). If $X\in J(\Q_2)$, then $35X$ lies in the kernel of reduction, that is, in the formal group with parameters in $2\Z_2$. In the formal logarithm of an integral invariant differential, the coefficients of the homogeneous part of degree $n$ lie in $n^{-1}\Z_2$ (apply the Euler operator to the closed integral differential), so its value has valuation at least $n-v_2(n)\ge1$. Since 35 is odd, $\log J(\Q_2)\subseteq2\Z_2^3$. If $X\in J(\Q)$ and $2X=aD_0+bD_1$, then $a\log D_0/2+b\log D_1/2=\log X\in2\Z_2^3$. The matrix formed by the first two components of the two half-logarithms is invertible modulo 2, so $a,b$ are even, and $X-(a/2)D_0-(b/2)D_1\in J(\Q)[2]=0$. Hence $J(\Q)/H$ has no element of order 2, and the index $N=[J(\Q):H]$ is odd. The rest follows because the logarithm is a homomorphism and vanishes on torsion elements. The coefficients are obtained from the inverse over $\Z_2$ of the matrix of the first two logarithm components. The cross product of the two logarithms has common valuation 2, and after division by $2^2$ it is the primitive annihilator $\mathbf a$ (\code{1.1.6d}). \end{proof} \paragraph{Step 5: 2-adic zeros.} $C(\F_2)$ consists of the five reductions of $P_0,P_1,A,B,\infty$, so there are also five residue disks, and each center has a known rational point. On the finite disks we use $t=u-u_0$, at $\infty$ we use $t=z/u$, and we put $t=2s$, $s\in\Z_2$. The constant term of $G(s)=\int_0^{2s}\omega$ is 0, and by \cite{Coleman85,MP12} and Lemma~\ref{lem:1.1.6-sat} every rational point of a disk is a zero of $G$. \begin{lem}\label{lem:1.1.6-zeros} On the five disks, $G$ has 11 zeros in total, exactly one in each ball of the table below. Here $(c_0,c_1)$ are the coefficients of Lemma~\ref{lem:1.1.6-sat}. \[ \begin{array}{c|c|c|r|c|r|c} \text{Disk}&\text{Ball of }s&v&s\bmod2^{14}&(c_0,c_1)\bmod128&70c_0+c_1\bmod128&\text{Point}\\\hline P_0&1+2\Z_2&2&11613&(106,48)&44&-\\ P_0&2\Z_2&2&0&(1,0)&70&P_0\\ P_1&2+4\Z_2&4&10130&(124,35)&11&-\\ P_1&4\Z_2&4&0&(0,1)&1&P_1\\ A&7+8\Z_2&7&16383&(4,127)&23&R\\ A&3+8\Z_2&7&4595&(120,79)&31&-\\ A&2+4\Z_2&5&7586&(64,104)&104&-\\ A&4\Z_2&5&0&(126,0)&116&A\\ B&\Z_2&1&0&(0,126)&126&B\\ \infty&1+2\Z_2&2&529&(5,65)&31&-\\ \infty&2\Z_2&2&0&(0,0)&0&\infty \end{array} \] \end{lem} \begin{proof} We expand $G$ up to degree 40 over $\Z/2^{30}$ and divide each disk into balls $s=c+2^dr$. After dividing by the minimal valuation $v$ of the coefficients: if only the constant term is a unit, then Strassmann's theorem gives no zero; if the linear coefficient is a unit and all higher coefficients are even, then Hensel's lemma and Strassmann's theorem give exactly one zero. Every ball ends in one of these two cases within depth 3. The only balls outside the table are $1+2\Z_2$ in the $P_1$-disk and $1+4\Z_2$ in the $A$-disk, and neither has a zero (\code{1.1.6e}). The error of the annihilator is divisible by $2^{22}$, and after the substitution $t=2s$ the coefficient of degree $k$ acquires a factor of valuation $k-v_2(k)\ge1$, so the error of $G$ is divisible by $2^{23}$. The terms of degree above 40 have valuation at least 36. Hence the error of a root $s$ has valuation $23-v+d\ge19$, and the coefficients $(c_0,c_1)$ are determined modulo $2^{11}$. Below we use them only modulo $2^7$. $R$ is the point of the $A$-disk with $s=-1$. \end{proof} \paragraph{Step 6: a sieve modulo 23.} \begin{lem}\label{lem:1.1.6-sieve} $\#J(\F_{23})=11392=2^7\cdot89$. The element $H_1=89\overline{D_1}$ has order 128 and $H_0=89\overline{D_0}=70H_1$, so the 2-primary part of $J(\F_{23})$ is the cyclic group generated by $H_1$. Moreover, as $Q$ runs through $C(\F_{23})$, the set of $r\in\Z/128$ with $89[Q-\infty]=rH_1$ is \[ \mathcal R=\{0,1,2,4,6,8,23,24,33,35,52,58,70,99,101,116,117,126,127\} \] \end{lem} \begin{proof} We have $\#C(\F_{23})=22$, $\#C(\F_{23^2})=538$, $\#C(\F_{23^3})=12616$. Newton's identities and the functional equation give $L_{23}(T)=1-2T+6T^2+140T^3+138T^4-1058T^5+12167T^6$, and $\#J(\F_{23})=L_{23}(1)=11392$. The order, the relation $H_0=70H_1$, and the classes of the 22 points are computed exactly by Riemann--Roch divisor arithmetic over $\F_{23}$ (\code{1.1.6f}). \end{proof} \begin{proof}[Proof of Proposition~\ref{prop:1.1.6}] A point $X\in C(\Q)$ lies in one of the five disks and is a zero of $G$, so it is one of the 11 zeros in the table. A ball that contains a known point contains only one zero, so if it lies in such a ball, $X$ is that known point. Otherwise $X$ is one of the other five zeros. By Lemma~\ref{lem:1.1.6-sat}, $N[X-\infty]=aD_0+bD_1$ with $N$ odd. Reducing modulo 23 and multiplying by 89 gives $89N\,\overline{[X-\infty]}=aH_0+bH_1=(70a+b)H_1$, and $N$ is invertible modulo 128, so $89\,\overline{[X-\infty]}=(70c_0+c_1)H_1$. The reduction of $X$ is a point of $C(\F_{23})$, so by \cite{BS10}, we must have $70c_0+c_1\bmod128\in\mathcal R$. However, none of the values $44,11,31,104,31$ of the five zeros lies in $\mathcal R$. Therefore $C(\Q)$ consists only of the six known points. \end{proof} \subsection{\texorpdfstring{$x^4+y^3-y^2+x=0$ \ ($H=30$)}{x^4+y^3-y^2+x=0 (H=30)}}\label{sec:1.1.7} \begin{thm}\label{thm:1.1.7} The equation $x^4+y^3-y^2+x=0$ has only the following integer solutions: $(x,y)=(-4,-6),(-1,0),(-1,1),(0,0),(0,1),(1,-1)$. \end{thm} \paragraph{Setup.} Put $u=-y$, $z=-x$. Then $x^4+y^3-y^2+x=-(u^3+u^2-z^4+z)$, so the equation becomes \[ C:\quad F(u,z)=u^3+u^2-z^4+z=0 \] The projective completion $u^3w+u^2w^2-z^4+zw^3=0$ is a smooth plane quartic (genus 3), and the point at infinity $O=(1:0:0)$ is a rational hyperflex. With $t=z/u$, $w=1/u$, the equation near $O$ is $w+w^2+tw^3-t^4=0$. $C$ has good reduction at 2 ($F_z=1-4z^3$ and $F_w=1+2w+3tw^2$ are units), and an integral basis of holomorphic differentials is \[ \boldsymbol\omega=(1,z,u)\frac{dz}{3u^2+2u}=-(1,z,u)\frac{du}{1-4z^3}=-(w,t,1)\frac{dt}{1+2w+3tw^2} \] At 3 the curve has bad reduction, since modulo 3 the point $(u,z)=(0,1)$ is singular. \begin{prop}\label{prop:1.1.7} $C(\Q)=\{O,(0,0),(0,1),(-1,0),(-1,1),(1,-1),(6,4)\}$. \end{prop} Since $(x,y)=(-z,-u)$, the six affine points are $(0,0)$, $(-1,0)$, $(0,1)$, $(-1,1)$, $(1,-1)$ and $(-4,-6)$, in this order, and Theorem~\ref{thm:1.1.7} follows from Proposition~\ref{prop:1.1.7}. For $X\in C(\Q)$ we write $D_X=[X-O]$, and we put $G_1=D_{(0,1)}$, $G_2=D_{(1,-1)}$. The proof goes as follows. In Steps 1--3 we show $\operatorname{rank}J(\Q)\le2$; since 5 is a bad prime at which the component group has order 2, we determine the local image at 5 separately (Lemma~\ref{lem:1.1.7-five}). In Step 4 we show that the rank is 2 and that $\langle G_1,G_2\rangle$ has odd index. In Steps 5 and 6, 2-adic and 3-adic Chabauty each reduce the candidates for new rational points to two, and in Step 7 a comparison modulo 509 and 293 eliminates all of them. \paragraph{Step 1: the bitangent field and bad primes.} The 27 bitangents other than the tangent line at $O$ have the form $u=mz+\eta(m)$, where $m$ is a root of \[ \begin{aligned} f(m)={}&m^{27}-160m^{19}-1344m^{18}-1280m^{15}-3840m^{11}\\ &+23552m^{10}-40704m^9-4096m^7+8192m^6-4096m-4096 \end{aligned} \] The element $\eta\in\Q(m)$ is given explicitly, and we check exactly the two coefficient conditions which say that $F(mz+\eta,z)$ is $-1$ times the square of a quadratic (\code{1.1.7a}). The polynomial $f$ is irreducible, and $h=u-mz-\eta$ is $t^{-4}(1+\cdots)$ at $O$, so it is normalized. The field $K=\Q[m]/(f)$ has signature $(3,12)$ and discriminant $D_K=2^{24}\cdot3^6\cdot17^6\cdot1289^6$. There is one prime ideal above 2 ($e=9$, $f=3$), there are four above 3, and five above 5. The prime 5 is unramified in $K$, with residue degrees $3,6,6,6,6$. In characteristic different from 2 and 3, a singular point satisfies $F_u=u(3u+2)=0$ and $F_z=1-4z^3=0$. If $u=0$ then $F=3z/4\ne0$; if $u=-2/3$ then $F=4/27+3z/4$, so $z=-16/81$, and $|\operatorname{Res}(81z+16,4z^3-1)|=547825=5^2\cdot17\cdot1289$. At 17 and 1289 the only singular point is the reduction of $(-2/3,-16/81)$, the Hessian is a unit, and the critical value has valuation 1. Hence it is a node of thickness 1, and the component group is trivial. At 5, too, the only singular point is $(u,z)\equiv(1,4)$. However, at the approximate critical point $(-2/3,-16/81)$ the valuations of $F$ and $F_z$ are 2, $F_u=0$, and the Hessian is a unit, and at the true critical point obtained by Hensel lifting the value of $F$ changes only modulo $5^4$. Hence the node at 5 has thickness 2, and the geometric component group is $\Z/2$ (the singular fiber is geometrically irreducible). Therefore we take $S=\{2,3,5\}$ in Proposition~\ref{thm:A}. \paragraph{Step 2: the class group.} \begin{lem}\label{lem:1.1.7-class} $\Cl(K)[2]=0$. \end{lem} \begin{proof} As in the proof of Lemma~\ref{lem:1.1.6-class}, we apply Proposition~\ref{thm:B} to the one-cell window $f(x)=(1-|x|/T)_+$. Here $T=8$ and $e^8=2980.96\ldots<3000$. All 413 prime ideals of norm at most 3000 have explicit generators (\code{1.1.7b}), so $c_m(q)=a(q)$ in every term, and the same closed-form lower bound gives $B_0>0.0849899>0$ (\code{1.1.7c}). By Proposition~\ref{thm:B}, $\Cl(K)[2]=0$. \end{proof} \paragraph{Step 3: an upper bound for the rank.} \begin{lem}\label{lem:1.1.7-five} The local image $\delta_5\bigl(J(\Q_5)/2J(\Q_5)\bigr)$ at 5 is the line spanned by $\delta_5([X_5-O])$, where $X_5$ is the unique $\Q_5$-point with $u=1$, $z\equiv4\pmod{25}$. \end{lem} \begin{proof} The bitangent code embeds $J[2]$ Galois-equivariantly and injectively into $E=\F_2^{27}$, and $K$ is unramified at 5, so $J[2]$ is also unramified at 5. The Frobenius $\Phi$ at 5 preserves the quadratic form that the rational odd theta characteristic $2O$ defines on $J[2]$, so it lies in the orthogonal group $O_6^-(2)$ (of order 51840) of this form, and it has cycle type $(3,6,6,6,6)$ on the bitangents. Listing all such elements (\code{1.1.7d}), the only pairs consisting of $\dim J[2]^\Phi$ and the rank of the coinvariant map $J[2]/(\Phi-1)J[2]\to E/(\Phi-1)E$ are $(0,0)$ (720 elements), $(1,1)$ (4320 elements), and $(2,0)$ (1440 elements). If $p\ne2$, then $\dim J(\Q_p)/2J(\Q_p)=\dim J(\Q_p)[2]=\dim J[2]^\Phi$, so in the first two cases the local image has dimension at most 1. In the third case, let $J^0(\Q_5)$ be the subgroup of points that reduce to the identity component of the Néron model. The special fiber of the identity component is a connected commutative algebraic group, on which multiplication by 2 is surjective and étale. Hence, by Hensel's lemma, the points of $J^0(\Q_5)$ are divisible by 2 over the maximal unramified extension, and their Kummer classes are unramified classes, that is, elements of $J[2]/(\Phi-1)J[2]$. Since the coinvariant map has rank 0, $\delta_5(J^0(\Q_5))=0$. The quotient $J(\Q_5)/J^0(\Q_5)$ embeds into the component group $\Z/2$, so the local image has dimension at most 1. On the other hand, the image of $X_5$ is nonzero (checked exactly, including the margin for the approximation error, \code{1.1.7d}). Therefore the local image is that line. \end{proof} \begin{lem}\label{lem:1.1.7-rank} $\operatorname{rank}J(\Q)\le2$. \end{lem} \begin{proof} By Proposition~\ref{thm:A}, $J(\Q)[2]=0$ and $\delta$ is injective. By Lemma~\ref{lem:1.1.7-class}, $\dim K(S,2)=15+10=25$, and 25 explicit elements supported on the ten prime ideals above $S$ form a basis (rank 25 from valuation characters and quadratic characters at good primes). The conditions on the sign of the norm and on the parity of the valuations at 2, 3, 5 reduce the dimension to 21. The local conditions at the good primes $487$, $491$, $499$, $503$, $509$, $521$, $523$ (local images of dimensions $1,2,1,2,3,3,1$) reduce the dimension to $18$, $15$, $13$, $10$, $7$, $6$, $6$ (Proposition~\ref{thm:S}). At 2, as in the proof of Lemma~\ref{lem:1.1.6-rank}, $J(\Q_2)[2]=0$ and $\dim J(\Q_2)/2J(\Q_2)=3$, and the images of the three points obtained by Hensel lifting from $(u,z)\equiv(1,0),(2,0),(0,1)$ (the third is the rational point $(0,1)$) are independent (approximation modulo $2^{30}$; the relative error exceeds twice the ramification index). Hence the dimension drops to 3, and the condition of Lemma~\ref{lem:1.1.7-five} makes it 2 (\code{1.1.7a}). Therefore $\operatorname{rank}J(\Q)\le2$. \end{proof} \paragraph{Step 4: 2-adic logarithms and odd index.} Since $F(u,0)=F(u,1)=u^2(u+1)$ and $F(-z,z)=-z(z-1)(z+1)^2$, we have $\operatorname{div}(z)=2(0,0)+(-1,0)-3O$, $\operatorname{div}(z-1)=2(0,1)+(-1,1)-3O$, $\operatorname{div}(u+z)=(0,0)+(-1,1)+2(1,-1)-4O$, and \[ D_{(0,0)}=2G_1-2G_2,\qquad D_{(-1,0)}=-4G_1+4G_2,\qquad D_{(-1,1)}=-2G_1 \] \begin{lem}\label{lem:1.1.7-log2} \begin{enumerate}[label=(\arabic*)] \item For each of $X=(0,0),(0,1),(1,-1)$ there is a polynomial $h_X$ with $\operatorname{div}(h_X)=35X+E_X-38O$ ($E_X$ an effective divisor of degree 3), and modulo $2^{40}$ \[ \begin{array}{c|rrr} X&\log_0D_X&\log_1D_X&\log_2D_X\\\hline (0,0)&714153886116&445000321592&1048252576132\\ (0,1)&578889133504&570949811744&721623895966\\ (1,-1)&771568004334&348449650948&747253421788 \end{array} \] \item $\operatorname{rank}J(\Q)=2$, $\#J(\Q)_{\rm tors}$ divides 35, and $\langle G_1,G_2\rangle$ has odd index in $J(\Q)/J(\Q)_{\rm tors}$. Hence for each $Y\in J(\Q)$ we have $NY=n_1G_1+n_2G_2$ with $N$ odd and $n_1,n_2$ integers, and $(c_1,c_2)=(n_1/N,n_2/N)\in\Z_2^2$ is uniquely determined by the 0th and 2nd components of $\log Y$. \end{enumerate} \end{lem} \begin{proof} (1) $h_X$ is the combination of the $z^iu^j$ ($j<3$, $3i+4j\le38$) whose first 35 Taylor coefficients at $X$ vanish (the solution space is 1-dimensional). $\operatorname{Res}_u(F,h_X)$ is the product of a constant, $(z-z_X)^{35}$ and an irreducible cubic, and the order 35 and the pole of order 38 account for all zeros and poles. For $(0,1)$ the residual divisor lies in the $O$-disk (the roots have valuation $\ge1/2$). For the other two, as in the proof of Lemma~\ref{lem:1.1.6-log}, the cubic algebra becomes $\F_2[u]/(u^2(u+1))$ modulo 2, and the base divisor is a full fiber of $z$ or of $z-1$, so its class is 0. Hence $\log D_X$ is the trace of the tiny integrals divided by $-35$. The integrated term of degree $n$ has valuation at least $\lfloor n/2\rfloor-v_2(n)$; with 144 terms the tail has valuation at least 64, and the division in the computation modulo $2^{60}$ loses at most 7 digits, so the values modulo $2^{40}$ are determined (\code{1.1.7e}). The table agrees with $D_{(0,0)}=2G_1-2G_2$ modulo $2^{40}$. (2) From the point counts $5,9,5$ over $\F_2,\F_4,\F_8$ we get $\#J(\F_2)=35$. By \cite[Appendix]{Katz81}, torsion of odd order injects into $J(\F_2)$, and $J(\Q)[2]=0$, so $\#J(\Q)_{\rm tors}\mid35$. Since $\log G_1/2\equiv(0,0,1)$ and $\log G_2/2\equiv(1,0,0)\pmod2$ are independent, the rank is 2. As in the proof of Lemma~\ref{lem:1.1.6-sat}, $\log J(\Q_2)\subseteq2\Z_2^3$. The submatrix formed by the 0th and 2nd components of the two half-logarithms has unit determinant. If the index were even, then for a basis $B_1,B_2$ of the free part we would have $(G_1,G_2)=(B_1,B_2)M$ with $\det M$ even; since the half-logarithms of the $B_i$ are integral, the determinant of this submatrix would then be even, a contradiction. The coefficients are read off using the inverse of this submatrix over $\Z_2$. \end{proof} \paragraph{Step 5: 2-adic zeros.} For the five points of $C(\F_2)$ we take the rational points $O$, $(0,0)$, $(0,1)$, $(-1,0)$, $(-1,1)$ as the centers of the disks. On the finite disks we use $t=u-u_0$, at $O$ we use $t=z/u$, and we put $t=2s$. Let $\omega$ be the primitive annihilator obtained from the cross product of the two logarithms, normalized by its middle coefficient (which is odd). Then the zeros of $G(s)=\int_0^{2s}\omega$ include all rational points of the disk (\cite{Coleman85,MP12}). \begin{lem}\label{lem:1.1.7-zeros2} On the five disks, $G$ has exactly nine zeros, and each of them lies in a ball that contains only this zero. Seven of them are known points ($(6,4)$ is at $s=3$ in the $(0,0)$-disk, $(1,-1)$ is at $s=1$ in the $(-1,1)$-disk, and the others are at the centers). The other two are \[ \begin{array}{c|c|r|r|r} &\text{Disk}&s\bmod2^{20}&c_1\bmod2^{16}&c_2\bmod2^{16}\\\hline Q_2&(-1,0)&902770&18122&35366\\ R_2&O&607280&38704&39424 \end{array} \] $Q_2,R_2$ are only local zeros in $C(\Q_2)$; this does not mean that they are rational points. \end{lem} \begin{proof} We use the same ball decomposition as in the proof of Lemma~\ref{lem:1.1.6-zeros}. When we integrate up to degree 64, the discarded part is divisible by $2^{58}$ and the uncertainty of the annihilator by $2^{39}$; since we compute in $\Z/2^{50}$, the coefficients have an absolute precision of 39 digits. The valuation we divide by is at most 10 (depth 5), so the roots have a precision of at least $39-10+5=34$ digits, and since the half-logarithm submatrix is invertible, the coefficients are determined to 16 digits (\code{1.1.7f}). \end{proof} \paragraph{Step 6: 3-adic logarithms and 3-adic zeros.} \begin{lem}\label{lem:1.1.7-three} \begin{enumerate}[label=(\arabic*)] \item For $X=(0,0),(1,-1)$ we have $\operatorname{div}(h_X)=63X+E_X-66O$, and modulo $3^{24}$ \[ \begin{array}{c|rrr} X&3\log_0D_X&3\log_1D_X&3\log_2D_X\\\hline (0,0)&265295843469&256546487256&167647527990\\ (1,-1)&247088970603&249160065058&89859469233 \end{array} \] In the space of logarithms $G_1=G_2+D_{(0,0)}/2$; the cross product of $3\log G_1,3\log G_2$ has common valuation 2, and the primitive annihilator is $(16280295932,6711384369,10302306500)$ modulo $3^{22}$ and $(2,0,2)$ modulo 3. \item Integrate the differential made from this annihilator on each 3-adic chart, starting from the rational point at the center. The resulting function has exactly nine zeros on all of $C(\Q_3)$, and each of them lies in a ball that contains only this zero. Seven of them are known points, and the coefficients of the remaining two, $Q_3,R_3$, are $(c_1,c_2)\equiv(2705,832)$, $(1055,2942)\pmod{3^8}$. \item At all nine zeros, $c_1,c_2\in\Z_3$. \end{enumerate} \end{lem} \begin{proof} (1) As in the 2-adic case, $h_X$ is found from the $z^iu^j$ ($j<3$, $3i+4j\le66$) and checked with resultants. Both residual divisors lie in the $O$-disk, and the base divisor is $3O$. Since $t^3\in3\mathcal A$, the traced term of degree $n$ has valuation at least $\lfloor n/3\rfloor-v_3(n)$; with 120 terms the tail has valuation at least 35, and the division in the computation in $\Z/3^{40}$ loses at most 5 digits. Hence the values modulo $3^{24}$ are determined. The bad reduction at 3 is not used; we use only the fact that the $O$-disk is an analytic disk (\cite[Section 3]{KRZB16}; \code{1.1.7g}). (2) The rational points of the singular fiber modulo 3 are $O,(0,0),(1,-1),(-1,0),(-1,1),(0,1)$, and only $(0,1)$ is a singular point. At the five smooth points, $u-u_0$ ($z-1$ at $(-1,1)$, and $z/u$ at $O$) is a parameter that runs over $3\Z_3$. The points that reduce to $(0,1)$, written as $u=3a$, $z=1+3b$, satisfy $b+6b^2+12b^3+9b^4=a^2+3a^3$. For rational $01$, and this disk contains all $a\in\Z_3$ ($(0,1)$ is $a=0$, and $(6,4)$ is $a=2$). On it $\boldsymbol\omega=(1,1+3b,3a)\,da/(1+12b+36b^2+36b^3)$, and after integration the coefficient of $a^n$ has valuation at least $\lceil(n-1)/4\rceil-v_3(n)\ge0$, and at least 28 if $n\ge121$. By \cite[Section 3]{KRZB16}, the antiderivative starting at $a=0$ is the Abelian integral on this whole chart. When we decompose the six charts into balls (smooth disks: $t=3s$ with 120 terms, coefficient error $3^{23}$, tail $3^{100}$; singular chart: error $3^{22}$, tail $3^{28}$), every ball ends with either 0 zeros or 1 simple root. The zeros are the seven known points, $Q_3$ (in the chart $u=-1+3s$, $s\equiv273546\pmod{3^{12}}$), and $R_3$ (in the singular chart, $a\equiv397615\pmod{3^{12}}$) (\code{1.1.7g}). (3) Let $M$ be the matrix of the 1st and 2nd components of $3\log G_1,3\log G_2$. Then $\det M=9\varepsilon$ with $\varepsilon\in\Z_3^\times$. If $U=\varepsilon^{-1}\operatorname{adj}M$, then for $v=3\log D$ at a zero we have $U(v_1,v_2)^{\mathsf T}=9(c_1,c_2)^{\mathsf T}$. At all nine zeros both components of the left-hand side are divisible by 9 (the error has valuation at least 11, \code{1.1.7g}). \end{proof} \begin{lem}\label{lem:1.1.7-proj} If the 3-adic coefficients of $Y\in C(\Q)$ are $(c_1,c_2)\in\Z_3^2$, then at a good prime $\ell$, for the projection $\pi_3$ onto the 3-primary part of $J(\F_\ell)$, we have $\pi_3(\overline{D_Y})=\pi_3(c_1\overline{G_1}+c_2\overline{G_2})$. The same holds for the 2-primary part and the 2-adic coefficients of Lemma~\ref{lem:1.1.7-log2}(2). \end{lem} \begin{proof} The rank is 2 and the two 3-adic logarithms are independent, so $c_1,c_2$ are rational numbers, and by assumption their denominators are prime to 3. For a common denominator $d$ ($3\nmid d$), the element $T=dD_Y-dc_1G_1-dc_2G_2$ has logarithm 0, so it is a torsion element, and its order divides 35. Hence $\pi_3(\overline T)=0$, and $d$ is invertible on the 3-primary part. For the 2-primary part the same argument works with an odd $N$. \end{proof} \paragraph{Step 7: comparison modulo 509 and 293.} The point counts over three extensions determine the Weil polynomial and $\#J$ (\code{1.1.7h}). \[ \begin{array}{c|r|r|r|r} p&\#C(\F_p)&\#C(\F_{p^2})&\#C(\F_{p^3})&\#J(\F_p)\\\hline 293&284&85720&25164212&24294816=2^5\cdot3^4\cdot7\cdot13\cdot103\\ 509&510&259528&131838318&131974656=2^9\cdot3\cdot11\cdot73\cdot107 \end{array} \] Multiplication by $257763=\#J(\F_{509})/2^9$, $759213=\#J(\F_{293})/2^5$, $299936=\#J(\F_{293})/3^4$ is the projection onto the 2-primary or the 3-primary part. \begin{lem}\label{lem:1.1.7-sieve} \begin{enumerate}[label=(\arabic*)] \item $257763\cdot304\,\overline{G_1}$ is not equal to $257763\,[Q-O]$ for any $Q\in C(\F_{509})$. \item The points $Q\in C(\F_{293})$ with $759213\,[Q-O]=759213\,(10\overline{G_1}+6\overline{G_2})$ are the nine points $(u,z)=(0,0)$, $(1,20)$, $(47,88)$, $(60,82)$, $(68,67)$, $(119,277)$, $(124,164)$, $(162,64)$, $(197,60)$, and for none of them is $299936\,[Q-O]$ equal to $299936\,(32\overline{G_1}+22\overline{G_2})$ or to $299936\,(2\overline{G_1}+26\overline{G_2})$. \end{enumerate} \end{lem} \begin{proof} We compare all points by exact divisor arithmetic over $\F_\ell$ (\code{1.1.7h}). \end{proof} \begin{proof}[Proof of Proposition~\ref{prop:1.1.7}] Suppose that $Y\in C(\Q)$ is not a known point. By Lemma~\ref{lem:1.1.7-zeros2}, $Y$ is $Q_2$ or $R_2$ in $C(\Q_2)$. If $Y=R_2$, then its 2-adic coefficients are $(304,0)\bmod512$, so by Lemma~\ref{lem:1.1.7-proj}, $257763\,\overline{D_Y}=257763\cdot304\,\overline{G_1}$; but the reduction of $Y$ is a point of $C(\F_{509})$, so this contradicts Lemma~\ref{lem:1.1.7-sieve}(1) (\cite{BS10}). Hence $Y=Q_2$, its 2-adic coefficients are $(10,6)\bmod32$, and the reduction of $Y$ modulo 293 is one of the nine points of Lemma~\ref{lem:1.1.7-sieve}(2). On the other hand, by Lemma~\ref{lem:1.1.7-three}, $Y$ is $Q_3$ or $R_3$, and its 3-adic coefficients lie in $\Z_3$ and are $(32,22)$ or $(2,26)$ modulo 81. Applying Lemma~\ref{lem:1.1.7-proj} with $\ell=293$ gives a contradiction with Lemma~\ref{lem:1.1.7-sieve}(2). Therefore $C(\Q)$ consists only of the seven known points. \end{proof} \subsection{\texorpdfstring{$x^4+y^3+2x-y+1=0$ \ ($H=31$)}{x^4+y^3+2x-y+1=0 (H=31)}}\label{sec:1.1.8} \begin{thm}\label{thm:1.1.8} The only integer solutions of $x^4+y^3+2x-y+1=0$ are $(-1,-1)$, $(-1,0)$ and $(-1,1)$. \end{thm} \paragraph{Setup.} Put $u=-y$, $z=x$. Then the equation becomes \[ C:\quad F(u,z)=-u^3+u+z^4+2z+1=0 \] The projective completion $-U^3W+UW^3+Z^4+2ZW^3+W^4=0$ is a smooth plane quartic (genus 3) (Step 1), and it has only one point at infinity, $O=[1:0:0]$. The tangent line $W=0$ meets the curve at $O$ with multiplicity 4, since $Z^4=0$; so $O$ is a rational hyperflex. $J$ is the Jacobian of $C$. Theorem~\ref{thm:1.1.8} is equivalent to the following proposition. \begin{prop}\label{prop:1.1.8} The only integer solutions of $F(u,z)=0$ are $P_0=(0,-1)$, $P_1=(1,-1)$, $P_2=(-1,-1)$. \end{prop} Since $(x,y)=(z,-u)$, the three points are $(x,y)=(-1,0),(-1,-1),(-1,1)$, respectively. The proof has five steps. In Step 1 we fix the bitangent field $K$ and the bad primes. In Step 2 we show $\Cl(K)[2]=0$ by Proposition~\ref{thm:B}, and in Step 3 we show by 2-descent that the rank is at most 2. In Step 4 we construct an annihilating differential from 2-adic logarithms, and in Step 5 we count the zeros on the two 2-adic disks that cover the integral points. The curve has bad reduction at 2, so we use \cite[Section 3]{KRZB16} to identify the integrals on the disks with integrals of power series. \paragraph{Step 1: the bitangent field and bad primes.} The 27 bitangents other than the tangent line at $O$ have the form $u=mz+\eta$, and the slope $m$ is a root of \begin{align*} f(m)={}&m^{27}-480m^{19}+2688m^{18}+17280m^{15}-34560m^{11}-141312m^{10}\\ &-162816m^{9}+165888m^{7}+221184m^{6}-94208m^{3}-49152m+32768 \end{align*} The polynomial $f$ is irreducible. In $K=\Q[m]/(f)$ put \[ \eta=-\frac{11m^{13}-112m^5-224m^4+192m}{8(13m^9+16m-16)} \] Then the identity \[ F(mz+\eta,z)=\Bigl(z^2-\frac{m^3}{2}z-\frac{3m^2\eta}{2}-\frac{m^6}{8}\Bigr)^2 \] holds. $K$ has signature $(r_1,r_2)=(7,10)$ and discriminant $D=2^{30}\cdot5^{10}\cdot431^{6}\cdot659^{6}$. There are two prime ideals above 2, with $(e,f)=(8,3),(1,3)$ (\code{1.1.8a}). The bad primes are as follows. The point $O$ is smooth in every characteristic (at $[1:0:0]$ the derivative in the $W$ direction is $-1$). In characteristic 3, $F_u\equiv1$. In characteristic $p>3$, an affine singular point satisfies \[ u^2=1/3,\qquad z^3=-1/2,\qquad 4u+9z+6=0 \] The last equation comes from $12F-4uF_u-3zF_z=2(4u+9z+6)$. In the critical algebra defined by the first two equations, the norm of the critical value $(4u+9z+6)/6$ is $-5^2\cdot431\cdot659/(2^8\cdot3^9)$. Hence the only odd bad primes are $5$, $431$, $659$. The $z$-coordinates of the singular points are given by $z^2+3z+4\equiv0\pmod5$ (two conjugate points), $z\equiv-6\pmod{431}$, and $z\equiv-222\pmod{659}$, respectively. The Hessian determinant $-72uz^2$ is a unit, so all of them are ordinary double points (nodes), and since the critical value has valuation exactly 1, their thickness is 1. A plane quartic with at most two nodes is geometrically irreducible (if it were reducible, then by Bezout the components would meet in at least three points). Therefore, at these three primes the special fiber of the regular model has one component, and the component group of the Jacobian is trivial. Then $J(\Q_p^{\rm nr})$ is divisible by 2, so the descent image is unramified at $p$. Therefore $S=\{\mathfrak p\mid2\}$ suffices (\code{1.1.8a}). \paragraph{Step 2: the class group.} \begin{lem}\label{lem:1.1.8class} $\Cl(K)[2]=0$. \end{lem} \begin{proof} We apply Proposition~\ref{thm:B} with $n=27$, $(r_1,r_2)=(7,10)$, the $D$ above, and $T=\log10^7$. The window is a step function on 48 cells of equal width, and its heights are positive integers, symmetric from left to right (the input of \code{1.1.8d}). Two pieces of data are needed. \begin{enumerate}[label=(\roman*)] \item There are $665{,}664$ prime ideals of norm less than $10^7$ in total, and we count the number $a(q)$ for every norm (\code{1.1.8b}). \item All $9{,}432$ prime ideals of norm at most $10^5$ are principal ideals. For each of these ideals we give a witness $g$ for Lemma~\ref{lem:C} and check it in exact arithmetic (\code{1.1.8c}). Therefore we set $k(q)=a(q)$ if $q\le10^5$, and $k(q)=0$ otherwise. Since $(10^5)^2>10^7$, all terms with $m\ge2$ receive a positive sign. \end{enumerate} With these data the computation gives $B^-(f)>0.0930446107>0$ (\code{1.1.8d}). In detail, $C_\infty=-0.7891595062\ldots$, the integral term is $0.0842206841\ldots$, and the prime term is $0.9664248011\ldots$. The computation uses exact rational coefficients and directed interval arithmetic (224 bits), so rounding errors cannot change the conclusion. By Proposition~\ref{thm:B}, $\Cl(K)[2]=0$. \end{proof} \paragraph{Step 3: an upper bound for the rank.} \begin{lem}\label{lem:1.1.8rank} $\operatorname{rank}J(\Q)\le2$. \end{lem} \begin{proof} The hypotheses of Proposition~\ref{thm:A} (a rational hyperflex, irreducibility of $f$) were checked above. Hence $J(\Q)[2]=0$ and $\delta$ is injective. For an affine rational point $P$, $\delta(P-O)$ is the square class of $u(P)-mz(P)-\eta$. With $t=z/u$, $w=1/u$ at $O$, the value of $t^4(u-mz-\eta)$ is 1, so this normalization is correct. By Step 1 the image lies in $K(S,2)$, and by Lemma~\ref{lem:1.1.8class} \[ \dim_{\F_2}K(S,2)=(r_1+r_2)+\#S=17+2=19 \] Nineteen explicit elements form a basis of this space. We factor the principal divisors of 85 atomic elements exactly into 61 prime ideals to show that all 19 elements have even valuation outside 2, and we check that the matrix of quadratic characters has rank 19 (\code{1.1.8e}). We impose conditions on this 19-dimensional space one at a time. Imposing the condition that the norm is a square (the sign and the parity of the 2-adic valuation) gives dimension 17. At the good primes $7$, $11$, $59$, $83$, the polynomial $f$ splits into three irreducible factors of degree 9. A Frobenius-invariant codeword has weight divisible by 9, while the nonzero codewords have weights 12 and 16. Hence $J(\F_p)[2]=0$ and $J(\Q_p)/2J(\Q_p)=0$. That is, the quadratic characters at the three prime ideals above $p$ must all be 0. These conditions reduce the dimension to $15,13,11,9$ in turn. At 2, the bitangent orbits have sizes $24$ and $3$. The partial sums $0,3,24,27$ do not include 12 or 16, so $J(\Q_2)[2]=0$ and $\dim J(\Q_2)/2J(\Q_2)=g=3$. In the two components of $K\otimes\Q_2$, the spaces of square classes have dimensions $26$ and $5$. By Hensel's lemma every element of $1+\mathfrak p^{2e+1}$ is a square, so a finite computation in the unit group gives the whole space of square classes. We use three local points: the rational points $P_0$, $P_1$, and the point with $t=2$ in the disk at infinity. The last point is obtained by computing the root $w$ ($v_2(w)=4$) of $t^4-w+(1+2t)w^3+w^4=0$ modulo $2^{80}$. The relative error caused by the approximation has valuation at least 648 and at least 81 in the two local fields, respectively, which is more than the depths 19 and 5 needed to determine the square classes. The three images are independent, so they make up the whole local image. Keeping only the elements of the 9-dimensional space that map into this image leaves the 2-dimensional space spanned by (in the numbering of the 19 basis elements) \[ e_3+e_{14}+e_{19},\qquad e_5+e_{14} \] (\code{1.1.8e}). Since $J(\Q)[2]=0$, $\operatorname{rank}J(\Q)\le2$. \end{proof} \paragraph{Step 4: 2-adic logarithms and the annihilating differential.} Let $G_i=[P_i-O]$ ($i=0,1$), and take the basis of holomorphic differentials \[ \omega_\bullet=(\omega_0,\omega_1,\omega_2)=(1,z,u)\,\frac{dz}{1-3u^2} \] \begin{lem}\label{lem:1.1.8log} $\operatorname{rank}J(\Q)=2$. Moreover, there is a primitive vector $a=(a_0,a_1,a_2)\in\Z_2^3$ with $a\equiv(12,14,9)\pmod{16}$ such that $\omega=a_0\omega_0+a_1\omega_1+a_2\omega_2$ annihilates all 2-adic logarithms of $J(\Q)$. \end{lem} \begin{proof} There is a polynomial $h_i(u,z)$ of degree at most 2 in $u$ such that \[ \operatorname{div}(h_i)=16P_i+E_i-19O,\qquad \deg E_i=3 \] This is checked exactly: the Taylor order at $P_i$ is 16, the maximum of the pole orders $3a+4b$ of the terms $z^au^b$ is exactly 19, and $\operatorname{Res}_u(F,h_i)$ is a constant times $(z+1)^{16}$ times a squarefree cubic, and this cubic does not vanish at $z=-1$ (\code{1.1.8f}). At the point at infinity put $t=z/u$, $w=1/u$. Then the curve is $t^4-w+(1+2t)w^3+w^4=0$, and the derivative in the $w$ direction at $(0,0)$ is $-1$, so there is only one branch, $w=t^4+\cdots\in\Z_2[[t]]$. At the points of $E_i$, the characteristic polynomials of $t$ and of $w$ are monic with every coefficient except the leading coefficient even (equivalently, each reduces to $T^3$ modulo 2), so $E_i$ lies in the residue disk of $O$ ($v_2(t)=1/3,1/3,1/3$ on $E_0$, and $1,1,4$ on $E_1$). In this parameter the differential is $(w,t,1)(1-tw'/w)(w^2-3)^{-1}dt$, with 2-adically integral coefficients. Therefore \[ \log G_i=-\tfrac1{16}\operatorname{Tr}_{E_i}\int_0^{t}\omega_\bullet \] and the trace is computed exactly from the Newton sums of the characteristic polynomial of $t$. We compute up to 156 terms with 44 digits. The discarded term of degree $n$ has valuation at least $\lfloor n/3\rfloor-v_2(n)-4\ge44$ ($n\ge157$), and the finite computation loses at most 11 digits, so at least 33 digits are guaranteed (\code{1.1.8f}). Modulo $2^{12}$, \[ 4\log G_0\equiv(3044,\,434,\,1012),\qquad 4\log G_1\equiv(4064,\,481,\,1506) \] The two logarithms are linearly independent, so $\operatorname{rank}J(\Q)\ge2$, and together with Lemma~\ref{lem:1.1.8rank} this gives $\operatorname{rank}J(\Q)=2$. The minimal valuation of the components of the cross product of the two vectors is $-2$, and multiplying it by 4 gives the primitive vector $a\equiv(12,14,9)\pmod{16}$. The subgroup generated by $G_0,G_1$ has finite index in $J(\Q)$ modulo torsion, and the logarithm kills torsion, so $\omega$ kills all logarithms of $J(\Q)$. \end{proof} \paragraph{Step 5: two disks covering the integral points.} Modulo 2 we have $F\equiv z+1$, so $z$ is odd at an integral point. \begin{lem}\label{lem:1.1.8disk} Among the integer solutions of $F(u,z)=0$, the only one with $u$ even is $P_0$, and the only ones with $u$ odd are $P_1,P_2$. \end{lem} \begin{proof} \emph{The even disk.} If $u$ is even, substituting $z=-1+2t$ gives $4\mid u$. Writing $u=4a$, we get \[ a-16a^3=t-6t^2+8t^3-4t^4 \] For each $t\in\Z_2$ this equation has only one solution $a\in\Z_2$, by the contraction mapping principle. More generally, for a rational number $r$ with $03$, an affine singular point satisfies $x^3=-1/2$, $y^2=-1/3$, $9x+4y+6=0$ (the last equation comes from $12F-3xF_x-4yF_y=2(9x+4y+6)$). The norm of the critical value $(9x+4y+6)/6$ is $12527563/(2^8\cdot3^9)$, so the only odd bad prime is $p=12527563$. There the only singular point is $(x,y)\equiv(2189611,\,4469046)$, and the Hessian determinant $72x^2y\equiv4898948$ is a unit, so it is an ordinary node. The critical norm has valuation 1, so the thickness is 1. A plane quartic with only one node is geometrically irreducible, so the component group of the Jacobian is trivial, and the descent image is unramified at $p$. Therefore $S=\{\mathfrak p\mid2\}$ suffices (\code{1.1.9a}). \paragraph{Step 2: the class group.} \begin{lem}\label{lem:1.1.9class} $\Cl(K)[2]=0$. \end{lem} \begin{proof} We apply Proposition~\ref{thm:B} with $n=27$, $(r_1,r_2)=(3,12)$, the $D_K$ above, and $T=\log10^9$. The window is a step function on 96 cells of equal width, and its heights are positive integers, symmetric from left to right (the input of \code{1.1.9d}). Two pieces of data are needed. \begin{enumerate}[label=(\roman*)] \item There are $50{,}851{,}991$ prime ideals of norm less than $10^9$ in total, and we count the number $a(q)$ for every norm (\code{1.1.9b}). For the primes $p\le31622$ and the primes of the polynomial discriminant we factor into prime ideals directly in the maximal order; for the remaining primes we count the prime ideals of degree 1 by $\deg\gcd(f,X^p-X)$. \item All $879{,}491$ prime ideals of norm at most $1.35\cdot10^7$ are principal ideals (\code{1.1.9c}). For $879{,}341$ of them, $(g)=\mathfrak p$ is checked for a single element $g$ by an exact ideal equality. The remaining 150 are checked with a product $\prod_j\alpha_j^{e_j}$: we factor the principal divisor of each atomic element $\alpha_j$ exactly into prime ideals and add the exponents in integer arithmetic, and then the exponent of $\mathfrak p$ is 1 and all the others are 0. (\code{1.1.9c} constructs the witnesses anew, and in that case every ideal is verified by a single witness for Lemma~\ref{lem:C}, without product witnesses.) Therefore we set $k(q)=a(q)$ if $q\le1.35\cdot10^7$, and $k(q)=0$ otherwise. Since $(1.35\cdot10^7)^2>10^9$, all terms with $m\ge2$ receive a positive sign. \end{enumerate} With these data the computation gives the integral term $0.0631019954\ldots$ and \[ B^-(f)=3.3541015\ldots\cdot10^{-5}>0 \] (\code{1.1.9d}). The computation uses exact rational coefficients and directed interval arithmetic (288 bits), and the radius of the final interval is less than $10^{-61}$. Hence the sign is certain, even though the margin is small. By Proposition~\ref{thm:B}, $\Cl(K)[2]=0$. \end{proof} \paragraph{Step 3: the rank and the index.} \begin{lem}\label{lem:1.1.9rank} $J(\Q)[2]=0$ and $\operatorname{rank}J(\Q)=2$. Moreover, $\langle G_0,D\rangle$ has odd index in $J(\Q)$. \end{lem} \begin{proof} The hypotheses of Proposition~\ref{thm:A} were checked above. Hence $J(\Q)[2]=0$, $\delta$ is injective, and at an affine point $\delta(P-O)$ is the square class of $mx(P)+\eta-y(P)$ (the value of $t^4(mx+\eta-y)$ at $O$ is 1). By Step 1 and Lemma~\ref{lem:1.1.9class}, the image lies in the space of dimension \[ \dim_{\F_2}K(S,2)=(r_1+r_2)+\#S=15+2=17 \] Seventeen explicit elements form a basis of this space. We factor the principal divisors of 161 atomic elements exactly into 141 prime ideals to show that all valuations outside 2 are even, and we check that the matrix of quadratic characters has rank 17 (\code{1.1.9e}). We impose the conditions one at a time. The condition that the norm is a square reduces the dimension from 17 to 15. At a good odd prime $p$ the local image is the image of $V/(\mathrm{Frob}-1)V\to E/(\mathrm{Frob}-1)E$. Here $V=J[2]$, and $E=\F_2^{27}$ is the permutation module on the 27 bitangents. We bound its dimension from above as follows: among the elements of the orthogonal group (of order $51{,}840$) that preserves the quadratic form on $V$, we take all $T$ that induce on the bitangents the same cycle type as Frobenius, compute for each the dimension of the image of $V/(T-1)V\to E/(T-1)E$, and take the maximum (\code{1.1.9e}). The images of points obtained by Hensel lifting points of $C(\F_p)$ attain this upper bound, so we know the local image exactly. The results are as follows. \[ \begin{array}{c|c|c|c} p & \text{degrees of the factors of }f\bmod p & \text{dimension of local image} & \text{remaining dimension}\\\hline 13 & (9,9,9) & 0 & 13\\ 17 & (3,12,12) & 0 & 11\\ 43 & (9,9,9) & 0 & 9\\ 53 & (9,9,9) & 0 & 7\\ 59 & (3,3,3,3,3,3,3,3,3) & 2 & 4\\ 67 & (2,5,5,5,10) & 1 & 3 \end{array} \] At 2 the bitangent orbits have sizes $24$ and $3$, so $J(\Q_2)[2]=0$ and $\dim J(\Q_2)/2J(\Q_2)=3$. The spaces of square classes of the two local components have dimensions 26 and 5. The images of three local divisor classes are independent, so they make up the whole local image. The first is $P_0$ ($\delta=\eta-m$). The second is the point with $t=2$ in the disk at infinity, where $w$ is computed modulo $2^{80}$ ($v_2(w)=4$). The third is the trace divisor $P+P^\sigma-2O$ formed from the point $P=(d,y)$, $y\equiv d\pmod2$ (Hensel; $F_y$ is a unit), over the unramified quadratic extension $\Q_2(d)$, $d^2+d+1=0$. For the second and the third, the relative error of the approximation has valuation at least 640 and at least 80 in the two local fields, which is more than the required depths 19 and 5. Keeping only the elements of the 3-dimensional space that map into this local image leaves the 2-dimensional space spanned by (in the numbering of the 17 basis elements) \[ e_3+e_5,\qquad e_4+e_5+e_6+e_{17} \] Therefore $\operatorname{rank}J(\Q)\le2$. The lower bound and the index are as follows. We have $\delta(G_0)=\eta-m$, and at a point $Q$ of $D_{\rm eff}$, $mx+\eta-y=-(bx+a)$ ($a=-2-\eta$, $b=1-m$), so $\delta(D)=a^2+2ab+3b^2$. Computing the quadratic characters at the simple roots $m\equiv-1\pmod5$ and $m\equiv-21\pmod{37}$ of $f$, the rows, in the order $(G_0,D)$, are $(1,1)$ and $(1,0)$ (\code{1.1.9e}). The determinant is 1, so $G_0,D$ are independent in $J(\Q)/2J(\Q)$. Since $J(\Q)[2]=0$, we have $\dim J(\Q)/2J(\Q)=\operatorname{rank}J(\Q)\le2$; hence the rank is 2, and $G_0,D$ generate $J(\Q)/2J(\Q)$. That is, the index of $\langle G_0,D\rangle$ is odd. \end{proof} \paragraph{Step 4: 2-adic logarithms and the annihilating differential.} We take the basis of holomorphic differentials $\omega_\bullet=(\omega_0,\omega_1,\omega_2)=(1,x,y)\,dx/(1+3y^2)$. \begin{lem}\label{lem:1.1.9log} There is a primitive vector $a=(a_0,a_1,a_2)\in\Z_2^3$ with $a\equiv(6,10,9)\pmod{16}$ such that $\omega=a_0\omega_0+a_1\omega_1+a_2\omega_2$ annihilates all 2-adic logarithms of $J(\Q)$. The vector $a$ is known exactly to at least 50 digits. \end{lem} \begin{proof} There are polynomials $h_0,h_D$ of degree at most 2 in $y$ such that \[ \operatorname{div}(h_0)=16P_0+E_0-19O,\qquad \operatorname{div}(h_D)=8D_{\rm eff}+E_D-19O,\qquad \deg E_0=\deg E_D=3 \] This is checked by exact Taylor expansions at the given points, by the pole orders (the maximum of the weights $3i+4j$ of the terms $x^iy^j$ is 19, and the weights are all distinct), and by the factorization of the norm polynomials (\code{1.1.9f}). On $E_0$ and $E_D$, the valuations of the coefficients of the characteristic polynomial of $t$, from the lowest degree up, are $(1,3,3,0)$ and $(2,2,3,0)$, so $v_2(t)=1/3$ and $2/3$. For the characteristic polynomial of $w$ they are $(4,7,4,0)$ and $(8,9,5,0)$, so $w$ also has positive valuation. Hence $E_0,E_D$ lie in the residue disk of $O$. There the differential is $(w,t,1)(1-tw'/w)(w^2+3)^{-1}dt$, with 2-adically integral coefficients, so \[ \log G_0=-\tfrac1{16}\operatorname{Tr}_{E_0}\int_0^t\omega_\bullet,\qquad \log D=-\tfrac18\operatorname{Tr}_{E_D}\int_0^t\omega_\bullet \] We compute with 216 terms and 64 digits. The discarded term of degree $n$ satisfies $v_2(\operatorname{Tr}(t^n)/16n)\ge\lceil n/3\rceil-v_2(n)-4\ge65$ ($n\ge217$), and the finite computation loses at most 11 digits, so at least 53 digits are guaranteed (\code{1.1.9f}). Modulo $2^{12}$, \[ 4\log G_0\equiv(3988,\,2194,\,1012),\qquad 4\log D\equiv(1848,\,3542,\,3892) \] The two logarithms are independent, and the minimal valuation of the components of their cross product is $-1$. Multiplying it by 2 gives the primitive vector $a\equiv(6,10,9)\pmod{16}$. By Lemma~\ref{lem:1.1.9rank}, $\langle G_0,D\rangle$ has finite index in $J(\Q)$ modulo torsion, so $\omega$ kills all logarithms of $J(\Q)$. \end{proof} \paragraph{Step 5: the disk covering the integral points.} Modulo 2 we have $F\equiv x+1$, so $x$ is odd. Then $x^4+2x+1\equiv0\pmod4$. If $y$ is odd, then $y^3+y\equiv2\pmod4$, a contradiction; if $y$ is even, then $y^3+y\equiv0\pmod4$ gives $4\mid y$. Hence every integral point can be written as \[ x=-1+2t,\qquad y=4b,\qquad b+16b^3=t-6t^2+8t^3-4t^4\qquad(t,b\in\Z_2) \] \begin{lem}\label{lem:1.1.9disk} Let $H(t)$ be the antiderivative of $\omega$ starting at $P_0$ ($t=0$), divided by 8. The only zeros of $H$ in $\Z_2$ are $t=0$ and a single odd $t_*$, and $t_*\equiv3{,}366{,}055{,}599\pmod{2^{32}}$. The coordinates $(c_0,c_D)$ of the logarithm of the point $P(t_*)$ with respect to $\log G_0,\log D$ lie in $\Z_2$, and $(c_0,c_D)\equiv(11,10)\pmod{32}$. \end{lem} \begin{proof} For every $R$ with $1\le R<2^{2/3}$ and $|t|\le R$, the map $b\mapsto t-6t^2+8t^3-4t^4-16b^3$ is a contraction that preserves $|b|\le R$, so $b(t)$ is analytic on a disk of radius greater than 1, and $|b|\le R$. Pulling back $\omega$ gives \[ H'(t)=\frac{(a_0-a_1)/4+(a_1/2)t+a_2b(t)}{1+48b(t)^2},\qquad H(0)=0 \] On this disk $|H'|\le R$, so the Cauchy inequality gives \[ v_2([t^n]H)\ge\Bigl\lceil\frac{2(n-2)}3\Bigr\rceil-v_2(n) \] This bound is at least 1 if $n\ge3$ and $n\ne4$. The remaining coefficients are exactly \[ [t]H=\frac{a_0-a_1}4,\qquad [t^2]H=\frac{a_1+2a_2}4,\qquad [t^4]H=36a_0-42a_1-14a_2 \] and for $a\equiv(6,10,9)\pmod{16}$ the first two are odd and the last has valuation 1. Hence $H\equiv t+t^2\pmod2$, and by Strassmann's theorem there are at most two zeros. One of them is $t=0$, and since $H(t)/t\equiv1+t\pmod2$, Hensel's lemma gives exactly one odd zero $t_*$. Using $a$ to 50 digits and the antiderivative with 120 terms (the discarded terms have valuation at least 77), we get $t_*\equiv3{,}366{,}055{,}599\pmod{2^{32}}$ (\code{1.1.9g}). The vector $\log[P(t_*)-O]=\log G_0+\int_{P_0}^{P(t_*)}\omega_\bullet$ is annihilated by $\omega$, so it is a $\Q_2$-linear combination of $\log G_0,\log D$. When we solve for these coordinates with a minor of order 2, the entries of its inverse have valuation at least $-1$, and the $t$-derivative of the tiny integral is divisible by 2. Hence the error $2^{32}$ in $t_*$ changes the coordinates only by multiples of $2^{32}$. The result is $(c_0,c_D)\equiv(10{,}965{,}931,\,8{,}513{,}226)\pmod{2^{24}}$, that is, $(11,10)\pmod{32}$ (\code{1.1.9g}). \end{proof} \paragraph{Step 6: a sieve at 23.} \begin{lem}\label{lem:1.1.9sieve} $P(t_*)$ is not a rational point. \end{lem} \begin{proof} $C$ has good reduction at 23. The point counts are $\#C(\F_{23^k})=26,622,12044$ ($k=1,2,3$), and $\#J(\F_{23})=14432=451\cdot32$ (\code{1.1.9h}). Suppose that $P=P(t_*)$ is a rational point. By Lemma~\ref{lem:1.1.9rank} there is an odd $k$ with $k[P-O]=n_0G_0+n_DD+\tau$, where $\tau$ is torsion of odd order. Taking logarithms gives $n_0/k=c_0$ and $n_D/k=c_D$. In $J(\F_{23})$, multiplication by 451 kills the odd part and is invertible on the 2-part (of order 32), so we can divide by $k$ modulo 32. Therefore, in the form of \cite{BS10}, \[ 451\,[\bar P-\bar O]=451\,(11\,\bar G_0+10\,\bar D)\qquad(\text{in }J(\F_{23})) \] must hold. Comparing, by exact Jacobian arithmetic, the right-hand side with the value of the left-hand side for all 26 points of $C(\F_{23})$ ($O$ and 25 affine points), we find no matching point (\code{1.1.9h}). This is a contradiction. \end{proof} \begin{proof}[Proof of Proposition~\ref{prop:1.1.9}] An integral point $P$ lies in the disk of Step 5. Its parametrization is an analytic embedding of an open disk of radius greater than 1, so by \cite[Section 3]{KRZB16}, $8H(t(P))=\omega(\log[P-P_0])$. Since $[P-P_0]\in J(\Q)$, this value is 0 by Lemma~\ref{lem:1.1.9log}. By Lemma~\ref{lem:1.1.9disk}, $t(P)=0$ or $t(P)=t_*$, and by Lemma~\ref{lem:1.1.9sieve} the latter case does not occur. Therefore $P=P_0$. \end{proof} \subsection{\texorpdfstring{$x^4+y^3+2x+y-1=0$ \ ($H=31$)}{x^4+y^3+2x+y-1=0 (H=31)}}\label{sec:1.1.10} \begin{thm}\label{thm:1.1.10} The only integer solutions of $x^4+y^3+2x+y-1=0$ are $(x,y)=(-1,1),(1,-1)$. \end{thm} \paragraph{Setup.} Let $J$ denote the Jacobian of the smooth projective model of the curve. We do not change coordinates. $C$ is the smooth plane quartic (genus 3; Step 1) defined by \[ X^4+Y^3Z+2XZ^3+YZ^3-Z^4=0,\qquad F(x,y)=x^4+y^3+2x+y-1 \] and $J$ is its Jacobian. The only point at infinity is $O=[0:1:0]$. The tangent line $Z=0$ meets the curve there with multiplicity 4, since $X^4=0$, so $O$ is a rational hyperflex. Near $O$ we set $t=X/Y$, $w=Z/Y$; then the curve is $t^4+w+2tw^3+w^3-w^4=0$ and $w=-t^4+O(t^{12})$. We write $A=(-1,1)$, $B=(1,-1)$, and in $J(\Q)$ we use the same letters for $[A-O]$, $[B-O]$. The proof runs as follows. Unlike in other sections, we do not show $\Cl(K)[2]=0$. Instead we show that an element of the 2-Selmer group outside $\langle A,B\rangle$ produces a quadratic class character with special properties (Lemma~\ref{lem:1.1.10strict}), and we exclude this character with the explicit formula (Proposition~\ref{prop:1.1.10rank}). Then the rank is 2. Next, Chabauty--Coleman at $p=3,5$ reduces the possible new points to two residue classes (Lemmas~\ref{lem:1.1.10p3}, \ref{lem:1.1.10p5}), and the Mordell--Weil sieve excludes both (Lemma~\ref{lem:1.1.10sieve}). \paragraph{Step 1: the bitangent field and descent.} In characteristic $p>3$, an affine singular point satisfies $2x^3+1=0$ and $G(x)=243x^2-324x+124=0$, and $\operatorname{Res}(2x^3+1,G)=31\cdot433\cdot2341$. In characteristic 3 we have $F_y\equiv1$, and at $O$ the derivative in the $w$ direction is 1 in every characteristic. At each of the three primes there is exactly one singular point: $(x,y)\equiv(22,14)\pmod{31}$, $(6,421)\pmod{433}$, $(2236,823)\pmod{2341}$. The Hessian determinant $72x^2y$ is a unit, so the point is an ordinary node. Lifting the gradient equations modulo $p^4$ shows that the critical value has valuation exactly 1, so the thickness is 1. Since there is only one node, the special fiber is geometrically irreducible (if it were reducible, Bezout would give at least three intersection points), and the component group is trivial. Hence the descent image is unramified at the odd primes (\code{1.1.10a}). The 27 finite bitangents have the form $y=mx+\eta$, where $m$ is a root of \begin{align*} f(m)={}&m^{27}+480m^{19}-2688m^{18}-17280m^{15}-34560m^{11}-141312m^{10}\\ &-162816m^{9}+165888m^{7}+221184m^{6}-126976m^{3}+49152m-32768. \end{align*} $f$ is irreducible, and in $K=\Q[m]/(f)$ we have \[ \eta=\frac{m(11m^{12}+112m^4+224m^3-192)}{8(13m^9-16m+16)},\qquad F(x,mx+\eta)=\Bigl(x^2+\frac{m^3}2x+\frac{12m^2\eta-m^6}8\Bigr)^2. \] For an arbitrary line $y=mx+z$, put $a=m^3/2$ and $b=(12m^2z-m^6)/8$. The remaining coefficient conditions are $2ab=3mz^2+m+2$ and $b^2=z^3+z-1$. Their resultant with respect to $z$ (left side minus right side, in this order) is $f(m)/4096$, so these are all the finite bitangents. $K$ has $n=27$, $(r_1,r_2)=(3,12)$, $D=2^{30}\cdot31^6\cdot433^6\cdot2341^6$, and there are two prime ideals above 2, with $(e,f)=(8,3),(1,3)$ (\code{1.1.10a}). Equip $V=J[2]$ with the quadratic form $q$ determined by $O$. Then the 27 bitangents are indexed by $\Omega=\{w\in V:q(w)=0,\ w\ne0\}$, and the descent map is given by $\iota:V\to\F_2^\Omega$, $c\mapsto(\langle c,w\rangle)_{w\in\Omega}$. In suitable coordinates, $q(z)=z_1z_2+z_3z_4+z_5z_6+z_5+z_6$. \begin{lem}\label{lem:1.1.10descent} $J(\Q)[2]=0$, and $H^1(\Q,J[2])\to K^\times/K^{\times2}$ is injective. Composing with the Kummer map $\kappa:J(\Q)/2J(\Q)\to H^1(\Q,J[2])$, for a rational point $P\ne O$ the image of $\kappa([P-O])$ is the square class of $mx(P)+\eta-y(P)$. The image of the 2-Selmer group $\mathrm{Sel}$ lies in \[ K(\{2\},2)=\{[u]\in K^\times/K^{\times2}: v_{\mathfrak p}(u)\text{ is even for every }\mathfrak p\text{ not above 2}\}. \] \end{lem} \begin{proof} Since $f$ is irreducible, the Galois group acts transitively on $\Omega$. Proposition~\ref{thm:A} gives $J(\Q)[2]=0$. Its proof (the argument that a Sylow 3-subgroup containing a Sylow 3-subgroup of a point stabilizer is transitive on $\Omega$) gives injectivity not only on $J(\Q)/2J(\Q)$ but on all of $H^1(\Q,J[2])$. The value of $t^4(mx+\eta-y)$ at $O$ is 1, so the normalization is correct. A Selmer element maps to a local class that is unramified at every odd prime (at the bad primes, by the component group computation above), so its valuations away from 2 are even. \end{proof} \paragraph{Step 2: the 2-adic image and 17 known classes.} The square class spaces of the two components of $K\otimes\Q_2$ have dimensions 26 and 5, and their sum $\mathcal T_2$ has dimension 31. The 2-adic bitangent orbits have sizes $3$ and $24$, so there is no invariant codeword of weight 12 or 16. Hence $J(\Q_2)[2]=0$, and the image $\mathcal I_2\subset\mathcal T_2$ of $J(\Q_2)/2J(\Q_2)$ is 3-dimensional. \begin{lem}\label{lem:1.1.10dyadic} There are explicit elements $u_1,\dots,u_{17}$ of $K(\{2\},2)$ whose 2-adic images span a 17-dimensional space $\mathcal L_2$, and $\mathcal L_2\cap\mathcal I_2$ is the 2-dimensional space spanned by the images of $[A-O]$ and $[B-O]$. \end{lem} \begin{proof} Each $u_i$ is a product of some of 170 polynomials in $m$ with rational coefficients. We factor its principal ideal and check that all valuations away from 2 are even. The 2-adic images are computed from the parity of the valuations and the unit square class modulo $\mathfrak p^{2e+1}$ (elements of $1+\mathfrak p^{2e+1}$ are squares), and they have rank 17. $\mathcal I_2$ is spanned by the images of three actual 2-adic points: the point lifting $y\equiv0\pmod2$ at $x=-63$ ($F_y=3y^2+1$ is odd), and the points lifting $x\equiv1\pmod2$ at $y=-31,-29$ (the derivative $2x^3+1$ of $F(1+2u,y)/4\in\Z_2[u]$ is odd). We compute each point modulo $2^{48}$; the valuation of the approximation error is larger than the valuation of the image plus $2e$, so the square classes are exact. The three images have rank 3. The 17 and the 3 together have rank 18, so $\dim(\mathcal L_2\cap\mathcal I_2)=2$. The images of $A$, $B$ lie in both spaces and are independent (\code{1.1.10a}). \end{proof} \paragraph{Step 3: the localization image and strictification of Selmer classes.} Adding the three real places, we set $\mathcal T=\mathcal T_2\oplus\bigoplus_{v\ \text{real}}K_v^\times/K_v^{\times2}$ (dimension 34), and give it the $\F_2$-valued pairing equal to the sum of the Hilbert symbols at the 2-adic and real places. Each local Hilbert pairing is nondegenerate, so this pairing is nondegenerate too. \begin{lem}\label{lem:1.1.10loc} Let $\mathcal L$ be the span of the images of $u_1,\dots,u_{17}$ in $\mathcal T$. Then the image of $K(\{2\},2)$ in $\mathcal T$ is exactly $\mathcal L$. Moreover, the projection $\mathcal L\to\mathcal T_2$ is injective. \end{lem} \begin{proof} If $u,v\in K(\{2\},2)$, then at the odd finite places both are units modulo squares, so their Hilbert symbol is 1; it is also 1 at the complex places. By Hilbert reciprocity, the image of $K(\{2\},2)$ is isotropic in $\mathcal T$. By Lemma~\ref{lem:1.1.10dyadic}, $\mathcal L\to\mathcal T_2$ is injective and $\dim\mathcal L=17$. Since $\mathcal L$ is a 17-dimensional isotropic subspace of a 34-dimensional nondegenerate space, $\mathcal L=\mathcal L^\perp$. By reciprocity, the image of an element of $K(\{2\},2)$ is orthogonal to $\mathcal L$, so it lies in $\mathcal L$. \end{proof} \begin{lem}\label{lem:1.1.10strict} Suppose $\mathrm{Sel}\ne\langle\kappa(A),\kappa(B)\rangle$. Then there is a nontrivial character $\chi:\Cl(K)\to\{\pm1\}$ of the ordinary class group, and for every prime $p>15{,}000{,}000$ there are a Frobenius $g\in O(V,q)$ and $c\in V$ such that \[ \sum_{\mathfrak p\mid p,\ N\mathfrak p=p}\chi(\mathfrak p)=\sum_{w\in\Omega,\ gw=w}(-1)^{\langle c,w\rangle}. \] \end{lem} \begin{proof} Choose $\gamma\in\mathrm{Sel}\setminus\langle\kappa(A),\kappa(B)\rangle$. By Lemma~\ref{lem:1.1.10descent}, the image of $\gamma$ lies in $K(\{2\},2)$. Hence, by Lemma~\ref{lem:1.1.10loc}, its 2-adic image lies in $\mathcal L_2$, and by the Selmer condition it also lies in $\mathcal I_2$. By Lemma~\ref{lem:1.1.10dyadic}, there are $\epsilon,\epsilon'\in\F_2$ such that the 2-adic image of $\gamma'=\gamma-\epsilon\kappa(A)-\epsilon'\kappa(B)$ is 0. Then $\gamma'\ne0$, and it is still a Selmer element. By Lemma~\ref{lem:1.1.10loc}, the image of $\gamma'$ in $\mathcal T$ lies in $\mathcal L$, and its 2-adic component is 0, so its real components are also 0. Let $u$ be the image of $\gamma'$. Then $u\ne1$, and $K(\sqrt u)/K$ is unramified at the odd places (even valuations) and split at the 2-adic and real places. Therefore its character $\chi$ is a nontrivial quadratic character of the ordinary class group. If $p>15{,}000{,}000$, then $C$ has good reduction at $p$ and $K$ is unramified at $p$. $\gamma'$ is unramified at $p$, so it is determined by its cocycle value $c$ at Frobenius. A degree-one prime ideal $\mathfrak p\mid p$ of $K$ corresponds to a $w\in\Omega$ fixed by $g$, and the quadratic character of $u$ there is the $w$-component $\langle c,w\rangle$ of $\iota(c)$. \end{proof} \paragraph{Step 4: a lower bound for the character sum.} \begin{lem}\label{lem:1.1.10code} For a subspace $U\le V=\F_2^6$, let $r(U)=\#\{w\in U\setminus0:q(w)=0\}$. For every $c\in V$, \[ \sum_{w\in U,\ q(w)=0,\ w\ne0}(-1)^{\langle c,w\rangle}\ \ge\ \ell(r(U)), \qquad \begin{array}{c|cccccccccc} r&0&1&2&3&5&7&9&11&15&27\\\hline \ell(r)&0&-1&-2&-3&-3&-5&-3&-5&-5&-5 \end{array} \] In particular, this sum is always at least $-5$. \end{lem} \begin{proof} We examine all $2825$ subspaces of $V$ (by dimension: $1,63,651,1395,651,63,1$) and all 64 vectors $c$ (\code{1.1.10b}). \end{proof} On the right-hand side of Lemma~\ref{lem:1.1.10strict}, $U=\ker(g-1)$, and $r(U)$ is the number of roots of $f$ modulo $p$, $r(p)=\deg\gcd(f,z^p-z)$. The polynomial discriminant is \[ 2^{390}3^{54}17^2\,31^6\,101^2\,433^6\,2341^6\,2969^2\,5227^2\,924151^2, \] and all its prime factors are smaller than $15{,}000{,}000$. Hence for $p>15{,}000{,}000$, $r(p)$ is exactly the number of degree-one prime ideals. \paragraph{Step 5: principal ideals.} \begin{lem}\label{lem:1.1.10princ} All $971{,}856$ prime ideals of norm at most $M=15{,}000{,}000$ are principal. \end{lem} \begin{proof} We construct the prime ideals exhaustively as follows. For $p>\sqrt M$ not dividing the denominators of the integral basis, only degree-one prime ideals can have norm at most $M$, and for each root $r$ of $f$ modulo $p$ the kernel of $b_i\mapsto b_i(r)$ is a degree-one prime ideal. For the remaining $p$ we factor directly in the maximal order. For each $\mathfrak p$ we check an element $g\in\mathfrak p$ exactly. If $|N(g)|=N\mathfrak p$, then $(g)=\mathfrak p$. Otherwise we prove the prime factorization $|N(g)|/N\mathfrak p=\prod r^{e_r}$ and check that $r^{e_r}X$, there are no terms with $m\ge2$ for ideals of norm greater than $M$ (\code{1.1.10d}). \item For the $38{,}465{,}791$ primes with $MR(1-\frac1{5\log^2R})$ gives \[ \sum_{R0$ contradicts the inequality above. Hence $\mathrm{Sel}=\langle\kappa(A),\kappa(B)\rangle$. The 2-adic images of $\kappa(A),\kappa(B)$ are independent, so $\dim\mathrm{Sel}=2$. Since $J(\Q)[2]=0$, $\operatorname{rank}J(\Q)\le2$, and since $A,B$ are independent, the rank is 2. \end{proof} \paragraph{Step 7: Chabauty--Coleman at $p=3,5$.} Set $(\omega_0,\omega_1,\omega_2)=(1,x,y)\,dx/(3y^2+1)$, and use $L(P)=\int_O^P(\omega_0,\omega_1,\omega_2)$ as the abelian logarithm. At $O$ this basis is $-(w,t,1)H_w^{-1}dt$ ($H_w=1+6tw^2+3w^2-4w^3$), with integral coefficients. The plane model is smooth at 3 and 5. Since $\#C(\F_{3^k})=4,10,28$ and $\#C(\F_{5^k})=3,23,120$ ($k=1,2,3$), we have $\#J(\F_3)=28$, $\#J(\F_5)=64$ (\code{1.1.10g}). By \cite[Appendix]{Katz81}, the order of the rational torsion divides $\gcd(28,64)=4$, and since $J(\Q)[2]=0$, $J(\Q)$ is torsion-free. Since $\#J(\F_p)$ is prime to $p$, multiplying an element of $J(\Q_p)$ by $\#J(\F_p)$ puts it in the kernel of reduction, and by Lemma~\ref{lem:O} its logarithm lies in $p\Z_p^3$. Hence $L(J(\Q_p))\subset p\Z_p^3$. For integers $n,m$ set $N=n+m+5$, and let $h_{n,m}\in L(NO)$ be the primitive integral generator of the kernel (which is 1-dimensional) of the linear conditions $\operatorname{ord}_A\ge n+2$, $\operatorname{ord}_B\ge m$. The basis is $x^iy^j$ ($j\le2$, $3i+4j\le N$), and then $\operatorname{div}(h_{n,m})=(n+2)A+mB+E_{n,m}-NO$ with $\deg E_{n,m}=3$ (\code{1.1.10h}). \begin{lem}\label{lem:1.1.10p3} Modulo 81, $L(A)\equiv(0,78,51)$ and $L(B)\equiv(78,15,66)$. $A,B$ are independent, and the index of $\langle A,B\rangle$ is prime to 3. If there is a new rational point $P\notin\{O,A,B\}$, then the 3-adic coordinates of $[P-O]=nA+mB$ satisfy $(n,m)\equiv(6,3)$ or $(3,4)\pmod9$, and in the latter case $P\equiv(0,2)\pmod3$. \end{lem} \begin{proof} For $(n,m)=(3,1),(2,10)$, $E_{n,m}$ has one point in each of the residue disks of $A,B,O$ (the constant term of the local expansion is divisible by 3 and the linear term is a unit). Let $\tau_{n,m}$ be the sum of the three tiny integrals. Then $6L(A)+2L(B)=-\tau_{3,1}$, $5L(A)+11L(B)=-\tau_{2,10}$, and modulo 81, $\tau_{3,1}\equiv(6,69,48)$, $\tau_{2,10}\equiv(33,12,72)$. For a parameter $u\in3\Z_3$ the integral is $a_0u+a_1u^2/2+a_2u^3/3$ modulo 81, and the terms with $k\ge4$ have valuation $k-v_3(k)\ge4$. The determinant 56 is a unit, so $L(A),L(B)$ are as stated (\code{1.1.10h}). $L(A)/3,L(B)/3$ are independent modulo 3, so $A,B$ are independent. If $3x=nA+mB$ ($x\in J(\Q)$), then $L(x)\in3\Z_3^3$ gives $nL(A)/3+mL(B)/3\equiv0\pmod3$, so $n\equiv m\equiv0\pmod3$; since $J(\Q)$ is torsion-free, the index is prime to 3. Normalizing the cross product gives the annihilating differential $\alpha_3=(a+bx+y)\omega_0$ with $(a,b)\equiv(26,17)\pmod{27}$ (since the rank is 2, it kills all of $J(\Q)$). Among the points of $C(\F_3)=\{O,(2,1),(1,2),(0,2)\}$, $\alpha_3\bmod3=(2+2x+y)\omega_0$ is nonzero at $O$, $A\equiv(2,1)$ and $(0,2)$, and is zero at $B\equiv(1,2)$. The antiderivative of a differential with unit constant term is injective on $3\Z_3$, so the disks of $O$ and $A$ contain only the known points. In the disk of $B$, write $u=x-1$ and $\alpha_3=(f_0+f_1u+f_2u^2+\cdots)du$; then $(f_0,f_1,f_2)\equiv(24,14,12)\pmod{27}$. With $u=3v$, the antiderivative starting at $B$ is $u$ times an analytic function, and this analytic function divided by 3 is $8+7v+3v^2$ modulo 9 and $2+v$ modulo 3. Hence there is exactly one zero other than $B$, with $v\equiv1\pmod9$, $L(P)\equiv(18,0,18)\pmod{27}$, $(n,m)\equiv(6,3)\pmod9$. In the disk of $(0,2)$ we use the base point $C_0=(\xi,-1)$, where $\xi^3+\xi^2+\xi+3=0$, $\xi\in3\Z_3$ ($\xi\equiv42\pmod{81}$). The line $y=-1$ cuts out $B+C_0+R_1+R_2-4O$, and $R_1,R_2$, given by the roots of $u^2+(3+\xi)u+(3+2\xi+\xi^2)$ in $u=x-1$, lie in the disk of $B$ (Eisenstein, valuation $1/2$). The power sums $s_k=u(R_1)^k+u(R_2)^k$ satisfy $v_3(s_k)\ge\lceil k/2\rceil$, so the terms with $k\le6$ determine the trace modulo 27, and $L(C_0)=-3L(B)-\tau\equiv(0,24,9)\pmod{27}$. With $u=x-\xi=3v$ we get $\alpha_3(L(P))/3\equiv4+7v\pmod9$, so there is exactly one zero, with $v\equiv2\pmod9$, $L(P)\equiv(15,24,12)\pmod{27}$, $(n,m)\equiv(3,4)\pmod9$ (\code{1.1.10h}). A new rational point is one of these two zeros. \end{proof} \begin{lem}\label{lem:1.1.10p5} Modulo 125, $L(A)\equiv(80,60,120)$ and $L(B)\equiv(115,120,110)$, and the index of $\langle A,B\rangle$ is prime to 5. A new rational point $P$ reduces to $B$ modulo 5, and $(n,m)\equiv(3,3)\pmod5$. Therefore $(n,m)\equiv(33,3)$ or $(3,13)\pmod{45}$. \end{lem} \begin{proof} On the tangent line at $A$, $8F(x,(x+3)/2)=(x+1)^2(8x^2-15x+31)$. Let $Q,Q'$ be the two points with $x=(15\pm\sqrt{-767})/16$, $y=(x+3)/2$. Then $2L(A)+L(Q)+L(Q')=0$, and the quadratic field is unramified at 5. For $(n,m)=(3,13),(7,9)$, $E_{n,m}$ has one point near each of $Q,Q',O$, so $3L(A)+13L(B)=-\tau_{3,13}$, $7L(A)+9L(B)=-\tau_{7,9}$, and modulo 125, $\tau_{3,13}\equiv(15,10,85)$, $\tau_{7,9}\equiv(30,0,45)$ (the integrals are taken up to $a_0u+a_1u^2/2$; the terms with $k\ge3$ have valuation $k-v_5(k)\ge3$). The determinant $-64$ is a unit, so we get the values above. The normalized logarithms are independent modulo 5, so the index is prime to 5, and the annihilating differential is $\alpha_5=(a'+b'x+y)\omega_0$, $(a',b')\equiv(11,0)\pmod{25}$ (\code{1.1.10h}). We have $C(\F_5)=\{O,(4,1),(1,4)\}$. $\alpha_5$ is a unit at $O$ and $A$, so their disks contain only the known points. At $B$ the constant coefficient is 15 modulo 25 and the linear coefficient is 4 modulo 5. Setting $u=x-1=5v$ and dividing the antiderivative by $25v$ gives $3+2v$ modulo 5 (the terms with $k\ge3$ satisfy $k-2-v_5(k)\ge1$). Hence there is one more zero, with $v\equiv1\pmod5$, $L(P)\equiv(10,15,15)\pmod{25}$, $(n,m)\equiv(3,3)\pmod5$. Combining this with Lemma~\ref{lem:1.1.10p3}, $(6,3)\bmod9$ becomes $(33,3)\bmod45$, and $(3,4)\bmod9$ becomes $(3,13)\bmod45$. \end{proof} \paragraph{Step 8: Mordell--Weil sieve.} \begin{lem}\label{lem:1.1.10sieve} $C(\Q)=\{O,A,B\}$. \end{lem} \begin{proof} Suppose there is a new rational point $P$. The reduction is good at 3, 5, 11, 13, and $\#J(\F_{11})=2520$, $\#J(\F_{13})=2912$ ($\#C(\F_{11^k})=18,146,1296$, $\#C(\F_{13^k})=16,216,2128$). Arithmetic in $J(\F_p)$ is done exactly by solving Riemann--Roch with the basis $x^iy^j$ ($j\le2$, $3i+4j\le m\le9$) of $L(mO)$, and the equality of two classes is decided by whether the effective representative of their difference is empty (\code{1.1.10g}). Following \cite{BS10}, we show the following in order. \begin{enumerate}[label=(\arabic*)] \item In $J(\F_5)$, $\operatorname{ord}B=64$ and $A=17B$. The characters of order 2 obtained by multiplying by 32 at 5 and by 1260 at 11 give the rows $(1,1)$, $(0,1)$ on $(A,B)$, so $\langle A,B\rangle$ is saturated at 2. Together with Lemmas~\ref{lem:1.1.10p3}, \ref{lem:1.1.10p5}, the index is prime to 2, 3, 5, and the coordinates $(n,m)$ are integral at these primes. Since $P\equiv B\pmod5$, $17n+m\equiv1\pmod{64}$. \item At 11, multiplication by 56 leaves the $\{3,5\}$-part, of order 45. Of the two candidates, only $(3,13)\bmod45$ matches the image of one of the 18 points of $C(\F_{11})$, and that point is $(3,4)$. Hence $P\equiv(0,2)\pmod3$ and $P\equiv(3,4)\pmod{11}$. Multiplication by 7 leaves the $\{2,3,5\}$-part, of order 360. Among the 8 pairs modulo 360 satisfying $(3,13)\bmod45$ and $17n+m\equiv1\pmod8$, only $(228,13)$ has the same image as $(3,4)$. Hence $n\equiv4$, $m\equiv5\pmod8$. \item The 7-parts of $J(\F_3)$ and $J(\F_{11})$ have order 7. Projecting by multiplication by 4 and by 360, respectively, and normalizing the image of $A$ to 1, the images of $B$ are 4 and 3, and the images required for the point are 2 at $(0,2)$ and 1 at $(3,4)$. The determinant is $6\not\equiv0\pmod7$, so the group is saturated at 7 as well, and $n\equiv5$, $m\equiv1\pmod7$. \item In $J(\F_{13})$, multiplication by 13 leaves the $\{2,7\}$-part, of order 224. Reducing the conditions above modulo 32 and 7, there are four pairs modulo 224: $(68,29)$, $(12,85)$, $(180,141)$, $(124,197)$. None of their images equals the image of any of the 16 points of $C(\F_{13})$ (including $O$). \end{enumerate} This is a contradiction. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.1.10}] By Proposition~\ref{prop:1.1.10rank}, $\operatorname{rank}J(\Q)=2$, and as shown in Step 7, $J(\Q)$ is torsion-free. By Lemma~\ref{lem:1.1.10sieve}, $C(\Q)=\{O,A,B\}$. Since $O$ is the only point at infinity, the only integer solutions are $A=(-1,1)$ and $B=(1,-1)$. \end{proof} \subsection{\texorpdfstring{$x^4+y^3+xy-y-1=0$ \ ($H=31$)}{x^4+y^3+xy-y-1=0 (H=31)}}\label{sec:1.1.11} \begin{thm}\label{thm:1.1.11} The only integer solutions of $x^4+y^3+xy-y-1=0$ are $(x,y)=(-1,0),(1,0)$. \end{thm} \paragraph{Setup.} Setting $u=-x$, $v=y$, the equation becomes \[ C:\quad G(u,v):=u^4+v^3-(u+1)v-1=0. \] The projective completion $u^4+v^3w-(u+w)vw^2-w^4=0$ is a smooth plane quartic (genus 3), and its only bad primes are $2$ and $q_0=13839869$ (\code{1.1.11a}). At the point at infinity $O=(0:1:0)$ the line $w=0$ meets the curve with multiplicity 4, since $u^4=0$, so $O$ is a rational hyperflex. Near $O$ we use $t=u/v$, $w=1/v$. In this chart the curve is \[ t^4+w-tw^2-w^3-w^4=0, \] and the partial derivative with respect to $w$ is 1 at the origin, so $w=-t^4+t^9-t^{12}+\cdots\in\Z[[t]]$. Hence $u$ and $v$ have poles of order 3 and 4 at $O$, respectively. \begin{prop}\label{prop:1.1.11} $C(\Q)=\{O,P_+,P_-\}$, where $P_+=(1,0)$, $P_-=(-1,0)$. \end{prop} Since $(u,v)=(1,0),(-1,0)$ correspond to $(x,y)=(-1,0),(1,0)$, Theorem~\ref{thm:1.1.11} follows directly from this proposition. Since $G(-1,v)=v^3$, we have $\operatorname{div}(u+1)=3P_--3O$, and $[P_--O]$ is a torsion element. The proof has five steps. We fix the bitangent field $K$ (Step 1), show $\Cl(K)[2]=0$ with Proposition~\ref{thm:B} (Lemma~\ref{lem:1.1.11class}), and show by 2-descent that the rank is at most 1 (Lemma~\ref{lem:1.1.11rank}). Next we compute the 3-adic logarithm of $[P_+-O]$ from one principal divisor and show that the rank is exactly 1 (Lemma~\ref{lem:1.1.11log}). We finish with Chabauty--Coleman at $p=3$. \paragraph{Step 1: the bitangent field.} If the line $v=mu+b$ is a bitangent, then $u^4+(mu+b)^3-(u+1)(mu+b)-1=(u^2+c_1u+c_0)^2$ with $c_1^2\ne4c_0$. The slopes of the 27 bitangents other than the line at infinity are the roots of \[ \begin{aligned} f(m)={}&m^{27}+48m^{22}-480m^{19}+144m^{17}-17280m^{15}+2304m^{14}-832m^{12}-34560m^{11}\\ &+64512m^{10}+18944m^{9}-168448m^{7}-49152m^{6}-27648m^{5}+8192m^{4}-94208m^{3}\\ &+1024m^{2}-4096m+4096. \end{aligned} \] $f$ is irreducible, and $K=\Q[m]/(f)$ has degree 27, signature $(3,12)$ and absolute discriminant $D=2^{28}\cdot13839869^{6}$ (\code{1.1.11a}). There are four prime ideals above 2, with (ramification index, residue degree) equal to $(5,1),(10,1),(2,1),(5,2)$. We normalize the bitangent function as $h_m=mu+b-v$. Since $v=-t^{-4}+\cdots$ at $O$, the coefficient of $t^{-4}$ in $h_m$ is 1. At $q_0$ the reduction has only one singular point, $(u,v)\equiv(950455,\,6126477)\pmod{q_0}$. At the integral lift $(950455,\,4\cdot950455^3)$ the value of $G$ has $q_0$-adic valuation 1, and the Hessian $288u^5-1$ is a $q_0$-adic unit. So this point is an ordinary node of thickness 1. If the quartic split into several components, then by Bézout's theorem the components would have to meet at least three times, and if there were a multiple component, the singular points would not be isolated. Hence this fiber is geometrically irreducible, the Tamagawa number is 1, and the descent image is unramified at $q_0$. Therefore $S=\{2\}$ suffices. \paragraph{Step 2: the class group.} \begin{lem}\label{lem:1.1.11class} $\Cl(K)[2]=0$. \end{lem} \begin{proof} We apply Proposition~\ref{thm:B} with $n=27$, $(r_1,r_2)=(3,12)$, the $D$ above, and $T=\log(5\cdot10^8)$. The window is a step function with 64 cells of equal width, and its heights are 64 positive integers, symmetric left to right (input of \code{1.1.11d}). Two pieces of data are needed. \begin{enumerate}[label=(\roman*)] \item There are $26{,}350{,}721$ prime ideals of norm less than $5\cdot10^8$ in total, and we count the number $a(q)$ for every norm (\code{1.1.11b}). \item All $664{,}374$ prime ideals of norm at most $10^7$ have a witness for Lemma~\ref{lem:C} (\code{1.1.11c}). Hence we set $k(q)=a(q)$ if $q\le10^7$, and $k(q)=0$ otherwise. \end{enumerate} With these data we get $B^->0.00490937703090692\ldots>0$ (\code{1.1.11d}). The knot values of the autocorrelation are computed as exact rationals, and the archimedean term and the prime sum are computed with directed-rounding interval arithmetic. So rounding errors cannot change the conclusion. By Proposition~\ref{thm:B}, $\Cl(K)[2]=0$. \end{proof} \paragraph{Step 3: upper bound for the rank.} \begin{lem}\label{lem:1.1.11rank} $\operatorname{rank}J(\Q)\le1$. \end{lem} \begin{proof} The hypotheses of Proposition~\ref{thm:A} (smooth quartic, rational hyperflex $O$, irreducible $f$ of degree 27) were checked above. Hence $J(\Q)[2]=0$, the map $\delta:J(\Q)/2J(\Q)\to K^\times/K^{\times2}$ is injective, and its image has square norm and lies in $K(S,2)$, $S=\{2\}$. By Lemma~\ref{lem:1.1.11class}, \[ \dim_{\F_2}K(S,2)=(r_1+r_2)+\#\{\mathfrak p\mid2\}=15+4=19. \] There are 19 explicit elements supported on $S$ that form a basis of this space. Their independence is checked by the parities of the valuations at the prime ideals above 2 and by quadratic characters at auxiliary primes (\code{1.1.11e}). We impose conditions on this 19-dimensional space one after another. The condition that the norm be a square (sign and 2-adic valuation) cuts it down to 17 dimensions. Imposing the local conditions at the good primes $487$, $491$, $499$, $503$ in turn, the dimension becomes $10,7,5,4$. At each prime we use all points of $C(\F_p)$, and since $p>484$, the images of these points generate the whole local image (Proposition~\ref{thm:S}(3)). At 2 the situation is as follows. The decomposition group at 2 acts on the 27 bitangents with orbits of sizes $5,10,2,10$ (the values $ef$ above). Under $J[2]\hookrightarrow\F_2^{27}$, an element of $J(\Q_2)[2]$ maps to a codeword that is a union of these orbits. Nonzero codewords have weight 12 or 16. The only unions of orbits of size 12 or 16 are the two obtained by joining the orbit of size 2 with one orbit of size 10, and their sum has weight 20, so it is not a codeword. Hence $\dim J(\Q_2)[2]\le1$, and by Proposition~\ref{thm:S}(1), $\dim J(\Q_2)/2J(\Q_2)\le1+3=4$. On the other hand, the 2-adic images of the following four divisors are independent (\code{1.1.11e}). \begin{enumerate}[label=(\alph*)] \item $P_+-O$. \item $R_1-O$. $R_1$ is the 2-adic point with $u=5$, $v=8s$, where $s$ is a root of $\tfrac{1-u^4}{16}+\tfrac{u+1}2s-32s^3=0$. The derivative with respect to $s$ is odd, so by Hensel's lemma there is a unique such root in $\Z_2$. \item $R_2-O$. $R_2$ is the point in the disk of $O$ with $t=2$. In the equation of the chart the derivative with respect to $w$ is a unit, so $w\in2\Z_2$ is uniquely determined. \item $R_3+\bar R_3-2O$. $R_3$ is the point over $\Q_2(\sqrt2)$ with $t=\sqrt2$, and $\bar R_3$ is its conjugate. \end{enumerate} The coordinates of the points are approximated to precision $2^{60}$, and we check that the error of each image has valuation greater than $2e_{\mathfrak p}$ at each prime ideal $\mathfrak p$ above 2. So the square classes are determined exactly. Therefore the four images generate the whole 2-adic local image. The elements of the preceding 4-dimensional space that map into this local image form a 1-dimensional space. Since $\delta$ is injective and $J(\Q)[2]=0$, $\operatorname{rank}J(\Q)\le1$. \end{proof} \paragraph{Step 4: a principal divisor and the 3-adic logarithm.} 3 is a good prime. We take the integral basis of regular differentials \[ \boldsymbol\omega=(\omega_0,\omega_1,\omega_2)=(1,u,v)\frac{du}{u+1-3v^2}=(1,u,v)\frac{dv}{4u^3-v}. \] In the chart at $O$, \[ \boldsymbol\omega=(w,t,1)\frac{dt}{1-2tw-3w^2-4w^3}, \] and the three components are series with integer coefficients: \[ \omega_0=(-t^4+3t^9+\cdots)\,dt,\quad \omega_1=(t-2t^6+3t^9+\cdots)\,dt,\quad \omega_2=(1-2t^5+3t^8+\cdots)\,dt. \] Since $C(\F_3)=\{O,(1,0),(2,0)\}$, there are three residue disks, and each contains one of the known rational points $O$, $P_+$, $P_-$. \begin{lem}\label{lem:1.1.11div} There is a polynomial $h\in\Q[u,v]$ of degree at most 2 in $v$ such that \[ \operatorname{div}(h)=21P_++E-24O. \] Here $E$ is an effective divisor of degree 3 defined over $\Q$, and the $u$-coordinates of its three points are the roots of an irreducible cubic polynomial $U$. \end{lem} \begin{proof} $h$ is a linear combination of the 22 monomials $u^iv^j$ with $3i+4j\le24$, $j\le2$, and its coefficients are in the input of \code{1.1.11f}. Weight 24 occurs only for $u^8$, so $h$ has a pole of order exactly 24 at $O$. An exact computation verifies \[ \operatorname{Res}_v(G,h)=c\,(u-1)^{21}U(u)\quad(c\in\Q^\times),\qquad \gcd\bigl(G(1,v),h(1,v)\bigr)=v. \] Since $G_v(P_+)=-2\ne0$, $u-1$ is a local parameter at $P_+$. So $h$ vanishes to order exactly 21 at $P_+$ and does not vanish at the other points above $u=1$. The remaining three zeros lie above the roots of $U$, and their $v$-coordinates are given by a quadratic polynomial in $u$ with rational coefficients. Substituting these, we check $G=h=0$. The total order of the zeros, $21+3$, equals the order 24 of the pole, so there are no other zeros. \end{proof} \begin{lem}\label{lem:1.1.11log} Let $L=(L_0,L_1,L_2)$ be the 3-adic logarithm of $[P_+-O]$ with respect to $\boldsymbol\omega$. Then $L\equiv(0,3,6)\pmod9$. Hence $\operatorname{rank}J(\Q)=1$. \end{lem} \begin{proof} The characteristic polynomial $T(Z)$ of $t=u/v$ on $E$ is \[ T(Z)\equiv Z^3+36Z^2+9Z+60\pmod{81}. \] It is 3-Eisenstein, so all three points of $E$ have $v_3(t)=1/3$. The coefficients of the characteristic polynomial of $v$ on $E$ have 3-adic valuations $(-4,-1,0,0)$, and its Newton polygon is a single segment, so $v_3(v)=-4/3$, that is, $v_3(w)=4/3>0$ (\code{1.1.11f}). Therefore the points of $E$, defined over a totally ramified cubic extension of $\Q_3$, lie in the residue disk of $O$, on the branch $w=w(t)$. By Lemma~\ref{lem:1.1.11div}, $21[P_+-O]=-[E-3O]$, so \[ L=-\frac1{21}\operatorname{Tr}\int_0^{t(E)}\boldsymbol\omega. \] The right-hand side is the sum of the tiny integrals to the three points of $E$. The integrals up to degree 9 are \[ I_0=-\frac{t^5}5,\qquad I_1=\frac{t^2}2-\frac{2t^7}7,\qquad I_2=t-\frac{t^6}3+\frac{t^9}3. \] For $n\ge10$, the degree-$n$ term $a\,t^n/n$ ($a\in\Z$) of the integral has valuation at least $n/3-v_3(n)\ge3$ at $v_3(t)=1/3$. For $10\le n\le26$ this is checked directly, and for $n\ge27$, $v_3(n)\le\log_3n\le n/3-3$. Hence the trace of all the discarded terms lies in $27\Z_3$. The power sums $p_k$ of the roots of $T$ are \[ (p_1,\dots,p_9)\equiv(45,63,63,54,27,27,0,0,0)\pmod{81}. \] Therefore \[ \operatorname{Tr}(I_0,I_1,I_2)=\Bigl(-\tfrac{p_5}5,\ \tfrac{p_2}2-\tfrac{2p_7}7,\ p_1-\tfrac{p_6}3+\tfrac{p_9}3\Bigr)\equiv(0,18,9)\pmod{27}, \] and dividing by $-21$ gives $L\equiv(0,3,6)\pmod9$. Since $L\ne0$, $[P_+-O]$ has infinite order, and together with Lemma~\ref{lem:1.1.11rank} this gives $\operatorname{rank}J(\Q)=1$. \end{proof} \paragraph{Step 5: Chabauty--Coleman at $p=3$.} Since $L_1/3$ is a 3-adic unit, \[ \eta_0=\omega_0-\frac{L_0}{L_1}\omega_1,\qquad \eta_2=\omega_2-\frac{L_2}{L_1}\omega_1 \] are integral differentials that annihilate $L$. From $L\equiv(0,3,6)\pmod9$ we get $L_0/L_1\equiv0$, $L_2/L_1\equiv2\pmod3$, so \[ \eta_0\equiv\omega_0,\qquad \eta_2\equiv\omega_1+\omega_2\pmod3. \] Since the rank is 1 and $[P_+-O]$ has infinite order, every element of $J(\Q)$, after multiplication by a suitable integer, becomes the sum of an integer multiple of $[P_+-O]$ and a torsion element. Hence $\eta_0$ and $\eta_2$ annihilate the logarithms of all of $J(\Q)$. \begin{proof}[Proof of Proposition~\ref{prop:1.1.11}] For each disk we choose the center and a local parameter $s$ as follows. \begin{itemize} \item $P_+$: $s=u-1$. Since $G_v(P_+)=-2$ is a unit, $v\in\Z_3[[s]]$, and at the center $\boldsymbol\omega/ds=(1,1,0)/2$. The constant coefficient $1/2$ of $\eta_0$ is a unit. \item $P_-$: $s=v$. Since $G_u(P_-)=-4$ is a unit, $u\in\Z_3[[s]]$, and at the center $\boldsymbol\omega/ds=(1,-1,0)/(-4)$. The constant coefficient $-1/4$ of $\eta_0$ is a unit. \item $O$: $s=t$. At the center $\boldsymbol\omega/ds=(0,0,1)$, and the constant coefficient 1 of $\eta_2$ is a unit. \end{itemize} In all three cases the coefficients of the differentials lie in $\Z_3$, and the rational points of the disk correspond to $s\in3\Z_3$. Let $\eta=\sum_{k\ge0}c_ks^k\,ds$ ($c_0\in\Z_3^\times$) be the chosen differential. Integrating from the center gives \[ I(3z)=\sum_{n\ge1}\frac{c_{n-1}}n(3z)^n=3z\Bigl(c_0+\sum_{n\ge2}\frac{c_{n-1}3^{n-1}}{n}z^{n-1}\Bigr). \] For $n\ge2$ we have $n-1-v_3(n)\ge1$, so the expression in parentheses is a unit for every $z\in\Z_3$. So the only solution of $I(3z)=0$ is $z=0$. For a rational point $R$ in the disk, $[R-\text{center}]\in J(\Q)$, so $I(s(R))=0$ by \cite{Coleman85,MP12}. Hence the only rational point in each disk is its center. $C$ has good reduction at 3, so every rational point lies in one of the three disks. Therefore $C(\Q)=\{O,P_+,P_-\}$. \end{proof} \subsection{\texorpdfstring{$x^4-y^3+xy+x-1=0$ \ ($H=31$)}{x^4-y^3+xy+x-1=0 (H=31)}}\label{sec:1.1.12} \begin{thm}\label{thm:1.1.12} The only integer solution of $x^4-y^3+xy+x-1=0$ is $(x,y)=(0,-1)$. \end{thm} \paragraph{Setup.} Setting $u=-x$, $v=y$, the equation becomes \[ C:\quad G(u,v):=u^4-v^3-uv-u-1=0. \] The projective completion $u^4-v^3w-uvw^2-uw^3-w^4=0$ is a smooth plane quartic (genus 3). The bad primes are $3,5,313,14087$, and the curve has good reduction at 2 (\code{1.1.12a}). At the point at infinity $O=(0:1:0)$ the line $w=0$ meets the curve with multiplicity 4, since $u^4=0$, so $O$ is a rational hyperflex. The parameter near $O$ is $t=u/v$, and $v=t^{-4}+\cdots$. At the point $P=(0,-1)$ we also have $G(u,-1)=u^4$, so the line $v=-1$ meets $C$ with multiplicity 4. Hence \[ \operatorname{div}(v+1)=4P-4O. \] Since $G_v(P)=-3\ne0$, $u$ is a local parameter near $P$, and $v+1=u^4/3-u^5/9+\cdots$. \begin{prop}\label{prop:1.1.12} $J(\Q)=\langle[P-O]\rangle\cong\Z/4\Z$ and $C(\Q)=\{O,P\}$. \end{prop} Since $P=(0,-1)$ corresponds to $(x,y)=(0,-1)$, Theorem~\ref{thm:1.1.12} follows from this proposition. The proof has four steps. We determine the bitangent algebra and the Galois structure of $J[2]$ (Step 1), show that the class groups of two fields have no 2-torsion (Step 2), show by descent that the rank is 0 (Step 3), and then determine the torsion subgroup and the rational points (Step 4). \paragraph{Step 1: the bitangent algebra and $J[2]$.} If the line $v=mu+b$ is a bitangent, then $u^4-(mu+b)^3-u(mu+b)-u-1$ is the square of a quadratic polynomial. The slopes of the 27 bitangents other than the line at infinity are $m=0$ ($b=-1$, that is, the tangent line $v=-1$ at $P$) and the roots of \[ \begin{aligned} f_{10}(m)&=m^{10}-24m^6-48m^2-16,\\ f_{16}(m)&=m^{16}+24m^{12}+48m^{11}+624m^8-192m^7+160m^6\\ &\quad-1152m^4-2304m^3+960m^2-64m+2304. \end{aligned} \] Both polynomials are irreducible. Hence the bitangent algebra is $\mathcal L=\Q\times F_{10}\times F_{16}$, $F_d=\Q[m]/(f_d)$ (\code{1.1.12a}), and \[ F_{10}:\ (r_1,r_2)=(2,4),\ |D|=2^8(5\cdot313\cdot14087)^2;\qquad F_{16}:\ (r_1,r_2)=(0,8),\ |D|=2^{12}(5\cdot313\cdot14087)^4. \] We take the bitangent function $h_m=v-mu-b$. Its coefficient of $t^{-4}$ at $O$ is 1. We use the framework of Proposition~\ref{thm:T}. $V=J[2]$ carries the quadratic form determined by $2O$, and in a suitable basis $V=\F_2^6$ and \[ q(z)=z_1z_2+z_3z_4+z_5z_6+z_5+z_6,\qquad B(a,b)=q(a+b)-q(a)-q(b). \] The 27 bitangents other than the line at infinity correspond to the 27 nonzero isotropic vectors of $q$. The map $a\mapsto(B(a,w))_w$ is a Galois-equivariant injection $V\hookrightarrow E=\F_2^{27}$, and nonzero images have weight 12 or 16. Let $H$ be the subgroup of the orthogonal group fixing the vector $v_0$ corresponding to the rational bitangent $v=-1$. Then the Galois image $G$ is contained in $H$. \begin{lem}\label{lem:1.1.12gal} $G=H$ and $|H|=1920$. \end{lem} \begin{proof} First we do the following finite computation (\code{1.1.12b}). The transvections $z\mapsto z+B(z,a)a$ for $a$ with $q(a)=1$, $B(a,v_0)=0$ generate a group of order 1920, and this group has orbits of sizes $1,10,16$ on the 27 isotropic vectors. This group is all of $H$. Indeed, there are 16 isotropic vectors that form a hyperbolic pair with $v_0$, and the elements fixing one such pair act faithfully as permutations of the five nonzero isotropic vectors of the 4-dimensional orthogonal complement, so $|H|\le16\cdot120$. The orbit of size 10 splits into five pairs $\{w,w+v_0\}$. The action $H\to S_5$ on these pairs is surjective, and its kernel $N$ is an elementary abelian group of order 16. The $H$-normal closure of every nonzero element of $N$ is $N$. The orbits of $G$ are $\{v_0\}$, the 10 roots of $f_{10}$, and the 16 roots of $f_{16}$. The image $\bar G$ of $G$ in $S_5$ is transitive on the five pairs, so its order is a multiple of 5. 13 splits in $F_{10}$ with residue degrees $(1,1,2,6)$, so the Frobenius at 13 has order divisible by 3; since $N$ is a 2-group, the order of $\bar G$ is also divisible by 3. 3 splits in $F_{10}$, $F_{16}$ with residue degrees $(1,1,4,4)$, $(1,1,2,4,4,4)$, and every element of $H$ with this cycle type maps to a 4-cycle in $S_5$. Hence the order of $\bar G$ is a multiple of 60 and $\bar G$ contains an odd permutation, so $\bar G=S_5$. Since $N$ is abelian, the conjugation action on $N$ factors through $H/N\cong S_5$, and since $G\to S_5$ is surjective, $G\cap N$ is $H$-invariant. Hence $G\cap N$ is $0$ or $N$. If it were $0$, then $|G|=120$; but there is an orbit of size 16, so $16\mid|G|$, a contradiction. Therefore $N\subset G$ and $G=H$. \end{proof} \begin{lem}\label{lem:1.1.12tor2} $\dim_{\F_2}J(\Q)[2]=1$, and $H^1(\Q,J[2])\to H^1(\Q,E)$ is injective. Hence $\delta:J(\Q)/2J(\Q)\to\mathcal L^\times/\mathcal L^{\times2}$ is injective. \end{lem} \begin{proof} A finite computation gives $\dim V^H=1$, $\dim E^H=3$, $\dim(E/V)^H=2$ (\code{1.1.12b}). So $E^H\to(E/V)^H$ is surjective. In the cohomology exact sequence of $0\to V\to E\to E/V\to0$, the kernel of $H^1(\Q,V)\to H^1(\Q,E)$ is the quotient of $(E/V)^{G}$ by the image of $E^{G}$, and by Lemma~\ref{lem:1.1.12gal} this quotient is 0. Moreover, $J(\Q)[2]=V^G=V^H$ is 1-dimensional. $\delta$ is the composition of the Kummer map with this map. \end{proof} At the bad primes $5$, $313$, $14087$ the reduction has a single singular point, an ordinary node of thickness 1, and the fiber is geometrically irreducible (\code{1.1.12a}). So the Tamagawa number is 1, and the descent image is unramified at these primes. At 3 we do not use this criterion. Hence $S=\{2,3\}$. \paragraph{Step 2: class groups.} \begin{lem}\label{lem:1.1.12class} $\Cl(F_{10})[2]=\Cl(F_{16})[2]=0$. \end{lem} \begin{proof} We apply Proposition~\ref{thm:B} to both fields. The window is a step function with a single cell, and its autocorrelation is the triangle $f(x)=(1-|x|/T)_+$. In this case the archimedean term of Proposition~\ref{thm:B} has the closed form \[ C_\infty-\frac{\pi^2}{4T}-\frac{2\alpha\log2}T+\frac2T\sum_{j\ \text{odd}}\frac{e^{-jT}}{j^2}+\frac{2\alpha}T\log(1+e^{-T}). \] For $F_{10}$ we take $n=10$, $\alpha=1/5$, $T=6$. There are 94 prime ideals of norm less than $e^6$ (the largest norm is 401), and we verify exactly, with witnesses for Lemma~\ref{lem:C}, that all of them are principal. So $k=a$, and $B^->0.0809177>0$ (\code{1.1.12c}). For $F_{16}$ we take $n=16$, $\alpha=0$, $T=16$. There are $596{,}471$ prime ideals of norm at most $8.9\cdot10^6$ ($>e^{16}$) in total, and we count the number $a(q)$ for every norm (\code{1.1.12d}). For all $18{,}074$ prime ideals of norm at most $2\cdot10^5$ we verify exactly, with witnesses for Lemma~\ref{lem:C}, that they are principal (\code{1.1.12e}). Setting $k(q)=a(q)$ for $q\le2\cdot10^5$ and $k(q)=0$ otherwise, we get $B^->0.0745721>0$ (\code{1.1.12e}). Both computations are done with 60-digit precision, and the margin is at least $0.07$, so rounding errors do not affect the conclusion. By Proposition~\ref{thm:B}, neither class group has 2-torsion. \end{proof} \paragraph{Step 3: rank 0.} \begin{lem}\label{lem:1.1.12rank} $\operatorname{rank}J(\Q)=0$. \end{lem} \begin{proof} By Lemma~\ref{lem:1.1.12tor2}, $\delta$ is injective, and its image has square norm and lies in $\mathcal L(S,2)=\Q(S,2)\times F_{10}(S,2)\times F_{16}(S,2)$ (Proposition~\ref{thm:T}). We have $\Q(S,2)=\langle-1,2,3\rangle$. There are $2+4$ prime ideals above 2 and 3 in $F_{10}$, and $3+6$ in $F_{16}$. By Lemma~\ref{lem:1.1.12class}, \[ \dim F_{10}(S,2)=(2+4)+6=12,\qquad \dim F_{16}(S,2)=(0+8)+9=17, \] and the whole space has dimension $3+12+17=32$. That explicit elements supported on $S$ form a basis of each space is checked by the rank of their valuation parities and quadratic characters (\code{1.1.12f}). We impose conditions one after another. The condition that the norm be a square (sign, 2-adic valuation, 3-adic valuation) gives 29 dimensions. Imposing the local conditions at the good primes $487,491,499,503,509,521,523,541$, the dimension becomes $26,19,17,12,12,11,9,8$ in turn. At each prime we use all points of $C(\F_p)$, so this is the condition given by the whole local image (Proposition~\ref{thm:S}(3)). At 2 the extension is tame, and the orbit sizes on the bitangents in the $\Q$, $F_{10}$, $F_{16}$ components are $(1)$, $(5,5)$, $(1,5,10)$. The inertia group is cyclic of order 5, and Frobenius sends a generator of inertia to its square. Every subgroup of $H$ satisfying this condition and having these orbit sizes has order 20 and a 2-dimensional fixed space in $V$ (\code{1.1.12b}). Hence $\dim J(\Q_2)[2]=2$, and by Proposition~\ref{thm:S}(1), $\dim J(\Q_2)/2J(\Q_2)=2+3=5$. On the other hand, the following images are independent: those of the divisors obtained by subtracting $O$ from the point with $u=0$ ($=P$), from the 2-adic points with $u=2$ and $u=4$, and from the point with $t=2$ in the disk of $O$; and that of the divisor obtained by subtracting $2O$ from the sum of the point with $u=\sqrt2$ over $\Q_2(\sqrt2)$ and its conjugate. The coordinates of the points are determined by Hensel's lemma, and we check that the error of each image has valuation greater than $2e_{\mathfrak p}$ at each prime ideal $\mathfrak p$ above 2. Therefore these five images generate the whole 2-adic local image. At 3, $\mathcal L$ is unramified, so inertia at 3 acts trivially on $E$, and hence on $V$. The Frobenius at 3 has the cycle type above, and its fixed space in $V$ is 3-dimensional (\code{1.1.12b}). So $\dim J(\Q_3)/2J(\Q_3)=\dim J(\Q_3)[2]=3$. The three images of the 3-adic points with $u=0$ ($=P$), $u=1$, $u=3$ are independent, so they generate the whole 3-adic local image. The elements of the preceding 8-dimensional space that map into both local images form a 1-dimensional space (\code{1.1.12f}). Since $\delta$ is injective, $\operatorname{rank}J(\Q)+\dim J(\Q)[2]\le1$, and since $\dim J(\Q)[2]=1$, the rank is 0. \end{proof} \paragraph{Step 4: torsion and rational points.} \begin{proof}[Proof of Proposition~\ref{prop:1.1.12}] By Lemma~\ref{lem:1.1.12rank}, $J(\Q)$ is a finite group. The numbers of points of $C$ over $\F_2,\F_4,\F_8$ are $2,4,8$ (\code{1.1.12g}). Hence the numerator of the congruence zeta function is $1-T-4T^5+8T^6$, and $\#J(\F_2)=1-1-4+8=4$. Since 2 is a good prime, by \cite[Appendix]{Katz81} the odd-order torsion of $J(\Q)$ injects into $J(\F_2)$. Hence $J(\Q)$ is a 2-group. Since $J(\Q)[2]$ is 1-dimensional, $J(\Q)$ is cyclic. Since $\operatorname{div}(v+1)=4P-4O$, we have $4[P-O]=0$. If $2[P-O]=0$, then $2P\sim2O$, which contradicts \cite[Section IV.5]{Hartshorne77} since $P\ne O$. So $[P-O]$ has order 4. Moreover, $[P-O]\notin2J(\Q)$. Indeed, the $\Q$-component of $\delta([P-O])$ is computed with the rational bitangent function $h_0=v+1$. Since $h_0$ has a zero of order 4 at $P$, its value at $P$ is taken to be the first coefficient $1/3$ with respect to the parameter $u$, and its normalized value at $O$ is 1. So this component is the square class of $3$, which is not a square. If $J(\Q)$ were cyclic of order at least 8, every element of order 4 would lie in $2J(\Q)$; hence $J(\Q)=\langle[P-O]\rangle\cong\Z/4\Z$. Now let $R\in C(\Q)$. Since the genus is positive, $R\mapsto[R-O]$ is injective, and $[R-O]=j[P-O]$ with $0\le j\le3$. If $j=0,1$, then $R=O,P$. If $j=2$, then $R+O\sim2P$, and if $j=3$, then $R+P\sim2O$. Since $O\ne P$, in either case two distinct effective divisors of degree 2 are linearly equivalent, which contradicts \cite[Section IV.5]{Hartshorne77}. Therefore $C(\Q)=\{O,P\}$. \end{proof} \subsection{\texorpdfstring{$x^4+y^3+xy+y-1=0$ \ ($H=31$)}{x^4+y^3+xy+y-1=0 (H=31)}}\label{sec:1.1.13} \begin{thm}\label{thm:1.1.13} The only integer solutions of $x^4+y^3+xy+y-1=0$ are $(x,y)=(-1,0),(1,0),(3,-4)$. \end{thm} \paragraph{Setup.} Setting $v=-y$, the equation becomes \[ C:\quad G(x,v):=x^4-v^3-(x+1)v-1=0. \] The projective completion $x^4-v^3w-(x+w)vw^2-w^4=0$ is a smooth plane quartic (genus 3), and the bad primes are $2,13,17,43,257$ (\code{1.1.13a}). At the point at infinity $O=(0:1:0)$ the line $w=0$ meets the curve with multiplicity 4, since $x^4=0$, so $O$ is a rational hyperflex. Near $O$ we use $t=x/v$, $w=1/v$. In this chart the curve is \[ t^4-w-tw^2-w^3-w^4=0, \] and $w=t^4-t^9-\cdots\in\Z[[t]]$. Hence $x$ and $v$ have poles of order 3 and 4 at $O$, respectively. The known rational points are $O$ and \[ P_-=(-1,0),\qquad P_1=(1,0),\qquad P_3=(3,4). \] Since $G(-1,v)=-v^3$ and $G(x,2x-2)=((x-1)(x-3))^2$, \[ \operatorname{div}(x+1)=3P_--3O,\qquad \operatorname{div}(v-2x+2)=2P_1+2P_3-4O. \] So $[P_--O]$ is an element of order 3 and $[P_1+P_3-2O]$ is an element of order 2. That both are nonzero follows from \cite[Section IV.5]{Hartshorne77}. \begin{prop}\label{prop:1.1.13} $C(\Q)=\{O,P_-,P_1,P_3\}$. \end{prop} Since $(x,v)=(-1,0),(1,0),(3,4)$ correspond to $(x,y)=(-1,0),(1,0),(3,-4)$, Theorem~\ref{thm:1.1.13} follows directly from this proposition. The proof has five steps: the bitangent algebra and the Galois structure of $J[2]$ (Step 1), the class groups of two fields (Step 2), the upper bound 1 for the rank (Step 3), the 3-adic logarithm computed from three principal divisors (Step 4), and the six residue disks at $p=3$ (Step 5). \paragraph{Step 1: the bitangent algebra and $J[2]$.} If the line $v=mx+b$ is a bitangent, then $x^4-(mx+b)^3-(x+1)(mx+b)-1$ is the square of a quadratic polynomial. The slopes of the 27 bitangents other than the line at infinity are $m=2$ ($b=-2$, that is, the line $v=2x-2$ tangent at $P_1$ and $P_3$) and the roots of \[ \begin{aligned} f_{10}(m)={}&m^{10}+10m^9+48m^8+144m^7+288m^6+384m^5+344m^4+224m^3+96m^2-16,\\ f_{16}(m)={}&m^{16}-8m^{15}+36m^{14}-112m^{13}+272m^{12}-528m^{11}+872m^{10}-1264m^9+1920m^8\\ &-3008m^7+4256m^6-4896m^5+4224m^4-2560m^3+768m^2-64m+128. \end{aligned} \] Both polynomials are irreducible. Hence the bitangent algebra is $\mathcal L=\Q\times K_{10}\times K_{16}$, $K_d=\Q[m]/(f_d)$ (\code{1.1.13a}), and \[ K_{10}:\ (r_1,r_2)=(2,4),\ |D|=2^8\cdot17^2\cdot43^2\cdot257^2;\qquad K_{16}:\ (r_1,r_2)=(0,8),\ |D|=2^{12}\cdot17^4\cdot43^4\cdot257^4. \] We take the bitangent function $h_m=v-mx-b$. Its coefficient of $t^{-4}$ at $O$ is 1. As in Section~\ref{sec:1.1.12}, we use the framework of Proposition~\ref{thm:T}. $V=J[2]=\F_2^6$, the quadratic form $q(z)=z_1z_2+z_3z_4+z_5z_6+z_5+z_6$, the correspondence between isotropic vectors and bitangents, the injection $V\hookrightarrow E=\F_2^{27}$, and the stabilizer $H$ of the vector $v_0$ of the rational bitangent ($|H|=1920$, with $N$ the kernel of $H\to S_5$) are as in the proof of Lemma~\ref{lem:1.1.12gal}. \begin{lem}\label{lem:1.1.13gal} The Galois image $G$ equals $H$. Hence $\dim_{\F_2}J(\Q)[2]=1$ ($J(\Q)[2]$ is generated by $[P_1+P_3-2O]$), and $\delta:J(\Q)/2J(\Q)\to\mathcal L^\times/\mathcal L^{\times2}$ is injective. \end{lem} \begin{proof} Let $\bar G\subset S_5$ be the image of $G$. Since $f_{10}$ is irreducible, $\bar G$ is transitive on the five pairs, and its order is a multiple of 5. 3 splits in $K_{10}$, $K_{16}$ with residue degrees $(2,8)$, $(8,8)$, and every element of $H$ with this cycle type maps to a 4-cycle in $S_5$. 5 splits with residue degrees $(4,6)$, $(4,12)$, and every element of $H$ with this cycle type maps to an element of cycle type $(2,3)$ (\code{1.1.13b}). Hence the order of $\bar G$ is a multiple of 60 and $\bar G$ contains an odd permutation, so $\bar G=S_5$. As in the proof of Lemma~\ref{lem:1.1.12gal}, $G\cap N$ is $0$ or $N$, and $0$ contradicts the orbit of size 16. Therefore $G=H$. The rest is as in the proof of Lemma~\ref{lem:1.1.12tor2}: since $\dim V^H=1$, $\dim E^H=3$, $\dim(E/V)^H=2$, the group $J(\Q)[2]$ is 1-dimensional and $H^1(\Q,J[2])\to H^1(\Q,E)$ is injective. \end{proof} At the bad primes $17,43,257$ the reduction has a single singular point, an ordinary node of thickness 1, and the fiber is geometrically irreducible. So the Tamagawa number is 1, and the descent image is unramified at these primes. At 13 the node has thickness 2, so we keep 13 (\code{1.1.13a}). Hence $S=\{2,13\}$. \paragraph{Step 2: class groups.} \begin{lem}\label{lem:1.1.13class} $\Cl(K_{10})[2]=\Cl(K_{16})[2]=0$. \end{lem} \begin{proof} As in the proof of Lemma~\ref{lem:1.1.12class}, we apply Proposition~\ref{thm:B} with the triangular window $f(x)=(1-|x|/T)_+$. For $K_{10}$ we take $T=5$, and we verify exactly, with witnesses for Lemma~\ref{lem:C}, that all 32 prime ideals of norm less than $e^5$ (the largest norm is 131) are principal. For $K_{16}$ we take $T=8$, and the same holds for all 465 prime ideals of norm less than $e^8$ (the largest norm is 2957). In both cases $k=a$, and \[ B^-(K_{10})>0.3436681640333619,\qquad B^-(K_{16})>0.0228202034642712 \] (\code{1.1.13c}). The conclusion follows from Proposition~\ref{thm:B}. \end{proof} \paragraph{Step 3: upper bound for the rank.} \begin{lem}\label{lem:1.1.13rank} $\operatorname{rank}J(\Q)\le1$. \end{lem} \begin{proof} By Lemma~\ref{lem:1.1.13gal}, $\delta$ is injective, and its image has square norm and lies in $\mathcal L(S,2)$ (Proposition~\ref{thm:T}). We have $\Q(S,2)=\langle-1,2,13\rangle$. There are $2+2$ prime ideals above 2 and 13 in $K_{10}$, and $3+4$ in $K_{16}$. By Lemma~\ref{lem:1.1.13class}, \[ \dim K_{10}(S,2)=(2+4)+4=10,\qquad \dim K_{16}(S,2)=(0+8)+7=15, \] and the whole space has dimension $3+10+15=28$. The explicit bases are checked by valuation parities and quadratic characters (\code{1.1.13d}). The condition that the norm be a square (sign, 2-adic and 13-adic valuations) gives 25 dimensions, and imposing the local conditions given by the images of all points of $C(\F_p)$ at the good primes \[ 487,491,499,503,509,521,523,541,547,557,563,569,571,577,587,593 \] gives 7 dimensions (Proposition~\ref{thm:S}(3)). At 2 the extension is tame, and the orbit sizes on the bitangents in the $\Q$, $K_{10}$, $K_{16}$ components are $(1)$, $(5,5)$, $(5,1,10)$. By the same computation as in the proof of Lemma~\ref{lem:1.1.12rank}, the decomposition group has order 20 and $\dim J(\Q_2)[2]=2$ (\code{1.1.13b}). Hence $\dim J(\Q_2)/2J(\Q_2)=5$ (Proposition~\ref{thm:S}(1)). The 2-adic images of the following five divisors are independent (\code{1.1.13d}). \begin{enumerate}[label=(\alph*)] \item $P_1-O$. Since $P_1$ lies on the rational bitangent $v=2x-2$, its component is computed, using the identity \[ (v-2x+2)\bigl((v+x-1)^2+3x^2-5x+4\bigr)=((x-1)(x-3))^2 \] on $C$, as the value of $1/\bigl((v+x-1)^2+3x^2-5x+4\bigr)$. \item The divisor obtained by subtracting $O$ from the 2-adic point with $x=-3$. \item The divisor obtained by subtracting $O$ from the point with $t=2$ in the disk of $O$. \item $D_1-2O$, for a divisor $D_1$ of degree 2 on the line $v=-5x-7$. \item $D_2-2O$, for a divisor $D_2$ of degree 2 on the line $v=-5x-5$. \end{enumerate} In (d), (e), write the line as $v=kx+b_0$. Then $G(x,kx+b_0)$ splits in $\Z_2[x]$ as a product of two monic quadratic polynomials with odd resultant (we apply Hensel's lemma to the factorization modulo $2^{50}$). $D_i$ is the effective divisor of degree 2 over $\Q_2$ determined by one of these quadratics. We check that the approximation error has valuation greater than $2e_{\mathfrak p}$ at each prime ideal $\mathfrak p$ above 2, so the square classes are determined exactly. Therefore the five images generate the whole 2-adic local image. The elements of the preceding 7-dimensional space that map into this local image form a 2-dimensional space. Since $\delta$ is injective and $\dim J(\Q)[2]=1$, $\operatorname{rank}J(\Q)\le2-1=1$. \end{proof} \paragraph{Step 4: principal divisors and the 3-adic logarithm.} 3 is a good prime. We take the integral basis of regular differentials \[ \boldsymbol\omega=(\omega_0,\omega_1,\omega_2)=(1,x,v)\frac{dx}{3v^2+x+1}=(1,x,v)\frac{dv}{4x^3-v}. \] In the chart at $O$, $\boldsymbol\omega=-(w,t,1)\,dt/(1+2tw+3w^2+4w^3)$. Modulo 3, $C(\F_3)$ consists of the six points \[ O,\quad(2,0),\quad(1,0),\quad(0,1),\quad(1,1),\quad(1,2). \] Above the first four lie the rational points $O,P_-,P_1,P_3$. Above the last two lie the $\Q_3$-points $Q_\pm=(1,\alpha_\pm)$, where $\alpha_\pm^2=-2$, $\alpha_+\equiv1$, $\alpha_-\equiv2\pmod3$; since $G(1,v)=-v(v^2+2)$, $Q_\pm$ are points over $\Q(\sqrt{-2})$. \begin{lem}\label{lem:1.1.13div} There are $h_1\in\Q[x,v]$ and $h_\pm\in\Q(\sqrt{-2})[x,v]$ such that \[ \begin{aligned} \operatorname{div}(h_1)&=22P_1+E_1-25O,\\ \operatorname{div}(h_+)&=P_-+18P_1+Q_++E_+-23O,\\ \operatorname{div}(h_-)&=2P_-+5P_1+Q_-+E_--11O. \end{aligned} \] Here $E_1,E_\pm$ are effective divisors of degree 3. \end{lem} \begin{proof} The three functions are computed by \code{1.1.13e} directly, as the kernel of the vanishing conditions (the kernel is one-dimensional, so they are unique up to a constant). In each function there is only one monomial of largest weight ($x$ has weight 3, $v$ has weight 4), so the pole orders are exactly 25, 23, 11. We check the orders of the zeros by Taylor expansion at $P_-$ and $P_1$, and check by substitution that $h_\pm$ vanishes at $Q_\pm$ and not at its conjugate. The norm to the $x$-line is the product of the known factors and one irreducible cubic, and this cubic does not vanish at the known $x$-coordinates. The remaining three points are given by a cubic for the $v$-coordinate and by an expression of $x$ as a polynomial in $v$; substituting them, we check $G=h=0$. The sum of the orders of the zeros equals the order of the pole, so there are no other zeros. \end{proof} \begin{lem}\label{lem:1.1.13log} Let $L=\log[P_1-O]$. The 3-adic logarithms with respect to $\boldsymbol\omega$ are as follows modulo 27. \[ \begin{array}{c|ccc} &\omega_0&\omega_1&\omega_2\\\hline L&24&15&15\\ \log(Q_+-O)&12&9&12\\ \log(Q_--O)&18&3&0 \end{array} \] In particular, $v_3(L_0)=1$, so $\operatorname{rank}J(\Q)=1$. \end{lem} \begin{proof} $[P_--O]$ is a torsion element, so its logarithm is 0. From Lemma~\ref{lem:1.1.13div}, \[ \begin{aligned} L&=-\frac{\log(E_1-3O)}{22},\\ \log(Q_+-O)&=-\log(E_+-3O)-18L,\\ \log(Q_--O)&=-\log(E_--3O)-5L. \end{aligned} \] Here 22 is a 3-adic unit. The points of $E_1$ and $E_+$ are defined over the unramified cubic extension of $\Q_3$, and their $x$-coordinates are 1 and 0 modulo 3, respectively. Over $\F_3$ the fibers $x=1$ and $x=0$ consist of three distinct points, and $\operatorname{div}(x-c)=(\text{fiber }x=c)-3O$. So, pairing each point $e$ of $E$ with the point $e'$ of the fiber in the same residue disk, we get $\log(E-3O)=\sum_e\int_{e'}^{e}\boldsymbol\omega$. On these disks $3v^2+x+1$ is a unit, so $x-c$ is a parameter. Writing $\boldsymbol\omega(x_e+s)=\sum_{n\ge0}a_ns^n\,ds$ ($a_n$ integral) and setting $\delta=x_e-c\in3\mathcal O$, we have \[ \int_{e'}^{e}\boldsymbol\omega=\sum_{n\ge0}\frac{(-1)^na_n\delta^{n+1}}{n+1}. \] The terms with $n+1\ge6$ have valuation at least $(n+1)-v_3(n+1)\ge5$. So if we compute the terms with $n+1\le5$ exactly in the cubic field and take the trace, the result is exact modulo 81. The cubic satisfied by the $v$-coordinates of $E_-$ is 3-Eisenstein, so all three points have $v_3(v)=1/3$, and $x$ reduces to $-1$. So $E_-$ lies in the disk of $P_-$. Since $3P_-\sim3O$, we have $\log(E_--3O)=\operatorname{Tr}\int_{P_-}^{e}\boldsymbol\omega$, and with respect to the parameter $v$ the denominator of $\boldsymbol\omega=(1,x,v)\,dv/(4x^3-v)$ is a unit. We use 18 terms of the integral; the remaining terms, of degree $n\ge19$, have valuation at least $n/3-v_3(n)\ge6$. The values obtained in this way are, modulo 81, \[ \log(E_1-3O)\equiv(39,48,21),\quad \log(E_+-3O)\equiv(69,72,69),\quad \log(E_--3O)\equiv(51,57,33) \] (\code{1.1.13f}). Substituting them into the relations above gives the table. Since $L_0\equiv24\pmod{27}$, $L\ne0$, and together with Lemma~\ref{lem:1.1.13rank} this gives $\operatorname{rank}J(\Q)=1$. \end{proof} \paragraph{Step 5: Chabauty--Coleman at $p=3$.} Since $v_3(L_0)=1\le v_3(L_1),v_3(L_2)$, \[ A=\Bigl(-\frac{L_1}{L_0},1,0\Bigr),\qquad B=\Bigl(-\frac{L_2}{L_0},0,1\Bigr) \] are integral vectors that annihilate $L$. From the table, $A\equiv(5,1,0)\pmod9$, $A\equiv(2,1,0)$, $B\equiv(2,0,1)\pmod3$. Since the rank is 1 and $[P_1-O]$ has infinite order, the differentials $A\boldsymbol\omega$ and $B\boldsymbol\omega$ annihilate the logarithms of all of $J(\Q)$. Also, from the table, \[ A\cdot\log(Q_+-O)\equiv15,\qquad A\cdot\log(Q_--O)\equiv12\pmod{27}, \] and both values have 3-adic valuation exactly 1. \begin{proof}[Proof of Proposition~\ref{prop:1.1.13}] $C$ has good reduction at 3, so every rational point lies in one of the six residue disks. We write the points of each disk with a parameter $s\in3\Z_3$. The disks of $Q_\pm$. The parameter is $s=x-1$, and at $Q_\pm$, $3v^2+x+1=-4$ is a unit. If such a disk contains a rational point $R$, then, since $A\boldsymbol\omega$ annihilates $J(\Q)$, \[ 0=A\cdot\log(R-O)=A\cdot\log(Q_\pm-O)+\int_{Q_\pm}^{R}A\boldsymbol\omega. \] Modulo 3, $A\boldsymbol\omega\equiv(2+x)\,dx/(3v^2+x+1)$, which vanishes at $x=1$. So the linear term of the integral has valuation at least 2, and the terms of degree $n\ge2$ also have valuation at least 2, since $n-v_3(n)\ge2$. Hence the integral lies in $9\Z_3$. On the other hand, $A\cdot\log(Q_\pm-O)$ has valuation 1, so the equality above cannot hold. These two disks contain no rational points. The remaining four disks. The centers, the parameters and the differentials used are as follows. \[ \begin{array}{l|cccc} \text{center}&O&P_-&P_1&P_3\\\hline \text{parameter}&t&v&x-1&x-3\\ \text{differential}&B\boldsymbol\omega&A\boldsymbol\omega&B\boldsymbol\omega&A\boldsymbol\omega\\ \text{coefficient at the center (modulo 3)}&-1&2&1&2 \end{array} \] All these coefficients are units, so by the same computation as in the proof of Proposition~\ref{prop:1.1.11}, the integral $I(3z)$ from the center has the form $3z\cdot(\text{unit})$. By \cite{Coleman85,MP12}, the rational points of a disk are zeros of this integral, so the center is the only one. Therefore $C(\Q)=\{O,P_-,P_1,P_3\}$. \end{proof} \subsection{\texorpdfstring{$x^4+y^3+xy+x+1=0$ \ ($H=31$)}{x^4+y^3+xy+x+1=0 (H=31)}}\label{sec:1.1.14} \begin{thm}\label{thm:1.1.14} The only integer solution of $x^4+y^3+xy+x+1=0$ is $(x,y)=(0,-1)$. \end{thm} This equation has the same shape as those in Sections~\ref{sec:1.1.1}--\ref{sec:1.1.13}, but the method of solution is different. Instead of bitangent 2-descent, an elementary fourth-power descent turns it into a problem about the rational points of a genus 5 curve. From that curve we construct a Prym map to the Jacobian of a genus 2 curve, and we finish with a sieve modulo 7 and 7-adic Chabauty. \paragraph{Setup.} Setting $v=y+1$, the equation becomes \begin{equation}\label{eq:1.1.14-v} x^4+v\,(v^2-3v+x+3)=0 \end{equation} (\code{1.1.14a}). In projective coordinates $(x:v:z)$ we use the three quadratic forms \[ q_1=vz,\qquad q_2=x^2,\qquad q_3=-(v^2-3vz+xz+3z^2); \] then the projective completion of \eqref{eq:1.1.14-v} is the smooth plane quartic $C:\ q_2^2=q_1q_3$. Let $(x:v:z:r:s)$ be coordinates on $\mathbb P^4$, and let $D$ be the intersection of the three quadrics \begin{equation}\label{eq:1.1.14-D} Q_1=vz+3r^2,\qquad Q_2=x^2+3rs,\qquad Q_3=q_3+3s^2. \end{equation} On $D$ we have $q_1=-3r^2$, $q_2=-3rs$, $q_3=-3s^2$, so $q_1q_3=q_2^2$. Hence $\pi(x:v:z:r:s)=(x:v:z)$ is a morphism $D\to C$ (if $x=v=z=0$, then $r=s=0$). A fixed point of the involution $\sigma:(r,s)\mapsto(-r,-s)$ must have $r=s=0$, but then $x=0$, $vz=0$, $v^2-3vz+3z^2=0$ force all coordinates to be 0. Hence $\pi$ is an unramified double cover. $D$ is smooth modulo 7, and hence also smooth over $\Q$ (\code{1.1.14b}). Therefore $D$ is a curve of genus 5. We use the two rational points \[ P_\pm=(0:0:1:0:\pm1) \] of $D$. Theorem~\ref{thm:1.1.14} follows from the descent in Step 1 and the following proposition. \begin{prop}\label{prop:1.1.14} $D(\Q)=\{P_+,P_-\}$. \end{prop} \paragraph{Step 1: elementary descent.} \begin{lem}\label{lem:1.1.14-vneg} Integer solutions of \eqref{eq:1.1.14-v} satisfy $v\le0$, and if $v=0$, then $x=0$. \end{lem} \begin{proof} Suppose $v\ge1$. If $x\ge0$, then $v^2-3v+x+3=(v-1)(v-2)+x+1\ge1$, so the left-hand side is positive. If $x=-t$ with $t\ge1$, then the left-hand side is $t^4-tv+(v-1)^3+1$. If $1\le v\le t^3$, then $t^4-tv\ge0$ and $(v-1)^3+1\ge1$. If $v\ge t^3+1$, then $(v-1)^2\ge t^6\ge t$, so $(v-1)^3\ge t(v-1)$, and the left-hand side is at least $t^4-t+1>0$. In either case the left-hand side is nonzero. If $v=0$, then $x^4=0$. \end{proof} \begin{lem}\label{lem:1.1.14-lift} If $(x,v)$ is an integer solution of \eqref{eq:1.1.14-v} with $v<0$, then there are integers $r,s$ such that $(x:v:1:r:s)\in D(\Q)$. \end{lem} \begin{proof} Set $n=-v>0$; then $x^4=n\,(n^2+3n+x+3)$. If $x=0$, then $n^2+3n+3=0$, which is impossible; so $x\ne0$, and both factors are positive. If a prime $p$ divides both factors, then $p\mid x$, and looking at the second factor modulo $p$ gives $p\mid3$. Hence for $p\ne3$ the exponent of $p$ in $n$ is a multiple of 4, and we can write \[ n=du^4,\qquad d\in\{1,3,9,27\},\quad u\in\Z_{>0}. \] If $d=1$, then $u^4\mid x^4$, so we write $x=uw$. If $d=3,9,27$, then $3\mid x$. Writing $x=3x'$, we get $u^4\mid9x'^4$, so $u\mid x'$, and we write $x=3uw$. Cancelling the common fourth power gives the following equations. \[ \begin{array}{c|l} d & \text{integer equation}\\ \hline 1 & w^4=u^8+3u^4+uw+3\\ 3 & 9w^4=3u^8+3u^4+uw+1\\ 9 & 3w^4=27u^8+9u^4+uw+1\\ 27 & w^4=243u^8+27u^4+uw+1 \end{array} \] $d=1$: If $u,w$ are both prime to 3, then modulo 3 we get $1\equiv1+uw$, a contradiction. If one of them is a multiple of 3, then by the equation so is the other, and modulo 9 we get $0\equiv3$, a contradiction. $d=9$: With $A=2w^2-6u^4-1$, $B=2w^2+6u^4+1>0$, the equation is equivalent to $3AB=4uw+1$ (\code{1.1.14a}). From $u^4\ge u^2$ and $4|uw|\le2u^2+2w^2$ we get $3B-|4uw+1|\ge4w^2+16u^2+2>0$. If $A\ne0$, then $|3AB|\ge3B>|4uw+1|$; if $A=0$, then $4uw+1=0$ would be required. Both are impossible. $d=3,27$: Set $r=u^2$ ($d=3$), $r=3u^2$ ($d=27$); then $n=3r^2$. $s=-x^2/(3r)$ is an integer, namely $-3w^2$ and $-w^2$, respectively. Now at $z=1$ we have $Q_1=-n+3r^2=0$ and $Q_2=x^2+3rs=0$. Also $n^2+3n+x+3=x^4/n=x^4/(3r^2)=3s^2$, so $Q_3=-(n^2+3n+x+3)+3s^2=0$. \end{proof} \paragraph{Step 2: the Prym map.} Let $M_1,M_2,M_3$ be the $5\times5$ symmetric matrices of $Q_1,Q_2,Q_3$ in \eqref{eq:1.1.14-D} ($Q_i(X)=X^{\mathsf T}M_iX$). In the pencil of quadrics \[ Q_t=Q_1-2tQ_2+t^2Q_3=(q_1-2tq_2+t^2q_3)+3(r-ts)^2, \] each $Q_t$ is a quadric of rank at most 4 with vertex $(0:0:0:t:1)$. Its discriminant is determined by the $(x,v,z)$ part. Indeed, \[ 12\det\bigl(M(q_1)-2tM(q_2)+t^2M(q_3)\bigr)=f(t):=3t^6-18t^5+36t^3+6t \] (\code{1.1.14a}; $M(q_i)$ is the $3\times3$ symmetric matrix of $q_i$). We have $f=3t\,(t^5-6t^4+12t^2+2)$, and its discriminant is $-2^6\cdot3^{10}\cdot19\cdot149\cdot3461$. Let \[ H:\ \eta^2=f(t),\qquad J=\operatorname{Jac}(H) \] be the genus 2 curve and its Jacobian. $H$ has good reduction at 5 and 7. If $f(t)\ne0$, then $Q_t$ has rank exactly 4, and its two families of planes correspond to $\eta=\pm\sqrt{f(t)}$. $D_\infty$ denotes the sum of the two points at infinity of $H$ (a rational divisor). The Bruin construction of \cite{Bruin08,RR18} gives a morphism $\phi:D\to J$ over $\Q$. Up to an overall sign, $\phi$ is the composition of $p\mapsto[2\sigma(p)-2p]\in\operatorname{Prym}(D/C)$ with the isomorphism $\operatorname{Prym}(D/C)\cong J$; see \cite[Proposition 6.2 and Remark 6.3]{Bruin08}. What the proof uses is the following tangent formula. Represent $p\in D$ by a vector $P$, and choose a vector $w$ that is tangent to $D$ at $p$ and not proportional to $P$ ($P^{\mathsf T}M_iw=0$, $i=1,2,3$). Then \begin{equation}\label{eq:1.1.14-tan} a_p(t)=w^{\mathsf T}M_1w-2t\,w^{\mathsf T}M_2w+t^2\,w^{\mathsf T}M_3w \end{equation} is nonzero. Replacing $w$ by $aw+bP$ ($a\ne0$) only multiplies $a_p$ by $a^2$. At a $t$ with $a_p(t)=0$, the tangent line at $p$ is contained in $Q_t$, and the family of planes containing this tangent line determines the sign of $\eta$. Let $D_p$ be the effective divisor of degree 2 on $H$ obtained in this way. Then \[ \phi(p)=[D_p-D_\infty]. \] In particular, the $t$-coordinates of $D_p$ are the roots of $a_p$. This Prym identification, including the quadratic twist, is independently checked by $L_p(D)=L_p(C)\,L_p(H)$ at $p=5,7,11,13,17$ (\code{1.1.14b}). \begin{lem}\label{lem:1.1.14-G} $A_0=(0,0)$ and $B_0=(24,20748)$ are rational points of $H$, and $\phi(P_+)=\pm G$, where \[ G=[B_0-A_0]=\bigl(t^2-24t,\ \tfrac{1729}{2}t\bigr)\in J(\Q) \] (Mumford representation). \end{lem} \begin{proof} We have $f(0)=0$ and $f(24)=20748^2$ (\code{1.1.14a}). In the coordinate order $(x,v,z,r,s)$, the three gradient vectors at $P_+$ are $(0,1,0,0,0)$, $(0,0,0,3,0)$, $(-1,3,-6,0,6)$, and $w=(1,0,0,0,\tfrac16)$ is orthogonal to all three. Since $Q_1(w)=0$, $Q_2(w)=1$, $Q_3(w)=3\cdot\tfrac1{36}=\tfrac1{12}$, \[ a_{P_+}(t)=-2t+\tfrac1{12}t^2=\tfrac1{12}\,t\,(t-24). \] The only point of $H$ above the root $t=0$ is the Weierstrass point $A_0$, and the points above $t=24$ are $(24,\pm20748)$. Hence $D_{P_+}=A_0+(24,\varepsilon\cdot20748)$ with $\varepsilon=\pm1$. Since $2A_0\sim D_\infty$, $\phi(P_+)=[(24,\varepsilon\cdot20748)-A_0]$. The hyperelliptic involution fixes $A_0$ and acts on $J$ as $-1$, so this is $\varepsilon G$. The Mumford representation is checked from $t(t-24)$, $b(0)=0$, $b(24)=1729\cdot12=20748$. \end{proof} \paragraph{Step 3: $J(\Q)$.} \begin{lem}\label{lem:1.1.14-MW} $J(\Q)\cong\Z$. If $H_0$ is a generator and $G=mH_0$, then $m$ is odd. Moreover, the image of the reduction modulo 7, $\rho_7:J(\Q)\to J(\F_7)$, is $\{0,\bar G\}$. \end{lem} \begin{proof} \code{1.1.14c} verifies the following by exact computation: \[ \#\mathrm{Sel}^{(2)}(J/\Q)=2,\qquad \#J(\F_5)=23,\qquad \#J(\F_7)=32,\qquad \operatorname{ord}\bar G=2,\qquad \bar G\notin2J(\F_7). \] By \cite[Appendix]{Katz81}, torsion of order prime to 5 injects into $J(\F_5)$, and torsion of order prime to 7 injects into $J(\F_7)$. If there were a torsion point of prime order $\ell$, we would need $\ell\mid23$ when $\ell\ne5$ and $\ell\mid32$ when $\ell\ne7$; no such $\ell$ exists. Hence $J(\Q)$ has no torsion, and in particular $J(\Q)[2]=0$. By \cite{Stoll01,PS97,PS99}, $\#\mathrm{Sel}^{(2)}=2$ gives $\operatorname{rank}J(\Q)\le1$. Since $G\ne0$, the rank is 1 and $J(\Q)\cong\Z$. If $m$ were even, then $\bar G=m\bar H_0\in2J(\F_7)$, a contradiction. Since $\#J(\F_7)=32$ and $m$ is odd, $\bar H_0$ and $\bar G=m\bar H_0$ generate the same subgroup, which is $\{0,\bar G\}$. \end{proof} \paragraph{Step 4: sieve modulo 7.} \begin{lem}\label{lem:1.1.14-sieve} Every point of $D(\Q)$ reduces modulo 7 to $\bar P_+$ or $\bar P_-$. \end{lem} \begin{proof} $D(\F_7)$ has eight points, all with $z\ne0$ (\code{1.1.14c}). Their coordinates $(x,v,r,s)$ normalized by $z=1$, and the tangent polynomials $a_{\bar p}$ normalized to be monic, are as follows. Two points paired by $\sigma$ have the same $a_{\bar p}$. \[ \begin{array}{l|c} (x,v,r,s)\bmod7 & a_{\bar p}(t)\\ \hline (0,0,0,1),\ (0,0,0,6)\quad(=\bar P_+,\bar P_-) & t^2+4t\\ (0,4,1,0),\ (0,4,6,0) & t^2+t+4\\ (4,2,2,2),\ (4,2,5,5) & t^2+t+4\\ (5,2,2,4),\ (5,2,5,3) & t^2+t+4 \end{array} \] Take $p\in D(\Q)$ and represent it by a primitive vector $P\in\Z_7^5$. Since $D$ is smooth over $\Z_7$, the $3\times5$ matrix $(P^{\mathsf T}M_i)_i$ has rank 3 also modulo 7. Its kernel is a saturated submodule of rank 2 containing $P$, so there is $w\in\Z_7^5$ that, together with $P$, forms a basis of the kernel. Then $\bar w$ is a tangent vector at $\bar p$ and is independent of $\bar P$. Hence the reduction modulo 7 of $a_p\in\Z_7[t]$ is a nonzero constant multiple of $a_{\bar p}$. Since $H$ has good reduction at 7, the reduction of $\phi(p)=[D_p-D_\infty]$ is $[\bar D_p-\bar D_\infty]$, where $\bar D_p$ is the effective divisor of degree 2 whose $t$-coordinates are the roots of $a_{\bar p}$. If $a_{\bar p}=t^2+t+4$, its discriminant $1-16\equiv6$ is not a square modulo 7. So $\bar D_p$ consists of two conjugate points over $\F_{49}$ with distinct $t$-coordinates. $\bar D_p$ is not a fiber of $t$, so $[\bar D_p-\bar D_\infty]\ne0$, and the first polynomial of its Mumford representation is $t^2+t+4$. The first polynomial of $\bar G$ is $t^2-24t\equiv t^2+4t$, so this class is not $\bar G$ either. On the other hand, by Lemma~\ref{lem:1.1.14-MW}, $\rho_7(\phi(p))\in\{0,\bar G\}$ (\cite{BS10}). Hence $\bar p\in\{\bar P_+,\bar P_-\}$. \end{proof} \paragraph{Step 5: 7-adic Chabauty in the two residue disks.} The differentials of $H$ \[ \omega_0=\frac{dt}{2\eta},\qquad \omega_1=\frac{t\,dt}{2\eta} \] form a basis of the regular differentials on the $\Z_7$-model of $H$, and hence a basis of the invariant differentials of $J$ over $\Z_7$. Let $\lambda=(\lambda_0,\lambda_1):J(\Q_7)\to\Q_7^2$ be the logarithm with respect to this basis, and write $\lambda_{\omega'}$ for the component with respect to a differential $\omega'$. \begin{lem}\label{lem:1.1.14-beta} Write $\lambda(G)=(\ell_0,\ell_1)$. Then $v_7(\ell_0)=v_7(\ell_1)=1$. $\beta=\ell_1/\ell_0$ is a unit of $\Z_7$ with $\beta\equiv24\pmod{49}$, in particular $\beta\equiv3\pmod7$ (more precisely, $\beta\equiv53924\pmod{7^6}$). The differential \[ \omega=\omega_1-\beta\omega_0=\frac{(t-\beta)\,dt}{2\eta} \] annihilates the logarithms of all of $J(\Q)$. \end{lem} \begin{proof} Since $f(24)\equiv0$ and $f'(24)\equiv1\pmod7$, by Hensel's lemma $f$ has a 7-adic root $\alpha$ with $\alpha\equiv24\pmod7$ ($\alpha\equiv24\pmod{49}$, $\alpha\equiv1641\pmod{7^4}$). Let $W=(\alpha,0)$. $A_0$ and $W$ are Weierstrass points, so $[W-A_0]$ is 2-torsion and its logarithm is 0. Hence $\lambda(G)=\lambda([B_0-W])$. Since $20748=7\cdot2964$, $B_0$ lies in the same residue disk as $W$, and since $f'(\alpha)$ is a unit, $\eta$ is a local parameter on this disk. Since $t=\alpha+\eta^2/f'(\alpha)+O(\eta^4)$, \[ \omega_0=\Bigl(\frac1{f'(\alpha)}+O(\eta^2)\Bigr)d\eta,\qquad \omega_1=\Bigl(\frac{\alpha}{f'(\alpha)}+O(\eta^2)\Bigr)d\eta, \] and all coefficients lie in $\Z_7$. By the tiny integrals of \cite{Coleman85,MP12}, $\ell_i=\int_W^{B_0}\omega_i$. The terms $\eta^{2k+1}/(2k+1)$ ($k\ge1$) have valuation at least $2k+1-v_7(2k+1)\ge3$ at $\eta=20748$, so \[ \ell_0\equiv\frac{20748}{f'(\alpha)},\qquad \ell_1\equiv\frac{20748\,\alpha}{f'(\alpha)}\pmod{7^3}. \] Hence both have valuation 1, and $\beta\equiv\alpha\pmod{49}$. The precise value is computed by \code{1.1.14e}. Since $J(\Q)=\Z H_0$ and $G=mH_0$ with $m\ne0$, we get $\lambda_\omega(H_0)=(\ell_1-\beta\ell_0)/m=0$. \end{proof} \begin{lem}\label{lem:1.1.14-disk} On the residue disk of $P_+$ (the points of $D(\Q_7)$ that reduce to $\bar P_+$ modulo 7), $x$ is a local parameter, and the points of the disk correspond bijectively to $x\in7\Z_7$. On this disk, \[ \phi^*\omega=g(x)\,dx,\qquad g\in\Z_7[[x]],\qquad g(0)=\mp\tfrac{\beta}{3}\in\Z_7^\times. \] \end{lem} \begin{proof} In the chart $z=1$, the Jacobian determinant of $(Q_1,Q_2,Q_3)$ with respect to $(v,r,s)$ at $P_+$ is \[ \det\begin{pmatrix}1&0&0\\0&3&0\\3&0&6\end{pmatrix}=18, \] a 7-adic unit. Hence, by Hensel's lemma (the implicit function theorem), $v,r,s\in\Z_7[[x]]$, and $x\in7\Z_7$ corresponds bijectively to the points of the disk. The first few terms are as follows (\code{1.1.14a}, \code{1.1.14f}). \[ s=1+\tfrac{x}{6}-\tfrac{x^2}{72}+\tfrac{x^3}{432}+O(x^4),\quad r=-\tfrac{x^2}{3}+\tfrac{x^3}{18}-\tfrac{x^4}{72}+O(x^5),\quad v=-\tfrac{x^4}{3}+\tfrac{x^5}{9}+O(x^6). \] \emph{Integrality.} The $\Z_7$-model of $D$ is smooth, and $J$ is an abelian scheme over $\Z_7$. By the Néron mapping property (\cite[Section 1.2]{BLR90}), $\phi$ extends to a morphism over $\Z_7$. Since $\omega_0,\omega_1$ are invariant differentials of $J$ over $\Z_7$ and $\beta\in\Z_7$, $\phi^*\omega$ is a regular differential on the $\Z_7$-model of $D$. Since $x$ is a parameter at $\bar P_+$, $\phi^*\omega=g(x)\,dx$ with $g\in\Z_7[[x]]$. \emph{Constant term.} Substituting the tangent vector $w=(1,v',0,r',s')$ at the point $p(x)$ of the disk into \eqref{eq:1.1.14-tan}, the coefficients of $a_{p(x)}$ at $x=0$ are $(0,-2,\tfrac1{12})$, and the two roots are \[ t_1=\tfrac23x^2-\tfrac19x^3+O(x^4),\qquad t_2=24-\tfrac{1729}{9}x^3+O(x^4). \] Since $f(t_1)=4x^2+O(x^3)$, we have $\eta_1=\pm(2x+O(x^2))$ and $dt_1/(2\eta_1)=\pm(\tfrac13+O(x))\,dx$. Since $dt_2/dx=O(x^2)$, the contribution of the second point is $O(x^2)\,dx$. Hence \[ \phi^*\omega=(t_1-\beta)\frac{dt_1}{2\eta_1}+(t_2-\beta)\frac{dt_2}{2\eta_2}=\mp\frac{\beta}{3}\,dx+O(x)\,dx. \] By Lemma~\ref{lem:1.1.14-beta}, $\beta/3\equiv1\pmod7$ is a unit. \code{1.1.14f} computes this expansion exactly up to degree 18 and shows that all coefficients up to degree 14 are 7-adic integers. \end{proof} \begin{proof}[Proof of Proposition~\ref{prop:1.1.14}] Let $R\in D(\Q)$. By Lemma~\ref{lem:1.1.14-sieve}, $R$ lies in the residue disk of $P_+$ or of $P_-$. $\sigma$ preserves $D(\Q)$ and interchanges $\bar P_+$ and $\bar P_-$, so we may assume that $R$ lies in the disk of $P_+$. Let $\xi=x(R)\in7\Z_7$. An integral within the disk is a difference of logarithms (\cite{Coleman85,MP12}), so \[ \int_{P_+}^{R}\phi^*\omega=\lambda_\omega\bigl(\phi(R)-\phi(P_+)\bigr)=0. \] This is because $\phi(R)-\phi(P_+)\in J(\Q)$, and by Lemma~\ref{lem:1.1.14-beta}, $\omega$ annihilates the logarithms of $J(\Q)$. On the other hand, writing $g=\sum_{j\ge0}c_jx^j$ as in Lemma~\ref{lem:1.1.14-disk}, \[ \int_{P_+}^{R}\phi^*\omega=\sum_{j\ge1}\frac{c_{j-1}}{j}\,\xi^{j}. \] If $\xi\ne0$, the first term has valuation $v_7(\xi)$, and the terms with $j\ge2$ have valuation at least $jv_7(\xi)-v_7(j)>v_7(\xi)$ (since $(j-1)v_7(\xi)\ge j-1>v_7(j)$). Then the integral has finite valuation $v_7(\xi)$, so it cannot be 0. Hence $\xi=0$, that is, $R=P_+$. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.1.14}] By Lemma~\ref{lem:1.1.14-vneg}, integer solutions have $v\le0$, and if $v=0$, then $(x,y)=(0,-1)$. This is indeed a solution. If there were a solution with $v<0$, then by Lemma~\ref{lem:1.1.14-lift} it would give a rational point $(x:v:1:r:s)\in D(\Q)$ with nonzero $v$-coordinate. The $v$-coordinates of $P_\pm$ are 0, so this contradicts Proposition~\ref{prop:1.1.14}. \end{proof} \subsection{\texorpdfstring{$x^4-x^2+xy+y^3=0$ \ ($H=32$)}{x^4-x^2+xy+y^3=0 (H=32)}}\label{sec:1.1.15} \begin{thm}\label{thm:1.1.15} The equation $x^4-x^2+xy+y^3=0$ has only the following integer solutions: $(x,y)=(0,0)$, $(1,0)$, $(-1,0)$, $(-1,1)$, $(-1,-1)$, $(2,-2)$, $(4,-6)$. \end{thm} \paragraph{Setup.} Unlike the other equations of Section~\ref{sec:1.1}, the plane quartic is singular: its lowest-degree part $-x^2+xy=x(y-x)$ has two distinct factors, so the origin is a node, and the point $(0:1:0)$ at infinity is smooth. If $x=0$, the equation gives $y=0$. For $x\ne0$ put \[ u=-\frac yx,\qquad v=-x . \] Then $x^4-x^2+xy+y^3=v^2(v^2+u^3v-u-1)$, so the solutions with $x\ne0$ lie on \begin{equation}\label{eq:1.1.15:C} C:\quad v^2+u^3v=u+1,\qquad\text{equivalently}\qquad w^2=u^6+4u+4\quad(w=2v+u^3). \end{equation} Put $J=\Jac(C)$, where $C$ denotes the smooth projective model of \eqref{eq:1.1.15:C}. Let $O$ and $E$ be the two points at infinity of \eqref{eq:1.1.15:C}, labelled so that $v\sim-u^3$ at $O$ and $v\sim u^{-2}$ at $E$; $O$ corresponds to the point $(0:1:0)$ of the quartic, and $E$ to one of the two branches at the origin. The functions \[ x=-v,\qquad y=uv,\qquad z=-u^2v=y^2/x \] have poles of orders $3,4,5$ at $O$ and no other poles. The affine curve $\mathcal U=C\setminus\{O\}$ has the integral model \begin{equation}\label{eq:1.1.15:U} y^2=xz,\qquad x^3-x+y(z+1)=0,\qquad z^2+z=y(1-x^2). \end{equation} On it $x=0$ forces $y=0$ and $z\in\{0,-1\}$; these are the points $D=(0,0,0)$ and $E=(0,0,-1)$, the two branches at the origin. Over $\Q$, \eqref{eq:1.1.15:U} is covered by the chart $z+1\ne0$, on which $u=(x^2-1)/(z+1)$ and $v=-x$, and the chart $z\ne0$, on which $s=y/z=x/y$ is a parameter at $E$. Integrally, the proof uses only three facts: every integer solution lifts to an integral point of \eqref{eq:1.1.15:U} (Lemma~\ref{lem:1.1.15:lift}); an integral point of \eqref{eq:1.1.15:U} does not reduce to $O$ modulo any prime (Lemma~\ref{lem:1.1.15:away}); and the $\Z_2$-points of \eqref{eq:1.1.15:U} are described by four residue disks (Step~6). \begin{lem}\label{lem:1.1.15:lift} If $(x,y)$ is an integer solution with $x\ne0$, then $x\mid y^2$, and $(x,y,y^2/x)$ is an integral point of \eqref{eq:1.1.15:U}. \end{lem} \begin{proof} Let $\ell$ be a prime with $\alpha=v_\ell(x)>0$, and $\beta=v_\ell(y)$. If $\beta\ge\alpha$, then $\alpha\le2\beta$. If $\beta<\alpha$, the four terms $x^4,x^2,xy,y^3$ have valuations $4\alpha,2\alpha,\alpha+\beta,3\beta$, and the smallest one must occur at least twice. Since $\alpha+\beta<2\alpha<4\alpha$, this forces $\alpha+\beta=3\beta$, that is, $\alpha=2\beta$. So $\alpha\le2\beta$ at every prime, and $x\mid y^2$. The three relations follow from the equation by substituting $z=y^2/x$. \end{proof} The seven solutions of the theorem give the following eight integral points of \eqref{eq:1.1.15:U} (the origin gives $D$ and $E$). They are used as known points; that they satisfy \eqref{eq:1.1.15:U} is checked by substitution. \[ \begin{array}{@{}c|cccccccc@{}} &A&B&C_1&D&E&F_1&G&H\\\hline (x,y,z)&(-1,-1,-1)&(-1,0,0)&(-1,1,-1)&(0,0,0)&(0,0,-1)&(1,0,0)&(2,-2,2)&(4,-6,9)\\ (x,y)&(-1,-1)&(-1,0)&(-1,1)&(0,0)&(0,0)&(1,0)&(2,-2)&(4,-6) \end{array} \] \begin{lem}\label{lem:1.1.15:model} The discriminant of \eqref{eq:1.1.15:C}, normalized as usual for genus 2, is $2^{-12}\operatorname{disc}(u^6+4u+4)=-8539$, and $8539$ is prime. Modulo $8539$, $f=u^6+4u+4$ satisfies $\gcd(f,f')=u+1709$ and $\gcd(f,f',f'')=1$. Hence \eqref{eq:1.1.15:C} has good reduction at every prime $p\ne8539$ (in particular at $2$), and at $8539$ the fibre is geometrically irreducible with a single ordinary node of thickness 1. The model is regular, and integral sections do not pass through the node (\code{1.1.15a}). \end{lem} \begin{proof} The four other roots of $f$ modulo 8539 are simple, so the fibre at 8539 has one ordinary double point and is geometrically irreducible; since $8539$ divides the discriminant exactly once, the node has thickness 1. After an unramified extension the local equation at the node is $ab=8539\cdot(\text{unit})$; this is regular, and a section through the node would make both $a$ and $b$ divisible by $8539$, which is impossible. \end{proof} So all fibres are irreducible, and local heights of divisors of degree 0 need no vertical correction. We abbreviate $[P-O]\in J(\Q)$ to $P$. The proof runs as follows. A 2-descent bounds the rank by 2 (Step~1). With explicit differentials of the second and third kind we write down the Coleman--Gross local height at $p=2$ (Steps~2--3). Since the differential $\omega_1$ used to normalize the self-height vanishes at the non-Weierstrass point $E$, we add the cross-height with $E$; the local terms at odd primes of this combination vanish at every integral point (Step~4). All periods and the height matrix are obtained from two tiny integrals, and they show that the rank is exactly 2 (Step~5). Finally, on each of the four 2-adic residue disks the resulting quadratic Chabauty function has at most two zeros, and the two known points in each disk account for them (Step~6). This is the method of \cite{CG89,BBM16}, at $p=2$ and with the cross-height term. The literature quoted for \cite{CG89,BBM16} does not cover this setting verbatim (it is usually stated for odd $p$, and no second point at infinity enters); the facts about heights used here are those of \cite{CG89,BB12,BB15}, which hold at every prime of good reduction, and the local heights are identified in Steps~3 and~4. \paragraph{Step 1: upper bound for the rank.} \begin{lem}\label{lem:1.1.15:rank} $\operatorname{rank}J(\Q)\le2$. \end{lem} \begin{proof} The 2-descent of \cite{Stoll01,PS97,PS99} for $w^2=u^6+4u+4$, computed with Magma's \texttt{RankBound}, gives the upper bound 2 (\code{1.1.15a}). The class groups used in the descent are computed with the proof bound set to the Minkowski bound, so the bound does not depend on GRH. Equality is proved in Lemma~\ref{lem:1.1.15:M}. \end{proof} \paragraph{Step 2: differentials and principal divisors.} Put \[ \omega_1=\frac{dx}{1+3z},\qquad\omega_2=\frac yx\,\omega_1,\qquad\eta_1=2y\,\omega_1,\qquad\eta_2=x\,\omega_1 \] (where an apparent denominator vanishes, these are understood as their regular extensions). On \eqref{eq:1.1.15:C}, $\omega_1=-du/(2v+u^3)$ and $\omega_2=u\,du/(2v+u^3)$, so $\omega_1,\omega_2$ form a basis of the holomorphic differentials. As a section of the relative dualizing sheaf of the regular model, $\omega_1$ has divisor $E+O$, with no vertical component. At $O$ we use the parameter $t=-x/y=1/u$, fixed once and for all: \begin{equation}\label{eq:1.1.15:infty} x=t^{-3}\rho,\quad y=-t^{-4}\rho,\quad z=t^{-5}\rho,\qquad \rho^2-\rho=t^5+t^6,\quad\rho(0)=1. \end{equation} The leading terms of $\omega_1,\omega_2,\eta_1,\eta_2$ are $-t\,dt$, $dt$, $2t^{-3}dt$, $-t^{-2}dt$, and the next terms begin in degrees $6,5,2,3$. Hence the $\eta_i$ are of the second kind with poles only at $O$, and with the convention $\langle[\alpha],[\beta]\rangle=\operatorname{Res}_O((\int\alpha)\beta)$ the cup product matrix of $\omega_1,\omega_2,\eta_1,\eta_2$ is $\left(\begin{smallmatrix}0&-I_2\\I_2&0\end{smallmatrix}\right)$. So $W=\langle[\eta_1],[\eta_2]\rangle$ is an isotropic subspace of $H^1_{\mathrm{dR}}$ complementary to the holomorphic differentials. \begin{lem}\label{lem:1.1.15:div} The following divisors hold on $C$ (\code{1.1.15b}): \[ \begin{aligned} \operatorname{div}(x)&=D+2E-3O,&\operatorname{div}(y)&=B+F_1+D+E-4O,\\ \operatorname{div}(y-x)&=A+2D+E-4O,&\operatorname{div}(x+1)&=A+B+C_1-3O,\\ \operatorname{div}(x+y)&=C_1+G+D+E-4O,&\operatorname{div}(z-2x-1)&=3A+C_1+H-5O. \end{aligned} \] Hence, in $J(\Q)$, \[ D=-2E,\quad B=E-F_1,\quad A=3E,\quad C_1=F_1-4E,\quad G=5E-F_1,\quad H=-5E-F_1. \] \end{lem} \begin{proof} The orders of zero of the six functions at the eight points are computed from local expansions, and in each case their sum equals the pole order at $O$ (\code{1.1.15b}). The relations follow successively. \end{proof} \paragraph{Step 3: the local height at 2.} For $P=(a,b,c)$ and $Q=(x,y,z)$ put \[ \nu_P(Q)=\frac{z+c+1+by/x}{x-a}\,\omega_1(Q). \] This is a differential of the third kind with residue divisor $P-O$: on the other branches over $x=a$ the numerator vanishes, and with the normalized branches it extends to $\nu_D=\frac{z+1}x\omega_1$ and $\nu_E=\frac zx\omega_1$. At $O$, \[ \nu_P=\bigl(-t^{-1}-at^2+bt^3+\cdots\bigr)\,dt, \] so the coefficients of $t^0$ and $t^1$ are 0. The bidifferential identity \begin{equation}\label{eq:1.1.15:kernel} d_P\nu_P(Q)-d_Q\nu_Q(P)=\sum_{i=1}^2\bigl(\eta_i(P)\omega_i(Q)-\omega_i(P)\eta_i(Q)\bigr) \end{equation} holds: after division by $\omega_1(P)\omega_1(Q)$ both sides reduce to $2b-2y+ay/x-xb/a$ on $C\times C$ (\code{1.1.15b}). Take $p=2$ and let $\log_2$ be the 2-adic logarithm with $\log_22=0$ (so also $\log_2(-1)=0$); it is the 2-component $\chi_2$ of the cyclotomic idele class character of \cite{CG89,BB12,BB15}. Put \[ I(P)=\Bigl(\int_O^P\omega_1,\int_O^P\omega_2\Bigr),\qquad J(P)=\Bigl(\int_O^P\eta_1,\int_O^P\eta_2\Bigr),\qquad K(P,Q)=\int_O^Q\nu_P, \] with the following normalizations at the end points. $J_j(P)=\Phi_j(P)$, where $\Phi_j$ is the Coleman integral of $\eta_j$ whose Laurent expansion in $t$ at $O$ has constant term 0 (this depends on the parameter $t$, not only on $\partial_t$). $K(P,Q)=\Lambda_P(Q)$, where $\Lambda_P$ is the Coleman integral of $\nu_P$ with $\Lambda_P+\log_2t\to0$ at $O$ (this depends only on $\partial_t$). For $P\ne E$ let $v_P$ be the tangent vector at $P$ with $\omega_1(v_P)=1$, and put $K(P,P)=\lim_{Q\to P}\bigl(\Lambda_P(Q)-\log_2z_P(Q)\bigr)$ for a parameter $z_P$ at $P$ with $dz_P(v_P)=1$; at $E$ we use $\partial_s$ and $z_E=s$, with $s=x/y$ as in \eqref{eq:1.1.15:Echart} below. Define \[ \mathcal H(P,Q)=K(P,Q)-J(P)\,I(Q)^{\mathsf T}. \] \begin{lem}\label{lem:1.1.15:height} For $P,Q\in\mathcal U(\Q_2)$, with the tangent vector $\partial_t$ at $O$ and, if $Q=P$, the tangent vector $v_P$ (or $\partial_s$ if $P=E$) at $P$, \[ \mathcal H(P,Q)=h_2(P-O,Q-O), \] where $h_2$ is the local height at 2 of \cite{CG89,BB12,BB15} for $W$ and $\chi_2=\log_2$. \end{lem} \begin{proof} By \cite[Section 2]{BB12} we need the differential $\omega_{P-O}$ of the third kind with residue divisor $P-O$ and $\Psi(\omega_{P-O})\in W$. Write $\Psi(\nu_P)=\sum_ia_i[\omega_i]+w$ with $w\in W$; by the cup product matrix, $\langle[\eta_j],\Psi(\nu_P)\rangle=a_j$. The global symbol of \cite[Section 2]{BB12} with $\nu=\nu_P$, $\beta=\eta_j$, $\Phi_\beta=\Phi_j$ has two terms: at $P$ the residue is $\Phi_j(P)=J_j(P)$, and at $O$ it is 0, because $\nu_P=(-t^{-1}+0+0\cdot t+O(t^2))\,dt$ while $\Phi_j$ has a pole of order at most 2 and constant term 0. So $a_j=J_j(P)$, and $\omega_{P-O}=\nu_P-\sum_iJ_i(P)\omega_i$. Hence $\int_{Q-O}\omega_{P-O}=K(P,Q)-J(P)I(Q)^{\mathsf T}$, the integrals being regularized at $O$ (and at $P$ if $Q=P$) by the chosen tangent vectors, as in the definition of $K$. This regularized pairing is bi-additive, agrees with $h_2$ for disjoint supports, and satisfies $h_2(\operatorname{div}f,\mathfrak a)=\log_2f[\mathfrak a]$ for divisors $\mathfrak a$ of degree 0, because $\omega_{\operatorname{div}f}=df/f$; by \cite{Gross86,BB15} these properties determine $h_2$. \end{proof} Consistently with this, \eqref{eq:1.1.15:kernel} says that the mixed derivative $d_Pd_Q$ of $\mathcal H(P,Q)-\mathcal H(Q,P)$ vanishes, as it must since $h_2$ is symmetric for isotropic $W$; with the opposite sign of the term $J(P)I(Q)^{\mathsf T}$ it would not. Integrating \eqref{eq:1.1.15:kernel}, with $Q$ replaced by a variable point $R$, from $R=O$ to $R=Q$ gives \[ d_P\mathcal H(P,Q)=\nu_Q(P)-\sum_i\omega_i(P)J_i(Q). \] The boundary term at $O$ vanishes: if $R\to O$ and $t=t(R)$, then $\nu_R(P)=-t^{-2}\omega_1(P)+t^{-1}\omega_2(P)+O(t)$ by \eqref{eq:1.1.15:infty}, while $\Phi_1=-t^{-2}+O(t^3)$ and $\Phi_2=t^{-1}+O(t^4)$; and the normalization of $K(P,\cdot)$ at $O$ does not depend on $P$. \begin{lem}\label{lem:1.1.15:tau} Put $\tau_*(P)=\mathcal H(P,P)+\mathcal H(P,E)$. Then \[ d\mathcal H(P,P)=-2\sum_i\omega_i(P)J_i(P)-\nu_E(P),\qquad d\mathcal H(P,E)=\nu_E(P)-\sum_i\omega_i(P)J_i(E), \] so $d\tau_*=-\sum_i\omega_i\bigl(2J_i(P)+J_i(E)\bigr)$, and $\tau_*$ has no logarithmic singularity at $E$. Its value at $E$, defined by the regular extension, is $2\mathcal H(E,E)=2h_2(E-O,E-O)$ with respect to $\partial_s$. \end{lem} \begin{proof} On the diagonal the two regular parts of $\nu_P(Q)$ and $\nu_Q(P)$ combine with the term $d\log$ of the normalization by $v_P$. Let $\Delta=1+3c$ and $b'=db/da$. Where $x$ is a parameter, the regular coefficient of $\nu_P$ on the diagonal is $r=(-3bb'/a+c/a)/\Delta$, and \[ 2r+\frac d{da}\log\Delta=-\frac c{a\Delta}, \] which is the coefficient of $-\nu_E$; the identity extends to the other charts. This gives the first formula; the second is the formula for $d_P\mathcal H$ above. Near $E$ we use $s=x/y$: \begin{equation}\label{eq:1.1.15:Echart} z+s^5z^2+1-s=0,\quad z(0)=-1,\qquad x=s^2z,\quad y=sz,\quad\omega_1=\frac{s\,ds}{1+2s^5z}. \end{equation} Since $\omega_1$ vanishes to order 1 at $E$, $v_P$ degenerates as $P\to E$; the resulting term $-\log_2s$ in the self-height and the term $+\log_2s$ in the cross-height cancel, and since the leading coefficient is the unit 1, no constant arises. \end{proof} \paragraph{Step 4: local heights away from 2.} \begin{lem}\label{lem:1.1.15:away} Let $\ell$ be an odd prime and $P\in\mathcal U(\Z)$. With the tangent vectors $\partial_t$ at $O$, $v_P$ at $P$ and $\partial_s$ at $E$, \[ h_\ell(P-O,P-O)+h_\ell(P-O,E-O)=0, \] where for $P=E$ the left-hand side is $2h_\ell(E-O,E-O)$ with respect to $\partial_s$. \end{lem} \begin{proof} Near the points at infinity use $T=1/u$ and $R=vT^3$, with $R^2+R=T^5+T^6$; then $x=-R/T^3$ and $\omega_1=T\,dT/(2R+1)$. The sections $\bar O$ and $\bar E$ are $T=0$ together with $R=-1$, resp.\ $R=0$; they are disjoint at every prime, and $T$ is a local equation of each, since $2R+1=\mp1$ there. (i) Integral points do not meet $\bar O$: if $P\in\mathcal U(\Z)$ reduced to $\bar O$ modulo $\ell$, then $T(P)\in\ell\Z_\ell$ and $R(P)\in-1+\ell\Z_\ell$, so $v_\ell(x(P))\le-3$. They do not pass through the node at 8539 (Lemma~\ref{lem:1.1.15:model}), all fibres are irreducible, and $\chi_\infty=0$. (ii) By \cite[(1.3)]{CG89}, (3), $h_\ell(\mathfrak a_1,\mathfrak a_2)=\chi_\ell(\ell)(\mathfrak a_1\cdot\mathfrak a_2)_\ell=-(\mathfrak a_1\cdot\mathfrak a_2)_\ell\log_2\ell$, also for common support when the tangent vectors at the common points are integral (the self-intersection of a section then contributes 0), and replacing a tangent vector $t_x$ by $\alpha t_x$ adds $\chi_\ell(\alpha)\operatorname{ord}_x(\mathfrak a_1)\operatorname{ord}_x(\mathfrak a_2)$. (iii) The tangent vectors $\partial_t$ at $O$ and $\partial_s$ at $E$ ($s=-T$) are integral. Let $P\ne E$ and $n=(P\cdot E)_\ell$. Since $\operatorname{div}\omega_1=\bar E+\bar O$ with no vertical part and $\bar P$ meets the special fibre in a smooth point and avoids $\bar O$, the restriction of $\omega_1$ to $\bar P$ vanishes to order exactly $n$. Hence $\omega_1(t_P)\in\ell^n\Z_\ell^\times$ for an integral tangent vector $t_P$, and $v_P=\alpha t_P$ with $v_\ell(\alpha)=-n$ (the adjunction formula $(P\cdot P)_{v_P}=-(P\cdot\operatorname{div}\omega_1)_\ell$). By (i)--(iii), $h_\ell(P-O,P-O)=\chi_\ell(\alpha)=n\log_2\ell$ and $h_\ell(P-O,E-O)=-(P\cdot E)_\ell\log_2\ell=-n\log_2\ell$, while $h_\ell(E-O,E-O)=0$ with respect to $\partial_s$. \end{proof} For the eight known points, the resulting decomposition $\mathcal H(P,Q)-I(P)MI(Q)^{\mathsf T}=\sum_{\ell\ne2}(P\cdot Q)_\ell\log_2\ell$ ($P\ne Q$) and $\mathcal H(P,P)=I(P)MI(P)^{\mathsf T}$, with $M$ of Lemma~\ref{lem:1.1.15:M}, was confirmed numerically to 310 bits for all 64 ordered pairs, including $E$; the right-hand side is nonzero only for $\{A,H\}$ at $\ell=5$ and for $\{C_1,G\}$, $\{F_1,H\}$ at $\ell=3$. No known point meets $E$ at an odd prime, so the case $n\ne0$ of (iii) rests on the argument above. \begin{lem}\label{lem:1.1.15:global} Let $h$ be the global height pairing on $J(\Q)$ of \cite{CG89} for $W$. For every $P\in\mathcal U(\Z)$, \[ \tau_*(P)=h(P,P)+h(P,E)\qquad(\text{for }P=E\text{: }\tau_*(E)=2h(E,E)). \] \end{lem} \begin{proof} The tangent vectors $\partial_t$ at $O$, $v_P$ at $P\ne E$ and $\partial_s$ at $E$ are $\Q$-rational, so by \cite{CG89} $h(P,P)+h(P,E)=\sum_v\bigl(h_v(P-O,P-O)+h_v(P-O,E-O)\bigr)$. The term at 2 is $\tau_*(P)$ by Lemmas~\ref{lem:1.1.15:height} and \ref{lem:1.1.15:tau}, the terms at odd $\ell$ vanish by Lemma~\ref{lem:1.1.15:away}, and there is no archimedean term. \end{proof} \paragraph{Step 5: periods and the height matrix.} Write $\omega=(\omega_1,\omega_2)$, $\eta=(\eta_1,\eta_2)$, and $I_P=I(P)$, $J_P=J(P)$. Put \[ \delta_1=\int_B^{F_1}\omega,\quad\delta_2=\int_D^G\omega,\quad\epsilon_1=\int_B^{F_1}\eta,\quad\epsilon_2=\int_D^G\eta . \] Each of these is a tiny integral within one 2-adic residue disk (Step~6). \begin{lem}\label{lem:1.1.15:periods} \[ I_E=\frac{\delta_1+2\delta_2}{13},\quad I_{F_1}=\frac{7\delta_1+\delta_2}{13},\quad J_E=\frac{\epsilon_1+2\epsilon_2+(4,0)}{13},\quad J_{F_1}=\frac{7\epsilon_1+\epsilon_2+(2,0)}{13}, \] and \[ \begin{aligned} I_A&=3I_E,& I_B&=I_E-I_{F_1},& I_D&=-2I_E,\\ J_A&=3J_E-(1,1),&J_B&=J_E-J_{F_1},&J_D&=-2J_E,\\ J_{C_1}&=J_{F_1}-4J_E+(1,1),&J_G&=5J_E-J_{F_1}+(-2,0). \end{aligned} \] \end{lem} \begin{proof} The relations for $I$ follow from Lemma~\ref{lem:1.1.15:div} and the additivity of abelian integrals ($\sum_P\operatorname{ord}_P(f)I(P)=0$ for every rational function $f$). For $J$ the constant terms coming from the second kind must be kept: if $f=ct^k(1+O(t))$ at $O$ with $k=\operatorname{ord}_Of$, then \[ \sum_{P\ne O}\operatorname{ord}_P(f)\,J(P)=\operatorname{Res}_O\Bigl(\log_2\frac f{ct^k}\;\eta\Bigr). \] This is the global symbol identity $\sum_x\operatorname{Res}_x((df/f)\Phi_j)=\langle[\eta_j],\Psi(df/f)\rangle=0$ of \cite[Section 2]{BB12}, after an integration by parts at $O$ that uses the normalization of $\Phi_j$; the leading term $ct^k$ is divided out, since otherwise $\log_2f$ would contain $\log_2t$. For the six functions of Lemma~\ref{lem:1.1.15:div}, in the order listed there, the residue vectors are $(0,0),(0,0),(-1,-1),(0,0),(-1,1),(-4,0)$; only the coefficients of $t$ and $t^2$ of the normalized logarithms are needed, and they are read off from \eqref{eq:1.1.15:infty} (\code{1.1.15b}). Solving the resulting linear relations gives the formulas. \end{proof} The base divisors are $D-O$ and $F_1-O$. The integral sections of $D$, $F_1$ and $E$ are pairwise disjoint at every prime, and none of them meets $\bar O$ (Lemma~\ref{lem:1.1.15:away}(i)). So $\omega_1$ is primitive at $D$ and $F_1$ (that is, $n=0$ in Lemma~\ref{lem:1.1.15:away}(iii)), all local terms away from 2 of the heights among $D-O$ and $F_1-O$ vanish, and by \cite{CG89} the global heights among these two divisors equal $\mathcal H$. \begin{lem}\label{lem:1.1.15:K} For $P\in\{D,F_1\}$ put $k_1(P)=\int_B^{F_1}\nu_P$ and $k_2(P)=\int_D^G\nu_P$. Then \[ K(P,E)=\frac{k_1(P)+2k_2(P)}{13},\qquad K(P,F_1)=\frac{7k_1(P)+k_2(P)}{13},\qquad K(P,D)=-2K(P,E). \] \end{lem} \begin{proof} By \cite{Gross86,BB15} and the symmetry of $h_2$, $\sum_Q\operatorname{ord}_Q(f)K(P,Q)=h_2(P-O,\operatorname{div}f)=\log_2\bigl(f[P]/f[O]\bigr)$, where $f[\cdot]$ is the value normalized by the chosen tangent vectors (the first equality holds because $\sum_Q\operatorname{ord}_Q(f)I(Q)=0$). We use the functions $x$, $y/x$ and $(x+y)(y-x)y/((x+1)x^4)$, with divisors $D+2E-3O$, $B+F_1-E-O$ and $G+F_1-5E+3O$. For $P=D,F_1$ the ratio of the normalized value at $P$ to the leading coefficient at $O$ is always of the form $\pm2^m$, so all these logarithms are 0, and the formulas follow as in Lemma~\ref{lem:1.1.15:periods}. In the tiny integrals the poles are subtracted algebraically: on the disk of $D$, with $x=s$ and $y=sV$, $\nu_D-dx/x=-2V^2\omega_1$, and on the disk of $B$ and $F_1$, $\nu_{F_1}-dx/(x-1)=\frac{2y(x+1)}{z+1}\omega_1$; the subtracted logarithms take values $\log_2(\pm2^m)=0$. \end{proof} \begin{lem}\label{lem:1.1.15:M} Let $L$ be the matrix with rows $I_D,I_{F_1}$, let $B_h=\bigl(K(P,Q)-J(P)I(Q)^{\mathsf T}\bigr)_{P,Q\in\{D,F_1\}}$, and $M=L^{-1}B_h(L^{-1})^{\mathsf T}$. Then $v_2(\det L)=3$, $B_h$ is symmetric, and \[ I_E\equiv(2,0),\quad I_{F_1}\equiv(2,22),\quad J_E\equiv(60,44),\quad J_{F_1}\equiv(62,30),\quad M\equiv\begin{pmatrix}56&19\\19&53\end{pmatrix}\pmod{64}. \] In particular $M\in M_2(\Z_2)$, $I_E,I_{F_1}\in2\Z_2^2$ and $J_E\in4\Z_2^2$. Moreover $\operatorname{rank}J(\Q)=2$, and $h(P,Q)=I(P)\,M\,I(Q)^{\mathsf T}$ for all $P,Q\in J(\Q)$, where $I$ is extended additively to $J(\Q)$. \end{lem} \begin{proof} The values are computed by exact arithmetic modulo $2^{40}$ (\code{1.1.15c}); a 70-bit computation in Magma gives the same values (\code{1.1.15d}). Since $v_2(\det L)=3$, the logarithms of $D$ and $F_1$ are independent, so $\operatorname{rank}J(\Q)\ge2$, and with Lemma~\ref{lem:1.1.15:rank} the rank is 2. Then $\log\colon J(\Q)\otimes\Q_2\to\Q_2^2$ is an isomorphism, and by \cite{CG89} the bilinear form $h$ is determined by its values on $D,F_1$, which form the matrix $B_h$ by the remark before Lemma~\ref{lem:1.1.15:K}. No integral basis of $J(\Q)$ and no saturation is needed. \end{proof} \paragraph{Step 6: the quadratic Chabauty function on the four disks.} By Lemmas~\ref{lem:1.1.15:global} and \ref{lem:1.1.15:M}, the function \[ q(P)=\tau_*(P)-I(P)\,M\,I(P)^{\mathsf T}-I(P)\,M\,I_E^{\mathsf T} \] vanishes at every $P\in\mathcal U(\Z)$. It involves both the self-height and the cross-height with $E$. By Lemma~\ref{lem:1.1.15:tau}, \[ dq=-\sum_{i=1}^2\omega_i\Bigl(2J_i(P)+J_i(E)+\sum_{j=1}^2M_{ij}\bigl(2I_j(P)+I_j(E)\bigr)\Bigr). \] Solving \eqref{eq:1.1.15:U} over $\F_2$ gives the four points $(1,1,1)$, $(1,0,0)$, $(0,0,0)$, $(0,0,1)$; so $\mathcal U(\Z_2)$ is the union of four residue disks, with centres and parameters \[ \begin{array}{c|c|c|c} \text{centre }P_0&\text{parameter }s&\text{known points}&u\text{ of the second point }(s=2u)\\\hline A&y+1&A,\ C_1&1\\ B&x+1&B,\ F_1&1\\ D&x&D,\ G&1\\ E&x/y=y/z&E,\ H&-1/3 \end{array} \] Each parameter is primitive over $\Z_2$, and $s\in2\Z_2$ describes the whole disk. The disk of $D$ is given by $V+sV^3+s^2-1=0$, $V(0)=1$, $x=s$, $y=sV$, $z=sV^2$, and the disk of $E$ by \eqref{eq:1.1.15:Echart}. On the disks of $A$ and $B$ the derivative of the equation with respect to $x$, resp.\ $y$, is odd, so Newton's method gives a unique series in $\Z_2[[s]]$. On all four disks $\omega_i/ds$ and $\eta_i/ds$ lie in $\Z_2[[s]]$. Since the centres are integral points, $q(P_0)=0$, and integrating $dq$ with constant term 0 gives \[ q_{P_0}(2u)=\sum_{n\ge1}c_{P_0,n}u^n . \] \begin{lem}\label{lem:1.1.15:table} The first seven coefficients, modulo 32, are as follows (\code{1.1.15c}). \[ \begin{array}{c|c|rrrrrrr} P_0&e&c_1&c_2&c_3&c_4&c_5&c_6&c_7\\\hline A&2&20&28&16&16&16&0&0\\ B&3&24&24&0&16&0&0&0\\ D&2&20&12&0&16&16&0&0\\ E&2&20&4&8&0&16&0&0 \end{array} \] \end{lem} \begin{proof} \code{1.1.15c} computes these values by polynomial arithmetic over $\Z/2^{40}\Z$ with series of length 64; every series that is inverted has odd constant term, and since the $n$-th term of an integral is $a_{n-1}2^n/n$, the 2-part of $n$ is split off first and only odd parts are inverted. No floating-point approximation is used. Write $\lambda(n)$ for the integer with $2^{\lambda(n)}\le n<2^{\lambda(n)+1}$. The terms omitted from the tiny integrals are those with $n\ge62$, and their valuations are at least $n-v_2(n)\ge n-\lambda(n)\ge57$. The entries of $L^{-1}$ have valuation at least $-2$ and those of $B_h$ at least 2, so errors of valuation at least 40 in $L$ and $B_h$ change $M$ only in valuation at least 36: $L^{-1}(\delta B_h)(L^{-1})^{\mathsf T}$ has valuation at least $-2+40-2$, and $\delta(L^{-1})=-L^{-1}(\delta L)L^{-1}$ has valuation at least 36 while $B_h(L^{-1})^{\mathsf T}$ has valuation at least 0. So at most four bits are lost, and the five bits needed for the table are guaranteed. \end{proof} \begin{lem}\label{lem:1.1.15:tail} On each of the four disks, $v_2(c_{P_0,n})\ge n+1-v_2(n)-\lambda(n-1)$ for $n\ge2$. In particular this bound is at least 4 for $n\ge8$, and the series converges on $\Z_2$. \end{lem} \begin{proof} The series coefficients of $\omega_i,\eta_i$ are integers. At the centres, $I(P_0)\in2\Z_2^2$ and $J(P_0)\in\Z_2^2$, by Lemmas~\ref{lem:1.1.15:periods} and \ref{lem:1.1.15:M}; moreover $M$ is integral, $I_E\in2\Z_2^2$ and $J_E\in2\Z_2^2$. Hence the constant term of the bracket in $dq$ lies in $2\Z_2$, and the non-constant terms are integral combinations of twice the integrals of integral series. In such an integral the denominator of the coefficient of degree $k\ge1$ divides $k$, so the coefficient of $s^{n-1}$ in $dq/ds$ has valuation at least $1-\lambda(n-1)$. Integrating once more and substituting $s=2u$ adds $n-v_2(n)$. If $n=2^k$ with $k\ge3$, the bound is $2^k+2-2k\ge4$; if $2^k4$. \end{proof} \begin{lem}\label{lem:1.1.15:disks} Each of the four residue disks contains exactly two points of $\mathcal U(\Z)$, namely the two known points of the table. \end{lem} \begin{proof} By Lemmas~\ref{lem:1.1.15:table} and \ref{lem:1.1.15:tail}, for the whole series \[ 2^{-e}q_{P_0}(2u)\equiv u^2+u\pmod2,\qquad e_A=e_D=e_E=2,\quad e_B=3 . \] Write $2^{-e}q_{P_0}(2u)=\sum_{n\ge1}b_nu^n$. Then $b_n\in\Z_2$, $b_n\to0$, $b_1$ and $b_2$ are units and $b_n\in2\Z_2$ for $n\ge3$, so by \cite{Strassmann28} the series has at most two zeros in $\Z_2$. The two known points lie at $u\equiv0$ and $u\equiv1\pmod2$, and they are zeros of $q$. (The 70-bit computation \code{1.1.15d} gives the same exponents $e$ and reductions.) \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.1.15}] By Lemma~\ref{lem:1.1.15:disks}, $\mathcal U(\Z)=\{A,B,C_1,D,E,F_1,G,H\}$. By Lemma~\ref{lem:1.1.15:lift}, every integer solution with $x\ne0$ lifts to one of these points with $x\ne0$, which gives $(-1,-1)$, $(-1,0)$, $(-1,1)$, $(1,0)$, $(2,-2)$, $(4,-6)$, and the only solution with $x=0$ is $(0,0)$. The genus and the rank are Lemmas~\ref{lem:1.1.15:model} and \ref{lem:1.1.15:M}. \end{proof} \section{Primitive generalized Fermat equations}\label{sec:1.2} Primitivity means $\gcd(x,y,z)=1$. Each proof reduces to finitely many covering curves and then determines the relevant rational or number-field points. All signs in the lists below are independent. \subsection{\texorpdfstring{$x^4+3y^3+z^3=0$ \ ($H=48$)}{x^4+3y^3+z^3=0 (H=48)}}\label{sec:1.2.1} \begin{thm}\label{thm:1.2.1} The only primitive integer solutions of $x^4+3y^3+z^3=0$, that is, integer solutions with $\gcd(x,y,z)=1$, are \[ (x,y,z)=(\pm1,0,-1),\qquad(\pm895,-6202,4199). \] \end{thm} \paragraph{Setup.} Put $X=x$, $Y_0=-y$, $Z_0=-z$. Then the equation becomes \[ Z_0^3+3Y_0^3=X^4 \] For a primitive solution, $X\ne0$, $\gcd(Y_0,Z_0)=1$ and $3\nmid XZ_0$. Indeed, if $X=0$, then $Z_0^3=-3Y_0^3$ gives $Y_0=Z_0=0$. A common prime factor of $Y_0,Z_0$ also divides $X$. If 3 divides one of $X$ and $Z_0$, then by the equation it divides both; then $27\mid 3Y_0^3$, so $3\mid Y_0$, a contradiction. We use three fields. \begin{itemize} \item $K=\Q(\theta)$, $\theta^3=3$. The unit $\varepsilon=4+3\theta+2\theta^2$, $\varepsilon^{-1}=\theta^2-2$. \item $L=\Q(\theta,\xi)$, $\xi^2=3-2\theta$ (degree 6). $\xi$ is a root of $T^6-9T^4+27T^2-3$. \item $M=K(R_0,W)$, $R_0^2=9+6\theta+4\theta^2=\theta^2\varepsilon$, $W^2=6+2\theta+2R_0$ (degree 12). \end{itemize} The proof runs as follows. A fourth-power descent over $K$ gives four plane quartic curves $C_j$ (Step 1). The even $j$ are settled by elliptic curves over $K$ (Step 2). The case $j=1$ is eliminated by congruences, and $j=-1$ becomes the problem of finding the rational points of the genus 2 curve $v^2=f(q)$ (Step 3). A two-cover descent on this curve (Step 4) lifts all its points to an elliptic curve $E$ over the degree 12 field $M$ (Step 5). We show $\Cl(M)[2]=0$ (Step 6) and $\operatorname{rank}E(M)=6$ (Step 7), and we obtain $q\in\{0,-8/3\}$ by a sieve at 661 (Step 8) and elliptic Chabauty (Step 9). \paragraph{The disk argument.} We first record a fact that is used repeatedly in this section and in Sections~\ref{sec:1.2.3}--\ref{sec:1.2.4}. \begin{lem}\label{lem:1.2.1-disk} Let $p\ge3$, and let $\Lambda(T)=a_0T+\sum_{n\ge2}a_{n-1}T^n/n$ with $a_0\in\Z_p^\times$, $a_i\in\Z_p$. Then the only $T\in p\Z_p$ with $\Lambda(T)=0$ is $T=0$. \end{lem} \begin{proof} If $T\ne0$ and $n\ge2$, then $v_p(a_{n-1}T^n/n)\ge nv_p(T)-v_p(n)>v_p(T)=v_p(a_0T)$, because $(n-1)v_p(T)\ge n-1>v_p(n)$. These valuations tend to infinity, so the sum of the tail has larger valuation than the linear term and cannot cancel it. \end{proof} In elliptic Chabauty (Proposition~\ref{thm:G}), for a local parameter $T$ of a residue disk, we take $\Lambda$ to be the composition of a global annihilator with the elliptic logarithm. This $\Lambda$ vanishes at every rational point in the disk. Since $\Lambda'$ is the pullback differential, if the map and the differential are integral on the disk and $\Lambda'(0)$ is a unit, then by Lemma~\ref{lem:1.2.1-disk} the only rational point in the disk is its center. \paragraph{Step 1: descent over the cubic field.} \begin{lem}\label{lem:1.2.1-cubic} \begin{enumerate}[label=(\arabic*)] \item $\mathcal O_K=\Z[\theta]$, $\operatorname{disc}K=-243$, $h(K)=1$, $\mathcal O_K^\times=\{\pm\varepsilon^n\}$. \item For each primitive solution we have \[ \begin{gathered} Z_0+\theta Y_0=\varepsilon^j\beta^4,\qquad j\in\{0,1,2,-1\},\qquad \beta=a+b\theta+c\theta^2,\\ a,b,c\in\Z,\qquad\gcd(a,b,c)=1,\qquad 3\nmid N\beta. \end{gathered} \] and $|X|=|N\beta|$. Hence $(a:b:c)$ is a rational point of the plane quartic curve $C_j:\ [\varepsilon^j(a+b\theta+c\theta^2)^4]_{\theta^2}=0$. Here $[\,\cdot\,]_{\theta^2}$ denotes the coefficient of $\theta^2$ in the basis $1,\theta,\theta^2$. \end{enumerate} \end{lem} \begin{proof} (1) $T^3-3$ is Eisenstein at 3 and its discriminant is $-3^5$, so $\mathcal O_K=\Z[\theta]$. The Minkowski constant is $\frac4\pi\cdot\frac{3!}{3^3}\sqrt{243}<4.42$, and the only prime ideals of norm 2, 3, 4 are $(\theta-1)$, $(\theta)$, $(\theta^2+\theta+1)$, so $h(K)=1$. The signature is $(1,1)$, so the unit rank is 1. For a unit with $1<\eta<\varepsilon$, the other two conjugates have absolute value less than 1, and solving for the coefficients from the three embeddings gives $\eta=a+b\theta+c\theta^2$ with $0\le a\le4$, $0\le b\le3$, $0\le c\le2$. In this range the only elements whose norm $a^3+3b^3+9c^3-9abc$ equals 1 are 1 and $\varepsilon$ (\code{1.2.1a}). Hence $\varepsilon$ is the fundamental unit. (2) If a prime ideal $\mathfrak p$ above $p\ne3$ divides $Z_0+\theta Y_0$, then $\mathfrak p\nmid Y_0$ (otherwise $\mathfrak p\mid Z_0$), and $\theta\equiv-Z_0/Y_0\pmod{\mathfrak p}$. Since $T^3-3$ has no repeated root modulo $p$, such a $\mathfrak p$ is the only one among the prime ideals above $p$, and it has residue degree 1. Therefore $v_{\mathfrak p}(Z_0+\theta Y_0)=v_p(X^4)$ is a multiple of 4. Since $3\nmid X$, $(\theta)$ does not divide it. Hence $(Z_0+\theta Y_0)$ is the fourth power of an ideal, and by (1), $Z_0+\theta Y_0=\pm\varepsilon^j\beta^4$. The norm $X^4$ is positive, $N(-1)=-1$ and $N\varepsilon=1$, so the sign is $+$. A prime dividing all of $a,b,c$ would divide $Y_0,Z_0$, so $\gcd(a,b,c)=1$. We have $(N\beta)^4=X^4$ and $3\nmid X$. Finally, the $\theta^2$-coefficient of $Z_0+\theta Y_0$ is 0. \end{proof} \paragraph{Step 2: the even case.} For a point of $C_j$, define $Y_0,Z_0\in\Q$ by $\varepsilon^j\beta^4=Z_0+\theta Y_0$, and put $n=N\beta$, $\nu=n/\beta^3$, $\lambda=(\theta Y_0-Z_0)/(\theta Y_0+Z_0)$. From $(Z_0+\theta Y_0)(Z_0^2-\theta Y_0Z_0+\theta^2Y_0^2)=Z_0^3+3Y_0^3=n^4$ we get $4\nu^4=\varepsilon^{3j}(1+3\lambda^2)$, and hence \[ \varphi_j:\ C_j\longrightarrow E_j:\ V^2=U^3-36\varepsilon^jU,\qquad (U,V)=(12\varepsilon^{-j}\nu^2,\ 36\lambda\nu) \] is a map over $K$. Written in homogeneous coordinates, it has no base points at the primes used below. \begin{lem}\label{lem:1.2.1-even} $C_2(\Q)=\varnothing$ and $C_0(\Q)=\{(1:0:0),(0:1:0)\}$. The only primitive solutions coming from even $j$ are $(\pm1,0,-1)$. \end{lem} \begin{proof} If $j$ is even, then $U^3-36\varepsilon^jU=U(U-6\varepsilon^{j/2})(U+6\varepsilon^{j/2})$, so all 2-torsion points of $E_j$ are $K$-rational. We carry out the full 2-descent of \cite{Simon02,Silverman09} with the square class generators $-1,\varepsilon,\theta-1,\theta^2+\theta+1,\theta$. At the two places above 2 the local images have sizes 8 and 16, and at 3 the size is 4. Of the $32^2$ global candidate pairs, 16 survive for $E_0$ and 4 for $E_2$. On $E_0$, the 2-torsion and the two points \[ A=(-3,9),\qquad B=(6\theta-6\theta^2,\ 36-36\theta+12\theta^2) \] fill all 16. Hence $\operatorname{rank}E_0(K)=2$, $\operatorname{rank}E_2(K)=0$, and $\langle A,B,E_0[2]\rangle$ has odd index. Reduction at the primes 5, 11 ($E_0$) and 61, 67 ($E_2$) of residue degree 1 and \cite[Appendix]{Katz81} show that the torsion is just $E_j[2]$ (\code{1.2.1a}). Since $\nu\ne0$, a rational point of $C_2$ maps to a point with $U\ne0$. For $U=\pm6\varepsilon$ we would need $\pm\varepsilon^3/2$ to be a square in $K$, and it is not (\code{1.2.1a}). So the image is not a torsion point, which contradicts rank 0. Hence $C_2(\Q)=\varnothing$. For $C_0$ we apply \cite{BS10} with $p=577$. The prime $577$ splits completely with $\theta\equiv260,329,565$, and all three reductions have $E_0(\F_{577})\cong(\Z/24)^2$. The subgroup $H$ of the product generated by the reductions of $A,B,E_0[2]$ has order 576. Since $[8]A$, $[8]B$ are independent in the 3-torsion of the product, the index of $\langle A,B,E_0[2]\rangle$ is prime to 3; being odd, it is prime to the order $576^2=2^{12}3^4$ of the quotient. Hence the reduction of $E_0(K)$ is exactly $H$. A full scan of $\mathbb P^2(\F_{577})$ finds 572 points on $C_0$, and the only ones whose image under $\varphi_0$ lies in $H$ are the reductions of $(1:0:0)$ and $(0:1:0)$ (\code{1.2.1b}). We apply Proposition~\ref{thm:G} to the two remaining disks. Lifting the three roots $\theta$ modulo $577^3$ and computing the logarithm of $\omega=dU/(2V)$ (Lemma~\ref{lem:O}), we get, modulo $577^2$, \[ \tfrac{1}{577}\log A\equiv(4351,4351,4351),\qquad \tfrac{1}{577}\log B\equiv(55622,183530,93777) \] The cross product of the two normalized logarithms is an integral annihilator that kills all logarithms of $E_0(K)$; modulo 577 it is $(28,304,245)$. With the parameter $b/a$ at $(1:0:0)$ and $a/b$ at $(0:1:0)$, the coefficients of $\varphi_0^*\omega$ at the centers are $\theta$ and $-\theta^2/3$, respectively, and their combinations with the annihilator are 496 and 241 modulo 577 (\code{1.2.1b}). $C_0$ is smooth at 577, and the derivatives of the dependent coordinate are 4 and 12, which are units. By Lemma~\ref{lem:1.2.1-disk}, each disk contains only its center. The point $(0:1:0)$ gives $\beta=\theta$, $N\beta=3$, which violates $3\nmid N\beta$. The point $(1:0:0)$ gives $\beta=\pm1$, $Z_0+\theta Y_0=1$, so $(X,Y_0,Z_0)=(\pm1,0,1)$, that is, $(x,y,z)=(\pm1,0,-1)$. \end{proof} \paragraph{Step 3: the odd case and a genus 2 curve.} We use the following polynomials. \begin{align*} P(m,n)&=-720m^6+1944m^5n-1620m^4n^2-60m^3n^3+810m^2n^4-432mn^5+73n^6,\\ Y(m,n)&=84m^4-216m^3n+216m^2n^2-100mn^3+18n^4,\\ Z(m,n)&=-108m^4+312m^3n-324m^2n^2+144mn^3-23n^4,\\ f(q)&=16q^6+72q^5+180q^4+300q^3+270q^2+108q+9. \end{align*} We have $Z^3+3Y^3=P^2$, and $f$ is irreducible with discriminant $2^{18}3^{25}$ (\code{1.2.1c}). \begin{lem}\label{lem:1.2.1-odd} There is no primitive solution with $j=1$. For a primitive solution with $j=-1$, there are coprime integers $m,n$ ($n>0$, $2\mid m$, $\gcd(n,6)=1$) and $w>0$ such that \[ w^2=P(m,n),\qquad (x,y,z)=(\pm w,\,-Y(m,n),\,-Z(m,n)) \] Moreover, if we put $q=(2n-3m)/(2m-n)$ and $v=w/(2m-n)^3$, then $v^2=f(q)$, and $q$ has even numerator and odd denominator. \end{lem} \begin{proof} $j=1$: Modulo 2 the equation of $C_1$ is $b^4\equiv0$, so $b$ is even. If $b$ is even, then modulo 8 we would need $2a^4+6a^2c^2+4c^4\equiv0$, but in the three cases where $a,c$ are not both even the values are $2,4,4$ (\code{1.2.1c}). This contradicts $\gcd(a,b,c)=1$. $j=-1$: We have $\varepsilon^{-1}=\theta^2-2$. Writing $\beta^2=A+B\theta+C\theta^2$, the equation of $C_{-1}$ becomes the conic \[ A^2-4AC-2B^2+6BC=0 \] The integers $A,B,C$ are coprime. Indeed, if a prime $\ell$ other than 3 divides all of them, then $\beta$ is a nilpotent element of $\Z[\theta]/\ell$ (a reduced algebra), so $\ell\mid\beta$; and if 3 divides all of them, then $3\mid N\beta$. If $A=0$, then $B=0$ or $B=3C$, and then $N\beta^2$ is $\pm9$ or $\pm90$, which violates $3\nmid N\beta$. If $A\ne0$, the equation is linear in $C$, so putting $B/A=m/n$ (in lowest terms, $n>0$) we get \[ (A:B:C)=(h_A:h_B:h_C)=\bigl(2n(3m-2n):2m(3m-2n):2m^2-n^2\bigr) \] The gcd $g$ of the three forms is 1 if $n$ is odd and 2 if $n$ is even, and $N(h_A+h_B\theta+h_C\theta^2)=-P(m,n)$. We have $\beta^2=\pm g^{-1}(h_A+h_B\theta+h_C\theta^2)$, and $(N\beta)^2$ must be a perfect square. But if $n$ is even, then $\pm N(h/2)\equiv2\pmod4$, and if $n,m$ are both odd, then $\pm N(h)\equiv5,3\pmod8$. Hence $m$ is even, $n$ is odd, and \[ \beta^2=-(h_A+h_B\theta+h_C\theta^2),\qquad (N\beta)^2=P(m,n) \] From $P(m,n)\equiv n^6\pmod3$ and $3\nmid N\beta$ we get $3\nmid n$. Now $Z_0+\theta Y_0=(\theta^2-2)\beta^4=Z(m,n)+\theta Y(m,n)$ and $X^4=P(m,n)^2$, so with $w=|N\beta|$ we get $(x,y,z)=(X,-Y_0,-Z_0)=(\pm w,-Y(m,n),-Z(m,n))$. With $\ell=2m-n$, $k=2n-3m$ we have $(m,n)=(k+2\ell,2k+3\ell)$, and the identity $P(k+2\ell,2k+3\ell)=\ell^6f(k/\ell)$ gives $v^2=f(q)$. Since $n$ is odd and $m$ is even, $\ell$ is odd and $k$ is even. \end{proof} \paragraph{Step 4: two-cover descent.} Put \[ \begin{aligned} r&=\tfrac14\bigl(-3+\theta-\theta^2+(3+3\theta+\theta^2)\xi\bigr),& \delta&=\tfrac12(3-3\theta+\theta^2)(1+\xi),\\ \pi_0&=\theta-1,& \pi_1&=-1+\theta-\theta^2+2\theta\xi+\delta. \end{aligned} \] Then $f(r)=0$. \begin{lem}\label{lem:1.2.1-twocover} The pair $(q,v)$ of Lemma~\ref{lem:1.2.1-odd} satisfies \[ q-r=c\,\pi_1\gamma^2,\qquad c\in\Q^\times,\ \gamma\in L^\times \] \end{lem} \begin{proof} An integral basis of $L$ is $(1,\theta,\theta^2,\xi,\theta\xi,\delta)$, and $\operatorname{disc}L=2^43^{11}$, $h(L)=1$. Above 2 there are the prime ideals $(\pi_0)$, $(\pi_1)$ of norm 4, and $2/(\pi_0\pi_1^2)$ is a unit. The prime 3 is totally ramified as $(\xi)^6$. The square classes of units are generated by $-1$, $\theta^2-2$, $u_2=-1+\theta\xi+\delta$, $u_3=\theta-1-\xi$, and the identity $3/\xi^6=(\theta^2-2)(52+36\theta+25\theta^2)^2$ holds (\code{1.2.1d}). By \cite{BS09}, the class of $q-r$ in $L^\times/L^{\times2}\Q^\times$ is unramified outside the primes 2, 3 dividing the leading coefficient and the discriminant of $f$. Since $N(q-r)=f(q)/16=(v/4)^2$ and $(\xi)$ is the only prime ideal above 3 (of residue degree 1), $v_\xi(q-r)$ is even. Modulo $\Q^\times$ the classes of $-1$ and $\theta^2-2\equiv3$ disappear, and since $2\equiv\pi_0\cdot(\text{unit})$, the class of $\pi_0$ also becomes a unit class. Hence the global candidates are the eight classes $u_2^au_3^b\pi_1^c$ ($a,b,c\in\{0,1\}$). By Lemma~\ref{lem:1.2.1-odd}, $q\in2\Z_2$. For $q_0=0$ and for $q_0=2$, the $(\pi_0,\pi_1)$-valuations of $q_0-r$ are $(-1,-1)$, and those of $4/(q_0-r)$ are $(3,5)$, which exceed $2v(2)=(2,4)$. By Hensel's lemma, the square class of $q-r$ is constant on $4\Z_2$ and on $2+4\Z_2$. We now compare the local square classes of $-r$ and $2-r$ with the eight candidates multiplied by the scalars $\pm1,\pm2,\pm5,\pm10$ of $\Q_2$. The comparison is exact: we check against the 576 squares of units of $\mathcal O_L/8\mathcal O_L$. If $q\in4\Z_2$, the only matching candidate is $\pi_1$; if $q\in2+4\Z_2$, no candidate matches (\code{1.2.1d}). Hence $q\in4\Z_2$ and the class of $q-r$ is $\pi_1$. \end{proof} \paragraph{Step 5: an elliptic curve over the degree 12 field.} Let $\xi_1,\xi_2$ be the two roots of $T^2-WT+R_0$. Put \[ R(T)=\tfrac1{16}(T^5-T^4-12T^3+4T^2+39T-15),\qquad \Pi(T)=\tfrac18(T^5-T^4-8T^3+8T^2+27T-11) \] Then $r=R(\xi)$ and $\pi_1=\Pi(\xi)$. Put $r_i=R(\xi_i)$ and $D_*=\Pi(\xi_1)\Pi(\xi_2)$. Also put \[ d=-3-2\theta+3\theta^2+\bigl(\theta+\tfrac56\theta^2\bigr)W+\bigl(1-\tfrac12\theta^2\bigr)W^2-\tfrac{\theta}{12}W^3,\qquad E:\ y^2=x^3+2dx^2+2d^2x \] We have $c_4(E)=-32d^2$, $\Delta_E=-256d^6$, $j(E)=128$, $N_M(d)=-5184=-2^63^4$. \begin{lem}\label{lem:1.2.1-cover} \begin{enumerate}[label=(\arabic*)] \item $[M:\Q]=12$, and $M$ has signature $(2,5)$. \item All points of Lemma~\ref{lem:1.2.1-twocover} lift to $M$-points $(q,V)$, $q\in\Q$, of the quartic curve \[ \mathcal Q:\ V^2=Q(q):=D_*\,\frac{f(q)}{(q-r_1)(q-r_2)} \] Here $Q$ is a separable quartic over $M$, and $Q(0)=(2\theta^2)^2$. \item Write $Q(q)=a_4q^4+a_3q^3+a_2q^2+a_1q+e^2$ with $e=2\theta^2$, and put \[ X'=\frac{2e(V+e)+a_1q}{q^2},\quad Y'=\frac{(X'^2-4e^2a_4)q-(a_1X'+2e^2a_3)}{2e},\quad x=\frac{X'-h_0}{s_0},\quad y=\frac{Y'}{s_0e} \] ($s_0=12\theta$, $h_0=(2ds_0-a_2)/3$). Then $\varphi:(q,V)\mapsto(x,y)$ is a map $\mathcal Q\to E$, and $\varphi^*\bigl(dx/(2y)\bigr)=-\theta^2\,dq/V$. \item The two points $(0,e)$, $(0,-e)$ above $q=0$ map to $O$ and to a point $P$, and the two points above $q=-8/3$ map to $P_1$ and $P-P_1$. Here $P,P_1\in E(M)$. \end{enumerate} \end{lem} \begin{proof} (1) $\varepsilon$ is not a square in $K$, because its residue at the prime ideal above 5 with $\theta\equiv2$ is 3. $N_{K(R_0)/K}(W^2)=(6+2\theta)^2-4R_0^2=-12\theta^2$ is negative at the real embedding of $K$, so $W^2$ is not a square in $K(R_0)$. Since $\varepsilon>0$, $K(R_0)$ has two real embeddings, and the two values of $W^2$ at them have negative product, so exactly one of them is positive. Hence $M$ has 2 real embeddings. (2) By the identity $T^6-9T^4+27T^2-3=(T^2-3+2\theta)(T^2-WT+R_0)(T^2+WT+R_0)$, the maps $\xi\mapsto\xi_i$ are two embeddings $\sigma_i$ of $L$ into $\overline M$. Applying $\sigma_1,\sigma_2$ to the relation of Lemma~\ref{lem:1.2.1-twocover} and multiplying gives $(q-r_1)(q-r_2)=D_*H^2$ with $H=c\,\sigma_1(\gamma)\sigma_2(\gamma)$. The elements $H$, $r_1+r_2$, $r_1r_2$, $D_*$ are symmetric under $\xi_1\leftrightarrow\xi_2$, so they lie in $M$. Hence $Q(q)=(v/H)^2$. From $r_1r_2=\theta^2D_*/4$ we get $Q(0)=4\cdot9/\theta^2=(2\theta^2)^2$. (3), (4) By the standard Weierstrass formulas for a quartic, $Y'^2=X'^3+a_2X'^2+(a_3a_1-4a_4e^2)X'+a_3^2e^2+a_4a_1^2-4a_4a_2e^2$, and a change of coordinates gives $E$. All identities are checked by polynomial division in $M$. The coordinates of $P$, $P_1$ and the two limits are also checked by exact substitution (\code{1.2.1e}). \end{proof} \paragraph{Step 6: the class group of $M$.} \begin{lem}\label{lem:1.2.1-class} $\Cl(M)[2]=0$. \end{lem} \begin{proof} We use twice the ambiguous class number formula for a quadratic extension $B/A$ (\cite[Theorem 1]{Lemmermeyer13}) \[ |\Cl(B)^{\operatorname{Gal}(B/A)}|=\frac{h(A)\,2^{t-1}}{[\mathcal O_A^\times:\mathcal O_A^\times\cap N_{B/A}B^\times]} \] ($t$ is the number of ramified places, infinite places included). If this value is odd, then $h(B)$ is odd as well. Indeed, the generator $\sigma$ acts as an involution on the $\F_2$-space $\Cl(B)[2]$, and an involution on a nonzero space has a nonzero fixed vector. Put $F=K(\kappa)$, $\kappa=R_0/\theta$, $\kappa^2=\varepsilon$. Since $\varepsilon$ is a unit and is positive at the real embedding, $F/K$ can ramify only at places above 2. At the dyadic place of degree 1 we have $\theta\equiv3$ and $\varepsilon\equiv7\pmod8$, and at the dyadic place of degree 2 we have $\theta^2+3\theta+1\equiv0$ and $\varepsilon\equiv2+\theta\pmod4$, which is not a square; so both places ramify. Hence $t=2$, $\mathcal O_F=\Z[\theta,\kappa]$ and $\operatorname{disc}F=2^63^{10}$. At the dyadic place of degree 1, $-1$ is not a norm from $\Q_2(\sqrt7)$ ($a^2-7b^2\equiv-1\pmod4$ is impossible). So the unit norm index is at least 2, and the formula gives $1\cdot2/2=1$. Hence $h(F)$ is odd. We have $M=F(W)$. From an explicit integral basis of $M$ we get $\operatorname{disc}M=-2^{14}3^{21}$ (\code{1.2.1e}), so the norm of the relative discriminant of $M/F$ is $2^{14}3^{21}/(2^63^{10})^2=12$. A ramified dyadic quadratic extension has discriminant exponent at least 2, so the dyadic prime ideal of $F$ of norm 4 is unramified, and the dyadic place of norm 2 ramifies with exponent 2. The remaining factor 3 comes from the ramification of one place above 3. At the place above 3 with $\kappa\equiv1$ we have $(\kappa+2)(\kappa-2)=\varepsilon-4=3\theta+2\theta^2$, so $v_\theta(\kappa+2)=2$, and $W^2=6-2\theta+2\theta(\kappa+2)$ has valuation 1. So the ramified 3-adic place is this one. Of the two real places of $F$, only the one with $\kappa<0$ becomes complex (see the proof of Lemma~\ref{lem:1.2.1-cover}(1)). Therefore $t=3$. The units $-1$, $\kappa$ have norm obstruction vectors $(1,1)$, $(1,0)$ at the real place that becomes complex and at the ramified 3-adic place (a unit that is a local norm there has a square residue in the residue field $\F_3$), so the unit norm index is at least 4. Since squares of units are norms, this index is a power of 2; since the value $h(F)\cdot4/[\,\cdot\,]$ of the formula is an integer and $h(F)$ is odd, the index is exactly 4. Hence $|\Cl(M)^{\operatorname{Gal}}|=h(F)$ is odd, and $\Cl(M)[2]=0$ (\code{1.2.1f}). \end{proof} \paragraph{Step 7: the rank.} We use the 2-isogenous curve $E':\ y^2=x^3-4dx^2-4d^2x$ and the isogenies $\phi$, $\hat\phi$ between $E$ and $E'$. The maps $\alpha:E(M)\to M^\times/M^{\times2}$ and $\alpha':E'(M)\to M^\times/M^{\times2}$ send a point to the square class of its $x$-coordinate, and $\alpha(T)=2$, $\alpha'(T')=-1$ ($T=(0,0)$, $T'=(0,0)$). By \cite{Simon02,Silverman09}, \[ 2^{\operatorname{rank}E(M)}=\frac{|\alpha(E(M))|\,|\alpha'(E'(M))|}{4}. \] \begin{lem}\label{lem:1.2.1-rank} $\operatorname{rank}E(M)=6$. For points $G_1,\dots,G_4\in E(M)$ (their coordinates are in the input of \code{1.2.1g}) and $P,Q,T$, the subgroup $G=\langle G_1,G_2,G_3,G_4,P,Q,T\rangle$ has odd index. Moreover, $E(M)_{\mathrm{tors}}=\{O,T\}$. \end{lem} \begin{proof} Both images consist of classes with even valuation outside 2, 3 ($N_M(d)=-2^63^4$). By Lemma~\ref{lem:1.2.1-class}, an ideal whose square is principal is itself principal. So the candidates are the $2^{12}=4096$ classes generated by 7 unit square classes (unit rank 6 for the signature $(2,5)$, together with $\pm1$) and 5 generators of the prime ideals above 2 and 3. The decomposition of 2 is $(e,f)=(4,1),(2,2),(2,2)$, and that of 3 is $(6,1),(3,2)$. The sizes of the local images are as follows. \begin{center} \begin{tabular}{lccc} \toprule Places & Local degree & $|\alpha_v|$ & $|\alpha'_v|$\\ \midrule Three places above 2 & 4 & 16 & 4\\ Two places above 3 & 6 & 2 & 2\\ \bottomrule \end{tabular} \end{center} At each place the product of the two sizes equals the value of the local isogeny quotient formula ($2^{4+2}$ at a dyadic place, 4 at an odd place), so these images, generated by local point witnesses, are complete. At the real places, $x^3+2dx^2+2d^2x=x\bigl((x+d)^2+d^2\bigr)$, so $x>0$ on $E$. The numbers of candidates remaining after each condition are as follows. \begin{center} \begin{tabular}{lcc} \toprule Condition & $E$ & $E'$\\ \midrule First dyadic place & 1024 & 256\\ Second dyadic place & 256 & 16\\ Third dyadic place & 64 & 8\\ Two places above 3 & 64 & 8\\ Two real places & 32 & 8\\ \bottomrule \end{tabular} \end{center} On the other hand, $\alpha(T),\alpha(G_1),\dots,\alpha(G_4)$ are independent, and for the points $Q_0,D'$ on $E'$, $\alpha'(T'),\alpha'(Q_0),\alpha'(D')$ are also independent (\code{1.2.1g}). These match the upper bounds, so $|\alpha(E(M))|=32$, $|\alpha'(E'(M))|=8$, and $2^{\operatorname{rank}}=32\cdot8/4=64$. The dual isogeny $\hat\phi(x,y)=\bigl((x-4d-4d^2/x)/4,\ y(1+4d^2/x^2)/8\bigr)$ sends $Q_0\mapsto P$, $D'\mapsto Q$ and $T'\mapsto O$. Therefore $T,G_1,\dots,G_4$ generate $E(M)/\hat\phi E'(M)$, $P,Q$ generate $\hat\phi E'(M)/2E(M)$, and $G$ generates $E(M)/2E(M)$. Hence the index is odd. At the places $(p,\theta,d)=(23,12,9)$, $(83,11,12)$ of residue degree 1 the reductions have orders 28 and 90, so by \cite[Appendix]{Katz81} the order of the torsion divides 2. \end{proof} \paragraph{Step 8: a sieve at 661.} The prime $p=661$ splits completely in $M$. Listing the twelve embeddings as $(\theta,d)\bmod p$, we get \[ \begin{array}{llll} (261,34) & (261,334) & (261,454) & (261,500)\\ (481,390) & (481,441) & (481,513) & (481,639)\\ (580,17) & (580,153) & (580,517) & (580,635) \end{array} \] and in this order $\#E_i(\F_p)$ is $(628,696,628,696,696,696,628,628,696,696,696,696)$. Here $628=4\cdot157$ and $696=8\cdot3\cdot29$. \begin{lem}\label{lem:1.2.1-sieve} If $q\in\Q$ and $(q,V)\in\mathcal Q(M)$, then $q\equiv0$ or $q\equiv438\equiv-8/3\pmod{661}$. The sign patterns of $V$ at the twelve embeddings are restricted to 64. \end{lem} \begin{proof} The subgroup $H$ of the product generated by the reductions of $G$ has index $2^{22}3^229^2$. Its 2-primary part has order 1024 and index $2^{22}$. For odd primes $\ell$, the matrix of $\ell$-components of the free generators $G_1,\dots,G_4,P,Q$ is $8\times6$ of rank 6 for $\ell=3$, $8\times6$ of rank 6 for $\ell=29$, and $4\times6$ of rank 4 (the whole image) for $\ell=157$. Since the column rank is full for $\ell=3,29$, no nonzero combination of the generators modulo $\ell$ can lie in $\ell E(M)$ (multiplying by $\#E_i(\F_p)/\ell$ in each factor would kill that combination). The torsion has order 2, so $[E(M):G]$ is prime to 3 and 29; being odd, it is prime to the quotient order $2^{22}3^229^2$. Hence, by \cite{BS10}, the reduction of $E(M)$ is exactly $H$. For each $q\in\mathbb P^1(\F_p)$ we run over all signs of the ordinates of $V^2=Q(q)$ at the twelve embeddings (including zero ordinates and removable poles of the map). There are 27 values of $q$ with points in every factor, and 106512 sign patterns. After the 2-primary test 6146 remain, after the 3-primary test 1152, and after the 29-primary test 64. All 64 remaining patterns have $q\equiv0$ or $q\equiv438$ (\code{1.2.1h}). \end{proof} \paragraph{Step 9: 661-adic logarithms.} \begin{lem}\label{lem:1.2.1-chab} If $q\in\Q$ and $(q,V)\in\mathcal Q(M)$, then $q=0$ or $q=-8/3$. \end{lem} \begin{proof} At each embedding we use the logarithm of $\omega=dx/(2y)$. We pass to the short model $X=x+2d/3$, $Y^2=X^3+\frac23d^2X-\frac{20}{27}d^3$, and multiply a point $G$ by the order $N_i$ (prime to $p$) of its reduction to put it into the formal group. Since $\log\tau=\tau+O(\tau^5)$ for $\tau=-X/Y$, computing $\tau([N_i]G)$ modulo $p^3$ determines $\log(G)/p$ modulo $p^2$ (Lemma~\ref{lem:O}). We compute with two kinds of arithmetic (projective scalar multiplication and division polynomials). The $6\times12$ matrix of the normalized logarithms of $G_1,\dots,G_4,P,Q$ reduced modulo $p$ has rank 6, and the minor of its first six columns is a unit. Therefore we can choose an integral basis $c^{(1)},\dots,c^{(6)}$ of the kernel of the exact normalized logarithms whose last six columns form the identity matrix. The first row is \[ c^{(1)}\equiv(74,2,659,439,29,256,1,0,0,0,0,0)\pmod{661} \] (the full matrix and the logarithm of $P_1$ are in \code{1.2.1i}). Since $G$ has finite index, these six annihilators kill the logarithms of all points of $E(M)$. For each pattern of Lemma~\ref{lem:1.2.1-sieve} we take the center $q_0=0$ or $q_1=-8/3$. The image of the center in factor $i$ is $O$ or $P_i$ for $q_0$, and $(P_1)_i$ or $P_i-(P_1)_i$ for $q_1$, so the normalized logarithm vector $\ell$ is determined. Writing $q=q_j+pT$ with $T\in\Z_p$, Lemma~\ref{lem:1.2.1-cover}(3) gives the differential vector $D_i\equiv-\theta_i^2/V_i\pmod p$, and the ordinates of the centers are all units. If the disk contains a rational point, then for every $c^{(k)}$ \[ c^{(k)}\cdot\ell+T\,(c^{(k)}\cdot D)\equiv0\pmod p \] The terms of degree at least 2 in $pT$ are still 0 modulo $p$ after division by $p$, so this condition is necessary on the whole disk. For 60 of the 64 patterns, no $T$ satisfies these six equations simultaneously. The remaining four are, for each center, the two patterns with all signs equal (\code{1.2.1i}). The centers of the four remaining disks are preimages of the actual global points $O$, $P$, $P_1$, $P-P_1$, so the annihilating logarithms vanish exactly there. The derivative coefficients for $c^{(1)}$ are as follows. \begin{center} \begin{tabular}{lcc} \toprule Center & Sign on the side of $O$ or $P_1$ & Opposite sign\\ \midrule $q=0$ & 592 & 69\\ $q=-8/3$ & 266 & 395\\ \bottomrule \end{tabular} \end{center} All of them are nonzero. The map and the differential are integral on these smooth disks, so by Proposition~\ref{thm:G} and Lemma~\ref{lem:1.2.1-disk} each disk contains only its center. Points above parameter infinity were already eliminated by the sieve of Step 8. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.2.1}] By Lemma~\ref{lem:1.2.1-even}, even $j$ gives $(\pm1,0,-1)$. There is no solution with $j=1$. By Lemmas~\ref{lem:1.2.1-odd}, \ref{lem:1.2.1-twocover} and \ref{lem:1.2.1-cover}, a primitive solution with $j=-1$ gives a point of $\mathcal Q(M)$, and by Lemma~\ref{lem:1.2.1-chab}, $q\in\{0,-8/3\}$. If $q=0$, then $2n=3m$, so $3\mid n$, which contradicts $\gcd(n,6)=1$. If $q=-8/3$, then $7m=2n$, so $(m,n)=(2,7)$, and \[ P(2,7)=895^2,\qquad Y(2,7)=6202,\qquad Z(2,7)=-4199 \] gives $(x,y,z)=(\pm895,-6202,4199)$. Substituting these values, we get $895^4+3(-6202)^3+4199^3=0$, and $\gcd(895,6202,4199)=1$. \end{proof} \subsection{\texorpdfstring{$2x^4-y^4+z^3=0$ \ ($H=56$)}{2x^4-y^4+z^3=0 (H=56)}}\label{sec:1.2.3} \begin{thm}\label{thm:1.2.3} The only primitive integer solutions ($\gcd(x,y,z)=1$) of $2x^4-y^4+z^3=0$ are \[ (x,y,z)=(0,\pm1,1),\qquad(\pm1,\pm1,-1) \] (with independent signs). \end{thm} \paragraph{Setup.} The equation is $y^4-2x^4=z^3$. For a primitive solution, $\gcd(x,y)=1$ and $y,z$ are odd. Indeed, if $y$ were even, then $x$ would be odd and $v_2(y^4-2x^4)=1$, which is impossible for a cube. The ring $\Z[\sqrt2]$ is norm-Euclidean: rounding the two coefficients leaves a remainder $a+b\sqrt2$ with $|a|,|b|\le1/2$ and $|a^2-2b^2|\le1/2$. The units are $\pm(1+\sqrt2)^n$. A unit with $1<\eta<1+\sqrt2$ has a conjugate of absolute value less than 1, so writing $\eta=a+b\sqrt2$ we get $a=1$, $b\in\{0,1\}$, which are the two endpoints of the interval. We use the two polynomials \[ f(A,B)=A^3+6A^2B+6AB^2+4B^3,\qquad g(A,B)=A^3+3A^2B+6AB^2+2B^3 \] and $F(t)=f(t,1)g(t,1)=t^6+9t^5+30t^4+60t^3+60t^2+36t+8$. We have $\operatorname{disc}F=2^{26}3^6$. The two genus 2 curves \[ C_+:\ w^2=F(t),\qquad C_-:\ w^2=-F(t) \] are understood as their smooth projective models. $C_+$ has the points at infinity $\infty_+,\infty_-$. The proof runs as follows. A cube descent over $\Q(\sqrt2)$ gives $C_\pm$ (Step 1). Over the cubic field $K=\Q(h)$, a 2-adic two-cover descent eliminates $C_-$ and lifts the points of $C_+$ to an elliptic curve $E$ over $K$ (Steps 2--4). We show $\operatorname{rank}E(K)=2$ (Step 5), and a sieve at 29 (Step 6) and elliptic Chabauty (Step 7) give $C_+(\Q)=\{\infty_+,\infty_-\}$. \paragraph{Step 1: cube descent.} \begin{lem}\label{lem:1.2.3-red} For each primitive solution, $y^2+x^2\sqrt2=(1+\sqrt2)^j(a+b\sqrt2)^3$ with $j\in\{-1,0,1\}$ and $\gcd(a,b)=1$. If $j=0$, then $(x,y,z)=(0,\pm1,1)$. If $j=\pm1$ and $b\ne0$, we get a finite rational point of $C_{\pm}$. Hence, if \[ C_-(\Q)=\varnothing,\qquad C_+(\Q)=\{\infty_+,\infty_-\} \] then Theorem~\ref{thm:1.2.3} holds. \end{lem} \begin{proof} The norm of $y^2\pm x^2\sqrt2$ is the odd number $z^3$, so both elements are units above 2; a common odd prime ideal would divide both $x$ and $y$, so the two elements are coprime. By unique factorization, $y^2+x^2\sqrt2$ is a unit times a cube, and the sign of the unit and $(1+\sqrt2)^{3k}$ are absorbed into the cube. $j=0$: $y^2=a(a^2+6b^2)$ and $x^2=b(3a^2+2b^2)$. Since $y$ is odd, $a>0$ is odd and $b\ge0$. If $b=0$, primitivity gives $(x,y,z)=(0,\pm1,1)$. If $b>0$, then $\gcd(b,3a^2+2b^2)\mid3$. If $3\nmid b$, then $b$ and $3a^2+2b^2$ are both squares, but the latter is 2 modulo 3. If $3\mid b$, then $3a^2+2b^2$ has 3-adic valuation 1, so $b=3v^2$. From $\gcd(a,a^2+6b^2)=1$ we get $a=u^2$ and $(x/(3v))^2=u^4+6v^4$. Then $X=u^2/v^2$, $Y=u x/(3v^3)$ is a rational point of $E_6:\ Y^2=X^3+6X$ with $X\ne0$. But $E_6(\Q)=\{O,(0,0)\}$. In the 2-isogeny descent of \cite{Simon02,Silverman09}, the image for $E_6$ lies in the positive classes $\{1,2,3,6\}$; the class 2 is excluded because $W^2=2U^4+3V^4$ is impossible modulo 16, and 3 is excluded after multiplying by $6$, so the image is $\{1,6\}$. The image for the partner curve $Y^2=X^3-24X$ contains $\{1,-6\}$, and the representatives $-1,-2,2$ of the other three cosets are excluded because $W^2=-U^4+24V^4$, $-2U^4+12V^4$, $2U^4-12V^4$ are impossible modulo 16, 16, 32, respectively, for $(U,V)$ not both even. Hence $2^{\operatorname{rank}}=2\cdot2/4=1$. The reductions at 5, 7, 13 have orders 4, 8, 10, so the torsion has order 2 (\code{1.2.3a}). $j=1$: $y^2=f(a,b)$ and $x^2=g(a,b)$. $j=-1$: $y^2=f(-a,b)$ and $x^2=-g(-a,b)$. If $b\ne0$, then $t=a/b$ or $-a/b$, together with $w=xy/b^3$, is a finite rational point of $C_+$ or $C_-$. If $b=0$: for $j=-1$, $y^2=-a^3$, $x^2=a^3$ is incompatible with a primitive solution; for $j=1$, $y^2=x^2=a^3$, so $a=1$, that is, $(x,y,z)=(\pm1,\pm1,-1)$. \end{proof} \paragraph{Step 2: the cubic field.} Put \[ K=\Q(h),\quad h^3+3h-2=0,\quad e=1-h-h^2,\quad \rho=h-1,\quad \pi=-3-h-h^2, \] \[ r_f=-4-h-h^2=\frac{2}{h-1},\qquad r_g=h-1. \] The elements $r_f$, $r_g$ are roots of $f(t,1)$, $g(t,1)$. The two polynomials have discriminants $-864$, $-216$, and $\operatorname{Res}(f,g)=-512$. \begin{lem}\label{lem:1.2.3-K} $\mathcal O_K=\Z[h]$, $\operatorname{disc}K=-216$, $\Cl(K)=0$, $\mathcal O_K^\times=\{\pm e^n\}$. The prime ideals above 2 are $(h)$, $(\rho)$, and the prime ideal above 3 is $(\pi)$. The square classes of elements with even valuation outside 2, 3 are generated by the ordered basis $\mathcal B=(-1,e,\rho,h,\pi)$. \end{lem} \begin{proof} The polynomial $T^3-3T^2+6T-6$ obtained by the substitution $h=T-1$ is Eisenstein at 3, so $\Z[h]$ is maximal at 3. The polynomial discriminant is $-216=-2^33^3$, so the only other candidate index prime is 2; but if the index were 2, the field discriminant would be $-54$, which violates the condition that a discriminant is 0 or 1 modulo 4 (Stickelberger). The signature is $(1,1)$, and the Minkowski constant $\frac4\pi\cdot\frac{3!}{3^3}\sqrt{216}$ is less than 5. Modulo 2, $h^3+3h-2\equiv h(h+1)^2$, and modulo 3, $h^3+3h-2\equiv(h+1)^3$, so the only prime ideals of norm less than 5 are $(h)$, $(\rho)$, $(\pi)$, of norms 2, 2, 3, and $\Cl(K)=0$. We have $e^{-1}=17+3h+5h^2$ and $Ne=1$. Using the real root $0.592v_\omega(2)=4$, squares modulo 8 lift to squares. The seven-bit labels of the elements of $\mathcal B$ are $(41,49,122,13,11)$. For $t\in a+2^n\Z_2$, the class of the $\Q_2$-component of $t-r$ is constant if $n-v_2(a-r)>2$, and the class of the $\Q_2(\omega)$-component is constant if $2n-v_\omega(a-r)>4$. Splitting balls repeatedly, we cover $\Z_2$ by 42 disjoint balls. On 40 of them all four classes are constant; on the two root balls $10+64\Z_2$, $45+64\Z_2$ only the scalar class of the vanishing factor varies, so we test all eight values. The remaining chart $u=1/t\in2\Z_2$ is covered by $u\equiv0,2,4,6\pmod8$, and there we use $1-ur_f$, $1-ur_g$. The admissible local labels, modulo a common $\Q_2$ scalar, are $\{0,1032,9328,9715\}$ for $\sigma=+1$ and $\{1,1160,9329,9714\}$ for $\sigma=-1$. Comparing the 16 global candidate pairs with these sets, only the trivial pair survives for $\sigma=+1$, and nothing survives for $\sigma=-1$ (\code{1.2.3c}). Since $C_-$ has leading coefficient $-1$, it has no rational point at infinity either. \end{proof} \paragraph{Step 4: an elliptic curve over the cubic field.} By Lemma~\ref{lem:1.2.3-2adic}, $(t-r_f)(t-r_g)$ is a square in $K$. Let \[ q(t)=t^4+(4-h^2)t^3+(8+2h-2h^2)t^2+(8-2h^2)t+4 \] so that $(t-r_f)(t-r_g)q(t)=F(t)$. Then the finite rational points of $C_+$ lift to $K$-points ($t\in\Q$) of $v^2=q(t)$. Put \[ k=2-\tfrac12h^2,\quad m=2+\tfrac34h+\tfrac38h^2,\quad n=\tfrac32-\tfrac92h-\tfrac{21}8h^2,\quad r_1=6h+3h^2,\quad d=2-h+h^2, \] \[ X=2(v+t^2+kt+m),\qquad Y=(2t+k)X+n,\qquad x_E=4X-r_1,\qquad y_E=8Y. \] Then \[ E:\ y_E^2=x_E(x_E-12d)(x_E+4d),\qquad \phi^*\frac{dx_E}{2y_E}=\frac{dt}{4v} \] The two points at infinity with $v/t^2\to+1$, $-1$ map to $O$ and to \[ Q=(-6h-3h^2,\ -12+36h+21h^2) \] respectively. Writing $u=1/t$, the formal square root of $u^4q(1/u)$ begins with $1+ku+mu^2+\frac12nu^3+O(u^4)$. All identities are checked by polynomial division using $h^3+3h-2=0$ and $v^2=q(t)$ (\code{1.2.3d}). \paragraph{Step 5: the rank.} \begin{lem}\label{lem:1.2.3-rank} $\operatorname{rank}E(K)=2$ and $E(K)_{\mathrm{tors}}=E[2]$. The subgroup $G$ generated by \[ P=(-8h+4h^2,\ -16+56h),\qquad Q=(-6h-3h^2,\ -12+36h+21h^2) \] and $T_0=(0,0)$, $T_1=(12d,0)$ has index prime to 6. \end{lem} \begin{proof} We carry out the full 2-descent of \cite{Simon02,Silverman09} on the isomorphic model $y^2=x(x-3d)(x+d)$ with the Kummer map $\delta=(x,x-3d)$. Its values at $T_0,T_1$ are $(-3d^2,-3d)$, $(3d,12d^2)$. From the differences of the roots and $N(d)=16$, all Kummer values have even valuation outside 2, 3, so the global candidates are the $32^2=1024$ pairs written in $\mathcal B$. The local images have dimensions 3 and 4 at the two places above 2 and dimension 2 above 3, and they are filled by the torsion and three local points ($x=-6-7h-5h^2$, $-7-6h-7h^2$, $-7-6h+3h^2$). At the real place the two signs must agree. Of the 1024 pairs, 16 remain. The mask pairs of $T_0,T_1,P,Q$ are $(18,29)$, $(28,19)$, $(24,26)$, $(24,24)$; they are independent, so they fill all 16 pairs (\code{1.2.3e}). Hence the rank is 2 and the index of $G$ is odd. At the places $(13,h-9)$, $(29,h-6)$ of residue degree 1 the reductions have orders 16 and 36 (\cite[Appendix]{Katz81}), so the order of the torsion divides 4, and all 2-torsion points are rational. We check coprimality to 3 at 29. The prime 29 splits completely with $h\equiv6,9,14$, and in the first two factors the exponents of the groups are 18 and 12. Multiplying by one third of the exponent, we get $[6]P=[6]Q=(16,28)$ at $h\equiv6$, and $[4]P=(22,17)$, $[4]Q=(22,12)=-[4]P$ at $h\equiv9$; all of these have order 3. If $aP+bQ\in3E(K)$, then $a+b\equiv0$ and $a-b\equiv0\pmod3$, so $a\equiv b\equiv0$. The torsion has no 3-part, so the index is prime to 3 (\code{1.2.3f}). \end{proof} \paragraph{Step 6: a sieve at 29.} \begin{lem}\label{lem:1.2.3-sieve} If $(t,w)$ is a rational point of $C_+$ and $t$ is a 29-adic integer, we get a contradiction. Hence $u=1/t\in29\Z_{29}$, and the point lies in one of the two disks centered at the two points at infinity $\infty_\pm$. \end{lem} \begin{proof} In the three factors, $\#E(\F_{29})=36,24,36$, the exponents are $18,12,18$, the orders of $P$ are $18,12,3$, and the orders of $Q$ are $9,3,9$. Adding $P,Q,T_0,T_1$ one at a time, the generated subgroups have sizes $36,324,648,1296$, and the index in the product, of order $36\cdot24\cdot36$, is 24. By Lemma~\ref{lem:1.2.3-rank}, the index of $G$ is prime to 24, so by \cite{BS10} the reduction of $E(K)$ equals this subgroup $H$. Suppose that $t$ is a 29-adic integer. By Lemma~\ref{lem:1.2.3-2adic}, $t-r_f$ and $t-r_g$ are the same $c$ times squares at all three embeddings. If $v_{29}(c)$ is odd, all six values $t-r$ must be divisible by 29; but the residues of the six roots, $(r_f)_i=(12,22,18)$ and $(r_g)_i=(5,8,13)$, are distinct, so this is impossible. Hence $v_{29}(c)$ is even, and the quadratic characters of the nonzero residues $t-r$ must all be equal. Testing all $t=0,\dots,28$, none satisfies this condition (\code{1.2.3f}). Therefore $u=1/t\in29\Z_{29}$. In this chart, $\hat v=u^2v$ satisfies $\hat v^2=u^4q(1/u)\equiv1$, so it reduces to $\pm1$ at each of the three embeddings, and the corresponding elliptic reductions are the eight tuples whose components are $O$ or $Q_i$. Among them, only $(O,O,O)$ and $(Q_1,Q_2,Q_3)$ lie in $H$. That is, the three signs are all equal. \end{proof} \paragraph{Step 7: 29-adic logarithms.} \begin{lem}\label{lem:1.2.3-chab} $C_+(\Q)=\{\infty_+,\infty_-\}$. \end{lem} \begin{proof} Lifting $h$ at the three embeddings modulo $29^3$ gives $13172$, $14132$, $21474$. At each embedding we pass to a short model $y^2=x^3+Ax+B$, multiply points by the orders of their reductions to put them into the formal group, and use $\log\tau=\tau+O(\tau^5)$ with $\tau=-x/y$ (Lemma~\ref{lem:O}). Computing by two methods, projective arithmetic and division polynomials, we find that the normalized logarithms of $\omega=dx_E/(2y_E)$ are, modulo $29^2$, \[ \tfrac1{29}\log P\equiv(647,368,410),\qquad \tfrac1{29}\log Q\equiv(672,302,764) \] Modulo 29 they are $(9,20,4)$ and $(5,12,10)$, which are independent. The cross product $c$ of the two exact normalized logarithms is an integral annihilator, with $c\equiv(7,17,8)\pmod{29}$ and $c_1+c_2+c_3\equiv3$. Since $P,Q$ span the logarithm space and torsion points have logarithm 0, $c$ kills all logarithms of $E(K)$ (\code{1.2.3g}). On the disks of Lemma~\ref{lem:1.2.3-sieve}, $\hat v$ is an integral power series whose constant term at each embedding is $\varepsilon=\pm1$ (the same for all three embeddings), and $\phi^*(dx_E/2y_E)=-du/(4\hat v)$. The constant term of the pullback differential combined with the annihilator is $-\varepsilon(c_1+c_2+c_3)/4\equiv\mp3/4$, that is, 21 or 8 modulo 29. Both are units. The centers $\infty_\pm$ of the disks map to the global points $O$, $Q$, so the annihilating logarithm vanishes there. By Proposition~\ref{thm:G} and Lemma~\ref{lem:1.2.1-disk}, each disk contains only its center. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.2.3}] By Lemmas~\ref{lem:1.2.3-2adic} and \ref{lem:1.2.3-chab}, $C_-(\Q)=\varnothing$ and $C_+(\Q)=\{\infty_+,\infty_-\}$. By Lemma~\ref{lem:1.2.3-red}, the only primitive solutions are $(0,\pm1,1)$ and $(\pm1,\pm1,-1)$. Substituting gives $0-1+1=0$ and $2-1-1=0$. \end{proof} \subsection{\texorpdfstring{$x^5+y^4=2z^2$ \ ($H=56$)}{x^5+y^4=2z^2 (H=56)}}\label{sec:1.2.4} \begin{thm}\label{thm:1.2.4} The only primitive integer solutions ($\gcd(x,y,z)=1$) of $x^5+y^4=2z^2$ are \[ (x,y,z)=(-1,\pm1,0),\qquad(1,\pm1,\pm1) \] (with independent signs). \end{thm} \paragraph{Setup.} The three numbers of a primitive solution are pairwise coprime. Modulo 2, $x$ and $y$ have the same parity; if both were even, $2z^2$ would be a multiple of 16, so $z$ would be even as well. Hence $x,y$ are odd. If $z=0$, then $(x,y,z)=(-1,\pm1,0)$. From now on $z\ne0$. Put $d=\sqrt2$ and $\varepsilon=1+d$. The ring $\Z[d]$ is norm-Euclidean and its units are $\pm\varepsilon^n$ (for a unit with $1<\eta<\varepsilon$, the $d$-coefficient $(\eta-N(\eta)/\eta)/(2d)$ would have to lie between 0 and 1). We use the following forms. \[ F=a^5+20a^3b^2+20ab^4,\quad G=5a^4b+20a^2b^3+4b^5,\quad F_1=F+2G,\quad F_2=3F+4G. \] We have $(a+bd)^5=F+Gd$. The quintic field is \[ L=\Q(\theta),\qquad g(\theta)=\theta^5+5\theta^3+5\theta-2=0 \] and we use the following elements. \begin{align*} u_1&=1-\theta-\theta^3, & u_2&=1-3\theta+\theta^2-\theta^3,\\ q&=-1+2\theta+2\theta^2+\theta^3, & \ell&=1+\theta^2,\\ \eta&=-\theta^4-\theta^3-5\theta^2-4\theta-6, & \chi&=\theta^4+\theta^3+3\theta^2+6\theta-4,\\ \delta&=\theta^4+\theta^3+3\theta^2+3\theta-1.&& \end{align*} The proof runs as follows. A fifth-power descent over $\Q(\sqrt2)$ splits the problem into five cases according to the unit exponent $j$ (Step 1). The case $j=0$ is settled by a rank 0 elliptic curve over $\Q$ (Step 2). The case $j=2$ is eliminated by a computation modulo 8 in $L$, and $j=1$ lifts to an elliptic curve $E$ over $L$ (Steps 3--4). We show $\operatorname{rank}E(L)=2$ (Step 5) and finish with a sieve at 751 (Step 6) and elliptic Chabauty (Step 7). \paragraph{Step 1: fifth-power descent.} \begin{lem}\label{lem:1.2.4-descent} For each primitive solution ($z\ne0$) we have \[ y^2+zd=\varepsilon^j(a+bd)^5,\qquad j\in\{-2,-1,0,1,2\},\quad\gcd(a,b)=1,\quad a\ \text{odd},\qquad x=(-1)^{j+1}(a^2-2b^2) \] If $j=0,1,2$, then $y^2=F$, $y^2=F_1$, $y^2=F_2$, respectively. Taking conjugates changes $(j,a,b,z)$ into $(-j,(-1)^ja,(-1)^{j+1}b,-z)$, so it suffices to consider $j=0,1,2$. \end{lem} \begin{proof} The product of $y^2\pm zd$ is $-x^5$, and their norms are odd. A common prime ideal does not lie above 2, and if there were one, it would give a common prime factor of $y,z$. Hence the two elements are coprime, and by unique factorization $y^2+zd$ is a unit times a fifth power. The sign of the unit is absorbed into the fifth power, since 5 is odd. Taking norms gives $-x^5=(-1)^j(a^2-2b^2)^5$. Since $x$ is odd, $a$ is odd. The rational parts of $(1+d)(F+Gd)$ and $(3+2d)(F+Gd)$ are $F_1$ and $F_2$. Since $\bar\varepsilon=-\varepsilon^{-1}$, the rule for conjugation follows. \end{proof} \paragraph{Step 2: $j=0$.} \begin{lem}\label{lem:1.2.4-j0} If $j=0$, the only solutions are $(x,y,z)=(-1,\pm1,0)$. In particular, there is no solution with $z\ne0$. \end{lem} \begin{proof} We have $F=a(a^4+20a^2b^2+20b^4)=y^2$, and the second factor is positive, so $a>0$. The gcd of the two factors divides $20$, and $a$ is odd, so the gcd is 1 or 5. If $a=5u^2$, then $F\equiv5$ or $13\pmod{16}$, which cannot be an odd square. Hence $a=u^2$ ($u$ odd, $\gcd(u,5b)=1$) and $c^2=u^8+20u^4b^2+20b^4$. Suppose $b\ne0$. The positive integers $m,k=(u^4+10b^2\pm c)/2$ satisfy $mk=20b^4$ and are coprime. Indeed, a common prime factor divides $20b^4$ and $m+k=u^4+10b^2$; but it cannot be a prime factor of $b$, the sum is odd, and $5\nmid u$. Up to order, the two factors have the form $(r^4,20s^4)$ or $(5r^4,4s^4)$ ($|b|=rs$, $\gcd(r,s)=1$). In the second case, $u^4=5r^4-10r^2s^2+4s^4$ is impossible modulo 16. In the first case, $u^4=r^4-10r^2s^2+20s^4$, and we get a rational point \[ (X,Y)=\Bigl(\frac{r^2}{s^2},\frac{ru^2}{s^3}\Bigr),\quad X\ne0,\qquad Y^2=X^3-10X^2+20X \] But the only rational points of this curve are $O$ and $(0,0)$. In the 2-isogeny descent of \cite{Simon02,Silverman09}, the partner curve is $Y^2=X^3+20X^2+20X$, and the candidate classes for the two curves are $\pm1,\pm2,\pm5,\pm10$. The class 5 is given by $(0,0)$. On the first curve, the negative classes are excluded by the sign, and the class 2 is excluded because $W^2=2U^4-10U^2V^2+10V^4$ is impossible modulo 25 for pairs primitive at 5. On the partner curve, the representatives $2,-2$ of the other two cosets are excluded modulo 25, and $-1$ is excluded because $W^2=-U^4+20U^2V^2-20V^4$ is impossible modulo 16 for pairs primitive at 2. Both images are $\{1,5\}$, so $2^{\operatorname{rank}}=2\cdot2/4=1$. The reductions at 3 and 7 have orders 2 and 10 (\cite[Appendix]{Katz81}), so the torsion has order 2 (\code{1.2.4a}). Hence $b=0$, and $y^2=a^5$ together with primitivity gives $a=1$, $z=0$. \end{proof} \paragraph{Step 3: the quintic field and $j=1,2$.} \begin{lem}\label{lem:1.2.4-L} $\mathcal O_L=\Z[\theta]$, $\operatorname{disc}L=200000=2^65^5$, $\Cl(L)=0$, $\mathcal O_L^\times/\mathcal O_L^{\times2}=\langle-1,u_1,u_2\rangle$. \end{lem} \begin{proof} $g(T+2)=T^5+10T^4+45T^3+110T^2+145T+80$ is Eisenstein at 5. Modulo 2 we have $g=T(T^2+T+1)^2$, and $(g-T(T^2+T+1)^2)/2=-T^4+T^3-T^2+2T-1$ is coprime to $T^2+T+1$ modulo 2, so by Dedekind's criterion \cite[Section~6.1]{Cohen93} $\Z[\theta]$ is also maximal at 2. Since $\operatorname{disc}g=2^65^5$, we get $\mathcal O_L=\Z[\theta]$. Since $g'(T)=5(T^4+3T^2+1)>0$, the signature is $(1,2)$. The Minkowski constant $(4/\pi)^2(5!/5^5)\sqrt{200000}$ is less than 29. The prime ideals of norm less than 29 above primes less than 29 are generated by $\theta$ (2), $q$ (4), $3+\theta+2\theta^2$ (3), $\ell$ (5), $3+\theta^2$ (7), $-1+3\theta+2\theta^3$ (11), $1+\theta$ (13), $-1+\theta+4\theta^2+\theta^4$ (17), $-1+2\theta-\theta^2+\theta^3$ (23), and factoring $g$ modulo $p$ shows that this list is complete (\code{1.2.4b}). Hence $h(L)=1$. We have $Nu_1=Nu_2=1$, and at the places $(7,\theta-5)$, $(17,\theta-3)$ of residue degree 1 the rows of quadratic characters of $u_1,u_2$ are $(0,1)$, $(1,0)$. The unit rank is 2, and since $u_1,u_2$ are positive at the real embedding, the class of $-1$ is independent of them. \end{proof} The following norm identities hold (\code{1.2.4b}). \[ N(a-\eta b)=F_1(a,b),\qquad N(3a-\chi b)=81F_2(a,b). \] The minimal polynomials of $\eta$ and $\chi$ are $Z^5+10Z^4+20Z^3+40Z^2+20Z+8$ and $Z^5+20Z^4+180Z^3+720Z^2+1620Z+1296$, and their discriminants are $2^{22}5^5$ and $2^{22}3^{12}5^5$. \begin{lem}\label{lem:1.2.4-j12} There is no primitive solution with $j=2$. If $j=1$, then $a-\eta b=w^2$ with $w\in\mathcal O_L$. \end{lem} \begin{proof} First, $5\nmid y$. Indeed, $F_1\equiv(a-2b)^5$ and $F_2\equiv3(a-3b)^5\pmod5$; the values $F_1(2,1)=560$, $F_2(3,1)=4885$ at the roots have 5-adic valuation 1, and the partial derivatives with respect to $a$ are divisible by 5, so the valuation remains 1 when the root is lifted. This is incompatible with $y^2$. Put $\gamma_1=a-\eta b$ and $\gamma_2=3a-\chi b$. Outside the primes dividing the discriminant, a prime ideal dividing $\gamma_j$ has residue degree 1 and is the only one above its rational prime (its residue is $\eta=a/b$ or $\chi=3a/b$). By the norm identities its valuation is even. Since $y$ is odd, $\gamma_j$ is a unit above 2, and by the previous paragraph it is also a unit above 5. For $\gamma_1$ the argument above also applies above 3. For $\gamma_2$: modulo 3 we have $g\equiv(T-1)(T^4+T^3-1)$, the element $\delta$ ($N\delta=81$) generates the prime ideal $\mathfrak p_4$ of degree 4, and $\chi$ has $N\chi=-1296$, is a unit at the 3-adic place of degree 1, and has $v_{\mathfrak p_4}(\chi)=1$. If $3\mid b$, then $3\nmid a$ and $v_{\mathfrak p_4}(\gamma_2)=1$. If $3\nmid b$, then $F_2(a,b)$ is a 3-adic unit, so the norm gives $v_{\mathfrak p_4}(\gamma_2)=1$. In either case the valuation at the 3-adic place of degree 1 is even. Therefore, from the unit classes and the positivity of the norm ($N(-1)=-1$ in a field of odd degree), \[ \gamma_1=u_1^iu_2^kw^2,\qquad\gamma_2=\delta u_1^iu_2^kw^2,\qquad i,k\in\{0,1\},\ w\in\mathcal O_L \] The ring $R=(\Z/8\Z)[T]/(g)$ has 32768 elements and 336 distinct squares. For $a$ odd and $b\bmod8$, the residues of $\gamma_j$ take 32 values. Comparing them with the four sets $u_1^iu_2^kR^2$, $\delta u_1^iu_2^kR^2$, the sizes of the intersections in the order $(i,k)=(0,0),(1,0),(0,1),(1,1)$ are $2,0,0,0$ for $\gamma_1$ and $0,0,0,0$ for $\gamma_2$ (\code{1.2.4c}). The two residues of $\gamma_1$ have coefficients $(1,0,0,0,0)$ and $(1,0,4,4,4)$. \end{proof} \paragraph{Step 4: the elliptic cover.} Suppose $j=1$ and $b\ne0$. Putting $t=a/b$ and $v=y/(b^2w)$, we get \[ v^2=q(t)=t^4+At^3+Bt^2+Ct+D,\qquad q(t)=\frac{N(t-\eta)}{t-\eta} \] where $A=10+\eta$, $B=20+10\eta+\eta^2$, $C=40+20\eta+10\eta^2+\eta^3$, $D=20+40\eta+20\eta^2+10\eta^3+\eta^4$. Here $t\in\Q$ and $v\in L$. Put \[ U=2v+2t^2+At,\quad V=2\bigl(U+B-\tfrac{A^2}4\bigr)t+\tfrac A2U+C,\quad R_\theta=2\theta^3-4,\quad \xi=\frac{U-R_\theta}4,\quad \zeta=\frac V8. \] Then $(\xi,\zeta)$ is a point of \[ E:\ \zeta^2=\xi^3-5\theta\xi^2+5\theta^2\xi \] and for $\omega=d\xi/(2\zeta)$ we have $\phi^*\omega=dt/v$ ($dU/dt=V/v$). The two points at infinity with $v/t^2\to+1$, $-1$ map to $O$ and to \[ P=\Bigl(\frac{-5\theta^4-25\theta^2+10\theta}{8},\ \frac{35\theta^4-5\theta^3+195\theta^2-20}{32}\Bigr) \] respectively (\code{1.2.4d}). \paragraph{Step 5: the rank.} The partner curve is $E':\ \zeta^2=\xi^3+10\theta\xi^2+5\theta^2\xi$. We use the 2-isogeny descent of \cite{Simon02,Silverman09}. For $E_{a,b}:y^2=x^3+ax^2+bx$, the map $\alpha$ sends a point to the square class of $x$, and a class $\mu$ lies in the image if and only if $\mu W^2=\mu^2U^4+a\mu U^2V^2+bV^4$ has a suitable nonzero point. If both curves have only one rational 2-torsion point (which holds, since $\sqrt5\notin L$), then \[ 2^{\operatorname{rank}E(L)}=\frac{|\alpha(E(L))|\,|\alpha'(E'(L))|}{4}. \] \begin{lem}\label{lem:1.2.4-rank} $\operatorname{rank}E(L)=2$ and $E(L)_{\mathrm{tors}}=\{O,T_0\}$ ($T_0=(0,0)$). For \[ Q=(4\theta+4\theta^2+3\theta^3+2\theta^4,\ 12-24\theta-7\theta^2-21\theta^3-3\theta^4) \] the subgroup $\langle P,Q,T_0\rangle$ has odd index. \end{lem} \begin{proof} Outside the primes above 2 and 5, the $x$-coordinate has even valuation, so the candidates are the 64 classes generated by the ordered basis $(-1,u_1,u_2,\theta,q,\ell)$. We impose necessary conditions one after another. \begin{itemize} \item Real place: $x\ge0$ on $E$. \item 5: Put $\lambda=\theta-2$; then $v_\lambda(5)=5$ and $\lambda^5/5\equiv-1\pmod\lambda$. For both curves, $a,b$ have valuation 5, and the normalized leading unit of $b$ is a square. If we take $U,V$ integral with one of them a unit, and take the representative of $\mu$ of valuation 0 or 1, then a class of even valuation must have a square unit part, and a class of odd valuation must have the class of $b$. \item 2: The 2-adic completions of $L$ are one of degree 1 and one of degree 4 (the even root of $g$ is 178 modulo 256, and the quartic factor is $T^4+2T^3+9T^2+2T+9$ modulo 16). We scan the charts $V=1$ ($U$ arbitrary) and $U=1$ ($V$ a nonunit) completely, modulo 256 for the linear factor and modulo 16 for the quartic factor. The ring of the quartic factor has 65536 elements, 49152 units and 1648 distinct squares. \end{itemize} The numbers of remaining candidates, for $E$ and $E'$ respectively, are 16, 32 after the real place and 5; 16, 16 after the dyadic place of degree 1; and 4, 4 after the dyadic place of degree 4. In both cases the masks are $0,6,32,38$, that is, $1,u_1u_2,\ell,u_1u_2\ell$. On the other hand, the images of $T_0,Q$ have masks $32,38$, and the images of the point $T_0'=(0,0)$ of $E'$ and of \[ S=(50+5\theta+50\theta^2+10\theta^4,\ 380+100\theta+320\theta^2+20\theta^3+60\theta^4) \] are also $32,38$ ($5\theta^2/\ell$, $\xi(Q)/(u_1u_2\ell)$, $\xi(S)/(u_1u_2\ell)$ are squares in $L$). Hence both images have exactly four elements, and $\operatorname{rank}E(L)=2$ (\code{1.2.4e}). For the dual isogeny $\hat\phi(x',y')=\bigl((x'+10\theta+5\theta^2/x')/4,\ y'(5\theta^2-x'^2)/(8x'^2)\bigr)$ we have $\hat\phi(S)=P+T_0$ and $\hat\phi(T_0')=O$. Since $T_0,Q$ generate $E(L)/\hat\phi E'(L)$ and $T_0',S$ generate $E'(L)/\phi E(L)$, the points $T_0,Q,P+T_0$, that is, $P,Q,T_0$, generate $E(L)/2E(L)$, and the index is odd. At the places $(3,\theta-1)$, $(11,\theta-5)$ of residue degree 1 the reductions of the two curves have orders 6 and 16 (\cite[Appendix]{Katz81}), so the torsion subgroups are $\{O,T_0\}$ and $\{O,T_0'\}$. \end{proof} \paragraph{Step 6: a sieve at 751.} \begin{lem}\label{lem:1.2.4-sieve} For a point of $v^2=q(t)$ ($t\in\Q$, $v\in L$), $t$ is not a 751-adic integer. So $u=1/t\in751\Z_{751}$, and the signs of $\hat v_i=u^2v_i$ at the five embeddings are all equal. \end{lem} \begin{proof} The prime $p=751$ splits completely with $\theta\equiv65,168,348,386,535$. Each reduction is $\Z/384\times\Z/2$ (of order 768), and $T_0$ has coordinates $(192,0)$ in every factor. The subgroup $H$ of the product of the five factors generated by $P,Q,T_0$ satisfies \[ |H|=36864=2^{12}3^2,\qquad[\textstyle\prod_iE_i(\F_p):H]=2^{28}3^3 \] If $mP+nQ\in3E(L)$, then multiplying by 128 and looking at the 3-components in each factor, we find that $(m,n)$ modulo 3 lies in the kernel of the matrix \[ \begin{pmatrix}2&1&2&2&0\\2&1&0&1&1\end{pmatrix}^{\!\top} \] This matrix has rank 2, so $m\equiv n\equiv0\pmod3$. The torsion has no 3-part, so the index is prime to 3; being odd, it is prime to $2^{28}3^3$. Hence, by \cite{BS10}, the reduction of $E(L)$ is exactly $H$. For each $t\in\F_p$ we check whether the five conjugate quartics $q_i(t)$ are squares, and we apply the map for all choices of signs of the ordinates. At $t=\infty$ we choose $O$ or $P_i$ independently in each factor. There are 57 parameter residues (including $\infty$) that are squares in every factor, and 1674 product points are tested; the only ones in $H$ are $(O,O,O,O,O)$ and $(P_1,\dots,P_5)$ at $t=\infty$ (\code{1.2.4f}). If $t$ is a $p$-adic integer, the reduction of the point is one of the finite tuples tested, so it is excluded. Otherwise $\hat v_i\equiv\pm1$, and the sieve leaves only the two patterns with all signs equal. \end{proof} \paragraph{Step 7: 751-adic logarithms.} \begin{lem}\label{lem:1.2.4-chab} The curve $v^2=q(t)$ has no finite point with $t\in\Q$. \end{lem} \begin{proof} The five roots lift modulo $p^3=423564751$ to $30368252$, $51283705$, $229643381$, $365409950$, $170424214$. We pass to the short model $Y^2=X^3-\frac{10}3\theta^2X-\frac{25}{27}\theta^3$, and we multiply $P$ by $192,192,192,192,64$ and $Q$ by $384,384,128,384,384$ to put them into the formal group (all these numbers are prime to $p$). Since $\log\tau=\tau+O(\tau^5)$ (Lemma~\ref{lem:O}), we get, modulo $p^2$, \begin{align*} \tfrac1p\log P&\equiv(423722,191178,352807,294331,252200),\\ \tfrac1p\log Q&\equiv(330447,456587,117975,48293,396869) \end{align*} The first three coordinates are independent modulo $p$. Let $c$ be the cross product of the first three coordinates of the exact normalized logarithms, followed by two zeros. Then $c$ is an integral annihilator, and \[ c\equiv(626,131,473,0,0)\pmod p,\qquad \textstyle\sum_ic_i\equiv479\pmod p \] (\code{1.2.4g}). Since $P,Q$ span $E(L)\otimes\Q$, $c$ kills all logarithms of $E(L)$. On the remaining disks the common parameter is $u=1/t$, and $\hat v_i^2=1+A_iu+B_iu^2+C_iu^3+D_iu^4$. The sign at $u=0$ fixes an integral analytic square root. Since $\phi^*\omega=dt/v=-du/\hat v$, the combined differential is $-\sum_ic_i\,du/\hat v_i$, and on the two disks with uniform signs its constant coefficients are $-\sum c_i$ and $\sum c_i$, that is, 272 and 479 modulo 751. Both are units. The centers map to the global points $O$, $P$, so the annihilating logarithm vanishes there. By Proposition~\ref{thm:G} and Lemma~\ref{lem:1.2.1-disk}, the disks contain only the center $u=0$, and this is not a finite point. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.2.4}] By Lemmas~\ref{lem:1.2.4-j12} and \ref{lem:1.2.4-chab}, if $j=1$ then $b=0$. Then $y^2=a^5$, $z=a^5$, and primitivity gives $a=1$, so $(x,y,z)=(1,\pm1,1)$. By conjugation, $j=-1$ gives $(1,\pm1,-1)$. By Lemma~\ref{lem:1.2.4-j12} there is nothing for $j=\pm2$, and by Lemma~\ref{lem:1.2.4-j0}, $j=0$ gives only $(-1,\pm1,0)$. Substituting gives $-1+1=0$ and $1+1=2$. \end{proof} \subsection{\texorpdfstring{$x^5+y^4=3z^2$ \ ($H=60$)}{x^5+y^4=3z^2 (H=60)}}\label{sec:1.2.6} \begin{thm}\label{thm:1.2.6} The only primitive integer solutions ($\gcd(x,y,z)=1$) of $x^5+y^4=3z^2$ are \[ (x,y,z)=(-1,\pm1,0),\qquad (11,\pm29,\pm538) \] (independent signs). \end{thm} \paragraph{Setup.} The proof goes as follows. We perform a fifth-power descent in $\Z[\sqrt3]$ and eliminate the cases with unit exponent $j\ne0$ by a square-class obstruction in a quintic field (Steps 1--2). The remaining case leads to the plane quartic curve \[ \mathcal C:\quad u^4=r^4-15r^2s^2+45s^4 \] (Step 3). A fourth-power descent over $\Q(\sqrt5)$ turns the points with $s\ne0$ into points of the genus 2 curve \[ \mathcal H:\quad Y^2=f(X):=X(X^2+30X+45)(X^2-45) \] with $X\in\Q^{\times2}\cup-\Q^{\times2}$ (Step 4). We lift these points by pointwise 2-descent to three elliptic curves over $K=\Q(\sqrt5)$, and find them all using ranks, a coset sieve and elliptic Chabauty (Steps 5--8). Let $K=\Q(d)$, $d=\sqrt5$, $e=(1+d)/2$. Then $\mathcal O_K=\Z[e]$, $h_K=1$ (Minkowski bound $\sqrt5/2<2$), the unit group is $\{\pm e^n\}$, and $N(e)=-1$. The primes 2 and 3 are inert, and $d$ is a prime element above 5. Further, putting \[ \rho=-15+6d=-21+12e,\qquad \tau=3d=-3+6e, \] we have $X^2+30X+45=(X-\rho)(X-\bar\rho)$ and $X^2-45=(X-\tau)(X+\tau)$. \paragraph{Step 1: descent in $\Z[\sqrt3]$.} \begin{lem}\label{lem:1.2.6-basic} For a primitive solution, $x,y,z$ are pairwise coprime, $3\nmid xy$, and $x\equiv2\pmod3$. The parities of $y$ and $z$ differ, and $x$ is odd. Also $x,y\ne0$, and if $z=0$ then $(x,y)=(-1,\pm1)$. \end{lem} \begin{proof} If 3 divides $x$ and $y$, then $3^4\mid x^5+y^4=3z^2$, so $3\mid z$. If another prime divides two of the variables, then by the equation it also divides the third. Hence they are pairwise coprime. Modulo 3, $3\mid x$ implies $3\mid y$ and conversely, so $3\nmid xy$, and $x^5\equiv-y^4\equiv-1$ gives $x\equiv2\pmod3$. If $y,z$ are both odd, then $v_2(y^4-3z^2)=1$, which is impossible since $y^4-3z^2=-x^5$. Therefore the parities differ and $x$ is odd. If $x=0$, then $y^4=3z^2$ gives $y=z=0$. If $y=0$, then $x^5=3z^2$ gives $3\mid x$. If $z=0$, then $x^5=-y^4$ gives $x=-1$, $y=\pm1$. \end{proof} \begin{lem}\label{lem:1.2.6-sqrt3} For every primitive solution there exist $j\in\{-2,-1,0,1,2\}$ and integers $a,b$ such that \[ y^2+z\sqrt3=(2+\sqrt3)^j(a+b\sqrt3)^5,\qquad \gcd(a,b)=1,\quad a+b\ \text{odd},\quad 3\nmid a,\quad x=3b^2-a^2. \] \end{lem} \begin{proof} $\Z[\sqrt3]$ is norm-Euclidean: rounding the two coefficients gives a remainder $A+B\sqrt3$ with $|A|,|B|\le\frac12$ and $|A^2-3B^2|\le\frac34$. The unit group is $\{\pm(2+\sqrt3)^n\}$. There is no unit of norm $-1$, by reduction modulo 3, and a unit of norm 1 between $1$ and $2+\sqrt3$ would have to have $\sqrt3$-coefficient between 0 and 1. The product of $y^2\pm z\sqrt3$ is $y^4-3z^2=-x^5$. The norms of the two factors are units at 2 and 3, and any other common prime factor would divide both $y$ and $z$. Hence the two factors are coprime, and $y^2+z\sqrt3=\pm(2+\sqrt3)^m\gamma^5$. Absorbing $-1$ and $(2+\sqrt3)^5$ into $\gamma^5$, we get $j\in\{-2,\dots,2\}$. A common prime factor of the coefficients of $\gamma=a+b\sqrt3$ divides $y$ and $z$, so $\gcd(a,b)=1$. Since $N(2+\sqrt3)=1$, comparing norms gives $a^2-3b^2=-x$. Since $x$ is odd, $a+b$ is odd, and since $3\nmid x$, $3\nmid a$. \end{proof} \paragraph{Step 2: elimination of $j\ne0$.} Writing $(a+b\sqrt3)^5=F+G\sqrt3$, we have \[ F=a(a^4+30a^2b^2+45b^4),\qquad G=b(5a^4+30a^2b^2+9b^4). \] Multiplying by $(2+\sqrt3)^1$ and $(2+\sqrt3)^2=7+4\sqrt3$, we get, for $j=1,2$, \[ y^2=F_1(a,b):=2F+3G,\qquad y^2=F_2(a,b):=7F+12G. \] Taking conjugates changes $(j,b,z)$ into $(-j,-b,-z)$, so it suffices to eliminate $j=1,2$. We use the quintic field $L=\Q(\vartheta)$, $g(\vartheta)=0$, $g(T)=T^5-5T^3+5T-4$. The following facts are checked by \code{1.2.6a}. \begin{itemize} \item $\operatorname{disc}(g)=2^4\cdot3^2\cdot5^5$ and $\mathcal O_L=\Z[\vartheta]$ ($g(T-1)$ is Eisenstein at 5, and at 2 and 3 we use Dedekind's criterion). The signature is $(1,2)$. \item $h_L=1$. The Minkowski bound is less than 43, and every prime ideal of norm less than 43 is generated by an element of the same norm. \item $e_1=1-\vartheta+\vartheta^3$ and $e_2=1+2\vartheta^2+3\vartheta^3+\vartheta^4$ are units of norm 1, and $-1,e_1,e_2$ form a basis of $\mathcal O_L^\times/\mathcal O_L^{\times2}$. \item $\varpi=\vartheta-2$ and $\kappa=\vartheta^2+\vartheta-1$ have norms 2 and 4, and $(2)=(\varpi)(\kappa)^2$. There is one prime ideal above 5, and its residue degree is 1. \item For $\mu_1=\vartheta^4+2\vartheta^3-5\vartheta^2-8\vartheta+1$ and $\mu_2=-\vartheta^4-\vartheta^3+3\vartheta^2+2\vartheta$, \[ N(2a-\mu_1 b)=16F_1(a,b),\qquad N(2a-\mu_2(a+b))=8F_2(a,b). \] The discriminants of the minimal polynomials of $\mu_1$ and $\mu_2$ are $2^{24}3^{10}5^5$ and $2^{16}3^{10}5^5$. \end{itemize} \begin{lem}\label{lem:1.2.6-j} In Lemma~\ref{lem:1.2.6-sqrt3}, $j=0$. \end{lem} \begin{proof} Suppose $j\in\{1,2\}$, and put $\gamma=2a-\mu_1 b$ ($j=1$) or $\gamma=2a-\mu_2(a+b)$ ($j=2$). Then $N(\gamma)=16y^2$ or $8y^2$, and $y\ne0$. Suppose a prime ideal $\mathfrak P$ above $p\notin\{2,3,5\}$ divides $\gamma$. The coefficient of $\mu_1$ or $\mu_2$ ($b$ or $a+b$) is not a multiple of $p$; otherwise $p\mid2a$, and $p$ divides both $a$ and $b$. Thus the residue of $\mu_1$ (or $\mu_2$) is determined as an element of $\F_p$. Since $p$ does not divide the discriminants of $\Z[\mu_1]$ and $\Z[\mu_2]$, only one prime ideal above $p$ divides $\gamma$, and its residue degree is 1. So $v_{\mathfrak P}(\gamma)=v_p(N\gamma)$ is even. At 3 we have $F_1\equiv2a^5$ and $F_2\equiv a^5\pmod3$, so $N\gamma$ is a unit. There is one prime ideal above 5, of residue degree 1, so its valuation is also $v_5(N\gamma)$, which is even. At 2, $v_2(N\gamma)=v_\varpi(\gamma)+2v_{\kappa}(\gamma)$, so $v_\varpi(\gamma)\equiv v_2(N\gamma)\equiv j-1\pmod2$. Since $h_L=1$, \[ \gamma=\varpi^{\,j-1}e_1^ie_2^k\kappa^{\ell}w^2,\qquad i,k,\ell\in\{0,1\},\ w\in\mathcal O_L. \] The factor $-1$ does not appear, because $N(-1)=-1$ and the norms of all the other factors are positive. And $w$ is integral, because all the remaining valuations are even and nonnegative. Now we compare in $\mathcal O_L/8\mathcal O_L=R=(\Z/8)[T]/(g)$. $R$ has $8^5=32768$ elements, and exactly 336 of them are squares. For $a,b\bmod8$ with $a+b$ odd, the left side $\gamma\bmod8$ takes 16 values for each of $j=1,2$. The right side is, for each of the eight $(i,k,\ell)$, the set of elements $\varpi^{j-1}e_1^ie_2^k\kappa^\ell\cdot(\text{square})$. The intersection of the two sets is empty for $j=1,2$ and all eight $(i,k,\ell)$ (\code{1.2.6a}). This is a contradiction. \end{proof} \paragraph{Step 3: the quartic curve $\mathcal C$.} \begin{lem}\label{lem:1.2.6-quartic} The primitive solutions are obtained from integral points $(u,r,s)$ of $\mathcal C$ with $u,r>0$, $s\ge0$, $\gcd(r,s)=1$, by \[ x=3r^2s^2-u^4,\qquad y=\pm u(r^4-45s^4),\qquad z=\pm rs(5u^8+30u^4r^2s^2+9r^4s^4). \] The solutions with $z=0$ come from $(u,r,s)=(1,1,0)$. \end{lem} \begin{proof} By Lemma~\ref{lem:1.2.6-j}, $y^2=aP$ with $P=a^4+30a^2b^2+45b^4>0$, and $z=b(5a^4+30a^2b^2+9b^4)$. A common prime factor of $a$ and $P$ divides $45b^4$, and since $3\nmid a$, it can only be 5. Since $y\ne0$, $a>0$. If $5\nmid a$, then $a=u^2$. If $5\mid a$, then $v_5(P)=1$, so $a=5u^2$; but then $y^2=aP\equiv a^5\equiv a\equiv2u^2\equiv2\pmod3$, a contradiction. Hence $a=u^2$, $c^2:=P(u^2,b)=u^8+30u^4b^2+45b^4$, and $\gcd(u,b)=\gcd(u,15)=1$. If $u$ is even, then $b$ is odd and $c^2\equiv13\pmod{16}$, which is impossible (\code{1.2.6b}). Therefore $u$ is odd, $b$ is even and $c$ is odd. If $b=0$, then $u=1$. If $b\ne0$, then $m=(u^4+15b^2+c)/2$ and $k=(u^4+15b^2-c)/2$ are positive integers with $mk=45b^4$. A common prime factor would divide $45b^4$ and $u^4+15b^2$, which contradicts $\gcd(u,b)=\gcd(u,15)=1$; and the sum is odd. Distributing the fourth powers, we get, up to order, $(m,k)=(r^4,45s^4)$ or $(5r^4,9s^4)$, with $|b|=rs$ and $\gcd(r,s)=1$. In the latter case the sum is 2 modulo 3, which is impossible since $m+k\equiv u^4\equiv1$. In the former case, $r^4+45s^4=u^4+15r^2s^2$ is $\mathcal C$, $c=|r^4-45s^4|$, and $y=\pm uc$. The remaining formulas follow by substituting $a=u^2$, $b=\pm rs$. \end{proof} \begin{lem}\label{lem:1.2.6-props} Every primitive integral point ($\gcd(u,r,s)=1$) of $\mathcal C$ satisfies $\gcd(r,s)=1$, $\gcd(u,30)=1$, $\gcd(r,15)=1$, $\gcd(u,rs)=1$. Moreover, there is no point with $u=0$ or $r=0$. \end{lem} \begin{proof} A common prime factor of $r,s$ also divides $u$. Modulo 16, the right-hand side is $7$ or $15$ if $r,s$ are both odd, $13$ if $4\mid r$ and $s$ is odd, and $5$ if $r$ is odd and $s\equiv2\pmod4$. In the remaining cases ($r\equiv2\pmod4$ and $s$ odd, or $r$ odd and $4\mid s$) it is 1 (\code{1.2.6b}). Hence $u$ is odd. If $3\mid u$, then $3\mid r$, $3\nmid s$, and the 3-adic valuation of the right-hand side is exactly 2. If $5\mid u$, then in the same way the 5-adic valuation is exactly 1. If $3\mid r$ or $5\mid r$, then $u$ is also divisible by that prime. A common prime factor of $u$ and $r$ divides $45s^4$, so it is 3 or 5, or it divides $s$, a contradiction. A common prime factor of $u$ and $s$ divides $r$. If $u=0$, then $r^2/s^2$ is a root of a quadratic polynomial of discriminant 45, and if $r=0$, then $u^4=45s^4$; both are impossible. \end{proof} \paragraph{Step 4: descent over $\Q(\sqrt5)$.} \begin{lem}\label{lem:1.2.6-sqrt5} If a primitive integral point of $\mathcal C$ has $s\ne0$, then there exist $\sigma\in\{\pm1\}$ and $t\in\Q^\times$ such that $(X,Y)=(9\sigma t^2,\,Y)$ is a point of $\mathcal H(\Q)$. In particular, $X\ne0$ and $X\in\sigma\Q^{\times2}$. Moreover, if $X=-9$, then $(u,r,s)=(\pm1,\pm2,\pm1)$. \end{lem} \begin{proof} Put $\alpha=r^2+\frac{-15+3d}{2}s^2$. Then $N(\alpha)=u^4>0$, so the two real embeddings of $\alpha$ have the same sign. Call this sign $\sigma$. A common prime factor of $(\alpha)$ and $(\bar\alpha)$ divides $3ds^2$. Those above 3 and 5 cannot divide $N\alpha=u^4$ because $\gcd(u,15)=1$, and one above a prime factor of $s$ would also divide $r$. Hence all valuations of $(\alpha)$ are multiples of 4, and $\alpha=\sigma e^j\beta^4$ with $j\in\{0,2\}$. In the residue field $\F_9$ above 3 we have $\alpha\equiv r^2\equiv1$, the fourth powers are $\pm1$, and $e^2\notin\F_3$; hence $j=0$. Writing $\beta^2=A+Bd$, we get \[ \sigma(A^2+5B^2)=r^2-\tfrac{15}{2}s^2,\qquad 4AB=3\sigma s^2,\qquad A^2-5B^2=u^2. \] Then $B\ne0$, and putting $t=s/(2B)$ we get $A=3\sigma Bt^2$ and \[ (r/B)^2=\sigma(9t^4+5)+30t^2,\qquad (u/B)^2=9t^4-5. \] Then $X=9\sigma t^2$ and $Y=27tru/B^2$ satisfy $Y^2=f(X)$ (\code{1.2.6b}). If $X=-9$, then $\sigma=-1$, $t^2=1$, and $|r/B|=4$, $|u/B|=2$, $|s/B|=2$, so $(u,r,s)=(\pm1,\pm2,\pm1)$. \end{proof} \begin{prop}\label{prop:1.2.6} The only rational points of $\mathcal H$ with $X\in\Q^{\times2}\cup-\Q^{\times2}$ are $(X,Y)=(-9,\pm216)$. \end{prop} \paragraph{Step 5: pointwise descent and three elliptic curves.} \begin{lem}\label{lem:1.2.6-pointwise} Let $(X,Y)\in\mathcal H(\Q)$ with $X=\sigma q^2\ne0$. Then the class of $(X-\rho,\,X-\tau)$ in $K^\times/K^{\times2}$ is $(1,1)$ if $\sigma=1$, and $(-3e,-6)$ or $(3ed,-3d)$ if $\sigma=-1$. \end{lem} \begin{proof} We have $f(X)=X\cdot N(X-\rho)\cdot N(X-\tau)$. Since $\operatorname{disc}(f)=2^{14}3^{20}5^{10}$, the five roots are distinct at prime ideals coprime to 30. At such a prime ideal, if $X$ is integral, then at most one of the five linear factors of $f$ has positive valuation, and that valuation is even. If $X$ is not integral, then all factors have valuation $v(X)=2v(q)$, which is even. Therefore both classes are represented by subproducts of $(-1,e,2,3,d)$. Imposing the norm condition $\sigma N(X-\rho)N(X-\tau)\in\Q^{\times2}$ leaves 256 pairs for each sign. Now we compute the local classes of the two numbers for all $q\in\Q_2$ and $q\in\Q_3$. At 2, we test $q\in\Z_2$ modulo 16, and if $q\notin\Z_2$ we test $v=1/q\in2\Z_2$ modulo 16 (in the latter case we use the square-equivalent pair $(\sigma-\rho v^2,\sigma-\tau v^2)$). On each ball the valuations of the two factors are at most 2, and changing the representative by 16 changes the factors by multiples of 32. So the ratio of the normalized units is 1 modulo 8, and it is a square in the unramified quadratic extension of $\Q_2$. At 3, we test the balls $q\bmod3$ and $v\in3\Z_3$; a change by 3 does not change the normalized residues. As a result, after the condition at 2, 4 pairs remain if $\sigma=1$ and 8 pairs if $\sigma=-1$; after the condition at 3, 1 pair and 2 pairs remain, respectively. The remaining pairs are those above (\code{1.2.6c}). \end{proof} We define three elliptic curves \[ E_0:\ \eta^2=\xi(\xi-\rho)(\xi+\tau),\quad E_1:\ \eta^2=\xi(\xi+2e\rho)(\xi-2e\tau),\quad E_2:\ \eta^2=\xi(\xi-e\rho)(\xi-e\tau). \] The three cases of Lemma~\ref{lem:1.2.6-pointwise} give, respectively, the point $\xi=X$ of $E_0$, the point $\xi=-2eX$ of $E_1$, and the point $\xi=eX$ of $E_2$. For example, in the middle case, if $X-\rho=-3ea_1^2$ and $X-\tau=-6a_2^2$, then taking conjugates gives $X+\tau=-6\bar a_2^2$, and $X(X-\rho)(X+\tau)=-2e(3qa_1\bar a_2)^2$. In general, if $X(X-r_1')(X-r_2')=\lambda V^2$, then $(\xi,\eta)=(\lambda X,\lambda^2V)$ is a point on $\eta^2=\xi(\xi-\lambda r_1')(\xi-\lambda r_2')$; here $\lambda=-2e$, and it is a point of $E_1$. In the third case, $X(X-\rho)(X-\tau)=e(3dqa_1a_2)^2$. Under the 2-descent map $\delta(\xi,\eta)=(\xi,\xi-r_1)$ (roots $0,r_1,r_2$), the classes of these points are $(2e,6)$ on $E_1$ and $(-e,3d)$ on $E_2$. \paragraph{Step 6: rank.} \begin{lem}\label{lem:1.2.6-rank} The ranks of $E_0,E_1,E_2$ over $K$ are $0,1,1$, and their torsion subgroups are $E_i[2]$. For $P_1=(18e,72)\in E_1(K)$ and $P_2=(1-2e,5-5e)\in E_2(K)$, the subgroup $\langle P_i,E_i[2]\rangle$ has odd index. The points obtained in Step 5 lie in $P_1+2E_1(K)$ on $E_1$, and in $T_0+2E_2(K)$ on $E_2$ ($T_0=(0,0)$). \end{lem} \begin{proof} We use the complete 2-descent of \cite{Simon02,Silverman09}. The differences of the roots are supported only above 2, 3, 5, so the candidates are the $32^2=1024$ pairs of classes of $(-1,e,2,3,d)$. The local images have dimensions $4,2,2$ at 2, 3, 5. At 3 and 5, the classes of the two torsion points span the local image. At 2, we add 2-adic points with the following $\xi$: $-1-e$, $-4+4e$ on $E_0$; $-2-e$, $1+e$ on $E_1$; $e$, $1-2e$ on $E_2$. At each of these points the valuation of the right-hand side is even and the normalized unit is a square modulo 8, so by Hensel's lemma they are genuine local points. Imposing the conditions at 2, 3, 5 and at the two real places in turn, the numbers of remaining candidates are $64,16,4,4,4$ for $E_0$, and $64,16,8,8,8$ for $E_1$ and $E_2$. The remaining candidates are generated by the classes of the torsion points and of $P_1$, $P_2$ (\code{1.2.6d}). Since $\dim E_i(K)[2]=2$, the ranks are $0,1,1$. Also $\langle P_i,E_i[2]\rangle+2E_i(K)=E_i(K)$, so multiplication by 2 is surjective on the finitely generated abelian group $E_i(K)/\langle P_i,E_i[2]\rangle$, which is therefore finite of odd order. Torsion: at the two prime ideals above 11, the orders of the reduced groups are $(8,12)$ for $E_0,E_1$ and $(12,16)$ for $E_2$; above 19 they are $(16,12)$, $(16,28)$, $(28,24)$. By \cite[Appendix]{Katz81}, the order of the prime-to-11 part of the torsion divides $\gcd(8,12)=\gcd(12,16)=4$, and since the orders above 19 are not multiples of 11, the 11-part is trivial. Hence the torsion is $E_i[2]$ (\code{1.2.6d}). On $E_1$ the root is $r_1=-2e\rho=18e-24$, so $\delta(P_1)=(18e,24)=(2e\cdot3^2,\,6\cdot2^2)\equiv(2e,6)$. On $E_2$ we have $r_1=e\rho$, $r_2=e\tau$, and $\delta(T_0)=(r_1r_2,-r_1)=\big((-e)(3d/e)^2,\,3d(e-1)^2\big)\equiv(-e,3d)$. The kernel of $\delta$ is $2E_i(K)$, so the last claim follows. \end{proof} The case $\sigma=1$ does not occur, because $E_0(K)=E_0[2]$ and none of the $\xi$-coordinates $0,\rho,-\tau$ of the torsion points lies in $\Q^\times$. \paragraph{Step 7: coset sieve.} \begin{lem}\label{lem:1.2.6-sieve} The points of $E_1$ from Step 5 have $X\equiv-9\pmod{89}$ and map to $\pm P_1$ in the two reductions above 89. The points of $E_2$ have $X\equiv0\pmod{11}$ and map to $T_0$ in the two reductions above 11. \end{lem} \begin{proof} The roots of $e^2-e-1$ are $10,80$ modulo 89 and $4,8$ modulo 11. Let $H_i$ be the subgroup of the product of the two reductions generated by the images of $P_i$ and $E_i[2]$. For $E_1$, the two reduced groups above 89 have orders $80,96$, and $|H_1|=480$ (index 16); for $E_2$, the orders above 11 are $12,16$, and $|H_2|=96$ (index 2). Since these indices are powers of 2 and the global index is odd, the image of $E_i(K)$ under reduction is $H_i$. Therefore the reductions of the cosets of Lemma~\ref{lem:1.2.6-rank} are $P_1+2H_1$ (60 elements) and $T_0+2H_2$ (12 elements). The $\xi$-coordinate of the point is $\lambda_iX$ ($\lambda_1=-2e$, $\lambda_2=e$, $X\in\Q$). Testing all residues $X$ and all signs of $\eta$ at the two embeddings, the only ones that land in the coset are $X\equiv80\equiv-9$ with the points $((2,72),(16,72))$ or $((2,17),(16,17))$, that is, $\pm P_1$, on $E_1$, and $X\equiv0$ with the point $(T_0,T_0)$ on $E_2$ (\code{1.2.6e}). Since $\lambda_i$ is a unit, if $X$ is a pole then the point maps to $(O,O)$, which lies in neither coset. Therefore $X$ is integral at the prime in question. \end{proof} \paragraph{Step 8: elliptic Chabauty.} \begin{lem}\label{lem:1.2.6-chab} The only point from Step 5 on $E_1$ is the one with $X=-9$, and there is none on $E_2$. \end{lem} \begin{proof} We apply Proposition~\ref{thm:G} with $\omega=d\xi/(2\eta)$ and $p=89$ ($E_1$), $p=11$ ($E_2$). The two $p$-adic embeddings give a logarithm $\log:E_i(K)\to\Q_p^2$, and since the rank is 1, its image is generated by $\log P_i$. Put $\ell_i=\log(P_i)/p$ and $c_i=((\ell_i)_2,-(\ell_i)_1)$; then $c_i$ kills the logarithms of all global points. Multiplying $P_i$ by its finite orders $n$ ($20,24$ at 89, $6,8$ at 11) to bring it into the formal group, and computing $t=-x_{\rm short}/\eta$ modulo $p^3$ ($\log=t+O(t^3)$), we get \[ \begin{gathered} \ell_1\equiv(2490,2948)\pmod{89^2},\quad c_1\equiv(11,2)\pmod{89},\\ \ell_2\equiv(117,76)\pmod{11^2},\quad c_2\equiv(10,4)\pmod{11} \end{gathered} \] (\code{1.2.6f}). The $c_i$ are primitive. $E_1$: by Step 7, the centers of the disks are $\pm P_1$, and $T=X+9\in89\Z_{89}$. The pullback of $\omega$ is $\lambda_1\,dX/(\pm144)$, and its coefficients for the positive sign at the two embeddings are $(32,78)$ modulo 89. The dot product with $c_1$ is $63$ or $26$, a unit. Since $\eta\equiv\pm72$ is a unit, the map and the differential have integral coefficients on the disk. $E_2$: here $X=-q^2$ with $q\in11\Z_{11}$, and the point maps to $T_0$ in both reductions. Resolving this double contact by $\xi=-eq^2$, $\eta=qW$, we get \[ W^2=-e^3(q^2+\rho)(q^2+\tau)=45-(12+6e)q^2-(1+2e)q^4. \] The square roots of 45 modulo 11 are $\pm1$, and we allow all four combinations of signs at the two embeddings. The pullback is $\omega=-(e/W)\,dq$, and for the sign combinations $(1,1),(1,-1),(-1,1),(-1,-1)$ of $W$ its dot products with $c_2$ are $5,3,8,6$, all units (\code{1.2.6f}). Since $2W$ is a unit, each branch is an analytic function with integral coefficients, and the center is $T_0$, whose logarithm is 0. On every disk the annihilating logarithm has the form $L(T)=vT+\sum_{n\ge2}a_{n-1}T^n/n$ ($v\in\Z_p^\times$, $a_i\in\Z_p$, $T\in p\Z_p$), and it vanishes at the center and at all global points. If $T\ne0$, then $v_p(a_{n-1}T^n/n)\ge nv_p(T)-v_p(n)>v_p(T)$, so $L(T)\ne0$. Hence on $E_1$ only $X=-9$ remains, and on $E_2$ only $q=0$ remains, which contradicts $X\ne0$. \end{proof} \begin{proof}[Proof of Proposition~\ref{prop:1.2.6}] If there is a rational point with $X=\sigma q^2\ne0$, then by Lemma~\ref{lem:1.2.6-pointwise} it gives a point on one of $E_0,E_1,E_2$. The case $\sigma=1$ ($E_0$) was eliminated at the end of Step 6, and $E_2$ is eliminated by Lemma~\ref{lem:1.2.6-chab}. What remains is $X=-9$ on $E_1$, and $f(-9)=46656=216^2$. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.2.6}] If $z=0$, then the solution is $(-1,\pm1,0)$ by Lemma~\ref{lem:1.2.6-basic}. If $z\ne0$, then the point $(u,r,s)$ of Lemma~\ref{lem:1.2.6-quartic} has $s\ne0$, and it is primitive since $\gcd(r,s)=1$. By Lemma~\ref{lem:1.2.6-sqrt5} and Proposition~\ref{prop:1.2.6}, $X=-9$ and $(u,r,s)=(1,2,1)$. Substituting gives $x=11$, $y=\pm(16-45)=\mp29$, $z=\pm2(5+120+144)=\pm538$. The equation is invariant under $y\mapsto-y$ and $z\mapsto-z$, so all four sign combinations are solutions, and all of them are primitive. \end{proof} \subsection{\texorpdfstring{$x^5+y^4+3z^2=0$ \ ($H=60$)}{x^5+y^4+3z^2=0 (H=60)}}\label{sec:1.2.7} \begin{thm}\label{thm:1.2.7} The only primitive integer solutions ($\gcd(x,y,z)=1$) of $x^5+y^4+3z^2=0$ are $(x,y,z)=(-1,1,0),\ (-1,-1,0)$. \end{thm} \paragraph{Setup.} Put $n=-x$. Then the equation becomes \[ y^4+3z^2=n^5. \] The proof has three steps. A fifth-power descent in the Eisenstein integers leads to the plane quartic curve $u^4=r^4+15r^2s^2+45s^4$ (Step 1). A fourth-power descent over $\Q(\sqrt5)$ leads to rational points with $X\ne0$ on the genus 2 curve \[ \mathcal H:\quad Y^2=f(X):=X(X^2-30X+45)(X^2-45) \] (Step 2). Finally, we show $J(\Q)\cong(\Z/2)^2$ for the Jacobian $J$ of $\mathcal H$, which gives $\mathcal H(\Q)=\{\infty,(0,0)\}$ (Step 3). \paragraph{Step 1: descent in the Eisenstein integers.} \begin{lem}\label{lem:1.2.7-basic} For a primitive solution, $n>0$ and $n,y,z$ are pairwise coprime. Moreover, $3\nmid ny$, the parities of $y$ and $z$ differ, and $n$ is odd. Also $y\ne0$, and if $z=0$ then $n=1$, $y=\pm1$. \end{lem} \begin{proof} If $n\le0$, then $y=z=0$ and $n=0$, contradicting primitivity. If a prime $p$ divides $n$ and $y$, then $p\mid3z^2$. If $p\ne3$, then $p\mid z$. If $p=3$, then $3^4\mid n^5-y^4=3z^2$, so again $3\mid z$. If $p\mid n,z$, then $p\mid y^4$, and if $p\mid y,z$, then $p\mid n^5$. Hence they are pairwise coprime. Modulo 3, $3\mid n$ implies $3\mid y$, and $3\mid y$ implies $3\mid n$, so $3\nmid ny$. If $y,z$ are both odd, then $y^4+3z^2\equiv4\pmod8$, so $v_2(n^5)=2$, which is not a multiple of 5. Therefore the parities differ and $n$ is odd. If $y=0$, then $3z^2=n^5$ gives $3\mid n$, a contradiction. If $z=0$, then $y^4=n^5$ and $\gcd(n,y)=1$, so $n=1$, $y=\pm1$. \end{proof} Let $d_0=\sqrt{-3}$, $\omega=(-1+d_0)/2$, and put \[ F(a,b)=a^4-30a^2b^2+45b^4. \] \begin{lem}\label{lem:1.2.7-eis} For every primitive solution there exist coprime integers $a,b$ of different parity ($3\nmid a$) such that $y^2+zd_0=(a+bd_0)^5$. Hence \[ n=a^2+3b^2,\qquad y^2=a\,F(a,b),\qquad z=b(5a^4-30a^2b^2+9b^4). \] \end{lem} \begin{proof} $\Z[\omega]$ is norm-Euclidean (rounding the two coefficients gives a remainder of norm at most $3/4$). The product of $y^2\pm zd_0$ is $n^5$. A common prime factor of the two factors divides $2y^2$ and $2zd_0$; since the norm $n^5$ is coprime to 6, it would divide both $y$ and $z$, a contradiction. Hence the two factors are coprime, and since all six units are fifth powers of units, $y^2+zd_0=\gamma^5$ ($\gamma\in\Z[\omega]$). In $\Z[\omega]/2\Z[\omega]\cong\F_4$ we have $y^2+zd_0\equiv y^2+z\equiv1$, and $\F_4^\times$ has order 3, so $\gamma^5=\gamma^2=1$ gives $\gamma\equiv1\pmod2$. That is, the $\omega$-coefficient of $\gamma$ is even, $\gamma=a+bd_0$ ($a,b\in\Z$), and $a+b$ is odd. A common prime factor of $a,b$ divides $y^2$ and $z$, so $\gcd(a,b)=1$. Comparing norms gives $n=a^2+3b^2$, and since $3\nmid n$, $3\nmid a$. The last two formulas come from expanding $(a+bd_0)^5$. \end{proof} \begin{lem}\label{lem:1.2.7-square} In Lemma~\ref{lem:1.2.7-eis}, $a=u^2$ and $F(u^2,b)=c^2$, with $\gcd(u,b)=\gcd(u,15)=1$. Moreover, $u,c$ are odd and $b$ is even. \end{lem} \begin{proof} Since $y\ne0$, $F\ne0$. A common prime factor of $a$ and $F$ divides $45b^4$, and since $\gcd(a,b)=1$ and $3\nmid a$, it can only be 5. If $5\nmid a$, then $\gcd(a,F)=1$ and $aF=y^2$, so $a=\pm u^2$, $F=\pm c^2$ (same sign). Since $F\equiv a^4\equiv1\pmod3$, $F=-c^2$ is impossible, and $a=u^2$. We rule out the case $5\mid a$. In this case $v_5(F)=v_5(45b^4)=1$, so $a=\pm5u^2$, $F=\pm5c^2$ (same sign). Since $F\equiv1\pmod3$ and $5c^2\equiv2c^2\not\equiv1\pmod3$, the sign is negative: \[ a=-5u^2,\qquad F=-5c^2 . \] We compare these two equations modulo 16. If $a$ is odd, then $b$ is even, $45b^4\equiv0$, $a^4\equiv1$, and $30a^2b^2\equiv0$ or $8\pmod{16}$, so $F\equiv1$ or $9$. If $a$ is even, then $u$ is even by $a=-5u^2$, so $4\mid a$; and $b$ is odd, so $F\equiv45b^4\equiv13\pmod{16}$. Hence $F\bmod16\in\{1,9,13\}$. On the other hand, $c^2\bmod16\in\{0,1,4,9\}$, so $-5c^2\bmod16\in\{0,3,11,12\}$. The two sets are disjoint, so this case does not occur (\code{1.2.7a}). Therefore $a=u^2$, $c^2=u^8-30u^4b^2+45b^4$, $\gcd(u,b)=1$ and $\gcd(u,15)=1$. If $u$ is even, then $b$ is odd and $c^2\equiv45\equiv13\pmod{16}$, a contradiction. Hence $u$ is odd, $b$ is even and $c$ is odd. \end{proof} \begin{lem}\label{lem:1.2.7-quartic} If there is a primitive solution with $z\ne0$, then there are integers $u,r,s$ with $\gcd(r,s)=1$ and $urs\ne0$ satisfying \[ \mathcal C:\quad u^4=r^4+15r^2s^2+45s^4 . \] \end{lem} \begin{proof} If $z\ne0$, then $b\ne0$. For $u,b,c$ as in Lemma~\ref{lem:1.2.7-square}, \[ m=\frac{u^4-15b^2+c}{2},\qquad k=\frac{u^4-15b^2-c}{2} \] are integers with $mk=45b^4$ and $m+k=u^4-15b^2$. A common prime factor of $m,k$ divides $45b^4$ and $u^4-15b^2$; if it is a prime factor of $b$ it also divides $u$, and if it is 3 or 5 it divides $u$, so this is impossible. Hence $m,k$ are coprime. Distributing the fourth powers, up to swapping $m$ and $k$, $\{m,k\}$ is $\{\pm r^4,\pm45s^4\}$ or $\{\pm9r^4,\pm5s^4\}$ (same sign, $|b|=rs$, $\gcd(r,s)=1$). Since $m+k\equiv u^4\equiv1\pmod3$, only $(r^4,45s^4)$ and $(-9r^4,-5s^4)$ remain. In the latter case, after renaming $r,s$, we get $u^4=-5r^4+15r^2s^2-9s^4$, and since exactly one of $r,s$ is even, the right-hand side is one of $3,7,11$ modulo 16. This contradicts $u^4\equiv1\pmod{16}$ (\code{1.2.7a}). In the former case, $r^4+45s^4=u^4-15r^2s^2$ is exactly $\mathcal C$. \end{proof} \begin{lem}\label{lem:1.2.7-props} Every primitive integral point ($\gcd(u,r,s)=1$) of $\mathcal C$ satisfies $\gcd(r,s)=1$, $\gcd(ur,30)=1$ and $4\mid s$. \end{lem} \begin{proof} A common prime factor of $r,s$ also divides $u$, so $\gcd(r,s)=1$. Modulo 16, the right-hand side is $5$ or $13$ if $r,s$ are both odd, $9$ or $13$ if $r$ is even, $13$ if $r$ is odd and $s\equiv2\pmod4$, and $1$ if $r$ is odd and $4\mid s$ (\code{1.2.7a}). Fourth powers are 0 or 1 modulo 16, so $u,r$ are odd and $4\mid s$. If $3\mid u$, then modulo 3 we get $3\mid r$ and $3\nmid s$, and the 3-adic valuation of the right-hand side is exactly 2, so it is not a fourth power. If $5\mid u$, then in the same way the 5-adic valuation of the right-hand side is exactly 1. If $3\mid r$ or $5\mid r$, then, modulo 3 or 5, $u$ is also divisible by that prime, which gives the contradiction just seen. \end{proof} \paragraph{Step 2: descent over $\Q(\sqrt5)$.} Let $K=\Q(d)$, $d=\sqrt5$, $e=(1+d)/2$. Then $\mathcal O_K=\Z[e]$, and since the Minkowski bound is $\sqrt5/2<2$, $h_K=1$. The unit group is $\{\pm e^j\}$ (if there were a unit $\rho$ with $1<\rho0$ such that $x=\varepsilon a^4$ and $N(y-\alpha x)=\varepsilon b^4$. Moreover, there exist $s=\pm1$, $0\le k\le3$ and $\beta\in\mathcal O_F$ such that \[ y-\alpha x=\delta\beta^4,\qquad \delta=s\alpha^k,\qquad \varepsilon=N(\delta)=s(-1)^k. \] In particular, $t=y/x\in\Q$ and $Y=b/(a^2\beta)\in F^\times$ satisfy \[ Y^4=(-\alpha)^k\,g(t). \] \end{lem} \begin{proof} The equation is $z^4=x(x^3+x^2y+y^3)=x\,N(y-\alpha x)$, and $\gcd(x,N(y-\alpha x))=\gcd(x,y^3)=1$. Since $z\ne0$, the two factors are coprime integers of the same sign, hence $x=\varepsilon a^4$ and $N(y-\alpha x)=\varepsilon b^4$. We show that $(y-\alpha x)$ is the fourth power of an ideal. If a prime ideal $\mathfrak R$ divides both $y-\alpha x$ and $G(x,y)$, then $\mathfrak R\nmid x$ (if $\mathfrak R\mid x$, then $\mathfrak R\mid y$, contradicting $\gcd(x,y)=1$). But $G(x,y)\equiv G(x,\alpha x)=Ax^2\pmod{y-\alpha x}$, so $\mathfrak R\mid A$, that is, $\mathfrak R=\mathfrak P$. Hence, if $\mathfrak R\ne\mathfrak P$, then $v_{\mathfrak R}(y-\alpha x)=v_{\mathfrak R}(N(y-\alpha x))=4v_{\mathfrak R}(b)\equiv0\pmod 4$. Since $\mathfrak P$ and $\mathfrak Q$ have degree 1, $v_{\mathfrak P}(y-\alpha x)=4v_{31}(b)-v_{\mathfrak Q}(y-\alpha x)\equiv0\pmod4$. Since $h_F=1$, $y-\alpha x=\delta\beta^4$ with $\delta=s\alpha^k$, and comparing norms gives $\varepsilon=N(\delta)=s(-1)^k$. Since $x\ne0$, $t=y/x$ is defined, and $x^2g(t)=G(x,y)=N(y-\alpha x)/(y-\alpha x)=\varepsilon b^4/(\delta\beta^4)$. Since $x^2=a^8$, we get $Y^4=b^4/(a^8\beta^4)=(\delta/\varepsilon)g(t)$, and $\delta/\varepsilon=\alpha^k(-1)^k$. \end{proof} \paragraph{Step 2: local elimination of $k=2,3$.} The two lemmas below treat the system \[ x=\varepsilon a^4,\qquad y-\alpha x=\delta\beta^4,\qquad \gcd(x,y)=1,\qquad \varepsilon=N(\delta)\tag{$\ast$} \] locally. The curve $Y^4=(-\alpha)^kg(t)$ alone is not enough. For example, the curve $Y^4=(\alpha+1)g(t)$ for $k=3$ has the $F$-point $(t,Y)=(-1,1)$ ($(\alpha+1)(\alpha^2-\alpha+2)=\alpha^3+\alpha+2=1$). \begin{lem}\label{lem:1.3.1-k3} If $\delta=s\alpha^3$, then $(\ast)$ has no solution in $\Z_3$. \end{lem} \begin{proof} Modulo 3, $T^3+T+1=(T-1)(T^2+T+2)$. The residue field of $\mathfrak P_1=(3,\alpha-1)$ is $\F_3$, and the residue field of $\mathfrak P_2$, of degree 2, is $\F_9=\F_3[\alpha]$. We have $\F_9^{\times4}=\F_3^\times$ and $\varepsilon=-s$. If $3\mid x$, then $y$ is a unit and $y-\alpha x\equiv y\pmod{\mathfrak P_2}$, so we would need $y/(s\alpha^3)\in\F_3^\times$, that is, $\alpha^3=-\alpha-1\in\F_3$, which is false. If $3\nmid x$, then $x\equiv\varepsilon=-s\pmod3$. Modulo $\mathfrak P_2$ we have $y-\alpha x=\lambda s\alpha^3=-\lambda s(1+\alpha)$ ($\lambda\in\F_3^\times$), and comparing the coefficients of $1$ and $\alpha$ gives $y=-\lambda s$, $x=\lambda s$. Hence $\lambda=-1$ and $y\equiv s\pmod3$. Then at $\mathfrak P_1$ ($\alpha\equiv1$) we get $(y-\alpha x)/\delta\equiv(s+s)/s=2\equiv-1$, which is not a square in $\F_3$ and so cannot be a fourth power. \end{proof} \begin{lem}\label{lem:1.3.1-k2} If $\delta=s\alpha^2$, then $(\ast)$ has no solution in $\Z_{13}$. \end{lem} \begin{proof} Modulo 13, $T^3+T+1=(T-7)(T^2+7T+11)$. The residue field of $\mathfrak P_1=(13,\alpha-7)$ is $\F_{13}$, and that of $\mathfrak P_2$ is $\F_{169}$. We have $\F_{13}^{\times4}=\{1,3,9\}$, and $w\in\F_{169}^\times$ is a fourth power if and only if its norm to $\F_{13}$ lies in $\{1,3,9\}$. At $\mathfrak P_2$ the norm of $\alpha$ is 11, and $\varepsilon=s$. If $13\mid x$, then $y-\alpha x\equiv y$ at both prime ideals. At $\mathfrak P_1$, $y/(10s)\in\{1,3,9\}$, so $y/s$ is a square modulo 13. Taking norms at $\mathfrak P_2$ gives $(y/s)^2/4\in\{1,3,9\}$, so $y/s\in\{2,5,6,7,8,11\}$, that is, a nonsquare. This is a contradiction. If $13\nmid x$, then $t=y/x\in\Z_{13}$ and $t-\alpha=\alpha^2(\beta/a)^4$. Taking norms at $\mathfrak P_2$ gives $(t^2+7t+11)/4\in\{1,3,9\}$. At $\mathfrak P_1$, $t\equiv7$ or $(t-7)/10\in\{1,3,9\}$, so $t\in\{4,6,7,11\}$. For these four values, $(t^2+7t+11)/4\equiv4,6,11,10$, and none of them lies in $\{1,3,9\}$. \end{proof} The conclusions of the two lemmas are also confirmed by an exhaustive check over the residue fields (\code{1.3.1b}). Therefore a solution with $z\ne0$ and $\gcd(x,y)=1$ gives, for $t=y/x\in\Q$ and $Y\in F^\times$, a point on \[ G_0:\ Y^4=g(t)\quad(k=0)\qquad\text{or}\qquad G_1:\ Y^4=-\alpha\,g(t)\quad(k=1). \] \paragraph{Step 3: two elliptic curves.} \begin{lem}\label{lem:1.3.1-curves} Smooth models of $G_0$ and $G_1$ are isomorphic over $F$ to \[ \begin{gathered} E_0:\ y^2=x^3+(3\alpha^2+4)x,\qquad E_1:\ y^2=x^3+k_1x,\\ k_1=\frac{39936\alpha^2-205568\alpha+169216}{29791}=(3\alpha^2+\alpha+1)\Big(\frac{40\alpha^2-4\alpha+28}{31}\Big)^4, \end{gathered} \] respectively. Hence each $F$-point $(t,Y)$ of $G_k$ corresponds to exactly one $R\in E_k(F)$, with $t(R)=t$. On $E_0$, $t=(3\alpha^2+4-2\alpha x-x^2)/(4x)$, and the only poles of this function are $O$ and $T=(0,0)$. On $E_1$, $t=(a_P(x)+b_P(x)y)/c_P(x)$ ($c_P$ monic of degree 4, $\deg a_P=3$, $\deg b_P=2$), and the points with $t=0$ are $O$ and \[ R_1=\Big(\frac{264\alpha^2-324\alpha-212}{961},\ \frac{23104\alpha^2+33848\alpha+49008}{29791}\Big). \] \end{lem} \begin{proof} $G_0$: put $U=Y^2$, $m=U-t$, $x=2m-\alpha$. Then $U^2=g(t)$ gives $t=(3\alpha^2+4-2\alpha x-x^2)/(4x)$ and $Y^2=(x^2+3\alpha^2+4)/(4x)$, hence $(2xY)^2=x^3+(3\alpha^2+4)x$. Putting $y=2xY$ gives a point of $E_0$. $G_1$: the conic $U^2=-\alpha g(t)$ has the point $(t,U)=(0,1)$ ($-\alpha(\alpha^2+1)=1$). Parametrizing by lines through this point gives $t=-(\alpha^2+2m)/(m^2+\alpha)$, $U=(\alpha-\alpha^2m-m^2)/(m^2+\alpha)$. Then $Y^2=U$ becomes $V^2=(\alpha-\alpha^2m-m^2)(m^2+\alpha)$ ($V=Y(m^2+\alpha)$), and this quartic takes the square value $(m_0^2+\alpha)^2$ at $m_0=-\alpha^2/2$. Substituting $m=m_0+1/s$ and dividing by $(m_0^2+\alpha)^2$ gives a monic quartic whose $s^2$-coefficient is 0, and the standard Weierstrass map of a monic quartic gives $E_1$. The second expression for $k_1$ shows that $E_1$ is isomorphic to $y^2=x^3+(3\alpha^2+\alpha+1)x$. Each map in the chain is verified by identities in function fields: each map lands on the target curve, and the two compositions are the identity (\code{1.3.1c}). The function $t$ has degree 4 on $E_0$, $E_1$ and $G_k$, so each map $E_k\to G_k$ has degree 1; being a birational map between smooth projective curves, it is an isomorphism. The points of $G_1$ with $t=0$ are $(0,\pm1)$, and they map to $O$ and $R_1$. \end{proof} \paragraph{Step 4: Mordell--Weil groups.} \begin{lem}\label{lem:1.3.1-mw} For $E\in\{E_0,E_1\}$, $E(F)\cong\Z\oplus\{O,T\}$. For $P=(-\alpha,2)\in E_0(F)$ and $R_1\in E_1(F)$, the index $[E(F):\langle P,T\rangle]$ (or $[E_1(F):\langle R_1,T\rangle]$) is prime to 2, 3, 47. \end{lem} \begin{proof} Rank: in the 2-isogeny descent (\cite{Simon02,Silverman09}) between $E_A:y^2=x^3+Ax$ and $E'_A:y^2=x^3-4Ax$, we have $2^{\operatorname{rank}}=|\operatorname{Im}\alpha_A|\,|\operatorname{Im}\alpha'_A|/4$, and the two images lie in the Selmer groups. With $S=\{(2),\mathfrak Q\}$, the dimensions of the Selmer groups are $(2,1)$ for $E_0$ ($A=3\alpha^2+4$) and $(1,2)$ for $E_1$ ($A=3\alpha^2+\alpha+1$) (\code{1.3.1d}). Hence both curves have rank at most 1. On $E_0$, $\alpha_A(P)=-\alpha$ and $\alpha_A(T)=A$ fill $\operatorname{Sel}(\alpha)$. Torsion: $\#E(\F_v)=48$ at the three prime ideals above 47, and $\#E(\F_v)=68$ at the three prime ideals above 67 (for both curves). By \cite[Appendix]{Katz81}, the order of the torsion divides $\gcd(48,68)=4$. Torsion of order 4 occurs only if $A$ or $-A$ is a square in $F$, and this is not the case since $N(3\alpha^2+4)=31$ and $N(k_1)=2^{24}/31^3$. Hence the torsion is $\{O,T\}$, $P$ and $R_1$ have infinite order, and the rank is 1. Index: if 2 divides the index, then $P$ or $P+T$ lies in $2E(F)$. But the halving quartic $(x^2-A)^2-4x_R(x^3+Ax)$ has no root in $F$ for $R=P$, $P+T$ (for example, for $P$ on $E_0$ it has no root modulo $(13,\alpha-7)$, and for $P+T$ none modulo $(23,\alpha-4)$). For an odd prime $\ell$: if there is a place $v$ of degree 1 such that $\ell\mid N=\#E(\F_v)$, $N/\ell$ is even and $(N/\ell)\bar P\ne O$, then $\ell$ does not divide the index. For $\ell=3$ we take $v=(11,\alpha-2)$, $N=12$, and for $\ell=47$ we take $v=(1879,\alpha-342)$, $N=1880$. The same places work for both curves (\code{1.3.1d}). \end{proof} \paragraph{Step 5: Mordell--Weil sieve at 47.} \begin{lem}\label{lem:1.3.1-sieve} If $R\in E_k(F)$ and $t(R)\in\mathbb P^1(\Q)$, then there is $R_0$, with $R_0\in\{O,T\}$ on $E_0$ and $R_0\in\{O,R_1\}$ on $E_1$, such that $R-R_0$ reduces to $O$ at all three prime ideals above 47. \end{lem} \begin{proof} The prime 47 splits completely in $F$. At the three places $v_1,v_2,v_3$ we have $\alpha\equiv25,34,35$. Both curves have good reduction at the three places, and $\#E(\F_{v_j})=48$. The function $t$ is well defined as a function on the reduced curve at each $v_j$; that is, at every reduced point there is an expression for $t$ whose numerator and denominator do not vanish simultaneously (on $E_0$ the formula above works as it is, since $A$ is a 47-adic unit; \code{1.3.1e}). Therefore, if $t(R)\in\mathbb P^1(\Q)$, the values of $t$ at the three reductions are equal in $\mathbb P^1(\F_{47})$. The exponent of $\prod_jE(\F_{v_j})$ divides 48, and $[E(F):\langle P,T\rangle]$ is prime to 48. Hence $\bar R=n\bar P+e\bar T$ ($n\bmod24$, $e\in\{0,1\}$). The orders of $\bar P$ are $24,4,12$ on $E_0$ and $24,6,6$ on $E_1$ ($P=R_1$). Among the 48 pairs $(n,e)$, the only ones with three equal $t$-values are $O$ and $T$ ($t=\infty$) on $E_0$, and $O$ and $R_1$ ($t=0$) on $E_1$ (\code{1.3.1e}). This is the claim (\cite{BS10}). \end{proof} \paragraph{Step 6: Strassmann argument at 47.} \begin{lem}\label{lem:1.3.1-strass} In Lemma~\ref{lem:1.3.1-sieve}, $R=R_0$. \end{lem} \begin{proof} Let $Q=24P$. Since the order of $\bar P$ divides 24 at the three places, $Q$ lies in the group $K_1$ of points that reduce to $O$ at all three places. $K_1$ is infinite cyclic ($T\notin K_1$), and by Lemma~\ref{lem:1.3.1-mw}, $[K_1:\langle Q\rangle]$ is prime to 47. Therefore $R-R_0=sQ$ with $s\in\Z_{(47)}$, and for the three embeddings $\sigma_j:F\to\Q_{47}$ we have $\sigma_j(R)=\sigma_j(R_0)+[s]\sigma_j(Q)$ with the same $s$. We use the framework of Proposition~\ref{thm:G}. Put $f=1/t$ on $E_0$ and $f=t$ on $E_1$, and let $\lambda_j=\log z(\sigma_jQ)$ ($z$ the formal group parameter). A computation gives $v(\lambda_j)=1$. Put \[ h(s)=f\big(\sigma_1(R_0)+[s]\sigma_1Q\big)-f\big(\sigma_2(R_0)+[s]\sigma_2Q\big)=\sum_{n\ge0}a_ns^n. \] Then $h(s)=0$ when $t(R)\in\Q$. Near $R_0$, $f$ lies in $\Z_{47}[[z]]$: at $O$ and $T$ on $E_0$ because $A$ is a 47-adic unit, and at $R_1$ on $E_1$ because $R_1$ is 47-adically integral and $c_P(x(R_1))$ is a 47-adic unit. The coefficients $e_n$ of the formal exponential satisfy $n!e_n\in\Z_{47}$, so $v(a_n)\ge n-(n-1)/46$. $E_0$: $t$ is a function of $x$ alone, so $t(-R)=t(R)$, and $-T=T$. Hence $h(-s)=h(s)$, and since $f(O)=f(T)=0$, $a_0=a_1=0$. A computation gives $v(a_2)=2$, and by the inequality above $v(a_n)>2$ for $n\ge3$. By Strassmann's theorem, $h$ has at most two zeros in $\Z_{47}$ counted with multiplicity, and since $s=0$ is a double zero, there are no other zeros. $E_1$: $a_0=0$, $v(a_1)=1$, and $v(a_n)>1$ for $n\ge2$. Hence the only zero is $s=0$. In both cases $s=0$, that is, $R=R_0$ (\code{1.3.1f}). \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.3.1}] If $d=\gcd(x,y)>1$, then $d^4\mid z^4$, so $(x/d,y/d,z/d)$ is also a solution. Hence we may assume $\gcd(x,y)=1$. If $z=0$, then $x\,N(y-\alpha x)=0$, and since $\alpha\notin\Q$, $N(y-\alpha x)\ne0$ when $x\ne0$. Therefore $x=0$, and the solution is $(0,y,0)$. If $z\ne0$, then by Lemmas~\ref{lem:1.3.1-descent}, \ref{lem:1.3.1-k3}, \ref{lem:1.3.1-k2} we get a point of $G_0$ or $G_1$, and by Lemma~\ref{lem:1.3.1-curves} there is $R\in E_k(F)$ with $t(R)=y/x$. By Lemmas~\ref{lem:1.3.1-sieve} and \ref{lem:1.3.1-strass}, on $E_0$ we have $R\in\{O,T\}$ and $t(R)=\infty$, which is impossible since $x=\varepsilon a^4\ne0$. On $E_1$ we have $R\in\{O,R_1\}$ and $t=0$, that is, $y=0$, $x=\pm1$, $z=\pm1$. Returning to the original solution, we get $(x,0,\pm x)$. \end{proof} \subsection{\texorpdfstring{$x^4-y^4+x^2yz+yz^3=0$ \ ($H=64$)}{x^4-y^4+x^2yz+yz^3=0 (H=64)}}\label{sec:1.3.2} \begin{thm}\label{thm:1.3.2} The only integer solutions of $x^4-y^4+x^2yz+yz^3=0$ are $(0,0,n)$, $(0,n,n)$, $(n,n,0)$, $(-n,n,0)$ $(n\in\Z)$. \end{thm} \paragraph{Setup.} Let $C\subset\mathbb P^2$ be the projective curve defined by the equation in the theorem. The partial derivatives of $F=x^4-y^4+x^2yz+yz^3$ are \[ F_x=2x(2x^2+yz),\qquad F_y=-4y^3+x^2z+z^3,\qquad F_z=y(x^2+3z^2). \] $F_z=0$ gives $y=0$ or $x^2=-3z^2$. If $y=0$, then $F_x=F_y=0$ gives $x=z=0$. If $x^2=-3z^2$, then $F_x=0$ gives $x=0$ (and then $z=0$, $y=0$) or $y=6z$, and in the latter case $F_y=-866z^3=0$. Hence $C$ is a smooth plane quartic curve (genus 3). The points with $xyz=0$ are found directly. If $x=0$, then $y(z^3-y^3)=0$ gives $(0:0:1),(0:1:1)$; if $y=0$, then $x=0$; and if $z=0$, then $x^4=y^4$ gives $(\pm1:1:0)$. Hence the theorem is equivalent to the statement that there is no rational point with $xyz\ne0$. Such points lie in the chart $x\ne0$, so we work with \[ r=y/x,\qquad t=z/x,\qquad F(r,t):=1+rt+rt^3-r^4=0,\qquad rt\ne0. \] The map $\iota(x:y:z)=(-x:y:z)$ is an involution of $C$. The Ritzenthaler--Romagny construction of \cite{Bruin08,RR18} maps the chart $x\ne0$ into the intersection of the three quadrics in $\mathbb P^4$ (coordinates $(x_1:x_2:x_3:u:v)$) \begin{equation}\label{eq:1.3.2-Q} \begin{aligned} Q_1&=-x_1x_2-x_2^2-x_1x_3-u^2,\\ Q_2&=-x_2^2-x_1x_3-uv,\\ Q_3&=x_1x_2+2x_2^2+2x_1x_3+x_2x_3-v^2 \end{aligned} \end{equation} by \[ V(r,t)=\bigl(r(t-r):1:r^2+rt+t^2:r:t\bigr). \] Indeed, $Q_1(V)=Q_2(V)=-F$ and $Q_3(V)=2F$. Let $M_i$ be the $5\times5$ symmetric matrix of $Q_i$, and $M_i'$ its upper left $3\times3$ block. Then \[ h(s):=-4\det(M_1'-2sM_2'+s^2M_3')=6s^6+22s^5+15s^4-14s^3-7s^2+6s-1 \] (\code{1.3.2a}), $h=(2s^2+2s-1)(3s^4+8s^3+s^2-4s+1)$, and the two factors are irreducible over $\Q$. By \cite{Bruin08,RR18}, the Prym of the cover $C\to C/\iota$ is isogenous to the Jacobian of the genus 2 curve \[ H:\ \eta^2=h(s),\qquad J=\operatorname{Jac}(H). \] Let $D_\infty$ be the sum of the two points at infinity of $H$ (a rational divisor), and let $\kappa:J\to\mathcal K\subset\mathbb P^3$ be the Kummer map. In Kummer coordinates $\kappa(0)=(0:0:0:1)$, and the first three coordinates of $[(s_1,\eta_1)+(s_2,\eta_2)-D_\infty]$ are $(1:s_1+s_2:s_1s_2)$. The proof below does not use the Prym identification itself. The required map $C\to J$ is obtained directly in Lemma~\ref{lem:1.3.2-map} by a computation over the function field. \paragraph{Step 1: the map $\phi:C\to J$.} At a point $P$ of $C$ ($rt\ne0$) we take the tangent direction \[ w=\frac1{2\Delta}\bigl(-N_1,\ 4r^3-t^3-t,\ -N_3,\ 2\Delta,\ 0\bigr),\qquad \Delta=r^4-rt^3+1, \] \[ \begin{aligned} N_1&=2r^4t-3r^2t^3+r^2t+rt^4-rt^2+4r-2t,\\ N_3&=2r^4t+4r^3t^2+3r^2t^3-r^2t+rt^4-rt^2-4r-t^5-t^3-2t. \end{aligned} \] The following hold (\code{1.3.2c}). \begin{enumerate}[label=(\roman*)] \item $V^{\mathsf T}M_iw=0$ ($i=1,2,3$), and the left $3\times3$ minor of the tangent matrix $(V^{\mathsf T}M_i)_i$ is $\Delta/4$. \item On $C$, $\Delta=rt+2$. If $rt=-2$, then multiplying $F=0$ by $-r^2$ gives $r^6+r^2+8=0$; so $\Delta\ne0$ at rational points. \item The fifth coordinate of $w$ is 0 and the fifth coordinate of $V$ is $t\ne0$, so $w$ is not proportional to $V$. \end{enumerate} Now put \[ q_i=w^{\mathsf T}M_iw\quad(i=1,2,3),\qquad A_P(s)=q_1(P)-2s\,q_2(P)+s^2q_3(P). \] The $q_i$ are rational functions on $C$ whose denominators are powers of $\Delta$. If we replace the tangent direction by $aw+bV$ ($a\ne0$), then, since $V^{\mathsf T}M_iV=0$ and $V^{\mathsf T}M_iw=0$, $A_P$ is only multiplied by $a^2$. \begin{lem}\label{lem:1.3.2-map} There is a map $\phi:C\to J$ over $\Q$ with the following properties. \begin{enumerate}[label=(\alph*)] \item For $P\in C(\Q)$ with $rt\ne0$, $(q_1,q_2,q_3)(P)\ne(0,0,0)$ and $\phi(P)=[D_P-D_\infty]$. Here $D_P$ is an effective divisor of degree 2 on $H$, and its $s$-coordinates are the roots of the quadratic form $A_P$ (if $q_3(P)=0$, one root is $\infty$). \item For the two quartics $\Phi_1,\Phi_2$ given in \code{1.3.2a}, \code{1.3.2b}, we have $\kappa(\phi(P))\in\Gamma:=\{\Phi_1=\Phi_2=0\}$ for all $P\in C$. \end{enumerate} \end{lem} \begin{proof} We compute over the function field $K=\Q(C)$ (\code{1.3.2b}). At the generic point, $A(s)=q_1-2sq_2+s^2q_3\in K[s]$ is an irreducible quadratic. There are exactly two elements $\pm\phi_K$ of $J(K)$ whose Mumford representation has first polynomial $A/q_3$. Their Kummer images coincide, and the coordinates $(1:2q_2/q_3:q_1/q_3:k_4)$ satisfy $\Phi_1$, $\Phi_2$ and the equation of the Kummer surface. We choose one $\phi_K$. Then $\phi_K\in J(K)$ is a rational map $C\dashrightarrow J$ over $\Q$. Since $C$ is a smooth projective curve and $J$ is a projective variety, it extends to a morphism $\phi:C\to J$. For the same reason, an effective divisor of degree 2 over $K$ representing $\phi_K$ gives a morphism $\psi:C\to\operatorname{Sym}^2H$, and $\phi(P)=[\psi(P)-D_\infty]$. The composition of $\psi$ with the $s$-projection $\operatorname{Sym}^2H\to\operatorname{Sym}^2\mathbb P^1=\mathbb P^2$ is $(q_1:-2q_2:q_3)$ at the generic point. Hence, at points where the three functions are regular and not all zero, it equals their value. (a) At rational points with $rt\ne0$ we have $\Delta\ne0$, so the $q_i$ are regular. If $q_2(P)=q_3(P)=0$, then the ideal memberships \[ (r-1)^2(r+1)^2(r^2+1)^2,\quad t-r^7+r^3\ \in\ \bigl(F,\ \operatorname{num}q_2,\ \operatorname{num}q_3\bigr)\subset\Q[r,t] \] (\code{1.3.2c}) give $r=\pm1$ and $t=r^7-r^3=0$, a contradiction. Hence the $s$-coordinates of $D_P:=\psi(P)$ are the roots of $A_P$. (b) $\Phi_i\circ\kappa\circ\phi$ is a section of a line bundle on $C$ that vanishes at the generic point, so it is identically 0. \end{proof} \paragraph{Step 2: $J(\Q)$.} \begin{lem}\label{lem:1.3.2-MW} $J(\Q)=\langle T\rangle\oplus\langle G\rangle\cong\Z/2\oplus\Z$, where, in Mumford representation, \[ T=\bigl(s^2+s-\tfrac12,\ 0\bigr),\qquad G=\bigl(s^2+2s-1,\ 2s-1\bigr). \] \end{lem} \begin{proof} Since $h$ is the product of an irreducible quadratic and an irreducible quartic over $\Q$, $J(\Q)[2]=\langle T\rangle\cong\Z/2$. The 2-Selmer group has order 8, and 2 is the only place at which $H$ is deficient. Since the number of deficient places is odd, $\mathrm{Sha}(J)[2]\ne0$, and by \cite{Stoll01,PS97,PS99}, $\operatorname{rank}J(\Q)\le3-1-1=1$. Since $G$ has infinite order, the rank is 1. That the torsion is only $\langle T\rangle$ and that $T,G$ generate all of $J(\Q)$ is proved by a saturation computation using height constants (\cite{CF96,Stoll99,Stoll02}). All of this is checked separately by \code{1.3.2a} and \code{1.3.2d}. \end{proof} \paragraph{Step 3: Mordell--Weil sieve.} For $P\in C(\Q)$ we write $\phi(P)=nG+eT$ ($n\in\Z$, $e\in\{0,1\}$). By Lemma~\ref{lem:1.3.2-map}(b), $\Phi_i(\kappa(nG+eT))=0$. Kummer coordinates are normalized to primitive integer vectors. $H$ has good reduction at 5, 29, 113. \begin{lem}\label{lem:1.3.2-sieve} $e=0$ and $n\equiv0,\pm2,\pm14\pmod{31}$. \end{lem} \begin{proof} Reduction is compatible with the group law and with the Kummer map, so $\kappa(n\bar G+e\bar T)\in\Gamma(\F_{29})$ (\cite{BS10}). The order of $\bar G\in J(\F_{29})$ is 31. Testing all 62 pairs $(n\bmod31,e)$ leaves only $(0,0),(2,0),(14,0),(17,0),(29,0)$ (\code{1.3.2a}). Here $17\equiv-14$ and $29\equiv-2$. (At $p=5$ the order of $\bar G$ is also 31, and 13 classes remain. They contain all five classes above, so nothing more is removed.) \end{proof} \begin{lem}\label{lem:1.3.2-lift} $n\not\equiv\pm14\pmod{31}$. \end{lem} \begin{proof} Put $Q=31G$; then $\bar Q=0$. Let $J^k$ be the kernel of reduction modulo $29^k$. Then $Q\in J^1$, and $z([29^k]Q)\equiv29^kz(Q)\pmod{29^{k+2}}$ from Lemma~\ref{lem:O} gives $[29^k]Q\in J^{k+1}$. Hence, if $n\equiv n'\pmod{31\cdot29^k}$, then $nG-n'G\in J^{k+1}$. Since the Kummer map is defined over $\Z_{29}$, the primitively normalized vectors $\kappa(nG)$ and $\kappa(n'G)$ agree modulo $29^{k+1}$ up to a unit multiple. Therefore $\Phi_i(\kappa(n'G))\equiv0\pmod{29^{k+1}}$ must hold. $k=1$: if $n\equiv14\pmod{31}$, then among $n'=14+31d$ ($0\le d\le28$) only $231$ satisfies this condition. Similarly, from $17$ only $668$ remains. $k=2$: among $231+899d$ ($0\le d\le28$) only $18211$ remains, and among $668+899d$ only $7860$ remains (\code{1.3.2a}). Hence $n\equiv18211$ or $7860\pmod{31\cdot29^2}$, and modulo $29^2$, $n\equiv550$ or $291$. At $p=113$ the order of $\bar G$ is $6728=2^3\cdot29^2$. For the 16 classes of $n\bmod6728$ with $n\bmod841\in\{291,550\}$, none of the points $\kappa(n\bar G)$ lies in $\Gamma(\F_{113})$ (\code{1.3.2a}). This is a contradiction. \end{proof} \paragraph{Step 4: 29-adic isolation of the three remaining disks.} \begin{lem}\label{lem:1.3.2-iso} $n\in\{0,2,-2\}$. \end{lem} \begin{proof} Let $z=(z_1,z_2)$ be the 29-adic formal group parameter of $J$; then $J^1$ corresponds to $z\in(29\Z_{29})^2$. Near the origin, writing the Kummer coordinates as $(\kappa_1:\kappa_2:\kappa_3:1)$, the $\kappa_j\in\Z_{29}[[z]]$ have no constant term. Since $\kappa\circ[-1]=\kappa$ and $[-1]z=-z+O(z^2)$, they have no linear term either. Hence $\kappa_j=B_j(z)+O(|z|^3)$ ($B_j$ quadratic forms). The primitive Kummer coordinates $(k_1,k_2,k_3,k_4)$ of $Q=31G$ have valuations $(2,2,2,0)$, and $(k_1,k_2,k_3)\equiv29^2\cdot(9,10,6)\pmod{29^3}$ (\code{1.3.2a}). Therefore $z(Q)=29\zeta$ with $\zeta\in\Z_{29}^2$, and $B(\zeta)\equiv c\,(9,10,6)\pmod{29}$ ($c$ a unit). (a) $n\equiv0\pmod{31}$: write $n=31m$, $m=29^eu\ne0$ ($u$ a unit). By Lemma~\ref{lem:O}, $z([m]Q)=29^{1+e}(u\zeta+O(29))$, so \[ (\kappa_1,\kappa_2,\kappa_3)([m]Q)=29^{2+2e}\bigl(u^2B(\zeta)+O(29)\bigr). \] Since $\Phi_i(k_1,k_2,k_3,1)$ has no terms $k_4^4$, $k_4^3k_j$, its lowest-order part is the quadratic form given by the coefficient of $k_4^2$: \[ L_1=-336k_1k_2+540k_1k_3-135k_2^2+672k_3^2,\qquad L_2=-192k_1k_2+388k_1k_3-97k_2^2+384k_3^2. \] Hence $\Phi_1(\kappa([m]Q))=29^{4+4e}\bigl(u^4c^2L_1(9,10,6)+O(29)\bigr)$. Since $L_1(9,10,6)\equiv13\not\equiv0\pmod{29}$ (and $L_2(9,10,6)\equiv24$), the valuation is $4+4e$, which is finite. That is, if $m\ne0$, then $\kappa(nG)\notin\Gamma$; so $n=0$. (b) $n\equiv2\pmod{31}$: write $n=2+31m$, $m=29^eu\ne0$. In the patch $k_1=1$, $\kappa(2G)\equiv b=(1,25,2,27)\pmod{29}$. Translation by $2G$ is an automorphism over $\Z_{29}$, so for $Y\in J^1$ we have $\kappa(2G+Y)=\kappa(2G)+\Lambda z(Y)+O(|z(Y)|^2)$ ($\Lambda$ a matrix over $\Z_{29}$). From the difference for $m=1$ we get $\Lambda\zeta\equiv d=(0,17,9,15)\pmod{29}$ (\code{1.3.2a}). Hence $\kappa(2G+[m]Q)=\kappa(2G)+29^{1+e}\bigl(u\,d+O(29)\bigr)$. Since $\Phi_i(\kappa(2G))=0$, \[ \Phi_i\bigl(\kappa(2G+[m]Q)\bigr)=29^{1+e}\bigl(u\,D_d\Phi_i(b)+O(29)\bigr), \] and since $(D_d\Phi_1(b),D_d\Phi_2(b))\equiv(23,9)\pmod{29}$, the valuation is $1+e$, which is finite. Hence $n=2$. (c) If $n\equiv-2\pmod{31}$, then $\kappa(nG)=\kappa(-nG)$ and $-n\equiv2\pmod{31}$, so $-n=2$ by (b). \end{proof} \paragraph{Step 5: fibers over the three classes.} \begin{lem}\label{lem:1.3.2-fib} For $P\in C(\Q)$ with $rt\ne0$, $\phi(P)\notin\{0,2G,-2G\}$. \end{lem} \begin{proof} (a) Suppose $\phi(P)=\pm2G$. Since $\kappa(\pm2G)=(32:-128:-52:-441)$, the first Mumford polynomial of $\pm2G$ is $s^2+4s-\tfrac{13}{8}$. In genus 2, the effective divisor $E$ of degree 2 in a nonzero class $[E-D_\infty]$ is unique, so $D_P$ of Lemma~\ref{lem:1.3.2-map}(a) is that divisor. Then $A_P$ is a constant multiple of $s^2+4s-\tfrac{13}{8}$, and at $P$ \[ 2q_2+4q_3=0,\qquad 8q_1+13q_3=0. \] The lexicographic Gröbner basis of the ideal of $\Q[r,t]$ generated by these two equations (with denominators cleared) and $F$ is \[ r^4-rt-1,\qquad (r^2-1)t^2,\qquad t^3 \] (\code{1.3.2e}). Hence $t=0$, contradicting $rt\ne0$. (b) Suppose $\phi(P)=0$. Then $D_P\sim D_\infty$. The complete linear system of $D_\infty$ is the unique $g^1_2$ of $H$, that is, the fibers of $s$, so the $s$-image of $D_P$ is a single point. Hence the discriminant $q_2^2-q_1q_3$ of $A_P$ vanishes at $P$. The resultant with respect to $t$ of $F$ and the numerator of this discriminant is \[ \tfrac1{64}\,r^4(r-1)^4(r+1)^4(r^2+1)^4A_8(r)B_8(r)C_{24}(r), \] where $A_8$, $B_8$, $C_{24}$ are irreducible polynomials over $\Q$ of degrees 8, 8, 24 (\code{1.3.2c}). For example, $A_8=27r^8+16r^6-19r^4-6r^2-1$ and $B_8=28r^8+16r^6-18r^4+2r^2-1$. Therefore $r\in\{0,\pm1\}$. But $r\ne0$, and $F(\pm1,t)=\pm t(t^2+1)$ gives $t=0$, a contradiction. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.3.2}] Suppose there is $P\in C(\Q)$ with $xyz\ne0$. Then $rt\ne0$. By Lemma~\ref{lem:1.3.2-MW} we can write $\phi(P)=nG+eT$. By Lemmas~\ref{lem:1.3.2-sieve}, \ref{lem:1.3.2-lift}, \ref{lem:1.3.2-iso}, $\phi(P)\in\{0,\pm2G\}$, which contradicts Lemma~\ref{lem:1.3.2-fib}. The rational points with $xyz=0$ are the four points found in the setup, and the integer solutions above them are the four forms in the theorem. \end{proof} At $(\pm1:1:0)$, the tangent quadratic computed with a different tangent direction is proportional to $s^2+4s-\tfrac{13}{8}$, and the only elements of $J(\Q)$ with this first polynomial are $\pm2G$ (\code{1.3.2a}). So $\pm2G$ necessarily survive the sieve; they are indeed the images of these two points. \subsection{\texorpdfstring{$x^4+x^2y^2+y^3z-yz^3=0$ \ ($H=64$)}{x^4+x^2y^2+y^3z-yz^3=0 (H=64)}}\label{sec:1.3.3} \begin{thm}\label{thm:1.3.3} The only integer solutions of $x^4+x^2y^2+y^3z-yz^3=0$ are $(0,0,n)$, $(0,n,0)$, $(0,n,n)$, $(0,n,-n)$ $(n\in\Z)$. \end{thm} \paragraph{Setup.} Let $C\subset\mathbb P^2$ be the projective curve defined by the equation in the theorem, and put $J_C=\Jac(C)$. $C$ is a smooth plane quartic curve (\code{1.3.3a}). Hence it has genus 3 and is not hyperelliptic. The map $\iota(x:y:z)=(-x:y:z)$ is an involution of $C$. Putting $x^2=X$, we get $(2X+y^2)^2=y^4-4y^3z+4yz^3$, so the quotient curve $E=C/\langle\iota\rangle$ is $W^2=y^4-4y^3z+4yz^3$. $E$ has the rational point $(y:z:W)=(0:1:0)$ and is isomorphic over $\Q$ to the elliptic curve $Y^2=X^3-16X+16$ of conductor 37. The point $Q_0=(0:0:1)$ is a rational fixed point of $\iota$. For $P\in C(\Q)$ put \[ \delta_P=[P-\iota(P)]\in J_C(\Q). \] Then $\delta_P=(1-\iota_*)[P-Q_0]$, so $\delta_P$ is a rational point of the Prym variety $\operatorname{Prym}(C/E)=(1-\iota_*)J_C$. In the form of Ritzenthaler--Romagny, \[ C:\ x^4-h_0x^2+f_0g_0=0,\qquad f_0=yz,\quad h_0=-y^2,\quad g_0=y^2-z^2. \] The matrix of coefficients of $(f_0,h_0,g_0)$ with respect to $(y^2,yz,z^2)$ and its inverse are \[ A=\begin{pmatrix}0&1&0\\-1&0&0\\1&0&-1\end{pmatrix},\qquad A^{-1}=\begin{pmatrix}0&-1&0\\1&0&0\\0&-1&-1\end{pmatrix}. \] Let $(a_i,b_i,c_i)$ be the rows of $A^{-1}$, and put $a(s)=a_1+2a_2s+a_3s^2$, and $b(s)$, $c(s)$ in the same way. Then $a=2s$, $b=-(s^2+1)$, $c=-s^2$. By \cite{Bruin08,RR18}, $\operatorname{Prym}(C/E)$ is isogenous over $\Q$ to the Jacobian $J_H$ of \[ H:\ \eta^2=b(b^2-ac)=-(s^2+1)(s^4+2s^3+2s^2+1) \] (\code{1.3.3a}). This identification, including the quadratic twist, is confirmed by $L_p(C)=L_p(E)\,L_p(H)$ at all good primes $3\le p\le79$ (\code{1.3.3b}). \paragraph{Step 1: the rank of the Prym.} \begin{lem}\label{lem:1.3.3-rank} $\operatorname{rank}J_H(\Q)=0$. Hence $\operatorname{Prym}(C/E)(\Q)$ is finite, and for every $P\in C(\Q)$, $\delta_P$ is a torsion point of $J_C(\Q)$. \end{lem} \begin{proof} The 2-descent of \cite{Stoll01,PS97,PS99} gives the upper bound 0 for $\operatorname{rank}J_H(\Q)$ (\code{1.3.3a}: the lower and upper bounds for the rank are both 0). In fact, $J_H(\Q)=\langle(s^2+1,\,0)\rangle\cong\Z/2$ (\code{1.3.3c}). Isogenies preserve the rank, so $\operatorname{Prym}(C/E)(\Q)$ is a finite group, and its element $\delta_P$ is torsion. \end{proof} \paragraph{Step 2: the order of $\delta_P$.} \begin{lem}\label{lem:1.3.3-ord} For every $P\in C(\Q)$, $2\delta_P=0$. \end{lem} \begin{proof} $C$ has good reduction at 3 and 5, and \[ J_C(\F_3)\cong\Z/2\oplus\Z/14,\qquad J_C(\F_5)\cong\Z/2\oplus\Z/176 \] (\code{1.3.3a}). The orders are $L_3(1)=28$ and $L_5(1)=352$, respectively, where \[ L_3(T)=1+27T^6,\qquad L_5(T)=1+4T+15T^2+32T^3+75T^4+100T^5+125T^6. \] Let $\ell$ be a prime dividing the order of $\delta_P$. By \cite[Appendix]{Katz81}, if $\ell\ne3$ then $\ell\mid28$, and if $\ell\ne5$ then $\ell\mid352$. If $\ell=3$ we would need $3\mid352$, and if $\ell=5$ we would need $5\mid28$; both are impossible. In the remaining cases $\ell\mid\gcd(28,352)=4$, so $\ell=2$. Hence the order of $\delta_P$ is a power of 2, and by \cite[Appendix]{Katz81} the group generated by $\delta_P$ maps injectively into $J_C(\F_3)$. The 2-part of $J_C(\F_3)$ is $(\Z/2)^2$, which has exponent 2, so $2\delta_P=0$. \end{proof} \paragraph{Step 3: fixed points.} \begin{proof}[Proof of Theorem~\ref{thm:1.3.3}] Let $P\in C(\Q)$ with $P\ne\iota(P)$. By Lemma~\ref{lem:1.3.3-ord}, $2[P-\iota(P)]=0$. This contradicts \cite[Section IV.5]{Hartshorne77} ($C$ is a smooth plane quartic curve, so it is not hyperelliptic). Hence $P=\iota(P)$. The fixed points of $\iota$ are the points on the line $x=0$ and $(1:0:0)$, and $(1:0:0)$ does not lie on $C$. If $x=0$, then $yz(y-z)(y+z)=0$, so $C(\Q)$ consists of the four points $(0:0:1)$, $(0:1:0)$, $(0:1:1)$, $(0:-1:1)$. The integer solutions above these four points are the four forms in the theorem. \end{proof} \subsection{\texorpdfstring{$x^4+y^4+x^2yz-yz^3=0$ \ ($H=64$)}{x^4+y^4+x^2yz-yz^3=0 (H=64)}}\label{sec:1.3.4} \begin{thm}\label{thm:1.3.4} The only integer solutions of $x^4+y^4+x^2yz-yz^3=0$ are $(0,0,n)$ and $(0,n,n)$ $(n\in\Z)$. \end{thm} \paragraph{Setup.} Since the equation is homogeneous, it suffices to determine the rational points of the projective plane quartic curve \[ C:\quad X^4+Y^4+X^2YZ-YZ^3=0. \] $C$ is smooth (\code{1.3.4a}). Hence it has genus 3 and, by \cite[Section IV.5]{Hartshorne77}, it is not hyperelliptic. Theorem~\ref{thm:1.3.4} follows directly from the next proposition, because a nonzero integer solution gives a rational point of $C$, and the integer solutions above the two points $(0:0:1)$, $(0:1:1)$ are $(0,0,n)$, $(0,n,n)$. \begin{prop}\label{prop:1.3.4} $C(\Q)=\{(0:0:1),(0:1:1)\}$. \end{prop} The involution $\iota(X:Y:Z)=(-X:Y:Z)$ preserves $C$. The fixed points of $\iota$ in the plane are the line $X=0$ and the point $(1:0:0)$, and $(1:0:0)$ does not lie on $C$. In the chart $Y=1$, with $u=X/Y$, $t=Z/Y$, the curve $C$ is $u^4+1+u^2t-t^3=0$. The functions $w=4t$, $v=4(2u^2+t)$ satisfy \[ v^2-(w^3+w^2-64)=64\,(u^4+1+u^2t-t^3), \] so the quotient curve $E=C/\iota$ is the elliptic curve $v^2=w^3+w^2-64$ (conductor 431). Let $P_C=\operatorname{im}(1-\iota_*)\subset J_C$ be the Prym variety of the cover $C\to E$. It is an abelian variety of dimension 2, and $J_C$ is isogenous to $E\times P_C$. Writing $C$ in the form $X^4-hX^2+fg=0$, we have $f=Y(Y-Z)$, $h=-YZ$, $g=Y^2+YZ+Z^2$. The matrix whose rows are the coefficients of $(Y^2,YZ,Z^2)$ in $f,h,g$, in this order, and its inverse are \[ A=\begin{pmatrix}1&-1&0\\0&-1&0\\1&1&1\end{pmatrix},\qquad A^{-1}=\begin{pmatrix}1&-1&0\\0&-1&0\\-1&2&1\end{pmatrix}. \] Forming $\alpha_1+2\alpha_2s+\alpha_3s^2$ from each column $(\alpha_1,\alpha_2,\alpha_3)^{\mathsf T}$ of $A^{-1}$ gives, in order, $a(s)=1-s^2$, $b(s)=2s^2-2s-1$, $c(s)=s^2$. By \cite{Bruin08,RR18}, $P_C$ is isogenous over $\Q$ to the Jacobian $J_H$ of the genus 2 curve \[ H:\quad y^2=f_H(s),\qquad \begin{aligned} f_H&=b(b^2-ac)=10s^6-26s^5+9s^4+18s^3-5s^2-6s-1\\ &=(2s^2-2s-1)(5s^4-8s^3-s^2+4s+1). \end{aligned} \] These polynomial identities are checked by \code{1.3.4a}. The same code also checks $L_p(C,T)=L_p(E,T)\,L_p(H,T)$ at the good primes $p=5,7,11$. These finite comparisons check the formula and twist; the isogeny itself follows from the cited construction. The proof has three steps. We show $\operatorname{rank}J_H(\Q)=0$ (Lemma~\ref{lem:1.3.4-rank}), show that the torsion of $J_C(\Q)$ has order at most 2 (Lemma~\ref{lem:1.3.4-tors}), and use these two facts to show that every rational point is a fixed point of $\iota$ (Lemma~\ref{lem:1.3.4-fix}). \paragraph{Step 1: the rank of $J_H$.} \begin{lem}\label{lem:1.3.4-rank} $\operatorname{rank}J_H(\Q)=0$. Hence $P_C(\Q)$ is a finite group. \end{lem} \begin{proof} Let $D_\infty$ be the sum of the two points at infinity of $H$ (a rational divisor). For the two roots $\alpha,\bar\alpha$ of $2s^2-2s-1$, the class $T=[(\alpha,0)+(\bar\alpha,0)-D_\infty]$, with Mumford representation $(s^2-s-\tfrac12,\,0)$, is an element of order 2 of $J_H(\Q)$. Hence $\dim_{\F_2}J_H(\Q)[2]\ge1$. The 2-Selmer group has 4 elements, that is, $\dim_{\F_2}\operatorname{Sel}^{2}(\Q,J_H)=2$ (\code{1.3.4b}). Moreover, the only place where $H$ is deficient, that is, where it has no $\Q_v$-rational divisor of degree 1, is $v=2$ (\code{1.3.4b}). In particular, $H(\Q_2)=\emptyset$, because $f_H(s)\equiv3,7\pmod8$ if $s\in\Z_2$, and $s^{-6}f_H(s)\equiv2\pmod4$ if $s\notin\Z_2$. On the other hand, at $\infty$ there are real points since the leading coefficient is $10>0$, and at the odd bad primes $3$, $431$ there are $\Q_p$-points. Since the number of deficient places is odd, the Poonen--Stoll version of \cite{Stoll01,PS97,PS99} gives $\mathrm{Sha}(J_H)[2]\ne0$ and \[ \dim\operatorname{Sel}^{2}\ \ge\ \operatorname{rank}J_H(\Q)+\dim J_H(\Q)[2]+1\ \ge\ \operatorname{rank}J_H(\Q)+2. \] Therefore $\operatorname{rank}J_H(\Q)=0$. Magma's \texttt{RankBounds} gives the same upper bound $0$ (\code{1.3.4a}). The rank does not change under isogeny, so $\operatorname{rank}P_C(\Q)=0$, and $P_C(\Q)$ is a finite group. \end{proof} \paragraph{Step 2: torsion of $J_C(\Q)$.} \begin{lem}\label{lem:1.3.4-tors} $J_C(\Q)_{\mathrm{tors}}$ has order at most 2. In particular, every torsion element $x\in J_C(\Q)$ satisfies $2x=0$. \end{lem} \begin{proof} $C$ has good reduction at $5$, $7$, $11$, and the Jacobians of the reductions are \[ J_C(\F_5)\simeq\Z/110,\qquad J_C(\F_7)\simeq\Z/2\oplus\Z/140,\qquad J_C(\F_{11})\simeq\Z/17\oplus\Z/68 \] (\code{1.3.4a}). Their orders are $110=2\cdot5\cdot11$, $280=2^3\cdot5\cdot7$, $1156=2^2\cdot17^2$, respectively. Suppose an odd prime $\ell$ divides $\#J_C(\Q)_{\mathrm{tors}}$. If $\ell\ne5$, then $\ell\mid110$ by \cite[Appendix]{Katz81}, so $\ell=11$. But $11\nmid280$, a contradiction at $p=7$. If $\ell=5$, then $5\nmid1156$ gives a contradiction at $p=11$. Hence $J_C(\Q)_{\mathrm{tors}}$ is a 2-group, and by \cite[Appendix]{Katz81} it injects into the 2-part $\Z/2$ of $J_C(\F_5)$. \end{proof} \paragraph{Step 3: rational points are fixed points of $\iota$.} \begin{lem}\label{lem:1.3.4-fix} For every $P\in C(\Q)$, $\iota(P)=P$. \end{lem} \begin{proof} $Q_0=(0:0:1)\in C(\Q)$ is a fixed point of $\iota$. For $P\in C(\Q)$, \[ \delta_P=[P-\iota(P)]=(1-\iota_*)[P-Q_0]\in P_C(\Q). \] By Lemma~\ref{lem:1.3.4-rank}, $\delta_P$ is a torsion element, and by Lemma~\ref{lem:1.3.4-tors}, $2\delta_P=0$, that is, $2P\sim2\iota(P)$. If $P\ne\iota(P)$, this contradicts \cite[Section IV.5]{Hartshorne77}. Hence $P=\iota(P)$. \end{proof} \begin{proof}[Proof of Proposition~\ref{prop:1.3.4}] By Lemma~\ref{lem:1.3.4-fix}, rational points are fixed points of $\iota$, and since $(1:0:0)\notin C$, we have $X=0$. Substituting $X=0$ gives \[ Y^4-YZ^3=Y(Y-Z)(Y^2+YZ+Z^2)=0. \] Since $Y^2+YZ+Z^2$ has discriminant $-3$, it has no rational root with $(Y,Z)\ne(0,0)$. The remaining points are $(0:0:1)$ and $(0:1:1)$, and both lie on $C$. \end{proof} \subsection{\texorpdfstring{$x^4-x^2y^2+y^3z+yz^3=0$ \ ($H=64$)}{x^4-x^2y^2+y^3z+yz^3=0 (H=64)}}\label{sec:1.3.5} \begin{thm}\label{thm:1.3.5} The only integer solutions of $x^4-x^2y^2+y^3z+yz^3=0$ are $(0,0,n)$, $(0,n,0)$, $(n,n,0)$, $(-n,n,0)$ $(n\in\Z)$. \end{thm} \paragraph{Setup.} The projective plane quartic curve \[ C:\quad F(X,Y,Z)=X^4-X^2Y^2+Y^3Z+YZ^3=0 \] is smooth modulo 3 (\code{1.3.5a}). Hence it is smooth over $\Q$ as well, has genus 3, and by \cite[Section IV.5]{Hartshorne77} is not hyperelliptic. Theorem~\ref{thm:1.3.5} is equivalent to the following proposition. \begin{prop}\label{prop:1.3.5} $C(\Q)=\{(0:0:1),(0:1:0),(1:1:0),(-1:1:0)\}$. \end{prop} \emph{Involution and Prym model.} Let $\iota(X:Y:Z)=(-X:Y:Z)$. In the chart $Y=1$, with $r=X/Y$, $t=Z/Y$, $R=r^2$, the curve $C$ is $r^4-r^2+t+t^3=0$, and the quotient curve $E=C/\iota$ is $R^2-R+t+t^3=0$ (an elliptic curve of conductor 91). Let $P_C=\operatorname{im}(1-\iota_*)\subset J_C$ be the Prym variety of the cover $C\to E$. $C$ has the form $X^4-hX^2+fg=0$ with $f=YZ$, $h=Y^2$, $g=Y^2+Z^2$. Applying the construction of \cite{Bruin08,RR18} to these data, we obtain three quadratic forms in $(a,b,c)$ \[ q_1=b^2+ac,\qquad q_2=ab,\qquad q_3=-b^2-ac+bc \] and the curve $D:\ q_1=u^2,\ q_2=uv,\ q_3=v^2$ in $\mathbf P^4$. The map \[ \pi:\ C\to D,\qquad (X:Y:Z)\mapsto(a:b:c:u:v)=(YZ:X^2:Y^2+Z^2:XY:XZ) \] satisfies $q_1-u^2=F$, $q_2-uv=0$, $q_3-v^2=-F$, so it indeed maps to $D$. On $X\ne0$, $\pi$ is an isomorphism onto its image, and the image lies in the smooth part $U$ of $D$ ($b=X^2\ne0$; \cite{Bruin08,RR18}). On $C$, $Y=0$ implies $X=0$, so the points with $X\ne0$ lie in the chart $Y=1$. Let $N_i$ be the $3\times3$ symmetric matrix of $q_i$, and let $M_i$ be the $5\times5$ symmetric matrices of the three equations $q_1-u^2$, $q_2-uv$, $q_3-v^2$ of $D$. The genus 2 curve is \[ \begin{aligned}H:\quad y^2&=-4\det(N_1-2sN_2+s^2N_3)\\&=-s^6-2s^5+3s^4+2s^3-3s^2+1\\&=-(s^2-1)(s^4+2s^3-2s^2+1).\end{aligned} \] (\code{1.3.5b}), and the prime factors of its discriminant are 2, 7, 13. $D_\infty$ is the sum of the two points at infinity of $H$ (a rational divisor). At a smooth point $\mathbf v$ of $D$ (viewed as a vector), the three rows $\mathbf v^{\mathsf T}M_i$ have rank 3, and their common kernel $W_{\mathbf v}$ is a 2-dimensional space containing $\mathbf v$. For $\mathbf w\in W_{\mathbf v}\setminus\Q\mathbf v$ we call \[ \tau_{\mathbf v}(s)=\mathbf w^{\mathsf T}(M_1-2sM_2+s^2M_3)\,\mathbf w \] the \emph{tangent quadratic} at $\mathbf v$. Since $\mathbf v^{\mathsf T}M_i\mathbf v=\mathbf v^{\mathsf T}M_i\mathbf w=0$, replacing $\mathbf w$ by $\alpha\mathbf w+\beta\mathbf v$ only multiplies $\tau_{\mathbf v}$ by $\alpha^2$. In the chart $Y=1$, at the image $\mathbf v_{\mathrm{gen}}=(t:R:1+t^2:r:rt)$ of the generic point of $C$, the tangent quadratic is proportional to $A_2s^2+Bs+A_0$, formed from \[ \begin{aligned} A_2&=-R^4+4R^3t-2R^3-R^2+12Rt^2-4Rt-4R-4t^2,\\ B&=2(3R^2t^2+7R^2+3Rt^2-5R+4t),\\ A_0&=-4R^3t+13R^2t-12Rt^2-Rt+4R+3t^2-1 \end{aligned} \] (\code{1.3.5b}). \emph{Kummer surface.} Let $\kappa:J_H\to K_H\subset\mathbf P^3$ be the Kummer map in Cassels--Flynn coordinates $(k_1:k_2:k_3:k_4)$. Then $\kappa(0)=(0:0:0:1)$, and $\kappa(x)=\kappa(y)$ is equivalent to $x=\pm y$. For an effective divisor $P_1+P_2$, $P_i=(x_i,y_i)$, that is not a hyperelliptic fiber, $\kappa([P_1+P_2-D_\infty])=(1:x_1+x_2:x_1x_2:\ast)$. $\Phi_1,\Phi_2\in\Z[k_1,\dots,k_4]$ are the two quartic forms given in \code{1.3.5b} (28 terms each). In the chart $k_4=1$, $\Phi_1$ has no constant or linear term, and \[ \Phi_1=512k_1k_2-992k_1k_3+248k_2^2+512k_2k_3+(\text{terms of degree 3 and 4}) \] (\code{1.3.5d}). The proof goes as follows. We construct a map $\varphi:C\to J_H$ over $\Q$ from the tangent quadratic (Lemma~\ref{lem:1.3.5-map}), determine $J_H(\Q)$ (Lemma~\ref{lem:1.3.5-mw}), and show $\varphi(C(\Q))\subset\{0,\pm4G\}$ by a sieve modulo 3 and 3-adic analysis (Lemma~\ref{lem:1.3.5-image}). Finally, we determine the two fibers (Lemmas~\ref{lem:1.3.5-fib4}, \ref{lem:1.3.5-fib0}). \paragraph{Step 1: the map $\varphi$.} \begin{lem}\label{lem:1.3.5-map} There is a map $\varphi:C\to J_H$ defined over $\Q$ such that the following hold. \begin{enumerate}[label=(\roman*)] \item For all $P\in C$, $\Phi_1(\kappa(\varphi(P)))=\Phi_2(\kappa(\varphi(P)))=0$. \item If $Y(P)\ne0$ and $(A_2,B,A_0)(P)\ne(0,0,0)$, then $\varphi(P)=[D_P-D_\infty]$. Here $D_P$ is an effective divisor of degree 2 on $H$, and its $x$-image is the zero set in $\mathbf P^1$ of $A_2(P)s^2+B(P)s+A_0(P)$. \item If $X(P)\ne0$ and $\varphi(P)=0$, then $2[P-\iota(P)]=0$. \end{enumerate} \end{lem} \begin{proof} Over the function field $\Q(C)=\Q(r,t)$, \code{1.3.5b} checks the following. The generic point $\mathbf v_{\mathrm{gen}}=(t:R:1+t^2:r:rt)$ lies on $D$, and $\dim W_{\mathbf v_{\mathrm{gen}}}=2$. The tangent quadratic has degree 2, is irreducible over $\Q(C)$, and is proportional to $A_2s^2+Bs+A_0$. There are exactly two points $\pm\varphi_{\mathrm{gen}}$ of $J_H(\Q(C))$ whose Mumford polynomial is the monic version of this quadratic, and their common Kummer point satisfies $\Phi_1=\Phi_2=0$ and $k_2A_2+k_1B=0$, $k_3A_2-k_1A_0=0$. $\varphi_{\mathrm{gen}}\in J_H(\Q(C))$ is a rational map $C\dashrightarrow J_H$ defined over $\Q$. Since $C$ is a smooth projective curve and $J_H$ is a projective variety, it extends to a morphism $\varphi:C\to J_H$. Likewise, the effective divisor of degree 2 determined by the Mumford representation of $\varphi_{\mathrm{gen}}$ is a $\Q(C)$-point of $\operatorname{Sym}^2H$, and it extends to a morphism $\psi:C\to\operatorname{Sym}^2H$. The two maps $\varphi$ and $P\mapsto[\psi(P)-D_\infty]$ agree at the generic point, hence everywhere. Put $D_P=\psi(P)$. (i) $\kappa\circ\varphi$ is a morphism, and $\Phi_i\circ\kappa\circ\varphi$ vanishes at the generic point. Since zero sets are closed, it vanishes on all of $C$. (ii) The composition of $\psi$ with $\operatorname{Sym}^2H\to\operatorname{Sym}^2\mathbf P^1\simeq\mathbf P^2$ (the coefficients of the quadratic form of the $x$-image) is a morphism $C\to\mathbf P^2$, and it equals $(A_2:B:A_0)$ at the generic point. In the chart $Y\ne0$, $A_2,B,A_0$ are polynomials in $R,t$, so at points where the three values are not all 0, the value of this morphism is $(A_2(P):B(P):A_0(P))$. (iii) By \cite{Bruin08,RR18}, on the smooth part $U$ of $D$ the map given by the tangent construction is $\lambda\circ d$. Here $d(Q)=[2\iota(Q)-2Q]$ is the map to the Prym scheme $P'$ of $D$ (the identity component of the kernel of the norm in $\operatorname{Pic}^0D$), and $\lambda:P'\to J_H$ is an isomorphism over $\Q$. At the generic point, the value of this map is the class whose Mumford polynomial is the tangent quadratic, so it is $\pm\varphi_{\mathrm{gen}}$. Hence $\varphi=\pm\lambda\circ d\circ\pi$ on $\pi^{-1}(U)\supset\{X\ne0\}$. Moreover, the pullback $\nu:P'\to P_C\subset J_C$ under the normalization $C\to D$ is an isogeny, and $\nu(d(\pi(P)))=2[\iota(P)-P]$. Therefore, if $X(P)\ne0$ and $\varphi(P)=0$, then $d(\pi(P))=0$ and $2[P-\iota(P)]=0$. \end{proof} \paragraph{Step 2: $J_H(\Q)$.} \begin{lem}\label{lem:1.3.5-mw} $J_H(\Q)=\langle T\rangle\oplus\langle G\rangle\simeq\Z/2\oplus\Z$, where, in Mumford representation, \[ T=(s^2-1,\ 0)=[(1,0)+(-1,0)-D_\infty],\qquad G=(s^2+s,\ -s-1)=[(0,-1)+(-1,0)-D_\infty], \] and $\kappa(4G)=(32:16:-12:73)$. \end{lem} \begin{proof} \code{1.3.5b} computes this with Magma's \texttt{MordellWeilGroupGenus2}. The rank upper bound from 2-descent (\cite{Stoll01,PS97,PS99}) is 1, the torsion subgroup is $\langle T\rangle$, and $G$ has infinite order. That $\langle T,G\rangle$ is saturated is proved using bounds for the canonical height (\cite{CF96,Stoll99,Stoll02}), and both proof flags are true. The same code also computes $\kappa(4G)$. \end{proof} \paragraph{Step 3: sieve modulo 3.} $H$ has good reduction at 3. The order of $\bar G$ in $J_H(\F_3)$ is 5. Put $S=5G$; then $S$ lies in the kernel $J_1$ of reduction modulo 3. \begin{lem}\label{lem:1.3.5-sieve} If $P\in C(\Q)$ and $\varphi(P)=eT+nG$ $(e\in\{0,1\},\ n\in\Z)$, then $e=0$ and $n\equiv0,1,4\pmod5$. Hence, for some $m\in\Z$, $\kappa(\varphi(P))=\kappa([m]S)$ or $\kappa(\varphi(P))=\kappa(4G+[m]S)$. \end{lem} \begin{proof} By Lemma~\ref{lem:1.3.5-map}(i), the primitive integer coordinates of $\kappa(\varphi(P))$ satisfy $\Phi_1=\Phi_2=0$, and so do their reductions modulo 3. Reduction modulo 3 commutes with the Kummer map (\cite{BS10}), so the Kummer point of $\overline{\varphi(P)}=e\bar T+n\bar G$ must satisfy $\Phi_1\equiv\Phi_2\equiv0\pmod3$. Among the 10 classes $e\bar T+n\bar G$ $(e\in\{0,1\},\ 0\le n<5)$ of $J_H(\F_3)$, only $(e,n)=(0,0),(0,1),(0,4)$ satisfy this (\code{1.3.5c}). If $n=5m$, then $\varphi(P)=[m]S$; if $n=4+5m$, then $\varphi(P)=4G+[m]S$. If $n=1+5m'$, then $-\varphi(P)=4G+[-1-m']S$, and $\kappa(-x)=\kappa(x)$. \end{proof} \paragraph{Step 4: 3-adic analysis.} $J_H$ extends to an abelian scheme over $\Z_3$, and $J_1$ is identified with $(3\Z_3)^2$ through the formal group $\hat F$ with integral formal parameter $z=(z_1,z_2)$. By Lemma~\ref{lem:O}, the formal logarithm $L$ and its inverse $\mathrm{Ex}$ are mutually inverse bijections of $(3\Z_3)^2$ that preserve valuations. Moreover, $L([m]Q)=mL(Q)$ $(m\in\Z_3)$, and $\partial^\alpha\mathrm{Ex}(0)\in\Z_3^2$ for every multi-index $\alpha$. The Kummer map also extends over $\Z_3$ (\cite{BS10}). So, if $B\in J_H(\Q)$ and $k_j$ is a 3-adic unit in the primitive integer coordinates of $\kappa(B)$, then on the disk $B+J_1$ the function $Q\mapsto(k_i/k_j)(\kappa(B+Q))$ is a power series in $z(Q)$ with coefficients in $\Z_3$. This is because translation by $B$ is an automorphism of the abelian scheme over $\Z_3$. The next two lemmas hold for any odd prime $p$ and any abelian scheme of dimension $g$ over $\Z_p$. We use the same symbols $z=(z_1,\dots,z_g)$, $J_1$, $\hat F$, $L$, $\mathrm{Ex}$. \begin{lem}[Formal translation]\label{lem:1.3.5-mahler} Let $S\in J_1(\Q_p)$, $\mathcal F\in\Z_p[[z]]$, and $f(m)=\mathcal F(z([m]S))$ $(m\in\Z_p)$. Put $a_j=\sum_{k=0}^{j}(-1)^{j-k}\binom jk f(k)$. Then $v_p(a_j)\ge j$, and $f(m)=\sum_{j\ge0}a_j\binom mj$ for all $m\in\Z_p$. \end{lem} \begin{proof} We have $z_S=z(S)\in(p\Z_p)^g$. In the formal group law $\hat F(z,w)$, all terms other than $z$ have $w$ as a factor, so $\hat F(z,z_S)=z+p\,h(z)$ with $h\in\Z_p[[z]]^g$. For $\mathcal G\in\Z_p[[z]]$ put $\tau\mathcal G(z)=\mathcal G(\hat F(z,z_S))$. In the Taylor expansion \[ \tau\mathcal G-\mathcal G=\sum_{\beta\ne0}\frac{\partial^\beta\mathcal G}{\beta!}(z)\,p^{|\beta|}h(z)^\beta \] we have $\partial^\beta\mathcal G/\beta!\in\Z_p[[z]]$, so $\tau\mathcal G-\mathcal G\in p\,\Z_p[[z]]$. Iterating gives $(\tau-1)^j\mathcal F\in p^j\Z_p[[z]]$. Since $z([k]S)=\hat F(z([k-1]S),z_S)$, we have $(\tau^k\mathcal F)(0)=f(k)$, and hence $a_j=((\tau-1)^j\mathcal F)(0)\in p^j\Z_p$. The series $\sum a_j\binom mj$ converges in $\Z_p$ and equals $f(m)$ for $m\in\Z_{\ge0}$ (Newton's forward difference formula). Both sides are continuous in $m$ ($[p^k]S\to0$), so they agree on all of $\Z_p$. \end{proof} \begin{lem}[Logarithmic coordinates]\label{lem:1.3.5-log} Suppose $\kappa\in\Z_p[[z]]$ satisfies $\kappa(0)=0$ and $\kappa\circ[-1]=\kappa$. Then $\kappa(\mathrm{Ex}(\ell))=\sum_{n\ge2,\ n\text{ even}}\kappa_n(\ell)$, where $\kappa_n$ is homogeneous of degree $n$ and its coefficients have valuation at least $-(n-1)/(p-1)$. Hence, if $Q\in J_1(\Q_p)$, $\ell_Q=L(z(Q))$, $m\in\Z_p$ and $e=v_p(m)$, then \[ \kappa([m]Q)=m^2\kappa_2(\ell_Q)+\varepsilon,\qquad v_p(\varepsilon)\ge4(e+1)-\frac{3}{p-1}. \] \end{lem} \begin{proof} Since $L\circ[-1]=-L$, we have $\mathrm{Ex}(-\ell)=[-1]\mathrm{Ex}(\ell)$, so $\kappa\circ\mathrm{Ex}$ is an even function. Since $\kappa(0)=0$, it has no constant term. By the chain rule and $\partial^\alpha\mathrm{Ex}(0)\in\Z_p^g$, we have $\partial^\alpha(\kappa\circ\mathrm{Ex})(0)\in\Z_p$. Therefore the coefficient of $\ell^\alpha$ lies in $\Z_p/\alpha!$, and $v_p(\alpha!)\le(|\alpha|-1)/(p-1)$. Since $v_p(\ell_Q)\ge1$, the valuation of $L([m]Q)=m\ell_Q$ is at least $e+1$, and $z([m]Q)=\mathrm{Ex}(m\ell_Q)$. The term of degree $n$, $m^n\kappa_n(\ell_Q)$, has valuation at least $n(e+1)-(n-1)/(p-1)$. This bound increases with $n$, so all terms with $n\ge4$ have valuation at least $4(e+1)-3/(p-1)$. \end{proof} Now we use $\kappa_i=k_i/k_4$ $(i=1,2,3)$ on the disk of the origin. These functions are $[-1]$-invariant, vanish at the origin, and lie in $\Z_3[[z]]$. \begin{lem}\label{lem:1.3.5-id} If $m\in\Z$ and $\Phi_1(\kappa([m]S))=0$, then $m=0$. \end{lem} \begin{proof} Put $f(m)=\Phi_1(\kappa_1,\kappa_2,\kappa_3,1)([m]S)$. Since $\Phi_1$ has integer coefficients, Lemma~\ref{lem:1.3.5-mahler} applies. The case $3\nmid m$. By \code{1.3.5d}, the Mahler coefficients are $a_0=0$, $v_3(a_1)=v_3(a_2)=5$, $v_3(a_3)=7$, $v_3(a_4)=8$, $v_3(a_5)=7$, with $a_1/3^5\equiv1$ and $a_2/3^5\equiv2\pmod3$. For $j\ge6$, $v_3(a_j)\ge j\ge6$. Hence \[ 3^{-5}f(m)\equiv\binom m1+2\binom m2=m^2\pmod3, \] and $f(m)\ne0$ if $3\nmid m$. The case $m=3^ea$, $e\ge1$, $3\nmid a$. We apply Lemma~\ref{lem:1.3.5-log} with $p=3$ to the $\kappa_i$, and write $\kappa_{i,2}$ for the quadratic part of $\kappa_i\circ\mathrm{Ex}$. The error has valuation at least $4e+\tfrac52$, and hence, since valuations are integers, at least $4e+3$. By \code{1.3.5d}, the valuations of $\kappa_1(3S),\kappa_2(3S),\kappa_3(3S)$ are $6,5,4$, and dividing them by $3^6,3^5,3^4$, respectively, gives $1$ modulo 3. For $m=3$ the error is a multiple of $3^7$, so $v_3(\kappa_{i,2}(\ell_S))=4,3,2$ $(i=1,2,3)$, and the normalized residues are all 1. For general $e\ge1$ we have $\kappa_i([m]S)=3^{2e}a^2\kappa_{i,2}(\ell_S)+O(3^{4e+3})$, and since $4e+3>2e+4$, \[ v_3(\kappa_1)=2e+4,\quad v_3(\kappa_2)=2e+3,\quad v_3(\kappa_3)=2e+2, \] with normalized residues $a^2\equiv1$. The numbers $512=2^9$, $992=2^5\cdot31$, $248=2^3\cdot31$ are 3-adic units. In the quadratic part of $\Phi_1$, the valuation of $512\kappa_2\kappa_3$ is exactly $4e+5$, those of $992\kappa_1\kappa_3$ and $248\kappa_2^2$ are $4e+6$, and that of $512\kappa_1\kappa_2$ is $4e+7$. The terms of degree 3 and higher have valuation at least $3(2e+2)>4e+5$. Hence $v_3(f(m))=4e+5$ and $f(m)\ne0$. \end{proof} \begin{lem}\label{lem:1.3.5-4G} If $m\in\Z$ and $\Phi_1(\kappa(4G+[m]S))=0$, then $m=0$. \end{lem} \begin{proof} In $\kappa(4G)=(32:16:-12:73)$ the coordinate $32$ is a 3-adic unit, so we use the chart $k_1=1$ on the whole disk $4G+J_1$. Then Lemma~\ref{lem:1.3.5-mahler} applies to $h(m)=\Phi_1(1,k_2/k_1,k_3/k_1,k_4/k_1)(\kappa(4G+[m]S))$. By \code{1.3.5d}, the Mahler coefficients satisfy \[ a_0=0,\qquad 3^{-2}a_1\equiv2,\qquad 3^{-2}a_2\equiv2,\qquad 3^{-3}a_3\equiv1\pmod3, \] and $v_3(a_j)\ge j$ for $j\ge3$. The Mahler coefficients $b_j=a_j/9$ of $g=h/9$ satisfy $b_1\equiv b_2\equiv2\pmod3$ and $v_3(b_j)\ge j-2$ $(j\ge3)$. (a) Modulo 3, $g(m)\equiv2\binom m1+2\binom m2=m(m+1)$, so $g(m)\ne0$ if $m\equiv1\pmod3$. (b) Suppose $m\equiv2\pmod3$. Choose $r\in\{2,5,8\}$ with $r\equiv m\pmod9$. By \code{1.3.5d}, $v_3(h(r))=3$, that is, $v_3(g(r))=1$. By the Vandermonde identity, \[ g(m)-g(r)=\sum_{j\ge1}b_j\sum_{i=1}^{j}\binom{m-r}{i}\binom{r}{j-i}, \] and $v_3\binom{m-r}{i}\ge2-v_3(i)$. If $j\le2$, then $v_3(i)=0$, and if $j\ge3$, then $v_3(b_j)\ge j-2\ge v_3(i)$. Hence $g(m)\equiv g(r)\not\equiv0\pmod9$. (c) Suppose $m\equiv0\pmod3$, $m\ne0$, and $e=v_3(m)$. Then \[ \frac{h(m)}{9m}=\frac{a_1}{9}+\frac{a_2}{9}\cdot\frac{m-1}{2}+\frac{a_3}{27}\cdot\frac{(m-1)(m-2)}{2}+\frac1{9m}\sum_{j\ge4}a_j\binom mj. \] Since $\binom mj=\frac mj\binom{m-1}{j-1}$, for $j\ge4$ we have $v_3(a_j\binom mj)\ge j+e-v_3(j)\ge e+3$, so the tail is 0 modulo 3. If $m\equiv0$, then $(m-1)/2\equiv1$ and $(m-1)(m-2)/2\equiv1\pmod3$, so $h(m)/(9m)\equiv2+2+1\equiv2\pmod3$. Hence $h(m)\ne0$. \end{proof} \begin{lem}\label{lem:1.3.5-image} For every $P\in C(\Q)$, $\varphi(P)\in\{0,\,4G,\,-4G\}$. \end{lem} \begin{proof} Applying Lemma~\ref{lem:1.3.5-map}(i) and Lemmas~\ref{lem:1.3.5-id}, \ref{lem:1.3.5-4G} to the two cases of Lemma~\ref{lem:1.3.5-sieve} gives $m=0$. Hence $\kappa(\varphi(P))$ is $\kappa(0)$ or $\kappa(4G)$, and $\varphi(P)=0$ or $\pm4G$. \end{proof} \paragraph{Step 5: the two fibers.} \begin{lem}\label{lem:1.3.5-fib4} If $P\in C(\Q)$, $Y(P)\ne0$ and $\varphi(P)=\pm4G$, then $P=(\pm1:1:0)$. \end{lem} \begin{proof} In the chart $Y\ne0$ of $C$, the only point with $A_2=B=0$ is $(R,t)=(0,0)$, and there $A_0=-1$ (\code{1.3.5e}). Therefore, by Lemma~\ref{lem:1.3.5-map}(ii), $\varphi(P)=[D_P-D_\infty]$, and the $x$-image of $D_P$ is the zero set of $A_2(P)s^2+B(P)s+A_0(P)$. Since the class $\pm4G$ is nonzero, by Riemann--Roch there is only one effective divisor in the class $\pm4G+[D_\infty]$. Since $\kappa(4G)=(32:16:-12:73)$, the two $x$-coordinates of that divisor satisfy $x_1+x_2=\tfrac12$, $x_1x_2=-\tfrac38$. That is, the $x$-image is the zero set of $8s^2-4s-3$, and the same holds for $-4G$. Hence $(A_2:B:A_0)(P)=(8:-4:-3)$, that is, \[ 2B+A_2=0,\qquad 8A_0+3A_2=0 \] hold at $(R,t)=(r^2,t)$. The Gröbner basis of the ideal $(R^2-R+t+t^3,\ 2B+A_2,\ 8A_0+3A_2)$ of $\Q[R,t]$ is $\{R-1,\ t\}$ (\code{1.3.5e}). Therefore $R=1$, $t=0$, $r=\pm1$, and $P=(\pm1:1:0)$. \end{proof} \begin{lem}\label{lem:1.3.5-fib0} If $P\in C(\Q)$ and $\varphi(P)=0$, then $X(P)=0$. \end{lem} \begin{proof} Suppose $X(P)\ne0$. By Lemma~\ref{lem:1.3.5-map}(iii), $2P\sim2\iota(P)$. On the other hand, $P\ne\iota(P)$, because the only fixed points of $\iota$ are the line $X=0$ and $(1:0:0)\notin C$. This contradicts \cite[Section IV.5]{Hartshorne77}. \end{proof} \begin{proof}[Proof of Proposition~\ref{prop:1.3.5}] Let $P\in C(\Q)$. If $X(P)\ne0$, then $Y(P)\ne0$, and by Lemma~\ref{lem:1.3.5-fib0}, $\varphi(P)\ne0$. By Lemma~\ref{lem:1.3.5-image}, $\varphi(P)=\pm4G$, and by Lemma~\ref{lem:1.3.5-fib4}, $P=(\pm1:1:0)$. If $X(P)=0$, then $YZ(Y^2+Z^2)=0$, so $P=(0:0:1)$ or $(0:1:0)$. All four points lie on $C$. \end{proof} \subsection{\texorpdfstring{$x^4+x^2y^2+y^3z+z^4=0$ \ ($H=64$)}{x^4+x^2y^2+y^3z+z^4=0 (H=64)}}\label{sec:1.3.6} \begin{thm}\label{thm:1.3.6} The only integer solutions of $x^4+x^2y^2+y^3z+z^4=0$ are $(0,n,0)$ and $(0,n,-n)$ $(n\in\Z)$. \end{thm} \paragraph{Setup.} The projective plane quartic curve \[ C:\quad F(X,Y,Z)=X^4+X^2Y^2+Y^3Z+Z^4=0 \] has good reduction at $5,7,11,13,47$ (\code{1.3.6a}). Hence it is smooth over $\Q$, has genus 3, and by \cite[Section IV.5]{Hartshorne77} is not hyperelliptic. Theorem~\ref{thm:1.3.6} is equivalent to the following proposition. \begin{prop}\label{prop:1.3.6} $C(\Q)=\{(0:1:0),(0:1:-1)\}$. \end{prop} \emph{Involution and Prym model.} The quotient curve $E=C/\iota$ by $\iota(X:Y:Z)=(-X:Y:Z)$ is $w^2=Y^4-4Y^3Z-4Z^4$, and it is isomorphic to the elliptic curve $y^2=x^3+16x-64$ (conductor 248). Let $P_C=\operatorname{im}(1-\iota_*)\subset J_C$ be the Prym variety of the cover $C\to E$. $C$ has the form $X^4-hX^2+fg=0$ with $f=Z(Y+Z)$, $h=-Y^2$, $g=Y^2-YZ+Z^2$. Applying the construction of \cite{Bruin08,RR18}, we obtain three quadratic forms in $(a,b,c)$ \[ q_1=-(b^2+ac),\qquad q_2=\tfrac12(ab-b^2-ac-bc),\qquad q_3=\tfrac12(ab+b^2+ac+bc) \] and the curve $D:\ q_1=u^2,\ q_2=uv,\ q_3=v^2$ ($\subset\mathbf P^4$). The map \[ \pi:\ C\to D,\qquad (X:Y:Z)\mapsto(a:b:c:u:v)=\bigl(Z(Y+Z):X^2:Y^2-YZ+Z^2:XY:XZ\bigr) \] satisfies $q_1-u^2=-F$, $q_2-uv=-\tfrac12F$, $q_3-v^2=\tfrac12F$, so it maps to $D$. On $X\ne0$, $\pi$ is an isomorphism onto its image, and the image lies in the smooth part $U$ of $D$ ($b=X^2\ne0$; \cite{Bruin08,RR18}). With $N_i$ the $3\times3$ symmetric matrix of $q_i$, the genus 2 curve is \[ \begin{aligned}H:\quad y^2&=-16\det(N_1-2sN_2+s^2N_3)\\&=2s^5+6s^4-4s^3-10s^2+12s-4\\&=2(s^2+2s-2)(s^3+s^2-2s+1).\end{aligned} \] (\code{1.3.6a}). The Ritzenthaler--Romagny model $y^2=-\tfrac12(s^2-2s-2)(s^3-s^2-2s-1)$ of \cite{Bruin08,RR18} becomes $H$ under $s\mapsto-s$, $y\mapsto y/2$. The prime factors of the discriminant of $H$ are 2, 3, 31. $H$ has one point at infinity $\infty$, and we put $D_\infty=2\infty$. \code{1.3.6a} also checks $L_p(C,T)=L_p(E,T)\,L_p(H,T)$ at $p=5,7,11,13,47$. The tangent quadratic is defined as in Section~\ref{sec:1.3.5}, using the $5\times5$ symmetric matrices $M_i$ of the three equations of $D$. \emph{Kummer surface.} $\kappa:J_H\to K_H\subset\mathbf P^3$ is the Kummer map in Cassels--Flynn coordinates $(k_1:k_2:k_3:k_4)$, and $\kappa(0)=(0:0:0:1)$. $\Phi_1,\Phi_2\in\Z[k_1,\dots,k_4]$ are the two primitive quartic forms printed by \code{1.3.6a} (28 terms each). In the chart $k_4=1$, neither has a constant or linear term, and their quadratic parts are \[ Q_1=48k_1^2-4k_1k_3+k_2^2+24k_2k_3,\qquad Q_2=-48k_1^2+240k_1k_3-60k_2^2-24k_2k_3 \] (\code{1.3.6a}). The proof goes as in Section~\ref{sec:1.3.5}. We construct a map $\varphi:C\to J_H$, determine $J_H(\Q)$, show $\varphi(C(\Q))=\{0\}$ by a sieve modulo 47 and 47-adic analysis, and then determine the zero fiber. \paragraph{Step 1: the map $\varphi$.} \begin{lem}\label{lem:1.3.6-map} There is a map $\varphi:C\to J_H$ defined over $\Q$ such that the following hold. \begin{enumerate}[label=(\roman*)] \item For all $P\in C$, $\Phi_1(\kappa(\varphi(P)))=\Phi_2(\kappa(\varphi(P)))=0$. \item If $X(P)\ne0$ and $\varphi(P)=0$, then $2[P-\iota(P)]=0$. \end{enumerate} \end{lem} \begin{proof} Over the function field $\Q(C)$, \code{1.3.6a} checks the following. At the image $\pi(X:Y:Z)$ of the generic point, the tangent quadratic has degree 2 and is irreducible. There are exactly two points $\pm\varphi_{\mathrm{gen}}$ of $J_H(\Q(C))$ whose Mumford polynomial is its monic version, and their common Kummer point satisfies $\Phi_1=\Phi_2=0$. The rest is as in the proof of Lemma~\ref{lem:1.3.5-map}. $\varphi_{\mathrm{gen}}$ extends to a morphism $\varphi:C\to J_H$, and (i) holds because a closed condition holds at the generic point. (ii) follows from \cite{Bruin08,RR18}. On the smooth part of $D$, the map given by the tangent construction is $\lambda\circ d$, where $d(Q)=[2\iota(Q)-2Q]$ is the map to the Prym scheme $P'$ and $\lambda:P'\to J_H$ is an isomorphism. Therefore $\varphi=\pm\lambda\circ d\circ\pi$ on $\{X\ne0\}$. The isogeny $\nu:P'\to P_C$ given by the normalization sends $d(\pi(P))$ to $2[\iota(P)-P]$. \end{proof} \paragraph{Step 2: $J_H(\Q)$.} \begin{lem}\label{lem:1.3.6-mw} $J_H(\Q)=\langle T\rangle\oplus\langle G\rangle\simeq\Z/2\oplus\Z$, where, in Mumford representation, \[ T=(s^2+2s-2,\ 0)=[(\alpha,0)+(\bar\alpha,0)-2\infty],\qquad G=(s+\tfrac52,\ -\tfrac94)=[(-\tfrac52,-\tfrac94)-\infty], \] and $\alpha,\bar\alpha$ are the two roots of $s^2+2s-2$. \end{lem} \begin{proof} \code{1.3.6a} computes this with Magma. The rank upper bound from 2-descent (\cite{Stoll01,PS97,PS99}) is 1, and $G$ has infinite order, so the rank is 1. The torsion subgroup is $\langle T\rangle\simeq\Z/2$. \texttt{MordellWeilGroupGenus2} proves, using bounds for the canonical height (\cite{CF96,Stoll99,Stoll02}), that $\langle T,G\rangle$ is all of $J_H(\Q)$, and both proof flags are true. \end{proof} \paragraph{Step 3: sieve modulo 47.} $H$ and $C$ have good reduction at 47. In $J_H(\F_{47})$, $\bar T$ has order 2 and $\bar G$ has order 28. Put $R=28G$; then $R$ lies in the kernel $J_1$ of reduction modulo 47. \begin{lem}\label{lem:1.3.6-sieve} For every $P\in C(\Q)$ there is $m\in\Z$ such that $\varphi(P)=[m]R$. \end{lem} \begin{proof} Write $\varphi(P)=eT+nG$ $(e\in\{0,1\},\ n\in\Z)$. By Lemma~\ref{lem:1.3.6-map}(i), the primitive integer coordinates of $\kappa(\varphi(P))$ satisfy $\Phi_1=\Phi_2=0$, and reduction modulo 47 commutes with the Kummer map (\cite{BS10}). Among the 56 classes $e\bar T+n\bar G$ $(e\in\{0,1\},\ 0\le n<28)$ of $J_H(\F_{47})$, the only one whose Kummer point satisfies $\Phi_1=\Phi_2=0$ modulo 47 is $(e,n)=(0,0)$ (\code{1.3.6a}). Hence $e=0$ and $28\mid n$. \end{proof} \paragraph{Step 4: 47-adic analysis.} $J_H$ extends to an abelian scheme over $\Z_{47}$, and $\kappa_i=k_i/k_4$ $(i=1,2,3)$ are power series with coefficients in $\Z_{47}$ in the integral formal parameters on the disk $J_1$ of the origin (\cite{BS10}). These functions are $[-1]$-invariant and vanish at the origin. By \code{1.3.6a}, the 47-adic valuations of the primitive integer coordinates of $\kappa(R)$ are $(2,2,2,0)$, and \[ \kappa_i(R)/47^2\equiv(24,\,7,\,1)_i\pmod{47},\qquad Q_1(24,7,1)\equiv39,\quad Q_2(24,7,1)\equiv8\pmod{47}. \] \begin{lem}\label{lem:1.3.6-padic} If $m\in\Z$ and $m\ne0$, then $\Phi_1(\kappa([m]R))\ne0$. More precisely, $v_{47}\bigl(\Phi_1(\kappa_1,\kappa_2,\kappa_3,1)([m]R)\bigr)=4v_{47}(m)+4$. \end{lem} \begin{proof} We apply Lemma~\ref{lem:1.3.5-log} with $p=47$ to the $\kappa_i$, and let $\kappa_{i,2}$ be the quadratic part of $\kappa_i\circ\mathrm{Ex}$. Writing $e=v_{47}(m)$ and $m=47^eu$, the error has valuation at least $4(e+1)-\tfrac3{46}$, that is, at least $4e+4$. For $m=1$ we have $\kappa_i(R)=\kappa_{i,2}(\ell_R)+O(47^4)$, so $\kappa_{i,2}(\ell_R)=47^2(d_i+O(47^2))$ with $(d_1,d_2,d_3)\equiv(24,7,1)\pmod{47}$. For general $m$, \[ \kappa_i([m]R)=m^2\kappa_{i,2}(\ell_R)+O(47^{4e+4})=47^{2e+2}\bigl(u^2d_i+O(47)\bigr). \] Since $\Phi_1$ has integer coefficients, \[ \Phi_1(\kappa_1,\kappa_2,\kappa_3,1)([m]R)=47^{4e+4}\bigl(u^4Q_1(d)+O(47)\bigr)+O(47^{6e+6}). \] Since $Q_1(d)\equiv39\not\equiv0\pmod{47}$ and $6e+6>4e+4$, the valuation is exactly $4e+4$. \end{proof} \paragraph{Step 5: the zero fiber.} \begin{lem}\label{lem:1.3.6-fix} For every $P\in C(\Q)$, $X(P)=0$. \end{lem} \begin{proof} By Lemma~\ref{lem:1.3.6-sieve}, $\varphi(P)=[m]R$, and by Lemma~\ref{lem:1.3.6-map}(i) and Lemma~\ref{lem:1.3.6-padic}, $m=0$, that is, $\varphi(P)=0$. Suppose $X(P)\ne0$. By Lemma~\ref{lem:1.3.6-map}(ii), $2P\sim2\iota(P)$. The only fixed points of $\iota$ are the line $X=0$ and $(1:0:0)\notin C$, so $P\ne\iota(P)$, and this contradicts \cite[Section IV.5]{Hartshorne77}. \end{proof} \begin{proof}[Proof of Proposition~\ref{prop:1.3.6}] By Lemma~\ref{lem:1.3.6-fix}, $X=0$. Substituting gives \[ Y^3Z+Z^4=Z(Y+Z)(Y^2-YZ+Z^2)=0. \] Since $Y^2-YZ+Z^2$ has discriminant $-3$, it has no rational root with $(Y,Z)\ne(0,0)$. The remaining points are $(0:1:0)$ and $(0:1:-1)$, and both lie on $C$. \end{proof} \subsection{\texorpdfstring{$x^3z+y^3x+y^2z^2-z^3y=0$ \ ($H=64$)}{x^3z+y^3x+y^2z^2-z^3y=0 (H=64)}}\label{sec:1.3.7} \begin{thm}\label{thm:1.3.7} The only integer solutions of $x^3z+y^3x+y^2z^2-z^3y=0$ are $(n,0,0)$, $(0,n,0)$, $(0,0,n)$, $(0,n,n)$ $(n\in\Z)$. \end{thm} \paragraph{Setup.} Let $C\subset\mathbb P^2$ be the curve defined by the equation in the theorem, and put $J=\Jac(C)$. Denote the following rational points by \[ P_1=(1:0:0),\quad P_2=(0:1:0),\quad P_3=(0:0:1),\quad P_4=(0:1:1), \] and set $G=[P_3-P_1]$. Write $F=x^3z+y^3x+y^2z^2-z^3y$. The proof runs as follows. $C$ is a smooth plane quartic with bad primes $2,11,41269$, $J(\Q)$ is torsion-free and $G$ has infinite order (Step~1). The 28 bitangents of $C$ are conjugate; their field of definition $K$ has degree 28, and their Galois group is $\operatorname{Sp}_6(\F_2)$ (Step~2). The descent map of Proposition~\ref{thm:AB} by the bitangents is then injective, with image in $K(2,S)$ modulo $\Q^\times$ (Step~3). The class number of $K$ is odd, proved without GRH with Proposition~\ref{thm:B} (Step~4), so $K(2,S)$ is spanned by 29 explicit $S$-units (Step~5). The local conditions at 2 and 11 cut the image down to dimension 1, so $\operatorname{rank}J(\Q)=1$ (Step~6), and Chabauty--Coleman at $p=3$ shows that each of the four residue disks of $C(\F_3)$ contains only one rational point (Step~7). Proposition~\ref{thm:A} does not apply here: no bitangent of $C$ is rational, so $C$ has no rational hyperflex. \paragraph{Step 1: the curve.} \begin{lem}\label{lem:1.3.7-bad} $C$ is smooth over $\Q$, so it has genus 3. The plane model $\mathcal C=\{F=0\}\subset\mathbb P^2_\Z$ is smooth over $\Z_p$ for every prime $p\notin\{2,11,41269\}$. For $p\in\{2,11,41269\}$ the reduction has exactly one singular point over $\overline\F_p$; it is $\F_p$-rational and an ordinary double point, and $v_p(F)=1$ at an integral lift (\code{1.3.7a}): \[ \begin{array}{c|c|c} p&\text{node}&\text{type}\\\hline 2&(1:1:1)&\text{split}\\ 11&(1:1:7)&\text{non-split}\\ 41269&(1:12101:2748)&\text{split} \end{array} \] For $p=11,41269$ the model $\mathcal C_{\Z_p}$ is regular with geometrically irreducible special fibre, and the Tamagawa number $c_p(J)$ is 1; at every prime $p\notin\{2,11,41269\}$ also $c_p(J)=1$. \end{lem} \begin{proof} In each of the three affine charts let $R_1,R_2,R_3$ be the resultants, with respect to one variable, of the pairs $(f,f_x)$, $(f,f_y)$, $(f_x,f_y)$ of the dehomogenized polynomial and its partial derivatives, and take the three pairwise resultants of the $R_i$ with respect to the other variable. In each chart some of these integers are 0 (in the charts $z=1$ and $x=1$, $R_1$ and $R_2$ have the common factor $x$, resp.\ $y^2$), but their greatest common divisor is $2\cdot11\cdot41269$. A singular point over $\overline\F_p$ in some chart makes $p$ divide all three integers of that chart, and a singular point over $\overline\Q$ would make them all 0. At $p=2,11,41269$ the singular points were found from the gcds of $f,f_x,f_y$; there is one, with nonzero discriminant of the quadratic part (\code{1.3.7a}). At $p=11,41269$, the linear part of $F$ at an integral lift of the node vanishes modulo $p$ and $F(\text{lift})\notin p^2\Z_p$, so $F\notin\mathfrak m^2$ and the model is regular there. The special fibre is reduced, and it is geometrically irreducible: two components $C',C''$ would meet only at the node, with $i(C',C'')=\deg C'\deg C''\ge3$ by Bézout, whereas each contains exactly one of the two branches there, which meet with multiplicity 1. By Raynaud's theorem \cite[9.6, Theorem~1]{BLR90} the component group of the Néron model is trivial. \end{proof} \begin{lem}\label{lem:1.3.7-points} (1) The points of $C$ with $xyz=0$ are $P_1,\dots,P_4$: $F(0,y,z)=yz^2(y-z)$, $F(x,0,z)=x^3z$, $F(x,y,0)=xy^3$. (2) $\operatorname{div}(z)=3P_1+P_2$, $\operatorname{div}(y)=P_1+3P_3$, $\operatorname{div}(x)=2P_2+P_3+P_4$. Hence $[P_2-P_1]=3G$ and $[P_4-P_1]=-4G$. (3) The numbers $N_k=\#C(\F_{p^k})$, the $L$-polynomials and $\#J(\F_p)=L_p(1)$ are (\code{1.3.7a}) \[ \begin{array}{c|c|l|r} p&(N_1,N_2,N_3)&L_p(T)&\#J(\F_p)\\\hline 3&(4,10,37)&1+3T^3+27T^6&31\\ 5&(6,26,171)&1+15T^3+125T^6&141=3\cdot47\\ 7&(10,66,313)&1+2T+10T^2+7T^3+70T^4+98T^5+343T^6&531=3^2\cdot59\\ 17&(30,308,4884)&1+12T+81T^2+386T^3+1377T^4+3468T^5+4913T^6&10238=2\cdot5119 \end{array} \] (4) $J(\Q)_{\mathrm{tors}}=0$, and $G$ has infinite order. \end{lem} \begin{proof} (2) $\operatorname{div}(z/y)=2P_1+P_2-3P_3$ and $\operatorname{div}(x/y)=2P_2+P_4-P_1-2P_3$. (3) With $s_k=p^k+1-N_k$, $L_p(T)=1+a_1T+a_2T^2+a_3T^3+pa_2T^4+p^2a_1T^5+p^3T^6$ with $a_1=-s_1$, $a_2=(s_1^2-s_2)/2$, $a_3=-(s_1^3-3s_1s_2+2s_3)/6$. (4) Since $\gcd(141,10238)=1$, \cite[Appendix]{Katz81} gives $J(\Q)_{\mathrm{tors}}=0$. Since $P_3\ne P_1$ and the genus is positive, $G\ne0$, so $G$ has infinite order. \end{proof} \paragraph{Step 2: the bitangents and the field $K$.} Let \begin{align*} B(m)&=m^{28}-208m^{26}-1024m^{25}+9344m^{24}-33792m^{22}-83636m^{21}+32256m^{20}+131816m^{19}\\ &\quad-44352m^{17}-23872m^{15}+87462m^{14}+28512m^{12}+5968m^{10}-3828m^{7}-504m^{5}+1,\\ g(w)&=w^{28}-288w^{21}-1028w^{19}-1440w^{17}-960w^{15}-346w^{14}-256w^{13}-748w^{12}\\ &\quad-528w^{10}-128w^{8}-56w^{7}-68w^{5}-16w^{3}+1, \end{align*} and $K=\Q(w)$ with $g(w)=0$. Put $n=-1/w^2$, and let $a\in K$ be the degree-27 polynomial in $w$ recorded in \texttt{data/a.txt} of \code{1.3.7b}. \begin{lem}\label{lem:1.3.7-bit} The following hold (\code{1.3.7b}). \begin{enumerate}[label=(\arabic*)] \item The line $y=mx+nz$ is a bitangent of $C$ only if $B(m)=0$. $B$ is irreducible over $\Q$ and squarefree. \item $B(a)=0$, so $K=\Q(a)$, and $F(x,ax+nz,z)=a^3(x^2+uxz+vz^2)^2$ for some $u,v\in K$. So $\ell=y-ax-nz$ defines a bitangent, and its 28 conjugates are the 28 bitangents of $C$. \item The maximal order of $K$ is certified without GRH; $d_K=2^{42}\cdot11^6\cdot41269^6$, and $K$ has signature $(4,12)$. $a$ is a unit, and $-n=w^{-2}$ is a square. \item The prime ideals of $K$ above $2$, $11$, $41269$ have $(e,f)$ \[ \begin{gathered} 2:\ (4,1),(2,3),(2,3),(4,3);\qquad 11:\ (1,1),(1,3),(1,6),(1,6),(2,6);\\ 41269:\ (2,2),(2,4),(1,8),(1,8). \end{gathered} \] Let $S$ be the set of these 13 prime ideals. \end{enumerate} \end{lem} \begin{proof} Substituting $y=mx+nz$ gives $F(x,mx+nz,z)=\sum_kc_kx^kz^{4-k}$ with $c_4=m^3$, $c_3=1+3m^2n$, $c_2=m^2+3mn^2$, $c_1=n^3+(2n-1)m$, $c_0=n^2-n$. This is $c_4(x^2+uxz+vz^2)^2$ if and only if $u=c_3/(2c_4)$, $v=(4c_2c_4-c_3^2)/(8c_4^2)$, $8c_1c_4^2=c_3(4c_2c_4-c_3^2)$ and $64c_0c_4^3=(4c_2c_4-c_3^2)^2$, and the resultant of the last two with respect to $n$ is $m^{24}B(m)$ up to a constant. For $m=0$ the line $y=nz$ meets $C$ in $z(x^3+n^3xz^2+(n^2-n)z^3)$, which vanishes to odd order at $P_1$. The rest is exact arithmetic in $K$; the 28 conjugates of $(a,n)$ have distinct slopes, and a smooth plane quartic has exactly 28 bitangents. \end{proof} \begin{lem}\label{lem:1.3.7-galois} The Galois group of $B$ is $\operatorname{Sp}_6(\F_2)$. In particular, $K$ has no quadratic subfield. \end{lem} \begin{proof} By Proposition~\ref{thm:AB}(1) the Galois group $\Gamma$ of $B$ is a subgroup of $\operatorname{Sp}_6(\F_2)$. $B$ is squarefree modulo $13$, $29$ and $3$, with factors of degrees $(3,5,5,15)$, $(7,7,7,7)$ and $(1,9,9,9)$ (\code{1.3.7b}), so $\Gamma$ contains elements of orders 15, 7 and 9. The maximal subgroups of $\operatorname{Sp}_6(\F_2)$ are $U_4(2){:}2$, $S_8$, $2^5{:}S_6$, $U_3(3){:}2$, $2^6{:}L_3(2)$, $2.[2^6]{:}(S_3\times S_3)$, $S_3\times S_6$ and $L_2(8){:}3$ \cite[p.~46]{ATLAS85}, of orders $2^73^45$, $2^73^25\cdot7$, $2^93^25$, $2^63^37$, $2^93\cdot7$, $2^93^2$, $2^53^35$, $2^33^37$. Only $S_8$ has order divisible by $5\cdot7$, and $S_8$ has no element of order 9. So $\Gamma=\operatorname{Sp}_6(\F_2)$, which is simple; hence it has no subgroup of index 2, and $K$ contains no quadratic field. \end{proof} \paragraph{Step 3: the descent map.} We use Proposition~\ref{thm:AB} with $\ell=y-ax-nz$. Since $\ell(1,0,0)=-a$, for $P\in C(k)$ not on a bitangent the class $\delta_k([P-P_1])$ is represented by $\ell(\tilde P)\,a$ for any coordinate vector $\tilde P$ of $P$. In particular $\delta(G)=\ell(P_3)/\ell(P_1)=(-n)/(-a)\equiv a$, because $-n$ is a square. \begin{lem}\label{lem:1.3.7-kernel} Let $V=\F_2^6$ with the standard symplectic form, $\Theta$ the 28 quadratic forms of Arf invariant 1, $\Gamma=\operatorname{Sp}_6(\F_2)$, $M=\F_2^\Theta/\F_2\mathbf1$ and $\iota(T)=(q(T))_{q\in\Theta}$. Then for $T\ne0$ exactly 12 forms satisfy $q(T)=0$, the 63 sets $\{q:q(T)=0\}$ are distinct, $\iota$ is injective, $V^\Gamma=0$ and $(M/\iota V)^\Gamma=0$ (\code{1.3.7c}). Hence $J(\Q)[2]=0$ and $\delta=\delta_\Q$ is injective. \end{lem} \begin{proof} The 63 transvections generate a permutation group of order $1451520=\#\operatorname{Sp}_6(\F_2)$ on $\Theta$, and the space of $m\in\F_2^{28}$ with $\gamma m-m\in\F_2\mathbf1+\iota V$ for all 63 generators $\gamma$ has dimension $7=\dim(\F_2\mathbf1+\iota V)$. The conclusion follows from Proposition~\ref{thm:AB}(1), (2) and Lemma~\ref{lem:1.3.7-galois}. \end{proof} Since $J(\Q)[2]=0$, injectivity gives $\operatorname{rank}J(\Q)\le\dim_{\F_2}\Sigma$ for every subgroup $\Sigma$ containing $\delta(J(\Q)/2J(\Q))$. \begin{lem}\label{lem:1.3.7-KS} Every class in $\delta(J(\Q)/2J(\Q))$ has a representative in \[ K(2,S)=\{\kappa\in K^\times/K^{\times2}:\ v_{\mathfrak p}(\kappa)\equiv0\pmod2\text{ for all prime ideals }\mathfrak p\notin S\}, \] and the image of $K(2,S)$ in $K^\times/K^{\times2}\Q^\times$ is $K(2,S)/\Q(2,S)$, where $\Q(2,S)=\langle-1,2,11,41269\rangle$ has dimension 4. Moreover, for such a representative, $(v_{\mathfrak p}(\kappa)\bmod2)_{\mathfrak p\mid11}\in\{0,(1,1,1,1,0)\}$ and $(v_{\mathfrak p}(\kappa)\bmod2)_{\mathfrak p\mid41269}\in\{0,(0,0,1,1)\}$. \end{lem} \begin{proof} At $p\notin\{2,11,41269\}$, $K/\Q$ is unramified and $c_p=1$, so by Proposition~\ref{thm:AB}(3) all $v_{\mathfrak p}(\kappa)$, $\mathfrak p\mid p$, have the same parity; multiplying $\kappa$ by suitable powers of these finitely many $p$ gives a representative in $K(2,S)$. At $p=11,41269$, $c_p=1$ (Lemma~\ref{lem:1.3.7-bad}), and Proposition~\ref{thm:AB}(3) gives the two conditions, with $(e_{\mathfrak p}\bmod2)$ read off from Lemma~\ref{lem:1.3.7-bit}(4). If $c\in\Q^\times$ lies in $K(2,S)K^{\times2}$, then $v_p(c)$ is even for $p\notin\{2,11,41269\}$, so $c\in\Q(2,S)\Q^{\times2}$; and $\Q(2,S)\to K^\times/K^{\times2}$ is injective because $K$ has no quadratic subfield. \end{proof} \paragraph{Step 4: the class group of $K$.} \begin{lem}\label{lem:1.3.7-principal} Every prime ideal of $\mathcal O_K$ of norm $0 \end{gathered} \] (\code{1.3.7d}; the GP script of \code{1.3.7d} gives the same value in floating point). \end{proof} \paragraph{Step 5: $S$-units.} \begin{lem}\label{lem:1.3.7-sunits} Let $u_1,\dots,u_{15}$ and $\pi_1,\dots,\pi_{13}$ be the elements recorded in \texttt{data/paper\_C2.json} of \code{1.3.7e}. Then $-1,u_1,\dots,u_{15},\pi_1,\dots,\pi_{13}$ is an $\F_2$-basis of $K(2,S)$, which has dimension 29. \end{lem} \begin{proof} Each $u_i$ is integral with norm $\pm1$, so it is a unit. Each $\pi_j$ is integral, and the ideals $(\pi_1),\dots,(\pi_{13})$ are the 13 prime ideals of $S$, in the order of Lemma~\ref{lem:1.3.7-bit}(4) (\code{1.3.7e}). By Dirichlet's $S$-unit theorem, $\mathcal O_{K,S}^\times\cong\{\pm1\}\times\Z^{r_1+r_2-1+\#S}=\{\pm1\}\times\Z^{28}$ ($K$ has a real place, so its only roots of unity are $\pm1$), hence $\dim\mathcal O_{K,S}^\times/\mathcal O_{K,S}^{\times2}=29$. The 29 elements are independent in $K^\times/K^{\times2}$: their quadratic characters at the 29 degree-one prime ideals $(q,w-r)$ with \[ \begin{gathered} (q,r)=(3,2),(5,1),(7,5),(19,10),(23,14),(23,16),(37,32),(41,10),(43,11),(47,15),\\ (47,18),(59,29),(67,18),(71,20),(79,38),(89,54),(103,29),(103,94),(107,92),(109,5),\\ (109,39),(113,76),(131,89),(137,16),(157,84),(157,116),(167,98),(173,9),(181,38) \end{gathered} \] form an invertible $29\times29$ matrix over $\F_2$. So they form a basis of $\mathcal O_{K,S}^\times/\mathcal O_{K,S}^{\times2}$. Finally $0\to\mathcal O_{K,S}^\times/\mathcal O_{K,S}^{\times2}\to K(2,S)\to\Cl(\mathcal O_{K,S})[2]\to0$ is exact, and $\Cl(\mathcal O_{K,S})$ is a quotient of $\Cl(K)$, which has odd order (Lemma~\ref{lem:1.3.7-cl}). \end{proof} \paragraph{Step 6: local images and the rank.} \begin{lem}\label{lem:1.3.7-dim2} $\dim_{\F_2}J(\Q_2)/2J(\Q_2)\le4$. \end{lem} \begin{proof} By Proposition~\ref{thm:A}(5), $\dim J(\Q_2)/2J(\Q_2)=3+\dim J(\Q_2)[2]$. The orbits of the image $\Delta$ of $\operatorname{Gal}(\overline\Q_2/\Q_2)$ on $\Theta$ correspond to the prime ideals above 2, so they have sizes $4,6,6,12$. By Proposition~\ref{thm:AB}(1) and Lemma~\ref{lem:1.3.7-kernel}, a nonzero $T\in J[2]$ is fixed by $\Delta$ if and only if the 12-element set $\{\theta:q_\theta(T)=0\}$ is $\Delta$-stable (these sets determine $T$), that is, a union of orbits: the 12-orbit or the union of the two 6-orbits. So $J(\Q_2)[2]=J[2]^\Delta$ has at most two nonzero elements, hence at most one, and $\dim J(\Q_2)[2]\le1$. \end{proof} To compute in $(K\otimes\Q_2)^\times/(K\otimes\Q_2)^{\times2}=\prod_{\mathfrak p\mid2}K_{\mathfrak p}^\times/K_{\mathfrak p}^{\times2}$ we use, for each $\mathfrak p\mid2$ with ramification index $e$, a uniformizer $\pi_{\mathfrak p}$ and record $v_{\mathfrak p}(x)$ modulo $2$, together with the discrete-logarithm coordinates modulo $2$ of the unit $x\pi_{\mathfrak p}^{-v_{\mathfrak p}(x)}$ in $(\mathcal O_K/\mathfrak p^{2e+1})^\times$. This is an isomorphism $K_{\mathfrak p}^\times/K_{\mathfrak p}^{\times2}\to\F_2^{2+ef}$, because a unit that is a square modulo $\mathfrak p^{2e+1}$ is a square (Hensel's lemma). Together they give an isomorphism $\mathrm{loc}_2:(K\otimes\Q_2)^\times/(\text{squares})\to\F_2^{36}$, under which the images of $-1,2,5$ are independent. \begin{lem}\label{lem:1.3.7-W2} Let $Q_1,\dots,Q_4\in C(\Q_2)$ be the points $(x_0:y_0:1)$ with \[ \begin{gathered} (x_0,y_0)=(-40,\ 591131412711573423326720+O(2^{84})),\\ (-38,\ 864734735839869549293544+O(2^{80})),\\ (-38,\ 273263295199333829574543+O(2^{81})),\\ (-36,\ 648823393722996630116800+O(2^{80})). \end{gathered} \] Let $W_2^+\subseteq\F_2^{36}$ be the span of $\mathrm{loc}_2(\ell(Q_i)a)$ ($i=1,\dots,4$) and of $\mathrm{loc}_2(-1),\mathrm{loc}_2(2),\mathrm{loc}_2(5)$. Then $\dim W_2^+=7$, and $W_2^+$ is the full preimage in $\F_2^{36}$ of $\delta_{\Q_2}(J(\Q_2)/2J(\Q_2))$ (\code{1.3.7f}). \end{lem} \begin{proof} For an integer $x_0$ and $h(Y)=F(x_0,Y,1)$, an integer $Y$ with $v_2(h(Y))>2v_2(h'(Y))$ determines a unique root $y_0\in\Z_2$ with $v_2(y_0-Y)\ge N=v_2(h(Y))-v_2(h'(Y))$. The exact element $(Y-ax_0-n)a$ is used only when $Ne_{\mathfrak p}\ge v_{\mathfrak p}((Y-ax_0-n)a)+2e_{\mathfrak p}+1$ for every $\mathfrak p\mid2$, so that it has the same square class as $\ell(Q)a$. The seven vectors span a space of dimension 7, and the last three span $\mathrm{loc}_2(\Q_2^\times)$. So the classes $\delta_{\Q_2}([Q_i-P_1])$ span a space of dimension 4 in the image of $\delta_{\Q_2}$, which has dimension at most 4 by Lemma~\ref{lem:1.3.7-dim2}. All 82 points of $C(\Q_2)$ with $x_0\in[-40,40]$, and the four displayed points, give the same space. \end{proof} Let $\Sigma\subseteq K(2,S)/\Q(2,S)$ consist of the classes $\kappa$ satisfying only \[ (v_{\mathfrak p}(\kappa)\bmod2)_{\mathfrak p\mid11}\in\{0,(1,1,1,1,0)\}, \qquad \mathrm{loc}_2(\kappa)\in W_2^+. \] These conditions are independent of the representative modulo $\Q(2,S)$. By Lemmas~\ref{lem:1.3.7-KS} and~\ref{lem:1.3.7-W2}, this space contains $\delta(J(\Q)/2J(\Q))$. \begin{lem}\label{lem:1.3.7-sel} $\dim_{\F_2}\Sigma=1$, and $\Sigma$ is spanned by $\delta(G)=[a]$. Hence $\operatorname{rank}J(\Q)=1$ and $J(\Q)\cong\Z$. \end{lem} \begin{proof} In the 29-element basis of Lemma~\ref{lem:1.3.7-sunits}, the two local conditions give an $\F_2$-linear system whose kernel has dimension 5. It contains the four-dimensional space $\Q(2,S)$ and the class of $a$, which is independent of that space (\code{1.3.7h}, the computation using only $p=2$ and $p=11$). Thus $\Sigma=\langle[a]\rangle$ has dimension 1. The descent map is injective and $J(\Q)$ is torsion-free, so the rank is at most 1; the nonzero class $G$ has infinite order, so the rank is exactly 1. \end{proof} \paragraph{Step 7: Chabauty--Coleman at $p=3$.} By Lemma~\ref{lem:1.3.7-bad}, $C$ has good reduction at 3, and $C(\F_3)=\{\bar P_1,\bar P_2,\bar P_3,\bar P_4\}$. $\#J(\F_3)=31$, so $31G$ reduces to 0. Since $J(\Q)\cong\Z$, every element of $J(\Q)$ has a nonzero multiple in $\Z\cdot31G$; so if $\eta$ is a regular differential with $\int_{31G}\eta=0$, then $\int_{P_i}^P\eta=0$ for all $P\in C(\Q)$ (\cite[Sections 4--5]{MP12}). The differential $\Omega=(x\,dy-y\,dx)/F_z=(y\,dz-z\,dy)/F_x=(z\,dx-x\,dz)/F_y$ is well defined on $C$, and $\omega_x=x\Omega$, $\omega_y=y\Omega$, $\omega_z=z\Omega$ form a basis of the regular differentials, regular on the plane model at every good prime. We use the local parameters $t=y/x$ at $P_1$, $t=z/y$ at $P_2$, and $t=x/z$ at $P_3$ and $P_4$. At these points the relevant partial derivative ($F_z$, $F_x$, $F_y$, $F_y$) is $\pm1$, so the remaining affine coordinate is a power series in $t$ with integer coefficients ($z=-t^3+\cdots$ at $P_1$, $x=-t^2+\cdots$ at $P_2$, $y=t^3+\cdots$ at $P_3$, $y=1-t+\cdots$ at $P_4$), and $\Omega=w_\Omega(t)\,dt$ with $w_\Omega$ a unit of $\Z[[t]]$. All series were computed exactly to $O(t^{140})$; after the substitution $t=3s$ the coefficient of $s^j$ in $\int_0^t\omega$ has valuation at least $j-\log_3j$. \begin{lem}\label{lem:1.3.7-E} Let $E$ be the effective divisor of degree 3 consisting of the three points $(x:V(x):1)$, where $x$ runs over the roots of \[ 80621568x^3+19551871x^2-576258756x+551007360,\qquad V(x)=\tfrac{27648}{10517}x^2+\tfrac{154341055}{30667572}x-\tfrac{31989800}{2555631}. \] Then $E\sim P_1+P_2+P_3+31G$. The three points of $E$ are $\Q_3$-rational and lie in the residue disks of $P_1$, $P_3$, $P_2$, with $v_3(t)=3,1,5$ respectively (\code{1.3.7i}). \end{lem} \begin{proof} The forms of degree 12 in the 46 monomials not divisible by $xy^3$ (a basis of $\Q[x,y,z]_{12}/(F)$) that vanish to order at least 44 at $P_1$ and at least 1 at $P_3$ form a space of dimension 1; let $\Gamma$ be a generator. The resultant $\operatorname{Res}_y(F(x,y,1),\Gamma(x,y,1))$ is $x$ times the cubic above (up to a constant), and the gcd of $F(x,y,1)$ and $\Gamma(x,y,1)$ over $\Q[x]/(\text{cubic})$ is $y-V(x)$. Since $\deg\operatorname{div}\Gamma=48=44+1+3$, $\operatorname{div}\Gamma=44P_1+P_3+E$, and $\operatorname{div}(\Gamma/(y^{11}z))=E+30P_1-32P_3-P_2$, that is, $E\sim P_2+32P_3-30P_1=P_1+P_2+P_3+31(P_3-P_1)$. The cubic has three roots in $\Q_3$, and the points reduce to $P_1,P_3,P_2$. \end{proof} \begin{lem}\label{lem:1.3.7-logs} With $L_\bullet=\int_{31G}\omega_\bullet$, which is a sum of three tiny integrals by Lemma~\ref{lem:1.3.7-E}, \[ L_x\equiv642322739790,\qquad L_y\equiv657631671201,\qquad L_z\equiv233976154908\pmod{3^{25}}, \] with valuations $2,4,1$; the values are known modulo $3^{55}$. The same computation for $62G$ (forms of degree 22 vanishing to order 85 at $P_1$) gives $2(L_x,L_y,L_z)$ (\code{1.3.7i}). \end{lem} The differentials $\eta_1=L_z\omega_x-L_x\omega_z$ and $\eta_2=L_z\omega_y-L_y\omega_z$ annihilate $J(\Q)$ and are independent ($L_z\ne0$). For $Q$ in the disk of $P_i$ with $t(Q)=3s$ put $I_k(s)=\int_{P_i}^Q\eta_k=\sum_jc_j^{(k)}s^j$. \begin{lem}\label{lem:1.3.7-disks} (1) In the disks of $P_1$, $P_2$, $P_4$ one of $I_1,I_2$ has Strassmann bound 1, so $P_i$ is the only rational point in its disk. (2) In the disk of $P_3$, $I_1$ has Strassmann bound 2; its zeros in $\Z_3$ are $s=0$ and $s=U^*\equiv589926054506\pmod{3^{25}}$, and $v_3(I_2(U^*))=6$. So $P_3$ is the only rational point in its disk (\code{1.3.7i}). \end{lem} \begin{proof} The valuations of $c_1,\dots,c_8$ ($\infty$ means that the coefficient is 0) are: \[ \begin{array}{c|c|c|c} \text{disk}&\text{differential}&v_3(c_1),\dots,v_3(c_8)&\text{bound}\\\hline P_1&\eta_1&2,\infty,\infty,6,\infty,6,\infty,10&1\\ P_2&\eta_2&2,6,\infty,\infty,\infty,7,9,10&1\\ P_3&\eta_1&3,3,\infty,6,6,\infty,10,10&2\\ P_3&\eta_2&5,\infty,\infty,5,\infty,\infty,9,13&4\\ P_4&\eta_2&2,3,4,5,6,7,8,12&1 \end{array} \] All $c_j$ with $j\le140$ were computed, and the bounds take all of them into account; for $j>140$ the tail bound gives $v_3(c_j)\ge j-\log_3j>130$. The errors from knowing $L_\bullet$ modulo $3^{55}$ have valuation at least 55. This proves (1) and the bound in (2), by \cite{Strassmann28}. For (2), $J_1(s)=I_1(s)/s$ satisfies $v_3(J_1(s_0))=43>2v_3(J_1'(s_0))=6$ at the Newton approximation $s_0$, so by Hensel's lemma it has a unique zero $U^*\equiv s_0\pmod{3^{40}}$, and $U^*\not\equiv0\pmod3$. Since the coefficients of $I_2$ are integral, $I_2(U^*)\equiv I_2(s_0)\equiv168750803736\pmod{3^{25}}$, of valuation 6. So $U^*$ is not a common zero of $I_1,I_2$, while every rational point of the disk satisfies $I_1=I_2=0$. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.3.7}] Every point of $C(\Q)$ lies in one of the four residue disks of $C(\F_3)$, and by Lemma~\ref{lem:1.3.7-disks} (which uses Lemma~\ref{lem:1.3.7-sel}) each disk contains only its centre. So $C(\Q)=\{P_1,P_2,P_3,P_4\}$. An integer solution $(x,y,z)\ne(0,0,0)$ is $d$ times a primitive coordinate vector of one of these points, $d=\gcd(x,y,z)$, which gives the four forms in the theorem. $J(\Q)\cong\Z$ is Lemma~\ref{lem:1.3.7-sel}. \end{proof} \subsection{\texorpdfstring{$x^4-y^3z+xyz^2+xz^3=0$ \ ($H=64$)}{x^4-y^3z+xyz^2+xz^3=0 (H=64)}}\label{sec:1.3.8} \begin{thm}\label{thm:1.3.8} The only integer solutions of $x^4-y^3z+xyz^2+xz^3=0$ are $(0,n,0)$, $(0,0,n)$, $(-n,0,n)$ $(n\in\Z)$. \end{thm} \paragraph{Setup.} Let $C\subset\mathbb P^2$ be the curve defined by the equation in the theorem, and put $J=\Jac(C)$. Write \[ \infty=(0:1:0),\qquad P_0=(0:0:1),\qquad P_1=(-1:0:1). \] Write $F=x^4-y^3z+xyz^2+xz^3$ and, in the chart $z=1$, $f=F(x,y,1)=x^4-y^3+xy+x$, so that $C$ is $y^3=x^4+xy+x$ there. Put $T=[P_0-\infty]$ and $G_1=[P_1-\infty]$ in $J(\Q)$. The line $z=0$ meets $C$ only at $\infty$, with multiplicity 4, so $\infty$ is a rational hyperflex and the 2-descent of Proposition~\ref{thm:A} applies. The proof runs as follows. $C$ is smooth, its odd bad primes are $11$ and $652081$, all odd Tamagawa numbers are 1, $J(\Q)_{\mathrm{tors}}=\langle T\rangle\cong\Z/3$, and $G_1$ has infinite order (Step~1). The other 27 bitangents are conjugate, with field of definition $K$ of degree 27 (Step~2). The descent map is injective, with image in the totally positive part of $K(\{2\},2)$ (Step~3). The class group of $K$ has odd order, proved without GRH with Proposition~\ref{thm:B} (Step~4); so $K(\{2\},2)$ is spanned by 19 explicit $S$-units, and the condition at 2 cuts the image down to dimension 1, so $\operatorname{rank}J(\Q)=1$ (Step~5). Chabauty--Coleman at $p=3$ shows that each of the three residue disks of $C(\F_3)$ contains only one rational point (Step~6). \paragraph{Step 1: the curve.} \begin{lem}\label{lem:1.3.8-zero} The points of $C$ with $x=0$ are $\infty$ and $P_0$; with $y=0$ they are $P_0$, $P_1$ and the two points $(\zeta:0:1)$ with $\zeta^2-\zeta+1=0$; with $z=0$ only $\infty$. So $\infty,P_0,P_1$ are the only rational points of $C$ with a zero coordinate. \end{lem} \begin{proof} $F(0,y,z)=-y^3z$, $F(x,0,z)=x(x+z)(x^2-xz+z^2)$, $F(x,y,0)=x^4$. \end{proof} \begin{lem}\label{lem:1.3.8-bad} (1) $C$ is smooth over $\Q$, so it has genus 3. (2) For an odd prime $p$, $C$ has good reduction at $p$ unless $p\in\{11,652081\}$. (3) For $p\in\{11,652081\}$ the reduction has exactly one singular point over $\overline\F_p$, namely $(x_0:y_0:1)$ with $y_0\equiv-9/5$, $x_0\equiv3y_0^2\pmod p$; it is an ordinary node, and the scheme $\{F=0\}\subset\mathbb P^2_{\Z_p}$ is regular. (4) For every odd prime $p$ the Tamagawa number $c_p(J)$ is 1 (\code{1.3.8a}). \end{lem} \begin{proof} On $z=0$ the only point is $\infty$, and $F_z(0,1,0)=-1$, so $\infty$ is a smooth point of every reduction. For affine points, $f_x=4x^3+y+1$ and $f_y=x-3y^2$, and for odd $p$ Euler's identity shows that an affine point is singular if and only if $f=f_x=f_y=0$. Over $\F_3$ and $\F_5$ the ideal $(f,f_x,f_y)$ has Gröbner basis $\{1\}$. For $p\ne3$, $f_y=0$ gives $x=3y^2$, and then $f_x=108y^6+y+1$ and $f=y^2(81y^6+2y+3)$. If $y=0$, then $x=0$ and $f_x=1$. Otherwise $4(81y^6+2y+3)-3(108y^6+y+1)=5y+9$, so for $p\ne5$ a singular point has $y=-9/5$, and then $108y^6+y+1=57383128/5^6=2^3\cdot11\cdot652081/5^6$. This proves (1), (2) and the first part of (3). At $p=11$ the node is $(4,7)$, the Hessian determinant $f_{xx}f_{yy}-f_{xy}^2$ is $9\not\equiv0$, and $f(4,7)\equiv66\not\equiv0\pmod{11^2}$; at $p=652081$ the node is $(573841,521663)$, the Hessian determinant is $217362\not\equiv0$, and $f(x_0,y_0)\equiv91909512788\not\equiv0\pmod{p^2}$. Since $f_x,f_y$ vanish at the node modulo $p$, the Taylor expansion shows that $f\in\mathfrak m^2$ ($\mathfrak m=(p,x-x_0,y-y_0)$) if and only if $f(x_0,y_0)\equiv0\pmod{p^2}$; so $f\notin\mathfrak m^2$, and the local ring at the node is regular. (4) The special fibre at $p=11,652081$ is reduced and geometrically irreducible: components would meet with total intersection number at least 3 (Bézout), only at singular points, but there is only one singular point, an ordinary node, where two branches meet with multiplicity 1. By Raynaud's theorem \cite[9.6, Theorem~1]{BLR90} the component group of the Néron model is trivial. \end{proof} \begin{lem}\label{lem:1.3.8-counts} The numbers $N_k=\#C(\F_{p^k})$, the coefficients of the $L$-polynomials and $\#J(\F_p)=L_p(1)$ are as follows (\code{1.3.8a}). \[ \begin{array}{r|c|l|r} p&(N_1,N_2,N_3)&\text{coefficients of }L_p(t)&\#J(\F_p)\\\hline 3&(3,9,27)&1,-1,0,0,0,-9,27&18=2\cdot3^2\\ 5&(5,29,119)&1,-1,2,-4,10,-25,125&108=2^2\cdot3^3\\ 7&(7,67,352)&1,-1,9,-6,63,-49,343&360=2^3\cdot3^2\cdot5\\ 17&(29,313,4856)&1,11,72,329,1224,3179,4913&9729=3^2\cdot23\cdot47\\ 29&(27,871,23904)&1,-3,19,-210,551,-2523,24389&22224=2^4\cdot3\cdot463 \end{array} \] Moreover $C(\F_3)=\{\bar\infty,\bar P_0,\bar P_1\}$. \end{lem} \begin{lem}\label{lem:1.3.8-tors} $J(\Q)_{\mathrm{tors}}=\langle T\rangle\cong\Z/3$, and $G_1$ has infinite order. \end{lem} \begin{proof} By \cite[Appendix]{Katz81}, the torsion subgroup $J(\Q)_{\mathrm{tors}}$ injects into $J(\F_p)$ for every odd prime $p$ of good reduction; $\#J(\F_{17})$ is odd and $\gcd(108,360,9729,22224)=3$, so $\#J(\Q)_{\mathrm{tors}}$ divides 3. The line $x=0$ meets $C$ in $3P_0+\infty$ and $z=0$ in $4\infty$, so $\operatorname{div}(x/z)=3P_0-3\infty$ and $3T=0$; $T\ne0$ since the genus is positive. The function $x/z$ has a pole of order exactly 3 at $\infty$, and $\dim L(3\infty)\le2$, so $L(3\infty)=\langle1,x/z\rangle$. If $G_1$ were torsion, then $3G_1=0$ and some $a+b\,x/z$ would have divisor $3P_1-3\infty$; vanishing at $P_1$ forces $a=b$, but $F(-z,y,z)=-yz(y^2+z^2)$ shows that $(x+z)/z$ has only a simple zero at $P_1$. \end{proof} \begin{lem}\label{lem:1.3.8-real} $C(\R)$ is connected. Hence $J(\R)/2J(\R)=0$ (\code{1.3.8a}). \end{lem} \begin{proof} For real $x$, the cubic $y^3-xy-(x^4+x)$ has discriminant $x^2\bigl(4x-27(x^3+1)^2\bigr)$. The second factor has no real root and is $-27$ at $x=0$, so the discriminant is negative for $x\ne0$, and for $x=0$ the cubic is $y^3$. So for every real $x$ there is exactly one real $y$, depending continuously on $x$; the affine real locus is a graph over $\R$, and $C(\R)$ is its closure. The rest is \cite{GH81}. \end{proof} \paragraph{Step 2: the bitangents and the field $K$.} Let \[ \begin{aligned} B(m)={}&m^{27}-48m^{22}+1344m^{18}+144m^{17}-3264m^{13}+832m^{12}-40704m^9\\ &-4608m^8-2560m^7-39936m^4-15360m^3-1024m^2+4096, \end{aligned} \] let $\theta$ be a root, and $K=\Q(\theta)$. \begin{lem}\label{lem:1.3.8-bit} (1) $B$ is irreducible, and \[ \operatorname{disc}(B)=2^{318}\cdot11^6\cdot41^2\cdot198839^2\cdot652081^6\cdot17743299257028546229859^2 \] (all factors prime). (2) For the line $y=mx+k$ write $f(x,mx+k)=x^4+c_3x^3+c_2x^2+c_1x+c_0$. It is the square of $x^2+ux+v$ if and only if $u=c_3/2=-m^3/2$, $v=(c_2-u^2)/2=(m-3m^2k-u^2)/2$, $E_1:=c_1-2uv=0$ and $E_2:=c_0-v^2=0$. The resultant of $E_1,E_2$ with respect to $k$ is $B(m)/4096$ (equivalently, that of the primitive integer polynomials $8E_1$, $64E_2$ is $512B(m)$), and the leading coefficient of $E_2$ in $k$ is $-1$; so $E_1(m_0,k)$ and $E_2(m_0,k)$ have a common root if and only if $B(m_0)=0$. Over $K$ the gcd of $E_1(\theta,k)$ and $E_2(\theta,k)$ is linear in $k$, with root $k(\theta)\in K$ (recorded in \texttt{descent/k\_paper.gp} of \code{1.3.8b}), and $f(x,\theta x+k(\theta))=(x^2+u(\theta)x+v(\theta))^2$ in $K[x]$. (3) The 28 bitangents of $C$ are $z=0$ and the 27 lines $\ell_\theta:\ y=\theta x+k(\theta)z$, $\theta$ running over the roots of $B$ (\code{1.3.8b}). \end{lem} \begin{proof} (3) A line through $\infty$ other than $z=0$ is $x=cz$; it meets $C$ at $\infty$ with multiplicity 1, so it is not a bitangent. Every other line is $y=mx+kz$, and by (2) it is a bitangent if and only if $B(m)=0$ and $k=k(m)$. These are 27 distinct lines. \end{proof} \begin{lem}\label{lem:1.3.8-K} $K$ has signature $(3,12)$ and discriminant $d_K=2^{34}\cdot11^6\cdot652081^6$; its maximal order is certified without GRH. The prime 2 factors as $2\mathcal O_K=\mathfrak p_1^2\mathfrak p_2^{10}\mathfrak p_3^5\mathfrak p_4^5$ with residue degrees $1,1,1,2$; equivalently, $B$ factors over $\Q_2$ into irreducible factors of degrees $2,10,5,10$ (\code{1.3.8b}, \code{1.3.8c}). \end{lem} \paragraph{Step 3: the descent map.} For each root $\theta$ of $B$ put $h_\theta=(y-\theta x-k(\theta)z)/z$. At $\infty$ the function $X=x/y$ is a uniformizer, $z/y=X^4+O(X^9)$ and $h_\theta=X^{-4}(1+O(X))$, so $h_\theta$ is normalized as in Proposition~\ref{thm:A}(2), and for an affine point $P=(x:y:1)$ that is not a point of contact, \[ \delta([P-\infty])=y-\theta x-k(\theta)\in K^\times/K^{\times2}. \] In particular $\delta(G_1)=\theta-k(\theta)$ and $\delta(T)=-k(\theta)$. \begin{lem}\label{lem:1.3.8-theta} For a root $\theta$ of $B$ let $D_\theta$ be the contact divisor of $\ell_\theta$ (so $\operatorname{div}h_\theta=2D_\theta-4\infty$) and $w_\theta=[D_\theta-2\infty]\in J[2]$. Then $\iota:J[2]\to\F_2^{27}$, $\varepsilon\mapsto(e_2(\varepsilon,w_\theta))_\theta$, is injective, and the nonzero vectors of $V=\iota(J[2])$ have weight 12 (36 vectors) or 16 (27 vectors); in particular $\mathbf 1\notin V$ (\code{1.3.8d}). \end{lem} \begin{proof} $q(w)=h^0(2\infty+w)+h^0(2\infty)\bmod2$ is a quadratic form on $J[2]$ with polar form $e_2$, whose zeros are $0$ and the 27 classes $w_\theta$; so it has Arf invariant 1, and $\iota(\varepsilon)_\theta=e_2(\varepsilon,w_\theta)$. All quadratic forms of Arf invariant 1 on $\F_2^6$ are equivalent, so it suffices to enumerate the 64 vectors for $q(a)=a_1a_2+a_3a_4+a_5^2+a_5a_6+a_6^2$. \end{proof} This is the computation behind Proposition~\ref{thm:A}(1), (2). \begin{lem}\label{lem:1.3.8-desc} $J(\Q)[2]=0$, and $\delta:J(\Q)/2J(\Q)\to K^\times/K^{\times2}$ is injective. Every element of $\delta(J(\Q))$ lies in \[ K(2,S)=\{a\in K^\times/K^{\times2}:\ v_{\mathfrak p}(a)\text{ even for all }\mathfrak p\notin S\},\qquad S=\{\mathfrak p_1,\mathfrak p_2,\mathfrak p_3,\mathfrak p_4\}, \] and is totally positive. \end{lem} \begin{proof} $B$ is irreducible of degree 27, so Proposition~\ref{thm:A}(1), (2) give the first statement. Since $c_p(J)=1$ for all odd $p$ (Lemma~\ref{lem:1.3.8-bad}), the set $S$ of Proposition~\ref{thm:A}(3) is $\{2\}$. At the real place, $\delta_\R$ is defined on $J(\R)/2J(\R)=0$ (Lemma~\ref{lem:1.3.8-real}) and is compatible with restriction (Proposition~\ref{thm:A}(3)). \end{proof} \paragraph{Step 4: the class group of $K$.} \begin{lem}\label{lem:1.3.8-principal} Every prime ideal of $K$ of norm $\le10^6$ is principal. There are $77{,}813$ of them ($77{,}583$ of degree 1, $193$ of degree 2, $18$ of degree 3, $15$ of degree 5, $3$ of degree 6, $1$ of degree 10) (\code{1.3.8e}). \end{lem} \begin{proof} For each prime ideal $\mathfrak p$ a generator $g$ was proposed by \texttt{bnfisprincipal} (used only as a proposer). The checker, which does not use PARI's class group functions, enumerates the prime ideals itself (for $p>1000$ not dividing $\operatorname{disc}B$ through the roots $r$ of $B$ modulo $p$, $\mathfrak p=(p,\theta-r)$; otherwise with \texttt{idealprimedec} in the certified maximal order), and verifies for each $\mathfrak p$: $g\in\mathcal O_K$ ($g$ is stored as a polynomial in $\theta$ divided by an integer, and its coordinates on the certified integral basis computed by the running PARI are integers; the check was run with PARI 2.15.4 and with PARI 2.17.4), $g\in\mathfrak p$, and $|N_{K/\Q}(g)|=N\mathfrak p$, where $N(g)$ is computed exactly as a resultant. Then $(g)\subseteq\mathfrak p$ and both have the same norm, so $(g)=\mathfrak p$. Each generator is used for exactly one prime ideal. \end{proof} \begin{lem}\label{lem:1.3.8-counts2} Let $R=5\cdot10^8$. There are $26{,}365{,}548$ prime ideals of $K$ of norm $0$ and $\Cl(K)[2]=0$. $\Cl(\mathcal O_{K,S})$ is a quotient of $\Cl(K)$. \end{proof} The window was chosen as the top eigenvector of the symmetric Toeplitz matrix $\Lambda$ with $B^-(f)=\mathbf h^{\mathsf T}\Lambda\mathbf h/\mathbf h^{\mathsf T}\mathbf h$ for the vector $\mathbf h$ of heights (\code{1.3.8g}, floating point; this only selects the window). If $f$ has knots $f_k=f(k\delta)$ ($\delta=T/64$), then $f=\sum_kw_k(1-|x|/k\delta)_+$ with $w_k=k(f_{k-1}-2f_k+f_{k+1})$, and $C_\infty-\int_0^\infty(1-f)W=\sum_kw_kA(k\delta)$ with \[ A(a)=C_\infty-\frac{\pi^2}{4a}-\frac{2\alpha\log2}a+\frac2a\Bigl(\operatorname{Li}_2(e^{-a})-\frac{\operatorname{Li}_2(e^{-2a})}4\Bigr)+\frac{2\alpha}a\log(1+e^{-a}). \] \paragraph{Step 5: $S$-units, the local image at 2, and the rank.} \begin{lem}\label{lem:1.3.8-sunits} Let $\varepsilon_1,\dots,\varepsilon_{14}$ and $\pi_1,\dots,\pi_4$ be the 14 units and 4 $S$-units recorded in \path{descent/paper_sunits.json} of \code{1.3.8i}. Then $-1,\varepsilon_1,\dots,\varepsilon_{14},\pi_1,\dots,\pi_4$ is a basis of $\mathcal O_{K,S}^\times/\mathcal O_{K,S}^{\times2}$ and of $K(2,S)$, and $\dim_{\F_2}K(2,S)=19$ (\code{1.3.8h}, \code{1.3.8i}). \end{lem} \begin{proof} Each $\varepsilon_i$ is integral with $N(\varepsilon_i)=\pm1$, and each $\pi_j$ is integral with $N(\pi_j)\in\{\pm2,\pm4\}$, so all are $S$-units. Their quadratic characters at the 400 degree-one prime ideals $(q,\theta-r)$ with $q\le2791$, $q\nmid\operatorname{disc}B$, form a matrix of rank 19 over $\F_2$, so the 19 elements are independent in $K^\times/K^{\times2}$. By Dirichlet's $S$-unit theorem, $\mathcal O_{K,S}^\times\cong\{\pm1\}\times\Z^{r_1+r_2-1+\#S}=\{\pm1\}\times\Z^{18}$, so $\dim\mathcal O_{K,S}^\times/\mathcal O_{K,S}^{\times2}=19$. The exact sequence $0\to\mathcal O_{K,S}^\times/\mathcal O_{K,S}^{\times2}\to K(2,S)\to\Cl(\mathcal O_{K,S})[2]\to0$ and Lemma~\ref{lem:1.3.8-cl} give $K(2,S)=\mathcal O_{K,S}^\times/\mathcal O_{K,S}^{\times2}$. \end{proof} \begin{lem}\label{lem:1.3.8-two} $\dim_{\F_2}J(\Q_2)[2]\le1$, so $\dim_{\F_2}\delta_{\Q_2}(J(\Q_2))\le4$. The four points $P=(x,y)\in C(\Q_2)$ with \[ \begin{array}{r|r} x&y\bmod2^{60}\\\hline -40&177036671367769442\\ -39&271477316397691054\\ -37&408928458825745180\\ -35&1073474074041454754 \end{array} \] have classes $\delta_{\Q_2}([P-\infty])$ that span a subspace $W_2\subset\prod_{\mathfrak p\mid2}K_{\mathfrak p}^\times/K_{\mathfrak p}^{\times2}$ of dimension 4. Hence $\delta_{\Q_2}(J(\Q_2))=W_2$ (\code{1.3.8d}, \code{1.3.8h}). \end{lem} \begin{proof} $J(\Q_2)[2]\cong V^D$ (Lemma~\ref{lem:1.3.8-theta}), where $D$ is the image of $\operatorname{Gal}(\overline\Q_2/\Q_2)$; its orbits on the 27 coordinates have sizes $2,5,10,10$ (Lemma~\ref{lem:1.3.8-K}). A $D$-invariant vector is a union of orbits, of weight in $\{0,2,5,7,10,12,15,17,20,22,25,27\}$, so by Lemma~\ref{lem:1.3.8-theta} a nonzero invariant vector of $V$ has weight $12=2+10$; there are at most two such unions, and their sum has weight 20, which does not occur. So $\dim V^D\le1$, and by Proposition~\ref{thm:A}(5) $\dim J(\Q_2)/2J(\Q_2)=3+\dim J(\Q_2)[2]\le4$. The values $y$ are the roots in $\Z_2$ of $y^3-xy-(x^4+x)$, computed to precision $2^{80}$. For each $\mathfrak p\mid2$ with ramification index $e$, the class of $a\in K_{\mathfrak p}^\times$ modulo squares is determined by $v_{\mathfrak p}(a)\bmod2$ and by the class of the unit $a\pi^{-v(a)}$ in $(\mathcal O_K/\mathfrak p^{2e+1})^\times$ modulo squares, read off from the discrete logarithm; the number of even cyclic factors of $(\mathcal O_K/\mathfrak p^{2e+1})^\times$ is $1+[K_{\mathfrak p}:\Q_2]$ in all four cases, as it must be. This gives coordinates in $\F_2^{35}$. The integer $Y\equiv y\pmod{2^{80}}$ satisfies $v_{\mathfrak p}(Y-y)-v_{\mathfrak p}(Y-\theta x-k)\ge2e+1$ for all $\mathfrak p\mid2$, so $Y-\theta x-k$ has the same square class as $y-\theta x-k$. The four class vectors have rank 4. (All 47 points of $C(\Q_2)$ with $x\in\Z\cap[-40,40]$, and the four points with $x=-24,-23,-21,-19$, give classes in $W_2$.) \end{proof} Let $\Sigma=\{a\in K(2,S):\ a\text{ is totally positive and }(a)_{\mathfrak p\mid2}\in W_2\}$. By Lemmas~\ref{lem:1.3.8-desc} and \ref{lem:1.3.8-two}, $\operatorname{rank}J(\Q)=\dim_{\F_2}J(\Q)/2J(\Q)\le\dim_{\F_2}\Sigma$. \begin{lem}\label{lem:1.3.8-sel} $\dim_{\F_2}\Sigma=1$, and $\Sigma$ is generated by $\theta-k(\theta)=\delta(G_1)$. Hence $\operatorname{rank}J(\Q)=1$, and $J(\Q)\cong\Z\oplus\Z/3$. \end{lem} \begin{proof} In the basis of Lemma~\ref{lem:1.3.8-sunits}, $\Sigma$ is the kernel of an explicit $\F_2$-linear map: the three real signs (computed with rigorous isolation of the real roots of $B$ in ball arithmetic, and confirmed by PARI's \texttt{nfeltsign}) and the 35 coordinates at $\mathfrak p\mid2$ taken modulo $W_2$. The kernel has dimension 1 (\code{1.3.8h}); in this basis it is generated by $\varepsilon_1\varepsilon_2\varepsilon_3\varepsilon_4$, which equals $\theta-k(\theta)$ modulo squares: both lie in $K(2,S)$ ($\theta-k(\theta)$ has even valuation at every prime ideal outside $S$), and their quadratic characters agree at 399 prime ideals (\code{1.3.8i}). So $\operatorname{rank}J(\Q)\le1$, and by Lemma~\ref{lem:1.3.8-tors} $\operatorname{rank}J(\Q)\ge1$ and $J(\Q)_{\mathrm{tors}}\cong\Z/3$. (Consistently, $\theta-k(\theta)$ is totally positive with local class in $W_2$, and $\delta(T)=-k(\theta)$ is a square in $K$.) \end{proof} The condition at 2 alone already gives the bound: in the coordinates of Lemma~\ref{lem:1.3.8-sunits} and on the basis of $W_2$ given by the four points of Lemma~\ref{lem:1.3.8-two}, the condition $(a)_{\mathfrak p\mid2}\in W_2$ is a homogeneous system of 35 linear equations over $\F_2$ in $19+4$ unknowns of rank 22, and the four basis vectors of $W_2$ are independent; so $\{a\in K(2,S):(a)_{\mathfrak p\mid2}\in W_2\}$ has dimension 1 (\code{1.3.8h}). Hence Lemma~\ref{lem:1.3.8-sel} holds without the sign condition, and without the norm condition of Proposition~\ref{thm:A}(4). \paragraph{Step 6: Chabauty--Coleman at $p=3$.} By Lemma~\ref{lem:1.3.8-bad}, $C$ has good reduction at 3, and by Lemma~\ref{lem:1.3.8-counts}, $C(\Q_3)$ is the union of the three residue disks $U_\infty,U_{P_0},U_{P_1}$ of the points of $C(\F_3)$, where $U_R$ is the set of points of $C(\Q_3)$ that reduce to $\bar R$; for a finite extension $M$ of $\Q_3$, $U_R(M)$ denotes the set of points of $C(M)$ that reduce to $\bar R$. Let $\Omega=dx/f_y(x,y)$ and $w_L=L\,\Omega$ for $L\in\{z,x,y\}$. These form a $\Z_3$-basis of the regular differentials of the smooth model over $\Z_3$, and $w_L$ vanishes exactly where $L$ does. In the chart $y=1$, with $X=x/y$ and $Z=z/y$, $F(X,1,Z)=X^4-Z+XZ^2+XZ^3$, and the equation $Z=X^4+XZ^2+XZ^3$ has a unique solution $Z(X)\in X^4\Z[[X]]$; so $w_L=\phi_L(X)\,dX$ with $\phi_L\in\Z[[X]]$: \[ \phi_z=X^4+\cdots,\qquad\phi_x=X+2X^6+\cdots,\qquad\phi_y=1+2X^5+\cdots . \] \begin{lem}\label{lem:1.3.8-log} Let $\lambda=\log(-9G_1)\in\Q_3^3$ be the vector of the abelian integrals of $w_z,w_x,w_y$ over $-9G_1$. Then $v_3(\lambda)=(5,2,6)$, and $3^{-2}\lambda$ is, up to a unit, congruent to $(432,1,81)$ modulo $3^6$. In particular, modulo 3, $\log G_1$ is proportional to $(0,1,0)$ (\code{1.3.8j}). \end{lem} \begin{proof} The functions $x^iy^j$ ($j\le2$, $3i+4j\le14$; $x,y$ have poles of orders 3 and 4 at $\infty$) form a basis of $L(14\infty)$. With the parameter $s=x+1$ at $P_1$ (where $f_y=-1$) and $y(s)\in\Z[[s]]$, the condition $\operatorname{ord}_{P_1}\ge9$ cuts out a 3-dimensional subspace, which contains \[ \begin{aligned} g={}&49412858216+191259723096x+291125394027x^2\\ &+206187126854x^3+56908597707x^4+37163261480y\\ &+102426117246xy+100459295010x^2y+35217797652x^3y+836804043y^2+36xy^2, \end{aligned} \] with $\operatorname{ord}_{P_1}g=9$ exactly and pole order 13 at $\infty$; its reduction modulo 3 is nonzero with pole order exactly 9 at $\bar\infty$. So $\operatorname{div}g=9P_1+\mathcal E-13\infty$ with $\mathcal E$ effective of degree 4, and $\operatorname{div}\bar g=9\bar P_1-9\bar\infty$; since divisors specialize on the smooth model, $\mathcal E$ reduces to $4\bar\infty$. Thus $[\mathcal E-4\infty]=-9G_1$. The $X$-coordinates of the points of $\mathcal E$ are the roots of the irreducible polynomial $\operatorname{Res}_Z(F(X,1,Z),Z^4g(X/Z,1/Z))/X^3$, \[ \begin{aligned} \mathcal E(X)={}&268895426627707157987116133388416X^4+320337638116517237204091843045891X^3\\ &+219672860573908519668631358279748X^2+157894577001790646570719666475904X\\ &+85988850899990436755428562963712, \end{aligned} \] whose four roots all have 3-adic valuation $3/4$ (the coefficients of $X^0,\dots,X^4$ have valuations $3,3,2,5,0$); their $Z$-coordinates have valuation 3. Since $3/4$ has denominator 4, $\mathcal E(X)$ is irreducible over $\Q_3$, and any of its roots generates a totally ramified extension $M$ of $\Q_3$ of degree 4. So the four points of $\mathcal E$ are conjugate over $\Q_3$ and lie in $U_\infty(M)$, where $X$ takes values in the maximal ideal of $\mathcal O_M$. The divisor $\mathcal E-4\infty$ is defined over $\Q_3$, and by \cite[Section 3]{KRZB16} \[ \lambda_L=\sum_{Q\in\mathcal E}\int_\infty^Qw_L=\operatorname{Tr}_{M/\Q_3}\int_\infty^{Q_1}w_L=\sum_{k\ge0}\frac{\phi_{L,k}}{k+1}\,p_{k+1},\qquad p_j=\sum_{Q\in\mathcal E}X(Q)^j, \] where $Q_1$ is any one of the four points and $p_j\in\Q$ is computed exactly by Newton's identities from $\mathcal E(X)$. Since $v_3(p_j)\ge3j/4$, the terms with $k\ge159$ have valuation greater than 115. As a check, the same computation with $\operatorname{ord}_{P_1}\ge18$ in $L(23\infty)$ (a function of pole order 22, again with $\deg\mathcal E=4$) gives $\log(-18G_1)=2\lambda$ to precision $3^{114}$. \end{proof} \begin{lem}\label{lem:1.3.8-disks} Each residue disk $U_\infty,U_{P_0},U_{P_1}$ contains exactly one rational point (\code{1.3.8j}). \end{lem} \begin{proof} Since $\operatorname{rank}J(\Q)=1$ and $G_1$ has infinite order (Lemma~\ref{lem:1.3.8-sel}), every $\omega=c_zw_z+c_xw_x+c_yw_y$ with $c\cdot\lambda=0$ satisfies $\int_R^P\omega=0$ for all $P,R\in C(\Q)$ (\cite[Sections 4--5]{MP12}). By Lemma~\ref{lem:1.3.8-log}, $\lambda/9$ has unit $x$-coordinate, so for any $c_z,c_y\in\Z_3$ there is a unique $c_x\in\Z_3$ with $c\cdot\lambda=0$, and then $c_x\equiv0\pmod3$. Let $R\in\{\infty,P_0,P_1\}$ and let $t$ be the local parameter $X$, $y$, $x+1$ respectively (at $P_0$, $f_x=1$ and $x(y)\in y^3\Z[[y]]$; at $P_1$, $f_y=-1$ and $y(s)\in\Z[[s]]$). The map $Q\mapsto t(Q)$ is a bijection $U_R\to3\Z_3$, and every regular differential is $(a_0+a_1t+\cdots)\,dt$ with $a_k\in\Z_3$ and $a_0\equiv(\text{unit})\cdot(c_z,c_x,c_y)\cdot(z,x,y)(R)\pmod3$. The reductions $(z,x,y)(R)\bmod3$ are $(0,0,1)$, $(1,0,0)$, $(1,2,0)$. Taking $(c_z,c_y)=(0,1)$ for $R=\infty$ and $(1,0)$ for $R=P_0,P_1$ gives annihilating differentials with $a_0\not\equiv0\pmod3$ (in fact $a_0\equiv1,-1,41849\pmod{3^{10}}$). For $Q\in U_R\cap C(\Q)$ write $t(Q)=3s$; then \[ 0=\frac13\int_R^Q\omega=a_0s+\sum_{k\ge1}a_k\frac{3^k}{k+1}s^{k+1}, \] and $v_3(3^k/(k+1))=k-v_3(k+1)\ge1$ for $k\ge1$. By \cite{Strassmann28} this power series has at most one zero in $\Z_3$, namely $s=0$. So $Q=R$. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.3.8}] $C(\Q)\subseteq U_\infty\cup U_{P_0}\cup U_{P_1}$, and by Lemma~\ref{lem:1.3.8-disks} each disk contains only its centre. So $C(\Q)=\{\infty,P_0,P_1\}$. An integer solution $(x,y,z)\ne(0,0,0)$ is $\gcd(x,y,z)$ times a primitive coordinate vector of one of these points, which gives the three forms in the theorem. $J(\Q)\cong\Z\oplus\Z/3$ is Lemma~\ref{lem:1.3.8-sel}. \end{proof} \section{Homogeneous surfaces and diagonal quintics}\label{sec:1.4} The first three cubic surfaces are excluded by cubic reciprocity, and the diagonal quartic by Hilbert reciprocity. The remaining six equations use the common Frey-curve framework below. \subsection{\texorpdfstring{$x^3+2x^2y+y^3+y^2z+xz^2+z^3+t^2x-t^2y-t^2z+tyz+t^3=0$ \ ($H=96$)}{x^3+2x^2y+y^3+y^2z+xz^2+z^3+t^2x-t^2y-t^2z+tyz+t^3=0 (H=96)}}\label{sec:1.4.1} \begin{thm}\label{thm:1.4.1} The only integer solution of \[ F(x,y,z,t):=x^3+2x^2y+y^3+y^2z+xz^2+z^3+t^2x-t^2y-t^2z+tyz+t^3=0 \] is $(0,0,0,0)$. \end{thm} \paragraph{Setup.} Let $S\subset\mathbb P^3$ be the cubic surface $F=0$. We prove that $S(\Q)=\varnothing$ by a Brauer--Manin obstruction of order 3 (Proposition~\ref{thm:AD}). Let \[ K=\Q(s),\quad s^2=-23,\qquad h(T)=T^3-T+1,\qquad L=K(a),\quad h(a)=0, \] let $\tau(a)=-\bigl(23a+s(6a^2+9a-4)\bigr)/46\in L$ (by Lemma~\ref{lem:1.4.1-field}(i), $a\mapsto\tau(a)$ defines a generator $\tau$ of $\operatorname{Gal}(L/K)$), and let $G$ be the cubic form \[ \begin{aligned} G={}&(102-18s)x^2y-(281+13s)x^2z-(119+19s)x^2t+288xy^2+(124-4s)xyz-(416-32s)xyt\\ &+(75-9s)y^3+(107-41s)y^2z+(112-16s)y^2t\\ &+(59-89s)(xz^2+xt^2+yzt-yt^2+z^3-zt^2+t^3). \end{aligned} \] We write $G=G_0+sG_1$ with $G_0,G_1\in\Z[x,y,z,t]$. If a prime $p$ splits in $K$, then for each of the two places $w\mid p$ of $K$ we have $K_w=\Q_p$, $s\mapsto\sigma_w\in\Z_p$ with $\sigma_w^2=-23$, and we put $G^{(w)}=G_0+\sigma_wG_1\in\Z_p[x,y,z,t]$; $\bar w$ denotes the other place above $p$. Points of $S(\Q_p)$ are always given by primitive coordinates in $\Z_p^4$, and $\bar P$ is the reduction of $P$ modulo $p$. Local invariants $\operatorname{inv}_w$ and Frobenius elements $\operatorname{Frob}_w\in\operatorname{Gal}(L/K)$ are those of \cite[Chapters V--VI]{Neukirch99}. The proof goes as follows. We construct a twisted cubic $C\subset S_L$ with $\divv_S(G)=C+\tau C+\tau^2C$ (Step 2), so that the cyclic algebra $\mathcal B=(L/K,\tau,G/(288x^3))$ lies in $\operatorname{Br}(S_K)$ (Step 3). After determining the reduction of $S$ modulo all primes (Step 4), we show that for every prime $p$ and every $P\in S(\Q_p)$ the sum of the local invariants of $\mathcal B(P)$ at the places above $p$ is $0$ if $p\ne3$ (Step 5) and $2/3$ if $p=3$ (Step 6). At a rational point the sum over all places would be $0$ (Step 7). \paragraph{Step 1: the fields.} \begin{lem}\label{lem:1.4.1-field} \begin{enumerate}[label=(\roman*)] \item $h$ is irreducible over $K$. Put $\tau^2(a):=-\bigl(23a-s(6a^2+9a-4)\bigr)/46$ (the expression for $\tau(a)$ with $s$ replaced by $-s$). Then $a,\tau(a),\tau^2(a)$ are the three roots of $h$ in $L$, and the $K$-automorphism $\tau$ of $L$ with $a\mapsto\tau(a)$ maps $\tau(a)$ to $\tau^2(a)$ and $\tau^2(a)$ to $a$. Hence $L/K$ is cyclic of degree 3 and $\operatorname{Gal}(L/K)=\langle\tau\rangle$. \item Let $w$ be a place of $K$ above a prime $p$. If $p$ is inert in $K$ or $p=23$, then $h$ splits into linear factors over $K_w$, so $L\otimes_KK_w\cong K_w^3$. If $p$ splits in $K$, then either $h\bmod p$ has three distinct roots and $L\otimes_KK_w\cong K_w^3$, or $h\bmod p$ is irreducible; in the latter case $L_w:=L\otimes_KK_w$ is the unramified cubic extension of $K_w=\Q_p$, $\operatorname{Frob}_w=\tau^{e}$ and $\operatorname{Frob}_{\bar w}=\tau^{-e}$ for some $e\in\{1,2\}$. In particular $L/K$ is unramified at all places. \item The primes $2,3,29,31$ split in $K$, and $h$ is irreducible modulo each of them; $59$ splits in $K$, and $h$ has the roots $17,46,55$ modulo 59; $7,337,9013$ are inert in $K$. Let $w_+$ and $w_-$ be the places of $K$ above 3 with $\sigma_{w_+}\equiv2$ and $\sigma_{w_-}\equiv-2\pmod{27}$. Then $\operatorname{Frob}_{w_+}=\tau$ and $\operatorname{Frob}_{w_-}=\tau^2$. \end{enumerate} \end{lem} \begin{proof} (i) $h(\pm1)\ne0$, so $h$ is irreducible over $\Q$, and a root in the quadratic field $K$ would generate a cubic subfield. That $\tau(a)$ is a root of $h$ and that $\tau(\tau(a))=\tau^2(a)$, $\tau(\tau^2(a))=a$ are identities in $L$ (\code{1.4.1a}). (ii) We have $\operatorname{disc}h=-23$. Let $p\ne2,23$, and let $\delta=(r_1-r_2)(r_1-r_3)(r_2-r_3)$ for the roots $r_i$ of $h\bmod p$ in $\bar\F_p$. Then $\delta^2=-23\ne0$, and the Frobenius $u\mapsto u^p$ maps $\delta$ to $\pm\delta$ according to the parity of the permutation it induces on the roots. Hence $h\bmod p$ has exactly one root if $\bigl(\frac{-23}p\bigr)=-1$, and three roots or none if $\bigl(\frac{-23}p\bigr)=1$. If $p$ is inert, Hensel's lemma gives $h=(T-r)g(T)$ over $\Z_p$, and $\operatorname{disc}g\cdot g(r)^2=\operatorname{disc}h=-23$ shows that $g$ splits over $K_w=\Q_p(\sqrt{-23})$. Modulo 23 we have $h\equiv(T-20)(T-13)^2$ with coprime factors, and the same argument applies, with $g(r)\equiv(20-13)^2\equiv3\pmod{23}$ a unit. If $p$ splits and $h\bmod p$ has three roots, they lift to $\Z_p$. If $h\bmod p$ is irreducible, $L\otimes_KK_w\cong\Q_p[T]/(h)$ ($a\mapsto T$, $s\mapsto\sigma_w$) is the unramified cubic extension of $\Q_p$, and $\tau$ acts on it as the automorphism $\alpha_w\colon T\mapsto P(T,\sigma_w)$, $P(T,s)=-(23T+s(6T^2+9T-4))/46$; $\alpha_w\ne1$ because $\tau(a)\ne a$. By (i), $P(P(a,s),s)=\tau^2(a)=P(a,-s)$, so $\alpha_{\bar w}=\alpha_w^2$ on $\Q_p[T]/(h)$. If $\alpha_w=\varphi^k$ for the Frobenius $\varphi$ of $\Q_p[T]/(h)$, then $\operatorname{Frob}_w=\tau^{e}$ with $e\equiv k^{-1}$ and $\operatorname{Frob}_{\bar w}=\tau^{e'}$ with $e'\equiv(2k)^{-1}\equiv-e\pmod 3$. For $p=2$ we have $-23\equiv1\pmod8$, so 2 splits in $K$, and $h\bmod2$ has no root, so it is irreducible. Finally, $L\otimes_KK_w$ is unramified over $K_w$ for $p\ne23$ because $h\bmod p$ is separable, and for $p=23$ because it is split. (iii) The splitting data are computed by \code{1.4.1a}. In $\F_{27}=\F_3[T]/(h)$ we have $P(T,2)\equiv T+2\equiv T^3$ and $P(T,-2)\equiv T+1\equiv T^9$ (\code{1.4.1a}), so $\alpha_{w_+}=\varphi$ and $\alpha_{w_-}=\varphi^2$. \end{proof} \paragraph{Step 2: the twisted cubic.} Let $q_0,q_1,q_2$ be the three quadratic forms over $L$ recorded in \mbox{\path{data/cubic_a_quadrics.json}} of \code{1.4.1b}. Every monomial of $q_0$ and of $q_1$ contains $x$ or $y$, so on the plane $(x,y)=(\lambda w,\mu w)$ we have \[ q_k(\lambda w,\mu w,z,t)=w\,\bigl(A_kw+B_kz+C_kt\bigr)\qquad(k=0,1), \] with binary forms $A_k$ of degree 2 and $B_k,C_k$ of degree 1 in $(\lambda,\mu)$. Put \[ \begin{gathered} W=B_0C_1-C_0B_1,\qquad Z=C_0A_1-A_0C_1,\qquad T=A_0B_1-B_0A_1,\\ \phi\colon\mathbb P^1_L\to\mathbb P^3_L,\qquad (\lambda:\mu)\mapsto(\lambda W:\mu W:Z:T), \end{gathered} \] so that $W$ is a quadratic and $Z,T$ are cubic binary forms; $(W:Z:T)$ is the common zero of the two linear forms $A_kw+B_kz+C_kt$ ($k=0,1$), and $\phi(\lambda:\mu)$ lies on the plane $\mu x=\lambda y$ through the line $x=y=0$. For $j=1,2$ let $\tau^j\phi$ be the map obtained by applying $\tau^j$ to the coefficients of $\phi$, so that $(\tau^j\phi)(\mathbb P^1)=\tau^j(\phi(\mathbb P^1))$. \begin{lem}\label{lem:1.4.1-curve} \begin{enumerate}[label=(\roman*)] \item $\phi^*q_0=\phi^*q_1=\phi^*q_2=0$, $\phi^*F=0$ and $\phi^*G=0$. \item The four binary cubics $\lambda W,\mu W,Z,T$ are linearly independent over $L$. Hence $C:=\phi(\mathbb P^1_L)$ is a twisted cubic: a smooth, geometrically integral curve of degree 3 and genus 0. It lies on $S_L$, on the cubic surface $G=0$ and on the three quadrics $q_k=0$. \item For $j=1,2$ the binary form $q_0(\tau^j\phi)$ is not $0$. Hence $\tau C$ and $\tau^2C$ do not lie on the quadric $q_0=0$, and $C$, $\tau C$, $\tau^2C$ are pairwise distinct. The line $x=y=0$ meets $C$ in $P_0=(0:0:a:1)$ and $P_1=(0:0:\tau(a):1)$, and $P_2=(0:0:\tau^2(a):1)\notin C$. \item $\divv_S(G)=C+\tau C+\tau^2C$. \end{enumerate} \end{lem} \begin{proof} (i)--(iii) are exact computations in $L[\lambda,\mu]$ (\code{1.4.1b}). In (ii), four linearly independent binary cubics form a basis of the space of binary cubics, so $\phi$ is the composition of $(\lambda:\mu)\mapsto(\lambda^3:\lambda^2\mu:\lambda\mu^2:\mu^3)$ with a projective linear transformation. In (iii), if $\tau C=C$ then $\tau^2C=C$, which is excluded, and $\tau^2C\ne\tau C$ follows by applying $\tau$. The points: $\operatorname{Res}(W,Z-aT)=\operatorname{Res}(W,Z-\tau(a)T)=0$, $\operatorname{Res}(W,Z-\tau^2(a)T)\ne0$ and $\operatorname{Res}(W,T)\ne0$ (\code{1.4.1b}). (iv) $F$ and $G$ have coefficients in $K$, so by (i) all three curves $\tau^jC$ lie on $S$ and on $G=0$. The coefficient of $x^3$ is 1 in $F$ and 0 in $G\ne0$, so $G$ is not a multiple of $F$ and does not vanish on the irreducible surface $S$. By Bézout's theorem \cite[Theorem I.7.7]{Hartshorne77}, $\divv_S(G)$ (the divisor of $G/\ell^3$ plus three times the plane section $\ell=0$, for any linear form $\ell$) is an effective divisor of degree $3\cdot3=9$. It contains the three distinct prime divisors $C,\tau C,\tau^2C$ of degree 3, so it is their sum. \end{proof} \paragraph{Step 3: the Brauer class.} For a linear form $\ell$ over $K$ let $H_\ell$ be the plane section $\ell=0$ of $S$. Put $\psi=G/(288x^3)\in K(S)^\times$. By Lemma~\ref{lem:1.4.1-curve}(iv), \[ \divv_S(\psi)=C+\tau C+\tau^2C-3H_x=N_{L/K}(C-H_x), \] so by Proposition~\ref{thm:AD}(1) the cyclic algebra $\mathcal B=(L/K,\tau,\psi)$ lies in $\operatorname{Br}(S_K)$. Since $G/(288\ell^3)=\psi\cdot(x/\ell)^3$, Proposition~\ref{thm:AD}(1) also gives: for every field $K'\supseteq K$, every linear form $\ell$ over $K$ and every $P\in S(K')$ with $G(P)\,\ell(P)\ne0$ (so that $P$ is not on $C\cup\tau C\cup\tau^2C\cup H_\ell$), \begin{equation}\label{eq:1.4.1-eval} \mathcal B(P)=\bigl(L\otimes_KK'/K',\ \tau,\ G(P)/(288\,\ell(P)^3)\bigr). \end{equation} \paragraph{Step 4: reduction modulo primes.} \begin{lem}\label{lem:1.4.1-bad} \begin{enumerate}[label=(\roman*)] \item $S$ is smooth. For every prime $p\notin\{2,3,23,31,59,9013\}$ the surface $\bar S_p\colon F\equiv0\pmod p$ is smooth over $\bar\F_p$, hence geometrically integral; $\bar S_3$ is smooth as well. \item Over $\bar\F_{31}$, the only singular point of $\bar S_{31}$ is $P_{31}=(1:23:28:12)$, and $\bar S_{31}$ is geometrically integral. The square roots of $-23$ modulo 31 are $15$ and $16$, and $G(P_{31})\equiv6$ for $s=15$, $G(P_{31})\equiv16$ for $s=16$ $\pmod{31}$. \item $\bar S_p$ is singular for $p=2,23,31,59,9013$. Hence these are exactly the primes of bad reduction of $S$. \end{enumerate} \end{lem} \begin{proof} (i) For an order of the variables, the Macaulay matrix in degree 5 of the four partial derivatives of $F$ has its 56 rows indexed by the monomials of degree 5, and the column of a monomial $m$ is the coefficient vector of $(m/v^2)\,\partial F/\partial v$, where $v$ is the first variable in the given order with $v^2\mid m$. If $\xi\in\bar\F_p^4\setminus\{0\}$ is a common zero of the partial derivatives modulo $p$, then the vector $(m(\xi))_m\ne0$ is in the left kernel of the matrix modulo $p$, so $p$ divides its determinant. For the orders $(x,y,z,t)$ and $(x,y,t,z)$ the determinants are \[ 2^6\cdot3^{17}\cdot7\cdot23\cdot29\cdot31\cdot59^2\cdot337\cdot9013^2\quad\text{and}\quad2^6\cdot3^{19}\cdot23\cdot31\cdot59^3\cdot73\cdot9013^2 \] (\code{1.4.1c}), with greatest common divisor $2^6\cdot3^{17}\cdot23\cdot31\cdot59^2\cdot9013^2$. For $p\ne3$, Euler's identity $3F=\sum_vv\,\partial F/\partial v$ shows that a singular point of $\bar S_p$ is a common zero of the partial derivatives, so $\bar S_p$ is smooth for $p\notin\{2,3,23,31,59,9013\}$; over $\bar\Q$ the same argument shows that $S$ is smooth. A cubic surface that is reducible or non-reduced over $\bar\F_p$ is singular along a curve (all partial derivatives of $F_1F_2$ vanish on $V(F_1)\cap V(F_2)$), so a cubic surface with finitely many singular points is geometrically integral. For $p\in\{3,31\}$ consider the $\F_p$-linear map $(g_0,g_1,\dots,g_4)\mapsto g_0F+\sum_ig_i\,\partial F/\partial v_i$ from (forms of degree 2)$\times$(forms of degree 3)$^4$ to the 56-dimensional space of forms of degree 5. A common zero $\xi\ne0$ of $F$ and its partial derivatives over $\bar\F_p$ gives the nonzero linear form ``evaluation at $\xi$'' on the forms of degree 5, which vanishes on the image, and two distinct points of $\mathbb P^3(\bar\F_p)$ give two linearly independent linear forms. The rank of the map is 56 for $p=3$ and 55 for $p=31$, and $F$ and its partial derivatives vanish at $P_{31}$ modulo 31 (\code{1.4.1c}). This proves the claims on $\bar S_3$ and the first claim of (ii); the values of $G$ are computed by \code{1.4.1c}. (iii) $F$ and its partial derivatives vanish modulo $p$ at $(1:18:3:10)$, $(1:23:28:12)$, $(1:47:12:14)$, $(1:7636:7996:3944)$ for $p=23,31,59,9013$, and modulo 2 at the three points $(1:r^3+r:r^5:r)$ over $\F_8$, $r^3+r^2+1=0$ (\code{1.4.1c}). \end{proof} \paragraph{Step 5: the places not above 3.} \begin{lem}\label{lem:1.4.1-zero} Let $p\ne3$ be a prime and $P\in S(\Q_p)$. Then $\sum_{w\mid p}\operatorname{inv}_w\mathcal B(P)=0$, and for $p\ne2$ each term is $0$. At the complex place $w_\infty$ of $K$, $\operatorname{inv}_{w_\infty}\mathcal B(P)=0$ for every $P\in S(\mathbb C)$. \end{lem} \begin{proof} $\operatorname{Br}(\mathbb C)=0$. If $L\otimes_KK_w\cong K_w^3$, which by Lemma~\ref{lem:1.4.1-field}(ii) is the case when $p$ is inert in $K$, when $p=23$, and when $p$ splits and $h\bmod p$ has three roots (for example $p=59$), then $\mathcal B(P)$ is split at $w$ (\cite[Chapters V--VI]{Neukirch99}). It remains to treat the primes $p$ that split in $K$ with $h\bmod p$ irreducible. Then $L_w/\Q_p$ is the unramified cubic extension, $\operatorname{Frob}_w=\tau^e$ and $\operatorname{Frob}_{\bar w}=\tau^{-e}$. $p=2$. Choose $\ell\in\{x,y,z,t\}$ with $\ell(P)\in\Z_2^\times$. For every primitive solution of $F\equiv0\pmod{32}$, $G^{(w)}$ has 2-adic valuation 1 if $\sigma_w\equiv5\pmod8$ and 4 if $\sigma_w\equiv3\pmod8$ (\code{1.4.1d}); since these valuations are less than 5, the same holds at $P$. In particular $G(P)\ne0$, and by~\eqref{eq:1.4.1-eval} the algebra $\mathcal B(P)$ at $w$ is $(L_w/\Q_2,\tau,c_w)$ with $v_2(c_w)=v_2(G^{(w)}(P))-v_2(288)\in\{-4,-1\}$, so $v_2(c_w)\equiv v_2(c_{\bar w})\equiv2\pmod3$. By \cite[Chapters V--VI]{Neukirch99}, $\operatorname{inv}_w\mathcal B(P)+\operatorname{inv}_{\bar w}\mathcal B(P)=\bigl(2e+2(-e)\bigr)/3=0$. $p\ge5$ (this covers $p=29$, $p=31$ and all primes of good reduction with $h\bmod p$ irreducible). By Lemma~\ref{lem:1.4.1-bad}, $\bar S_p$ is geometrically integral, and $\bar P$ is a smooth point of $\bar S_p$ unless $p=31$ and $\bar P=P_{31}$. The reduction of $G^{(w)}$ does not vanish on $\bar S_p$: otherwise it would be a multiple of $F\bmod p$, hence $0$ (compare the coefficients of $x^3$), but its coefficient of $xy^2$ is $288\not\equiv0$. If $\bar P$ is a smooth point, Proposition~\ref{thm:AD}(3) (with $\psi=G/(288\ell^3)$, $\ell$ a coordinate with $\ell(\bar P)\ne0$, and $v_p(288)=0$) gives $\operatorname{inv}_w\mathcal B(P)=0$. If $p=31$ and $\bar P=P_{31}$, then $x(P)$ is a unit and $G^{(w)}(P)\equiv G^{(w)}(P_{31})\equiv6$ or $16\pmod{31}$ is a unit (Lemma~\ref{lem:1.4.1-bad}(ii)), so by~\eqref{eq:1.4.1-eval} with $\ell=x$ and \cite[Chapters V--VI]{Neukirch99}, $\operatorname{inv}_w\mathcal B(P)=e\cdot0/3=0$. \end{proof} \paragraph{Step 6: the places above 3.} \begin{lem}\label{lem:1.4.1-three} For every $P\in S(\Q_3)$ we have $v_3(G^{(w_+)}(P))=0$, $v_3(G^{(w_-)}(P))=1$, and \[ \operatorname{inv}_{w_+}\mathcal B(P)+\operatorname{inv}_{w_-}\mathcal B(P)=\tfrac23 . \] \end{lem} \begin{proof} The primitive coordinates of $P$ reduce to one of the $10{,}206$ primitive solutions of $F\equiv0\pmod{27}$. For each of them, with $\sigma_{w_+}\equiv2$ and $\sigma_{w_-}\equiv25\pmod{27}$, the value of $G^{(w_+)}$ has 3-adic valuation 0 and that of $G^{(w_-)}$ has valuation 1 (\code{1.4.1e}; the same holds for the $275{,}562$ primitive solutions modulo 81). Valuations less than 3 are determined modulo 27, which proves the first claim. Choose $\ell\in\{x,y,z,t\}$ with $\ell(P)\in\Z_3^\times$, and put $c=G(P)/(288\ell(P)^3)$; since $v_3(288)=2$, $v_{w_+}(c)=-2$ and $v_{w_-}(c)=-1$. By~\eqref{eq:1.4.1-eval}, Lemma~\ref{lem:1.4.1-field}(iii) and \cite[Chapters V--VI]{Neukirch99}, \[ \operatorname{inv}_{w_+}\mathcal B(P)=\frac{1\cdot(-2)}3\equiv\frac13,\qquad \operatorname{inv}_{w_-}\mathcal B(P)=\frac{2\cdot(-1)}3\equiv\frac13\pmod1 . \] (With the opposite sign convention for the invariants the sum is $1/3$; it is nonzero in either case.) \end{proof} \paragraph{Step 7: conclusion.} \begin{proof}[Proof of Theorem~\ref{thm:1.4.1}] Suppose $P\in S(\Q)$. Then $P\in S(K)$, and by Proposition~\ref{thm:AD}(4), $\sum_w\operatorname{inv}_w\mathcal B(P)=0$, the sum over all places of $K$. Grouping the places by the primes below them, Lemmas~\ref{lem:1.4.1-zero} and~\ref{lem:1.4.1-three} give $\sum_w\operatorname{inv}_w\mathcal B(P)=2/3$, a contradiction. Hence $S(\Q)=\varnothing$. A nonzero integer solution would give a rational point of $S$. \end{proof} \begin{rem} $S$ has points over $\R$ and over every $\Q_p$: at $p=2,23,31,59,9013$ the points $(1:0:1:1)$, $(1:0:0:6)$, $(1:0:0:20)$, $(1:0:1:10)$, $(1:0:0:8363)$ are smooth points of $\bar S_p$, at $p=3$ the point $(1:0:0:1)$ is a smooth point of $\bar S_3$ (\code{1.4.1f}), at the other primes $\bar S_p$ is smooth and has an $\F_p$-point by the Chevalley--Warning theorem, and all of them lift by Hensel's lemma; $F(1,0,0,t)=t^3+t^2+1$ has a real root. So the obstruction is not local. In terms of $\mathcal A=\operatorname{Cor}_{K/\Q}\mathcal B\in\operatorname{Br}(S)$, whose local invariant at $P\in S(\Q_p)$ is $\sum_{w\mid p}\operatorname{inv}_w\mathcal B(P)$, Steps 5 and 6 say that $\sum_p\operatorname{inv}_p\mathcal A(P_p)=2/3$ for every adelic point $(P_p)$; so $S(\mathbb A_\Q)^{\operatorname{Br}}=\varnothing$, and the class of $\mathcal A$ in $\operatorname{Br}(S)/\operatorname{Br}(\Q)$ has order 3. \end{rem} \subsection{\texorpdfstring{$2x^3+x^2y+xy^2+y^3+y^2z-2xz^2-z^3+t^2y+tz^2-t^3=0$ \ ($H=96$)}{2x^3+x^2y+xy^2+y^3+y^2z-2xz^2-z^3+t^2y+tz^2-t^3=0 (H=96)}}\label{sec:1.4.2} \begin{thm}\label{thm:1.4.2} The only integer solution of \[ F(x,y,z,t)=2x^3+x^2y+xy^2+y^3+y^2z-2xz^2-z^3+t^2y+tz^2-t^3=0 \] is $(x,y,z,t)=(0,0,0,0)$. \end{thm} \paragraph{Setup.} Since $F$ is homogeneous, it suffices to show that the cubic surface $S:F=0$ in $\mathbb P^3$ has no rational point. Let $K=\Q(s)$ with $s^2=-23$, let $h(T)=T^3-T+1$, let $L$ be the splitting field of $h$ over $\Q$, and let $q\in L$ be a root of $h$. Put \[ \ell_\pm=(-1\pm s)x+4y,\qquad \rho=-\frac{(s-1)q^2+8}{4(3q^2-1)}. \] The plane $\ell_+=0$ meets $S$ in three lines that are defined over $L$ and permuted cyclically by $\operatorname{Gal}(L/K)$ (Step 2). For a rational point $P$ of $S$ we put $\psi=\ell_+(P)/\ell_-(P)\in K^\times$ and compute the local invariants $\operatorname{inv}_w(\psi)$ of the cyclic cubic extension $L/K$ (\cite[Chapters V--VI]{Neukirch99}). They vanish at every place of $K$ except the two places above 2 (Steps 3--4), and there they are equal and nonzero (Step 5), which contradicts the reciprocity law. For a place $w$ of $K$ we write $K_w$ for the completion and $v_w$ for the normalized valuation. Complex conjugation $c$ is the nontrivial automorphism of $K$, and $c(w)$ is the place with $v_{c(w)}(c(\alpha))=v_w(\alpha)$. Since $P$ is rational, $c(\ell_+(P))=\ell_-(P)$ and $c(\psi)=\psi^{-1}$. \paragraph{Step 1: the fields.} \begin{lem}\label{lem:1.4.2-field} \begin{enumerate}[label=(\roman*)] \item $h$ is irreducible over $\Q$ and $\operatorname{disc}(h)=-23$. $L=K(q)\cong K[T]/(h)$, $[L:K]=3$, and $\operatorname{Gal}(L/K)$ is cyclic, generated by the automorphism $\tau$ with $\tau(q)=\bigl(-q+s/h'(q)\bigr)/2$. Every element of $\operatorname{Gal}(L/\Q)$ outside $\operatorname{Gal}(L/K)$ conjugates $\tau$ to $\tau^{-1}$. \item Let $w$ be a place of $K$ above a prime $p$. If $h$ has a simple root modulo $p$, then $L\otimes_KK_w\cong K_w^3$, that is, $L/K$ is split at $w$. This is the case for $p=23$ ($h\equiv(T-13)^2(T-20)$), for every $p$ that is inert in $K$, and for every $p$ that splits in $K$ and modulo which $h$ has a root. \item If $p$ splits in $K$ and $h$ has no root modulo $p$, then $K_w=\Q_p$, $L\otimes_KK_w$ is the unramified extension of $\Q_p$ of degree 3, $w$ is unramified in $L$, and its Frobenius element generates $\operatorname{Gal}(L/K)$. This is the case for $p=2$ and $p=3$. \item Every prime $p$ falls under (ii) or (iii). \end{enumerate} \end{lem} \begin{proof} (i) $\pm1$ are not roots of $h$, so $h$ is irreducible, and $\operatorname{disc}(h)=-4(-1)^3-27=-23$. Let $q_1,q_2,q_3$ be the roots of $h$ in $L$ and $\delta=(q_1-q_2)(q_1-q_3)(q_2-q_3)$; then $\delta^2=-23$, so $K=\Q(\delta)\subseteq L$. The group $\operatorname{Gal}(L/\Q)$ acts faithfully and transitively on the roots, and the elements that fix $\delta$ are the even permutations; since $\delta\notin\Q$, $\operatorname{Gal}(L/\Q)\cong S_3$ and $\operatorname{Gal}(L/K)\cong A_3$ is cyclic of order 3. Hence $L=K(q)$, and $L\cong K[T]/(h)$ since $h$ is irreducible over $K$. The transpositions invert the 3-cycles. The element $(-q+s/h'(q))/2$ is a root of $h$ different from $q$ (we have $(q-q_2)(q-q_3)=h'(q)$ and $(q_2-q_3)^2=-23/h'(q)^2$), and the $K$-automorphism $\tau$ that maps $q$ to it has order 3 (\code{1.4.2b}). (ii) We have $L\otimes_KK_w\cong K_w[T]/(h)$, and this is the product of the completions of $L$ at the places above $w$, which are all isomorphic since $L/K$ is Galois. A simple root modulo $p$ lifts to a root in $\Z_p\subseteq K_w$ by Hensel's lemma, so one factor is $K_w$, and hence all of them are. Modulo 23, $h\equiv(T-13)^2(T-20)$ and 20 is a simple root. If $p$ is inert in $K$, then $p\ne2$ ($-23\equiv1\pmod 8$, so 2 splits) and $p\ne23$, so the roots $\theta_1,\theta_2,\theta_3$ of $h$ in $\overline{\F}_p$ are distinct, and $\bar\delta=\prod_{i0$, and put $y_0=(1-s)x/4$. Then $y-y_0=\ell_+(P)/4$ has positive valuation, so $F(x,y_0,z,t)\equiv F(x,y,z,t)=0$ modulo $\mathfrak W$. Put $m_i=t-\tau^i(\rho)x-\tau^i(q)z$. By Lemma~\ref{lem:1.4.2-plane}(i), (iii), $\prod_im_i$ has positive $\mathfrak W$-valuation and every $m_i$ is $\mathfrak W$-integral, so $v_{\mathfrak W}(m_i)>0$ for some $i$. Since $x,z,t\in\Q$, we have $m_i=\tau^i(m_0)$, and therefore $v_{\mathfrak W}(m_0)=v_{\mathfrak W}(m_1)=v_{\mathfrak W}(m_2)>0$. If $M$ is the matrix of Lemma~\ref{lem:1.4.2-plane}(ii), then $M\,(x,z,t)^{\mathsf T}=(m_0,m_1,m_2)^{\mathsf T}$, and $\det M=(s-1)/2$ is a unit at $\mathfrak W$ since $p\nmid6$. Hence $x,z,t\equiv0$ modulo $\mathfrak W$, that is, $p\mid x,z,t$, and then $4y=\ell_+(P)-(-1+s)x$ gives $p\mid y$. This contradicts the coprimality of the coordinates. The place $c(w)$ is of the same type, so also $v_{c(w)}(\ell_+(P))=0$, that is, $v_w(\ell_-(P))=v_w(c(\ell_+(P)))=0$. Hence $v_w(\psi)=0$, and $\operatorname{inv}_w(\psi)=0$ by \cite[Chapters V--VI]{Neukirch99}. \end{proof} For $5\le p<500$ the conclusion is also confirmed directly: for each of the 32 such primes and both square roots of $-23$ modulo $p$, the plane cubic $F(x,(1-s)x/4,z,t)$ has no zero in $\mathbb P^2(\F_p)$ (\code{1.4.2d}). \paragraph{Step 4: the places above 2 and 3.} Let $s_2\in\Z_2$ be the root of $s^2=-23$ with $s_2\equiv1\pmod4$; then $s_2\equiv45\pmod{256}$. Let $s_3\in\Z_3$ be the root with $s_3\equiv1\pmod3$; then $s_3\equiv25\pmod{81}$ ($s_3\equiv187\pmod{243}$). Let $w_2$ and $w_3$ be the places of $K$ given by $s\mapsto s_2$ and $s\mapsto s_3$. The other places above 2 and 3 are $c(w_2)$ and $c(w_3)$. \begin{lem}\label{lem:1.4.2-two} For every $(x,y,z,t)\in\Z_2^4$, not all even, with $F(x,y,z,t)=0$, the pair $(v_2(\ell_+),v_2(\ell_-))$ (with $s=s_2$) is $(2,1)$, $(5,1)$ or $(2,4)$, according as $(x,y,z,t)$ is congruent modulo 2 to $(1,0,0,0)$, $(1,1,1,1)$ or $(0,1,1,1)$. In particular $v_{w_2}(\psi)\equiv1\pmod3$. \end{lem} \begin{proof} Modulo 2, $F\equiv x^2y+xy^2+y^3+y^2z+z^3+t^2y+tz^2+t^3$, whose only projective zeros are the three residues of the statement. Moreover, modulo 16 every solution with $x,y,z,t$ odd has $y\equiv13x$, and every solution with $x$ even and $y,z,t$ odd has $x\equiv6y$ (\code{1.4.2c}). We have $(s_2-1)/4\equiv11\pmod{64}$ and $(s_2+1)/2\equiv23\pmod{128}$, so $v_2(s_2-1)=2$ and $v_2(s_2+1)=1$. If $x$ is odd and $y$ is even, then $v_2(\ell_+)=v_2((s_2-1)x)=2$ and $v_2(\ell_-)=v_2(-(s_2+1)x)=1$. If $x,y,z,t$ are odd, then $v_2(\ell_-)=1$, and $\ell_+/4=\frac{s_2-1}4x+y\equiv11x+13x=24x\pmod{16}$ has valuation 3, so $v_2(\ell_+)=5$. If $x$ is even and $y,z,t$ are odd, then $v_2(\ell_+)=2$, and $\ell_-/4=y-\frac{s_2+1}2\cdot\frac x2\equiv y-23\cdot3y=-68y\pmod 8$ has valuation 2, so $v_2(\ell_-)=4$. In each case $v_2(\ell_+)-v_2(\ell_-)\equiv1\pmod 3$, and $v_{w_2}(\psi)=v_2(\ell_+(P))-v_2(\ell_-(P))$. \end{proof} The lemma is also confirmed by the complete residue tree of all primitive 2-adic points modulo $2^8$ (49152 normalized classes; every pair is determined from $2^6$ on) and by a direct enumeration of all primitive solutions modulo $2^6$ (\code{1.4.2c}). \begin{lem}\label{lem:1.4.2-three} For every $(x,y,z,t)\in\Z_3^4$, not all divisible by 3, with $F(x,y,z,t)=0$, the pair $(v_3(\ell_+),v_3(\ell_-))$ (with $s=s_3$) is $(3,0)$ if $(x,y,z,t)$ is congruent to a multiple of $(1,0,0,-1)$ modulo 3, and $(0,0)$ otherwise. In particular $v_{w_3}(\psi)\equiv0\pmod3$. \end{lem} \begin{proof} Since $s_3\equiv1\pmod3$, we have $\ell_+\equiv y$ and $\ell_-\equiv x+y\pmod 3$. The projective zeros of $F$ modulo 3 are $(1,0,0,-1)$, $(1,1,1,0)$, $(1,1,-1,0)$, $(0,1,0,-1)$ (\code{1.4.2c}), and at the last three both forms are units. At the first one, scale to $x=1$ and write $y=3Y$, $z=3Z$, $t=-1+3T$. Then \[ \tfrac13F=1+2Y-3T+3Y^2-6TY+9\bigl(-T^3+T^2Y+T^2+TZ^2+Y^3+Y^2Z-Z^3-Z^2\bigr). \] Modulo 3 this gives $Y\equiv1$, and then modulo 9, with $Y=1+3Y_1$, it gives $6(1+Y_1)\equiv0$, so $Y\equiv7\pmod9$. With $Y=7+9Y_2$, the equation modulo 27 becomes $9(2Y_2+1+2T^2+TZ^2-Z^2)\equiv0$, that is, $Y_2\equiv1+2T^2+TZ^2+2Z^2\pmod 3$, and $2T^2+TZ^2+2Z^2\in\{0,2\}$ modulo 3 for all $T,Z$. Hence $Y\equiv7$ or $16\pmod{27}$. Now $\ell_+=(s_3-1)+12Y\equiv12(Y+2)\pmod{81}$ has valuation $1+2=3$, and $\ell_-\equiv1\pmod3$ is a unit. \end{proof} The lemma is also confirmed by the complete residue tree modulo $3^5$ (26244 classes; every pair is determined from $3^4$ on) and by a direct enumeration of all primitive solutions modulo $3^4$ (\code{1.4.2c}). \paragraph{Step 5: reciprocity.} \begin{proof}[Proof of Theorem~\ref{thm:1.4.2}] Suppose that $S$ has a rational point $P$, and let $\psi$ be as in Step 3. By \cite[Chapter VI, Section 5]{Neukirch99}, $\sum_w\operatorname{inv}_w(\psi)=0$, and by Lemma~\ref{lem:1.4.2-good} only the places $w_2,c(w_2),w_3,c(w_3)$ can contribute. They are unramified in $L$ with Frobenius elements generating $\operatorname{Gal}(L/K)$ (Lemma~\ref{lem:1.4.2-field}(iii)). Write $\operatorname{Frob}_{w_2}=\tau^m$ with $m\in\{1,2\}$. By Lemma~\ref{lem:1.4.2-field}(i), the hypothesis of \cite[Chapters V--VI]{Neukirch99} holds (with $k=\Q$), and $c(\psi)=\psi^{-1}$; so $\operatorname{inv}_{c(w)}(\psi)=\operatorname{inv}_w(\psi)$ for $w=w_2,w_3$. By \cite[Chapters V--VI]{Neukirch99} and Lemma~\ref{lem:1.4.2-three}, $\operatorname{inv}_{w_3}(\psi)=0$, since $v_{w_3}(\psi)\equiv0\pmod 3$. By \cite[Chapters V--VI]{Neukirch99} and Lemma~\ref{lem:1.4.2-two}, $\operatorname{inv}_{w_2}(\psi)=m\,v_{w_2}(\psi)/3\equiv m/3\pmod 1$. Hence \[ \sum_w\operatorname{inv}_w(\psi)=2\operatorname{inv}_{w_2}(\psi)=\frac{2m}3\not\equiv0\pmod1, \] a contradiction. Therefore $S(\Q)=\varnothing$. A nonzero integer solution would give a rational point of $S$, so the only integer solution is $(0,0,0,0)$. \end{proof} \subsection{\texorpdfstring{$2x^3+2x^2y+2y^3+y^2z-2xz^2-z^3+tz^2-t^3=0$ \ ($H=96$)}{2x^3+2x^2y+2y^3+y^2z-2xz^2-z^3+tz^2-t^3=0 (H=96)}}\label{sec:1.4.3} \begin{thm}\label{thm:1.4.3} The only integer solution of \[ F(x,y,z,t)=2x^3+2x^2y+2y^3+y^2z-2xz^2-z^3+tz^2-t^3=0 \] is $(x,y,z,t)=(0,0,0,0)$. \end{thm} \paragraph{Setup.} The proof is the same as that of Section~\ref{sec:1.4.2}, with a different decisive prime. It suffices to show that the cubic surface $S:F=0$ in $\mathbb P^3$ has no rational point. Let $K=\Q(s)$ with $s^2=-11$, let $g(U)=U^3-4U-4$, let $L$ be the splitting field of $g$ over $\Q$, let $u\in L$ be a root of $g$, and put $q=u/2$, so that $2q^3-2q-1=0$. (The field $\Q(u)$ is also generated by a root $v$ of $h_C(V)=V^3-V^2-V-1$: $u=v^2-1$ and $v=u^2/2-1$; and $L$ is the splitting field of $h_C$.) Put \[ \ell_\pm=(2\pm s)y+3t,\qquad \rho=\frac{s+2-6q^2}{6(3q^2-1)}. \] The plane $\ell_+=0$ meets $S$ in three lines permuted cyclically by $\operatorname{Gal}(L/K)$ (Step 2). For a rational point $P$ we put $\psi=\ell_+(P)/\ell_-(P)\in K^\times$; its local invariants for $L/K$ (\cite[Chapters V--VI]{Neukirch99}) vanish at every place of $K$ except the two places above 3 (Steps 3--4), where they are equal and nonzero (Step 5). The notation $K_w$, $v_w$, $c$, $c(w)$ is as in Section~\ref{sec:1.4.2}; again $c(\psi)=\psi^{-1}$. \paragraph{Step 1: the fields.} \begin{lem}\label{lem:1.4.3-field} \begin{enumerate}[label=(\roman*)] \item $g$ is irreducible over $\Q$ and $\operatorname{disc}(g)=-176=-11\cdot4^2$. $L=K(u)\cong K[U]/(g)$, $[L:K]=3$, and $\operatorname{Gal}(L/K)$ is cyclic, generated by the automorphism $\tau$ with $\tau(u)=\bigl(-u+4s/g'(u)\bigr)/2$. Every element of $\operatorname{Gal}(L/\Q)$ outside $\operatorname{Gal}(L/K)$ conjugates $\tau$ to $\tau^{-1}$. \item Let $w$ be a place of $K$ above a prime $p$. If $g$ has a simple root modulo $p$, then $L/K$ is split at $w$. This is the case for $p=11$ ($g\equiv(U-4)^2(U-3)$), for every odd $p$ that is inert in $K$, and for every $p$ that splits in $K$ and modulo which $g$ has a root. \item If $p$ splits in $K$ and $g$ has no root modulo $p$, then $K_w=\Q_p$, $L\otimes_KK_w$ is the unramified extension of $\Q_p$ of degree 3, $w$ is unramified in $L$, and its Frobenius element generates $\operatorname{Gal}(L/K)$. This is the case for $p=3$. \item The prime 2 is inert in $K$: $K_w=\Q_2(s)$ is the unramified quadratic extension of $\Q_2$, and with $\omega=(1+s)/2$ ($\omega^2-\omega+3=0$) we have $2+s=1+2\omega$ and $2-s=3-2\omega$. Every odd prime falls under (ii) or (iii). \end{enumerate} \end{lem} \begin{proof} As for Lemma~\ref{lem:1.4.2-field}. The rational roots of $g$ would divide 4, and $\pm1,\pm2,\pm4$ are not roots. $\operatorname{disc}(g)=-4(-4)^3-27\cdot16=-176$, and $K=\Q(\sqrt{-176})$. Since $u$ has two conjugates with $(u_2-u_3)^2=-176/g'(u)^2$, $\tau(u)$ is a root of $g$ different from $u$, and $\tau$ has order 3 (\code{1.4.3b}). Modulo 11, $g\equiv(U-4)^2(U-3)$. If $p$ is odd, $p\ne11$, the roots of $g$ modulo $p$ are distinct, and the argument with $\bar\delta^2=-176$ shows that $g$ has exactly one root modulo $p$ if $p$ is inert in $K$, and 0 or 3 roots if $p$ splits. The prime 3 splits ($-11\equiv1\pmod3$), and $g\equiv U^3-U-1$ has no root modulo 3. The prime 2 is inert since $-11\equiv5\pmod8$. These facts are checked by \code{1.4.3a}. \end{proof} In fact $h_C(V+1)=V^3+2V^2-2$ is Eisenstein at 2, so $L/K$ is totally (and tamely) ramified at the place above 2, and unramified at every other place; the field discriminant of $\Q(u)$ is $-44$ (\code{1.4.3a}). We do not use this: at 2 the argument of Lemma~\ref{lem:1.4.3-two} works for any cubic extension. \paragraph{Step 2: the plane section.} \begin{lem}\label{lem:1.4.3-plane} \begin{enumerate}[label=(\roman*)] \item As polynomials in $x,y,z$ with coefficients in $L$, \[ F\Bigl(x,y,z,-\frac{(2+s)y}{3}\Bigr)=2\prod_{i=0}^{2}\bigl(x-\tau^i(\rho)\,y-\tau^i(q)\,z\bigr). \] \item The determinant of the matrix with rows $\bigl(1,-\tau^i(\rho),-\tau^i(q)\bigr)$ ($i=0,1,2$) is 2. \item $u$ is an algebraic integer, $3q^2-1=(3u^2-4)/4$, and $N_{\Q(u)/\Q}(3u^2-4)=-\operatorname{disc}(g)=176$. Hence $\tau^i(\rho)$ and $\tau^i(q)$ are integral at every place of $L$ not above $2$, $3$ or $11$. \end{enumerate} \end{lem} \begin{proof} (i) and (ii) are identities in $\Q[u,s]/(g(u),s^2+11)$, with $\tau(u)=-u/2+s(-6u^2+9u+16)/22$ (\code{1.4.3b}); (i) is also verified through the resultant of $g$ with the numerator $6(3u^2-4)(x-uz/2)-4(s+2-\frac32u^2)y$. (iii) $\rho=4(s+2-\frac32u^2)/(6(3u^2-4))$. \end{proof} \begin{lem}\label{lem:1.4.3-nonzero} If $P$ is a rational point of $S$, then $\ell_\pm(P)\ne0$. \end{lem} \begin{proof} $\ell_\pm(P)=0$ implies $y=t=0$, and then $F(x,0,z,0)=2x^3-2xz^2-z^3=0$. If $x=0$, then $z=0$. Otherwise $r=z/x$ is a rational root of $r^3+2r^2-2$, but $\pm1,\pm2$ are not roots (\code{1.4.3b}). \end{proof} \paragraph{Step 3: the places not above 2 and 3.} From now on $P$ is a rational point of $S$ with coprime integer coordinates, and $\psi=\ell_+(P)/\ell_-(P)\in K^\times$. \begin{lem}\label{lem:1.4.3-good} $\operatorname{inv}_w(\psi)=0$ for every place $w$ of $K$ that does not lie above 2 or 3. \end{lem} \begin{proof} At the complex place and at the places of type (ii) of Lemma~\ref{lem:1.4.3-field}, $L/K$ is split. It remains to treat a place $w$ above a prime $p\ge5$ that splits in $K$ (so $p\ne11$) and modulo which $g$ has no root; there is a unique place $\mathfrak W$ of $L$ above $w$, and $v_{\mathfrak W}\circ\tau=v_{\mathfrak W}$. Suppose $v_w(\ell_+(P))>0$. With $t_0=-(2+s)y/3$ we have $t-t_0=\ell_+(P)/3$, so $F(x,y,z,t_0)\equiv0$ modulo $\mathfrak W$. By Lemma~\ref{lem:1.4.3-plane}(i), (iii), some, and hence every, $m_i=x-\tau^i(\rho)y-\tau^i(q)z=\tau^i(m_0)$ has positive $\mathfrak W$-valuation. The determinant 2 of Lemma~\ref{lem:1.4.3-plane}(ii) is a unit at $\mathfrak W$, so $p\mid x,y,z$, and then $3t=\ell_+(P)-(2+s)y$ gives $p\mid t$, a contradiction. Hence $v_w(\ell_+(P))=0$, and likewise $v_{c(w)}(\ell_+(P))=0$, that is, $v_w(\ell_-(P))=0$. So $v_w(\psi)=0$ and $\operatorname{inv}_w(\psi)=0$ by \cite[Chapters V--VI]{Neukirch99}. \end{proof} For the 32 such primes $5\le p<500$ the plane cubic $F(x,y,z,-(2+s)y/3)$ has no zero in $\mathbb P^2(\F_p)$ for either square root of $-11$ modulo $p$ (\code{1.4.3d}). \paragraph{Step 4: the places above 2 and 3.} Let $s_3\in\Z_3$ be the root of $s^2=-11$ with $s_3\equiv1\pmod3$; then $s_3\equiv31\pmod{81}$. Let $w_3$ be the place of $K$ given by $s\mapsto s_3$; the other place above 3 is $c(w_3)$. Let $w_2$ be the place above 2. \begin{lem}\label{lem:1.4.3-two} Every $(x,y,z,t)\in\Z_2^4$, not all even, with $F(x,y,z,t)=0$ has $y$ odd and $t$ even. Consequently $\ell_+(P)\equiv\ell_-(P)\equiv1$ modulo $2\mathcal O_{K_{w_2}}$, $\psi\in1+2\mathcal O_{K_{w_2}}$, and $\operatorname{inv}_{w_2}(\psi)=0$. \end{lem} \begin{proof} Modulo 2, $F\equiv y^2z+z^3+tz^2+t^3$. If $y$ is even and $t$ is odd, then $F\equiv z(y^2+z^2+tz)+t^3\equiv z^3+z^2+1\not\equiv0$; so if $y$ is even, then $t$ and $z$ are even and $x$ is odd, and then $F\equiv2x^2(x+y)\equiv2\pmod 4$. If $t$ is odd, then $F\equiv z^3+z^2+y^2z+1\pmod 2$ forces $z$ odd and then $y$ odd; with $y^2\equiv z^2\equiv t^2\equiv1\pmod8$ we get $F\equiv2(x^3+x^2y+y-x)\pmod8$, which is $\equiv4y\equiv4\pmod 8$ if $x$ is odd and $\equiv2(y-x)\equiv2\pmod4$ if $x$ is even. Hence $y$ is odd and $t$ is even (\code{1.4.3c} checks all primitive solutions modulo 8). By Lemma~\ref{lem:1.4.3-field}(iv), $\ell_\pm\equiv y+t\equiv1$ modulo $2\mathcal O_{K_{w_2}}$, so $\psi\equiv1$. By Hensel's lemma ($X^3-\psi$ at $X=1$; $3$ is a unit) $\psi=b^3$ with $b\in K_{w_2}$, and $b^3$ is the norm of $b$ from every cubic extension of $K_{w_2}$. Hence $\psi$ is a local norm and $\operatorname{inv}_{w_2}(\psi)=0$ by \cite[Chapters V--VI]{Neukirch99}. (Equivalently, by \cite[Chapters V--VI]{Neukirch99}: the residue of $\psi$ is $1$, a cube in $\F_4^\times$.) \end{proof} \begin{lem}\label{lem:1.4.3-three} For every $(x,y,z,t)\in\Z_3^4$, not all divisible by 3, with $F(x,y,z,t)=0$, the pair $(v_3(\ell_+),v_3(\ell_-))$ (with $s=s_3$) is $(1,3)$ if $(x,y,z,t)$ is congruent to a multiple of $(1,0,0,-1)$ modulo 3, and $(1,0)$ otherwise. In particular $v_{w_3}(\psi)\equiv1\pmod3$. \end{lem} \begin{proof} The projective zeros of $F$ modulo 3 are $(1,0,0,-1)$, $(1,1,0,0)$, $(1,-1,0,1)$, $(0,1,0,-1)$ (\code{1.4.3c}). Since $2+s_3\equiv33$ and $2-s_3\equiv-29\pmod{81}$, we have $\ell_+\equiv3(11y+t)$ modulo 81, and $\ell_-\equiv y\pmod3$. At the last three points, $3\nmid y$ and $2y+t\not\equiv0\pmod3$, so the pair is $(1,0)$. At the first point, scale to $t=1$ and write $x=-1+3X$, $y=3Y$, $z=3Z$. Then \[ \tfrac13F=18X^3+18X^2Y-18X^2-12XY-18XZ^2+6X+18Y^3+9Y^2Z+2Y-9Z^3+9Z^2-1 . \] Reducing modulo 3 and 9 gives $Y\equiv5\pmod 9$, and with $Y=5+9Y_2$ the equation modulo 27 gives $Y_2\equiv Z^2(X+1)-X^2-X-1\pmod3$. For $X\equiv0,1,-1$ this is $Z^2-1$, $-Z^2$, $-1$, so it is 0 or 2, never 1; hence $Y\equiv5$ or $23\pmod{27}$. Now $\ell_+=3\bigl((2+s_3)Y+1\bigr)\equiv3(33Y+1)$ has valuation 1, and $\ell_-=3\bigl((2-s_3)Y+1\bigr)\equiv3(1-29Y)\pmod{3^4}$, where $1-29Y\equiv18$ or $9\pmod{27}$ has valuation 2. So the pair is $(1,3)$. \end{proof} The lemma is also confirmed by the complete residue tree modulo $3^4$ (2916 normalized classes; the pairs are determined at $3^4$) and by a direct enumeration of all primitive solutions modulo $3^4$ (\code{1.4.3c}). \paragraph{Step 5: reciprocity.} \begin{proof}[Proof of Theorem~\ref{thm:1.4.3}] Suppose that $S$ has a rational point $P$, and let $\psi$ be as in Step 3. By Lemmas~\ref{lem:1.4.3-good} and~\ref{lem:1.4.3-two}, only $w_3$ and $c(w_3)$ can contribute to $\sum_w\operatorname{inv}_w(\psi)$, which is 0 by \cite[Chapter VI, Section 5]{Neukirch99}. These places are unramified in $L$, and $\operatorname{Frob}_{w_3}=\tau^m$ with $m\in\{1,2\}$ (Lemma~\ref{lem:1.4.3-field}(iii)). By \cite[Chapters V--VI]{Neukirch99} and $c(\psi)=\psi^{-1}$, $\operatorname{inv}_{c(w_3)}(\psi)=\operatorname{inv}_{w_3}(\psi)$, and by \cite[Chapters V--VI]{Neukirch99} and Lemma~\ref{lem:1.4.3-three}, $\operatorname{inv}_{w_3}(\psi)=m\,v_{w_3}(\psi)/3\equiv m/3\pmod1$. Hence $\sum_w\operatorname{inv}_w(\psi)=2m/3\not\equiv0\pmod1$, a contradiction. Therefore $S(\Q)=\varnothing$, and the only integer solution is $(0,0,0,0)$. \end{proof} \subsection{\texorpdfstring{$8x^4+7y^4+2z^4-t^4=0$ \ ($H=288$)}{8x^4+7y^4+2z^4-t^4=0 (H=288)}}\label{sec:1.4.4} \begin{thm}\label{thm:1.4.4} The only integer solution of $8x^4+7y^4+2z^4=t^4$ is $(x,y,z,t)=(0,0,0,0)$. \end{thm} \paragraph{Setup.} Suppose that there is a nonzero solution. Dividing by the greatest common divisor, we get a solution with $\gcd(x,y,z,t)=1$, which we fix. Put \[ f=7y^2+2z^2-3t^2,\qquad g=7y^2+2z^2+3t^2,\qquad A_\pm=2x^2\pm2xz+z^2 . \] We show that the Hilbert symbols $c_v=(-7,f)_v\,(7,A_+)_v$ satisfy $c_p=1$ for every prime $p$ and $c_\infty=-1$, which contradicts Hilbert's product formula (\cite[Chapter III, Sections 1--2]{Serre73}). For a prime $p$ and an integer $a$ prime to $p$, $(a/p)$ is the Legendre symbol. \paragraph{Step 1: identities and signs.} \begin{lem}\label{lem:1.4.4-id} \begin{enumerate}[label=(\roman*)] \item $-fg=72x^4+14(y^2-z^2)^2=2\bigl(36x^4+7(y^2-z^2)^2\bigr)$, which is twice the norm of $6x^2+(y^2-z^2)\sqrt{-7}$ from $\Q(\sqrt{-7})$. \item $t^4-7y^4=2A_+A_-$, where $A_+A_-=4x^4+z^4$, and $t^4-7y^4$ is the norm of $t^2+y^2\sqrt7$ from $\Q(\sqrt7)$. \item $g>0$, $A_+>0$, $f<0$. \end{enumerate} \end{lem} \begin{proof} As polynomial identities, $-fg-72x^4-14(y^2-z^2)^2=-9(8x^4+7y^4+2z^4-t^4)$ and $t^4-7y^4-2A_+A_-=-(8x^4+7y^4+2z^4-t^4)$ (\code{1.4.4a}); this gives (i) and (ii). (iii) $g=0$ forces $y=z=t=0$ and then $x=0$. $A_+=x^2+(x+z)^2=0$ forces $x=z=0$, and then $t^4=7y^4$ forces $y=t=0$. By (i), $f\le0$, and $f=0$ forces $x=0$ and $y^2=z^2$, so $t^4=9y^4$, $t^2=3y^2$, $y=t=0$. \end{proof} \paragraph{Step 2: odd primes other than 3 and 7.} \begin{lem}\label{lem:1.4.4-odd} Let $p\ne2,3,7$ be a prime. If $(-7/p)=-1$, then $v_p(f)$ is even. If $(7/p)=-1$, then $v_p(A_+)$ is even. \end{lem} \begin{proof} Suppose $(-7/p)=-1$ and $p\mid f$. If also $p\mid g$, then $p\mid g-f=6t^2$, so $p\mid t$, and $36x^4+7(y^2-z^2)^2\equiv0\pmod p$ by Lemma~\ref{lem:1.4.4-id}(i). Since $-7$ is not a square modulo $p$, this forces $y^2\equiv z^2$ and then $x\equiv0$. Then $f\equiv7y^2+2y^2=9y^2$, so $p\mid y$ and $p\mid z$, which contradicts $\gcd(x,y,z,t)=1$. Hence $p\nmid g$, and $v_p(f)=v_p(-fg)=v_p\bigl(N(6x^2+(y^2-z^2)\sqrt{-7})\bigr)$ is even, because $p$ is inert in $\Q(\sqrt{-7})$. Suppose $(7/p)=-1$ and $p\mid A_+$. Then $p\mid t^4-7y^4$, and since 7 is not a square modulo $p$, $p\mid y$ and $p\mid t$. If also $p\mid A_-$, then $p\mid A_+-A_-=4xz$ and $p\mid A_++A_-=4x^2+2z^2$, so $p\mid x$ and $p\mid z$, a contradiction. Hence $p\nmid A_-$, and $v_p(A_+)=v_p(t^4-7y^4)=v_p\bigl(N(t^2+y^2\sqrt7)\bigr)$ is even, because $p$ is inert in $\Q(\sqrt7)$. \end{proof} \paragraph{Step 3: the prime 3.} \begin{lem}\label{lem:1.4.4-three} $v_3(f)\in\{0,2\}$. \end{lem} \begin{proof} Modulo 3, $f\equiv y^2-z^2$ and $8x^4+7y^4+2z^4-t^4\equiv2x^4+y^4+2z^4-t^4$. If $3\mid f$, then $y^4\equiv z^4$, so $2x^4\equiv t^4$, which forces $3\mid x$ and $3\mid t$; then $y,z$ are units (if $3\mid y$, then $3\mid z$). In this case put $r=(z/y)^2\in\Z_3$, so $r\equiv1\pmod3$. Since $3^4$ divides $8x^4$ and $t^4$, we get $7+2r^2\equiv0\pmod{81}$, and hence \[ (7+2r)(7-2r)=49-4r^2=63-2(7+2r^2)\equiv63\pmod{81}. \] Since $7-2r\equiv2\pmod3$ is a unit, $v_3(7+2r)=v_3(63)=2$. Finally $f/y^2=7+2r-3(t/y)^2$ and $v_3(3(t/y)^2)\ge3$, so $v_3(f)=2$. The residue facts are checked by \code{1.4.4b}. \end{proof} \paragraph{Step 4: the prime 2.} \begin{lem}\label{lem:1.4.4-two} $x,y,z,t$ are odd, $A_+\equiv1$ or $5\pmod8$, and $f\equiv6\pmod8$. \end{lem} \begin{proof} A fourth power is $\equiv0$ or $1\pmod{16}$, according as the number is even or odd. Hence $8x^4+7y^4+2z^4-t^4\equiv8a+7b+2c-d\pmod{16}$, where $a,b,c,d\in\{0,1\}$ are the parities of $x,y,z,t$. Among the sixteen choices, only $a=b=c=d=0$ and $a=b=c=d=1$ give $0\pmod{16}$, and the first is excluded by $\gcd(x,y,z,t)=1$. For odd $x,z$ we have $A_+\equiv2+2xz+1\equiv1$ or $5\pmod 8$, and for odd $y,z,t$ we have $f\equiv7+2-3=6\pmod8$ (\code{1.4.4b}). \end{proof} \paragraph{Step 5: the prime 7.} \begin{lem}\label{lem:1.4.4-seven} $f$ and $A_+$ are prime to 7, and $(f/7)=(A_+/7)$. \end{lem} \begin{proof} If $7$ divides $x,z,t$, then $y$ is prime to 7 and $7y^4=t^4-8x^4-2z^4$ has 7-adic valuation 1 on the left and at least 4 on the right. Hence $(x,z,t)\not\equiv(0,0,0)\pmod7$, and modulo 7 the equation reads $x^4+2z^4\equiv t^4$, while $f\equiv2z^2-3t^2$. The fourth powers modulo 7 are $0,1,2,4$. If $z\equiv0$, then $x^4\equiv t^4\not\equiv0$, and $A_+\equiv2x^2$, $f\equiv-3t^2\equiv4t^2$ are nonzero squares ($2\equiv3^2$). If $z\not\equiv0$, put $r=x/z$, $s=t/z$; then $r^4+2\equiv s^4$, and the possibilities are \[ \begin{array}{c|c|c|c} r&s&A_+/z^2&f/z^2\\\hline 0&\pm2&1&4\\ 2&\pm3&6&3\\ -2&\pm3&5&3 \end{array} \] In each row $A_+/z^2$ and $f/z^2$ are both squares or both nonsquares modulo 7 (\code{1.4.4b}). \end{proof} \paragraph{Step 6: Hilbert symbols.} \begin{proof}[Proof of Theorem~\ref{thm:1.4.4}] By Lemma~\ref{lem:1.4.4-id}(iii), $f,A_+\ne0$, so the symbols are defined. We use the formulas of \cite[Chapter III, Sections 1--2]{Serre73}. $p\ne2,3,7$: $-7$ and $7$ are $p$-adic units, so $(-7,f)_p=(-7/p)^{v_p(f)}$ and $(7,A_+)_p=(7/p)^{v_p(A_+)}$. By Lemma~\ref{lem:1.4.4-odd}, both are 1. $p=3$: $(-7,f)_3=(-7/3)^{v_3(f)}=(-1)^{v_3(f)}=1$ by Lemma~\ref{lem:1.4.4-three}, and $(7,A_+)_3=(7/3)^{v_3(A_+)}=1$ since $7\equiv1\pmod3$. $p=2$: $-7\equiv1\pmod8$ is a square in $\Q_2$, so $(-7,f)_2=1$. Since $7$ and $A_+$ are 2-adic units and $A_+\equiv1\pmod4$ (Lemma~\ref{lem:1.4.4-two}), $(7,A_+)_2=(-1)^{\varepsilon(7)\varepsilon(A_+)}=1$. $p=7$: $f$ and $A_+$ are 7-adic units (Lemma~\ref{lem:1.4.4-seven}) and $v_7(-7)=v_7(7)=1$, so $(-7,f)_7=(f/7)$ and $(7,A_+)_7=(A_+/7)$, and $c_7=(f/7)(A_+/7)=1$ by Lemma~\ref{lem:1.4.4-seven}. $v=\infty$: $-7<0$ and $f<0$, so $(-7,f)_\infty=-1$, while $(7,A_+)_\infty=1$ since $7>0$. Hence $\prod_v(-7,f)_v\cdot\prod_v(7,A_+)_v=\prod_vc_v=-1$. By Hilbert's product formula (\cite[Chapter III, Sections 1--2]{Serre73}) both products are 1, a contradiction. Therefore there is no nonzero solution. \end{proof} The local conclusions $c_2=c_3=c_7=1$ are also confirmed by evaluating the two symbols, with the formulas of \cite[Chapter III, Sections 1--2]{Serre73}, on the complete residue trees of all primitive points over $\Z_2$, $\Z_3$, $\Z_7$ modulo $2^5$, $3^4$, $7^2$, and by a direct enumeration of all primitive solutions modulo $3^4$ and $7^2$ (\code{1.4.4c}). \subsection*{The common Frey-curve argument}\label{sec:frey} Each of these six equations has the form $Ax^5+By^5+Cz^5=0$, and we use the modular method with exponent $5$: a primitive solution gives a Frey curve $E$ whose $5$-torsion $E[5]$ (an $\F_5[G_\Q]$-module of dimension 2, $G_\Q=\operatorname{Gal}(\overline\Q/\Q)$) is irreducible, so by modularity and level lowering (\cite[Theorem 3 and Proposition 5.1]{Sik12}) it is congruent modulo a prime above 5 to a newform of weight 2 and of a small level; the newforms of that level are then excluded by comparing their Hecke eigenvalues with the traces that $E$ can have at a few auxiliary primes. The lemmas below are used by all six sections. We write $v_\ell$ for the $\ell$-adic valuation, $\Delta_{\min}$ for the minimal discriminant, $f_\ell(E)$ for the exponent of $\ell$ in the conductor $N_E$, $a_\ell(E)=\ell+1-\#\widetilde E(\F_\ell)$ for a prime $\ell$ of good reduction, and $\chi_5$ for the cyclotomic character modulo 5. For a newform $f=\sum_{n\ge1}a_n(f)q^n$ we write $K_f=\Q(a_n(f):n\ge1)$ for its coefficient field; the $a_n(f)$ are algebraic integers. \begin{lem}\label{lem:1.4:terms} Let $A,B,C$ be pairwise coprime nonzero integers with $v_p(A),v_p(B),v_p(C)\le4$ for every prime $p$, and let $(x,y,z)$ be a primitive solution ($\gcd(x,y,z)=1$) of $Ax^5+By^5+Cz^5=0$. \begin{enumerate}[label=(\arabic*)] \item If $z=0$, then $|A|=|B|=|x|=|y|=1$; similarly if $x=0$ or $y=0$. \item If $xyz\ne0$, the three integers $Ax^5$, $By^5$, $Cz^5$ are pairwise coprime, and exactly one of them is even. \end{enumerate} \end{lem} \begin{proof} (1) If $z=0$, then $\gcd(x,y)=1$ and $Ax^5=-By^5$. Since $\gcd(A,B)=\gcd(x,y)=1$, we get $|A|=|y|^5$ and $|B|=|x|^5$, and since no prime divides $A$ or $B$ to the fifth power, $|A|=|B|=1$ and $|x|=|y|=1$. (2) Suppose that a prime $p$ divides two of the three terms; then it divides the third as well, since their sum is 0. Since $A,B,C$ are pairwise coprime, $p$ divides at most one coefficient. If $p\nmid ABC$, then $p$ divides $x$, $y$ and $z$. If, say, $p\mid A$, then $p\nmid BC$, so $p\mid y$ and $p\mid z$; then $p^5$ divides $By^5+Cz^5=-Ax^5$, and since $v_p(A)\le4$, also $p\mid x$. The cases $p\mid B$ and $p\mid C$ are the same. Each case contradicts $\gcd(x,y,z)=1$. Hence the terms are pairwise coprime, so at most one of them is even, and they cannot all be odd, because their sum is 0. \end{proof} In the situation of Lemma~\ref{lem:1.4:terms}(2), let $t_j$ be the even term and $t_i$, $t_k$ the two odd terms, in an order fixed in each section, and put \[ (u,v,w)=\varepsilon\,(t_i,t_j,t_k),\qquad \varepsilon\in\{\pm1\}\ \text{such that}\ u\equiv-1\pmod4 . \] Then $u,v,w$ are pairwise coprime and nonzero, $u+v+w=0$, $u\equiv-1\pmod4$ and $v$ is even. The \emph{Frey curve} of the solution is \[ E=E_{u,v}:\quad Y^2=X(X-u)(X+v),\qquad c_4=16(u^2+uv+v^2),\qquad \Delta=16u^2v^2w^2 . \] Its 2-torsion points $(0,0)$, $(u,0)$, $(-v,0)$ are rational. The other possible choice, with $t_k$ in place of $t_i$, gives a curve isomorphic to $E$ if $v_2(v)\ge2$ (then it is $(-w,-v,-u)$), and the twist of $E$ by $-1$ if $v_2(v)=1$ (then it is $(w,v,u)$), by Lemma~\ref{lem:1.4:frey}(3); nothing below depends on this choice. \begin{lem}\label{lem:1.4:frey} Let $u,v,w$ be pairwise coprime nonzero integers with $u+v+w=0$, $u\equiv-1\pmod4$ and $v$ even, and put $r=v_2(v)$. \begin{enumerate}[label=(\arabic*)] \item At an odd prime $\ell$, $E_{u,v}$ has good reduction if $\ell\nmid uvw$, and multiplicative reduction with $v_\ell(\Delta_{\min})=2v_\ell(uvw)$ if $\ell\mid uvw$. \item At 2, the Kodaira type, $f_2(E_{u,v})$ and $v_2(\Delta_{\min})$ are \[ \begin{array}{c|ccccc} r&1&2&3&4&\ge5\\\hline \text{type}&\mathrm{III}&\mathrm I_1^*&\mathrm{III}^*&\mathrm I_0&\mathrm I_{2r-8}\\ f_2&5&3&3&0&1\\ v_2(\Delta_{\min})&6&8&10&0&2r-8 \end{array} \] \item The curves $E_{u',v'}$ obtained from the other orderings $(u',v',w')$ of $(u,v,w)$ and of $(-u,-v,-w)$ are isomorphic over $\Q$ to $E_{u,v}$ or to its quadratic twist by $-1$, and for $a\in\Q^\times$ the curve $E_{au,av}$ is the quadratic twist of $E_{u,v}$ by $a$. \end{enumerate} \end{lem} \begin{proof} The displayed $c_4$ and $\Delta$ give (1). Applying Tate's algorithm \cite{Tate75,Silverman94} to $a_2=v-u$, $a_4=-uv$ gives the table in (2), with the normalized model \[ S^2+RS=R^3+\frac{v-u-1}{4}R^2-\frac{uv}{16}R \] when $r\ge4$; \code{1.4qa} checks every residue class needed for the other rows. In (3), translation permutes the three roots, changing all their signs gives the twist by $-1$, and scaling them by $a$ gives the twist by $a$. \end{proof} The congruences and the model used in (2) are checked by \code{1.4qa}, which also compares the table with PARI's implementation of Tate's algorithm on all residue classes of $u$ and $v/2^r$ modulo $2^7$, for $r\le12$. For $u\equiv1\pmod4$ the exponents are different ($f_2=4$ for every $r\ge2$); this is why the sign is normalized. \begin{lem}\label{lem:1.4:level} In the situation of Lemma~\ref{lem:1.4:terms}(2), let $E$ be the Frey curve, $c_j$ the coefficient of the even term $t_j$ and $s_j$ its variable, so that $r=v_2(t_j)=v_2(c_j)+5v_2(s_j)$. Then $E$ is semistable at every odd prime, and its level of \cite[Theorem 3 and Proposition 5.1]{Sik12} for $p=5$, \[ N_5=N_E\Big/\prod_{\ell\,\parallel N_E,\ 5\mid v_\ell(\Delta_{\min})}\ell, \] is $N_5=2^\alpha\prod_{\ell\mid ABC,\ \ell\ \text{odd}}\ell$, where $\alpha=f_2(E)$ from Lemma~\ref{lem:1.4:frey}(2), except that $\alpha=0$ if $r\ge5$ and $r\equiv4\pmod5$. Moreover, if $5\nmid ABC$, then $E$ has good reduction at 5 or multiplicative reduction with $5\mid v_5(\Delta_{\min})$; if $5\mid ABC$, then $E$ has multiplicative reduction at 5 with $5\nmid v_5(\Delta_{\min})$. \end{lem} \begin{proof} Up to sign, $uvw=ABC(xyz)^5$, and every prime divides at most one of the three terms. Let $\ell$ be odd. By Lemma~\ref{lem:1.4:frey}(1), $E$ is good at $\ell$ if $\ell\nmid uvw$, and otherwise multiplicative with $v_\ell(\Delta_{\min})=2v_\ell(c)+10v_\ell(s)$, where $c$ and $s$ are the coefficient and the variable of the term divisible by $\ell$. If $\ell\mid ABC$, then $1\le v_\ell(c)\le4$, so $5\nmid v_\ell(\Delta_{\min})$ and $\ell$ divides $N_5$ exactly once. If $\ell\nmid ABC$, then $5\mid v_\ell(\Delta_{\min})$, so $\ell\nmid N_5$. At 2, Lemma~\ref{lem:1.4:frey}(2) gives $f_2(E)$; the prime 2 is removed only if $f_2(E)=1$, that is $r\ge5$, and $5\mid 2r-8$, that is $r\equiv4\pmod5$. The statement at 5 is the case $\ell=5$ above. \end{proof} Since $r\equiv v_2(c_j)\pmod5$, the exception $\alpha=0$ with $r\ge5$ occurs only when $v_2(c_j)=4$ (Section~\ref{sec:1.4.9}). The resulting levels for each equation are printed by \code{1.4.5a}--\code{1.4.10a}. \begin{lem}\label{lem:1.4:trace} Let $E/\Q$ be an elliptic curve which is semistable at every odd prime $\ell\ne5$, with $f_2(E)\le5$. If $E[5]$ is reducible, then $a_q(E)\equiv\pm2\pmod5$ for every prime $q\equiv1\pmod5$ at which $E$ has good reduction. \end{lem} \begin{proof} Let $L\subset E[5]$ be a $G_\Q$-stable line, with character $\psi\colon G_\Q\to\F_5^\times$; by the Weil pairing, $G_\Q$ acts on $E[5]/L$ by $\psi'=\chi_5\psi^{-1}$. At an odd prime $\ell\ne5$ of good reduction, $E[5]$ is unramified; at an odd prime $\ell\ne5$ of multiplicative reduction, the semisimplification of $E[5]$ is trivial on the inertia group (\cite[Chapter V]{Silverman94}). In both cases $\psi$ is unramified at $\ell$. At 5, the image of inertia under $\psi$ has order prime to 5, so $\psi$ is tamely ramified there, of conductor exponent at most 1. At 2, the Artin conductor exponents satisfy $a_2(\psi)+a_2(\psi')\le a_2(E[5])\le f_2(E)$. Indeed, the Swan conductor is additive in the exact sequence $0\to L\to E[5]\to E[5]/L\to0$, and the codimension of the inertia invariants of $E[5]$ is at least the sum of those of $L$ and $E[5]/L$; and $f_2(E)=\delta+\epsilon$, where $\delta$ is the Swan conductor of $E[5]$ and $\epsilon=0,1,2$ according as the reduction is good, multiplicative or additive \cite[Section IV.10]{Silverman94}, and $\epsilon$ is at least the codimension of the inertia invariants of $E[5]$ (in the multiplicative case because inertia fixes the line coming from $\mu_5$, \cite[Chapter V]{Silverman94}). Since $\chi_5$ is unramified at 2, $a_2(\psi')=a_2(\psi)$, so $a_2(\psi)\le2$. By the Kronecker--Weber theorem, $\psi$ is a Dirichlet character of conductor dividing $2^2\cdot5=20$. Let $q\equiv1\pmod5$ be a prime of good reduction. Then $q^2\equiv1\pmod{20}$, so $\psi(q)=\pm1$, and the trace of $\mathrm{Frob}_q$ on $E[5]$ is $\psi(q)+q\psi(q)^{-1}\equiv2\psi(q)\pmod5$. It equals $a_q(E)\bmod5$. \end{proof} \begin{lem}\label{lem:1.4:flat} Let $E/\Q$ be an elliptic curve with $E(\Q)[2]\cong(\Z/2)^2$, semistable at every odd prime, with $f_2(E)\le3$, and with good reduction at 5 or multiplicative reduction at 5 with $5\mid v_5(\Delta_{\min})$. Then $E[5]$ is irreducible. \end{lem} \begin{proof} Suppose that $L$, $\psi$, $\psi'=\chi_5\psi^{-1}$ are as in the proof of Lemma~\ref{lem:1.4:trace}. As there, $\psi$ and $\psi'$ are unramified at every odd prime $\ell\ne5$, and $2a_2(\psi)\le f_2(E)\le3$. A character of $G_{\Q_2}$ of conductor exponent at most 1 corresponds, by local class field theory, to a character of $\Q_2^\times$ trivial on $1+2\Z_2=\Z_2^\times$, so it is unramified; hence $\psi$ and $\psi'$ are unramified at 2. By \cite[Section 4.1]{Serre87}, $E[5]$ is finite flat at 5, and by \cite{Raynaud74,OT70} so are $L$ and $E[5]/L$, and the inertia group at 5 acts on each of them trivially or through $\chi_5$. Since $\psi\psi'=\chi_5$ is ramified at 5, one of $\psi,\psi'$ is unramified at 5, hence unramified everywhere, hence trivial, as $\Q$ has no nontrivial unramified extension. If $\psi=1$, then $L\subset E(\Q)$, and $E(\Q)$ contains $(\Z/2)^2\times\Z/5\cong\Z/2\times\Z/10$. If $\psi'=1$, let $\phi\colon E\to E'=E/L$ be the isogeny of degree 5 over $\Q$. Then $\phi(E[5])\cong E[5]/L$ is a subgroup of order 5 of $E'(\Q)$, and since $\deg\phi$ is odd, $\phi$ maps $E[2]$ isomorphically onto $E'[2]$, so $E'(\Q)$ contains $\Z/2\times\Z/10$. Both cases contradict \cite{Mazur77}. \end{proof} For a prime $q\nmid10ABC$ we define the \emph{trace set} $T_q=T_q(A,B,C)\subseteq\F_5$: it consists of the residues modulo 5 of $\pm a_q(E_{Aa^5,Bb^5})$ for every point $(a:b:c)\in\mathbb P^2(\F_q)$ of $Aa^5+Bb^5+Cc^5=0$ with $abc\ne0$, where \[ a_q(E_{u,v})=-\sum_{X\in\F_q}\Big(\frac{X(X-u)(X+v)}{q}\Big)\qquad(u,v,u+v\in\F_q^\times), \] together with the residues of $\pm(q+1)$ if the curve has a point over $\F_q$ with $abc=0$. The sets used in Sections~\ref{sec:1.4.5}--\ref{sec:1.4.10} are computed by \code{1.4.5b}, \code{1.4.6b}, \code{1.4.7b}, \code{1.4.8b}, \code{1.4.9a} and \code{1.4.10a}, each with two implementations: character sums as above, and point counting for all twelve signed orderings of the three terms and all scalings of $(a,b,c)$, which also confirms that these give only the two traces $\pm a_q$. \begin{lem}\label{lem:1.4:sieve} In the situation of Lemma~\ref{lem:1.4:terms}(2), let $E$ be the Frey curve and $N_5$ its level from Lemma~\ref{lem:1.4:level}, and let $q\nmid10ABC$ be a prime. \begin{enumerate}[label=(\arabic*)] \item If $q\nmid xyz$, then $E$ has good reduction at $q$ and $a_q(E)\bmod5\in T_q$. If $q\mid xyz$, then $q\parallel N_E$, $q\nmid N_5$, and $\pm(q+1)\bmod5\in T_q$. \item If $q\equiv1\pmod5$, $T_q\cap\{2,3\}=\emptyset$ and $f_2(E)\le5$, then $E[5]$ is irreducible. \item If $E[5]$ is irreducible, there are a newform $f\in S_2(\Gamma_0(N_5))$ and a prime ideal $\lambda\mid5$ of $\mathcal O_{K_f}$ such that $a_q(f)\bmod\lambda\in T_q$ for every prime $q\nmid10ABC$. \end{enumerate} \end{lem} \begin{proof} (1) Let $(a:b:c)$ be the reduction of $(x:y:z)$ modulo $q$, so $(x,y,z)\equiv\mu(a,b,c)$ with $\mu\in\F_q^\times$. If $q\nmid xyz$, then $q\nmid uvw$, so $E$ is good at $q$ (Lemma~\ref{lem:1.4:frey}(1)); its reduction is, by Lemma~\ref{lem:1.4:frey}(3), isomorphic to $E_{Ax^5,By^5}$ or its twist by $-1$, and $E_{Ax^5,By^5}\equiv E_{\mu^5Aa^5,\mu^5Bb^5}$ is the twist of $E_{Aa^5,Bb^5}$ by $\mu^5$. Quadratic twists over $\F_q$ change the trace at most by sign, so $a_q(E)=\pm a_q(E_{Aa^5,Bb^5})$. If $q\mid xyz$, then $q$ divides exactly one of $x,y,z$ (Lemma~\ref{lem:1.4:terms}(2)), $(a:b:c)$ has a zero coordinate, and by the proof of Lemma~\ref{lem:1.4:level}, $q\parallel N_E$ and $q\nmid N_5$. (2) By (1), $q\nmid xyz$, since $\pm(q+1)\equiv\pm2\pmod5$. So $E$ is good at $q$ and $a_q(E)\not\equiv\pm2\pmod5$. By Lemma~\ref{lem:1.4:level}, $E$ is semistable at every odd prime, and Lemma~\ref{lem:1.4:trace} applies. (3) By \cite[Theorem 3 and Proposition 5.1]{Sik12} with $p=5$ there are $f$ and $\lambda$ with $a_q(f)\equiv a_q(E)\pmod\lambda$ if $q\nmid5N_EN_5$, and $a_q(f)\equiv\pm(q+1)\pmod\lambda$ if $q\nmid5N_5$ and $q\parallel N_E$. Since $N_5$ divides $2^5ABC$, every prime $q\nmid10ABC$ is in one of these two cases, and (1) gives the claim. \end{proof} The newforms of the levels $N_5$ are computed with PARI/GP (\texttt{mfinit}, \texttt{mfeigenbasis}). For each Galois orbit with coefficient field $K_f=\Q[y]/(P)$ the codes compute the maximal order of $K_f$ (\texttt{nfinit}, certified by \texttt{nfcertify}) and test every prime ideal $\lambda\mid5$ of $\mathcal O_{K_f}$ separately: the pair (orbit, $\lambda$) is excluded at the first test prime $q$ for which $v_\lambda(a_q(f)-t)=0$ for every $t\in T_q$. No coefficient field is discarded for being non-rational. The dimensions of the new subspaces are confirmed by the genus formula for $X_0(N)$ (\code{1.4qb}), and the characteristic polynomials of the Hecke operators $T_q$ used, by modular symbols (\code{1.4qc}). \begin{lem}\label{lem:1.4:pairing} Let $p$ be an odd prime and $\ell$ a prime ($\ell=p$ is allowed). Let $E_1,E_2/\Q_\ell$ be elliptic curves with multiplicative reduction such that $p\nmid v_\ell(\Delta_{\min}(E_1))\,v_\ell(\Delta_{\min}(E_2))$, and let $\phi\colon E_1[p]\to E_2[p]$ be an isomorphism of $\F_p[G_{\Q_\ell}]$-modules. Let $d\in\F_p^\times$ be defined by $e_2(\phi P,\phi Q)=e_1(P,Q)^d$, where $e_1,e_2$ are the Weil pairings. Then $d$ and $v_\ell(\Delta_{\min}(E_2))/v_\ell(\Delta_{\min}(E_1))$ have the same class in $\F_p^\times/\F_p^{\times2}$. Consequently, if $E,G/\Q$ are elliptic curves with $E[p]\cong G[p]$ as $\F_p[G_\Q]$-modules, and $\ell_1,\ell_2$ are primes at which both curves have multiplicative reduction with minimal discriminant valuations prime to $p$, then $v_{\ell_1}(\Delta_{\min}(G))/v_{\ell_1}(\Delta_{\min}(E))$ and $v_{\ell_2}(\Delta_{\min}(G))/v_{\ell_2}(\Delta_{\min}(E))$ are both squares or both nonsquares modulo $p$. \end{lem} \begin{proof} Since $\phi^*e_2$ is again a nondegenerate alternating pairing with values in $\mu_p$ and the alternating forms on $E_1[p]$ form a space of dimension 1, $d$ is well defined. By \cite[Chapter V]{Silverman94}, $E_i$ is the twist by an unramified character $\varepsilon_i$ of the Tate curve $E_{q_i}$, with $v_\ell(q_i)=v_\ell(\Delta_{\min}(E_i))\not\equiv0\pmod p$. Then $q_i$ is not a $p$th power, so the extension $0\to\mu_p\to E_{q_i}[p]\to\Z/p\to0$ does not split, and $\varepsilon_i\mu_p$ is the only $G_{\Q_\ell}$-stable line of $E_i[p]$. Hence $\phi$ maps $\varepsilon_1\mu_p$ to $\varepsilon_2\mu_p$, so $\varepsilon_1=\varepsilon_2$ ($-1\ne1$ in $\F_p$), and twisting both curves by $\varepsilon_1$ changes neither $\phi$ nor the Weil pairings. So we may assume $E_i=E_{q_i}$. In the bases $(\zeta,Q_i)$ of \cite[Chapter V]{Silverman94}, written multiplicatively, $\phi(\zeta)=\zeta^a$ and $\phi(Q_1)=\zeta^bQ_2^{d'}$ with $a,d'\in\F_p^\times$. Comparing $\phi(\sigma Q_1)$ and $\sigma\phi(Q_1)$ gives $a\kappa_1(\sigma)+b=b\chi_p(\sigma)+d'\kappa_2(\sigma)$, so the cocycles $a\kappa_1$ and $d'\kappa_2$ differ by the coboundary of $\zeta^b$, and $q_1^aq_2^{-d'}\in\Q_\ell^{\times p}$. Taking valuations, $a\,v_\ell(q_1)\equiv d'v_\ell(q_2)\pmod p$. Finally $e_2(\phi\zeta,\phi Q_1)=e_2(\zeta,Q_2)^{ad'}=\zeta^{ad'}=e_1(\zeta,Q_1)^{ad'}$, so $d=ad'=(a/d')\,d'^2$, and $a/d'=v_\ell(q_2)/v_\ell(q_1)$. For the consequence, take a $G_\Q$-isomorphism $\phi\colon E[p]\to G[p]$; its multiplier $d$ is the same for the restrictions to $G_{\Q_{\ell_1}}$ and $G_{\Q_{\ell_2}}$, and the first part applies at $\ell_1$ and at $\ell_2$. \end{proof} \subsection{\texorpdfstring{$14x^5+5y^5+3z^5=0$ \ ($H=704$)}{14x^5+5y^5+3z^5=0 (H=704)}}\label{sec:1.4.5} \begin{thm}\label{thm:1.4.5} The only integer solution of $14x^5+5y^5+3z^5=0$ is $(x,y,z)=(0,0,0)$. \end{thm} \paragraph{Setup.} We use the common framework of Section~\ref{sec:1.4} (Lemmas~\ref{lem:1.4:terms}--\ref{lem:1.4:pairing}). A primitive solution gives a Frey curve $E$ with $E[5]$ irreducible, and a newform of level $210$ or $3360$ congruent to it (Steps 1--2). A sieve with twelve auxiliary primes leaves only two rational newforms of level 3360 (Step 3). They are the newforms of the elliptic curves \[ G_+:\ Y^2=X^3+X^2-890X-10200,\qquad G_-:\ Y^2=X^3-X^2-890X+10200 \] ($G_-$ is the twist of $G_+$ by $-1$), and they are excluded by comparing the Weil pairings at 3 and at 5 (Step 4). \paragraph{Step 1: the Frey curve.} \begin{lem}\label{lem:1.4.5-frey} Let $(x,y,z)$ be a primitive solution. Then $xyz\ne0$, the terms $14x^5$, $5y^5$, $3z^5$ are pairwise coprime, and $y$, $z$ are odd. Let $\varepsilon=\pm1$ with $3\varepsilon z^5\equiv-1\pmod4$, and let $E=E_{u,v}$ with $(u,v,w)=\varepsilon(3z^5,14x^5,5y^5)$. Then $E$ is semistable at every odd prime, it has multiplicative reduction at $3$, $5$, $7$ with \[ v_3(\Delta_{\min})=2+10v_3(z),\qquad v_5(\Delta_{\min})=2+10v_5(y),\qquad v_7(\Delta_{\min})=2+10v_7(x), \] and $f_2(E)=5$ if $x$ is odd, while $f_2(E)=1$ and $v_2(\Delta_{\min})=10v_2(x)-6$ if $x$ is even. The level of Lemma~\ref{lem:1.4:level} is $N_5=3360=2^5\cdot3\cdot5\cdot7$ if $x$ is odd and $N_5=210=2\cdot3\cdot5\cdot7$ if $x$ is even. \end{lem} \begin{proof} The coefficients $14=2\cdot7$, $5$, $3$ are pairwise coprime, every prime divides them at most once, and none is $\pm1$. By Lemma~\ref{lem:1.4:terms}, $xyz\ne0$ and the three terms are pairwise coprime; $14x^5$ is even, so it is the even term, and $y$, $z$ are odd. Here $r=v_2(14x^5)=1+5v_2(x)$. If $x$ is odd, then $r=1$, and Lemma~\ref{lem:1.4:frey}(2) gives $f_2=5$. If $x$ is even, then $r\ge6$, $f_2=1$ and $v_2(\Delta_{\min})=2r-8=10v_2(x)-6\equiv4\pmod5$, so $2$ divides $N_5$. The odd primes and $N_5$ are given by Lemma~\ref{lem:1.4:level}: the odd prime divisors of $14\cdot5\cdot3$ are $3,5,7$, and $v_\ell(\Delta_{\min})=2v_\ell(uvw)$ (\code{1.4.5a}; the 2-adic data are also checked with PARI's Tate algorithm on all residue classes of $x$ and $z$ modulo $2^6$ in \code{1.4.5c}). \end{proof} \paragraph{Step 2: irreducibility and level lowering.} The trace sets $T_q=T_q(14,5,3)$ of the common framework are, for the primes $q$ used below (\code{1.4.5b}), \[ \begin{array}{c|c@{\qquad}c|c@{\qquad}c|c} q&T_q&q&T_q&q&T_q\\\hline 11&\{0,2,3\}&41&\F_5^\times&131&\{0,2,3\}\\ 13&\F_5^\times&61&\F_5^\times&151&\F_5^\times\\ 17&\F_5^\times&71&\{2,3\}&181&\F_5^\times\\ 31&\{0,1,4\}&101&\F_5^\times&191&\F_5^\times \end{array} \] For example, modulo 31 the curve $14a^5+5b^5+3c^5=0$ has 50 points, none with a zero coordinate, and the traces of the curves $E_{14a^5,5b^5}$ at these points are $0$ and $\pm4$. \begin{lem}\label{lem:1.4.5-irr} $E[5]$ is irreducible. There are a newform $f\in S_2(\Gamma_0(N_5))$, with $N_5$ as in Lemma~\ref{lem:1.4.5-frey}, and a prime ideal $\lambda\mid5$ of $\mathcal O_{K_f}$ such that $a_q(f)\bmod\lambda\in T_q$ for the twelve primes $q$ of the table. \end{lem} \begin{proof} We have $31\equiv1\pmod5$, $T_{31}\cap\{2,3\}=\emptyset$ and $f_2(E)\le5$, so $E[5]$ is irreducible by Lemma~\ref{lem:1.4:sieve}(2). Since no $q$ in the table divides $10\cdot14\cdot5\cdot3$, Lemma~\ref{lem:1.4:sieve}(3) gives $f$ and $\lambda$. \end{proof} \paragraph{Step 3: the sieve.} \begin{lem}\label{lem:1.4.5-sieve} \begin{enumerate}[label=(\arabic*)] \item $S_2(\Gamma_0(210))^{\mathrm{new}}$ has dimension 5 and consists of five rational newforms. None of them satisfies the conditions of Lemma~\ref{lem:1.4.5-irr} (with $\lambda=5$). \item $S_2(\Gamma_0(3360))^{\mathrm{new}}$ has dimension 48 and 36 Galois orbits of newforms: 26 rational, 8 with quadratic and 2 with cubic coefficient fields, which give 38 pairs (orbit, $\lambda\mid5$). Exactly two pairs satisfy the conditions of Lemma~\ref{lem:1.4.5-irr}; they are two rational newforms $f_+$, $f_-$ with $\lambda=5$. \end{enumerate} \end{lem} \begin{proof} This is a finite computation (\code{1.4.5c}). At level 210, three forms are excluded at $q=11$, one at $q=61$ and one at $q=71$. At level 3360, the numbers of pairs that remain after $q=11,13,17,31$ are $12,8,6,2$, and the two remaining pairs satisfy the conditions at all twelve primes. The prime 5 is inert in the coefficient fields of six of the quadratic orbits and ramified in the other two (whose field is $\Q(\sqrt5)$), and it has two prime ideals, of residue degrees 1 and 2, in the two cubic fields; every one of these prime ideals is tested. The dimensions are also given by the genus formula (\code{1.4qb}), and the Hecke polynomials at the twelve primes agree with those computed by modular symbols (\code{1.4qc}). \end{proof} \paragraph{Step 4: the two remaining newforms.} \begin{lem}\label{lem:1.4.5-pairing} The newform $f$ of Lemma~\ref{lem:1.4.5-irr} is neither $f_+$ nor $f_-$. \end{lem} \begin{proof} The curves $G_\pm$ have conductor $3360$ and minimal discriminant $2^6\cdot3^6\cdot5^2\cdot7^4$, and $c_4(G_\pm)=2^4\cdot2671$ is prime to 15, so they have multiplicative reduction at 3 and 5, with $v_3(\Delta_{\min})=6$ and $v_5(\Delta_{\min})=2$. By \cite{BCDT01} their newforms lie in $S_2(\Gamma_0(3360))$. The index of $\Gamma_0(3360)$ in $\operatorname{SL}_2(\Z)$ is $9216$, so the Sturm bound is $9216/6=1536$, and the coefficients $a_0,\dots,a_{1536}$ of $f_+$ and $f_-$ are those of $G_+$ and $G_-$, respectively, that is, $a_0=0$ and the coefficients $a_n(G_\pm)$ of $L(G_\pm,s)$, which are those of the newforms of $G_\pm$ by \cite{BCDT01} (\code{1.4.5d}). By \cite{Sturm87}, $f_\pm$ is the newform of $G_\pm$. Suppose that $f=f_\pm$. By \cite[Section 5]{Sik12}, $E[5]\cong G_\pm[5]$ as $\F_5[G_\Q]$-modules. Both $E$ and $G_\pm$ have multiplicative reduction at 3 and at 5, with discriminant valuations prime to 5: by Lemma~\ref{lem:1.4.5-frey}, $v_3(\Delta_{\min}(E))\equiv v_5(\Delta_{\min}(E))\equiv2\pmod5$. The ratios are \[ \frac{v_3(\Delta_{\min}(G_\pm))}{v_3(\Delta_{\min}(E))}\equiv\frac62\equiv3,\qquad \frac{v_5(\Delta_{\min}(G_\pm))}{v_5(\Delta_{\min}(E))}\equiv\frac22=1\pmod5 , \] and 3 is not a square modulo 5, while 1 is. This contradicts Lemma~\ref{lem:1.4:pairing} with $p=5$, $\ell_1=3$, $\ell_2=5$. \end{proof} The ratio $v_3(\Delta_{\min})/v_5(\Delta_{\min})$ is a nonsquare modulo 5 for all four curves isogenous to $G_+$ (\code{1.4.5d}), as it must be, since the argument does not depend on the choice of the curve in its isogeny class. \begin{proof}[Proof of Theorem~\ref{thm:1.4.5}] If there is a solution other than $(0,0,0)$, dividing it by the greatest common divisor of its coordinates gives a primitive solution. By Lemmas~\ref{lem:1.4.5-frey} and~\ref{lem:1.4.5-irr} there is a newform $f$ of level 210 or 3360 with the conditions of Lemma~\ref{lem:1.4.5-irr}. By Lemma~\ref{lem:1.4.5-sieve}, $f$ is $f_+$ or $f_-$, which contradicts Lemma~\ref{lem:1.4.5-pairing}. \end{proof} \subsection{\texorpdfstring{$21x^5+5y^5+4z^5=0$ \ ($H=960$)}{21x^5+5y^5+4z^5=0 (H=960)}}\label{sec:1.4.6} \begin{thm}\label{thm:1.4.6} The only integer solution of $21x^5+5y^5+4z^5=0$ is $(x,y,z)=(0,0,0)$. \end{thm} \paragraph{Setup.} We use the common framework of Section~\ref{sec:1.4} (Lemmas~\ref{lem:1.4:terms}--\ref{lem:1.4:sieve}). A primitive solution gives a Frey curve $E$ with $E[5]$ irreducible and a newform of level $210$ or $840$ congruent to it (Steps 1--2); the sieve of Step 3 excludes all newforms of these levels, using only the auxiliary primes 11, 31, 41 and 61. \paragraph{Step 1: the Frey curve.} \begin{lem}\label{lem:1.4.6-frey} Let $(x,y,z)$ be a primitive solution. Then $xyz\ne0$, the terms $21x^5$, $5y^5$, $4z^5$ are pairwise coprime, and $x$, $y$ are odd. Let $\varepsilon=\pm1$ with $\varepsilon x\equiv3\pmod4$, and let $E=E_{u,v}$ with $(u,v,w)=\varepsilon(21x^5,4z^5,5y^5)$. Then $E$ is semistable at every odd prime, with multiplicative reduction and $v_\ell(\Delta_{\min})\equiv2\pmod5$ at $\ell=3,5,7$. If $z$ is odd, then $E$ has type $\mathrm I_1^*$ at 2, $f_2(E)=3$ and $N_5=840=2^3\cdot3\cdot5\cdot7$. If $z$ is even, then $f_2(E)=1$, $v_2(\Delta_{\min})=10v_2(z)-4$ and $N_5=210$. \end{lem} \begin{proof} The coefficients $21=3\cdot7$, $5$, $4=2^2$ are pairwise coprime, with exponents at most 2, and none is $\pm1$. By Lemma~\ref{lem:1.4:terms}, $xyz\ne0$, the terms are pairwise coprime, and since $4z^5$ is even, $x$ and $y$ are odd. As $x^5\equiv x$ and $21\equiv1\pmod4$, $u=21\varepsilon x^5\equiv\varepsilon x\equiv-1\pmod4$. Here $r=v_2(4z^5)=2+5v_2(z)$. If $z$ is odd, then $r=2$ and Lemma~\ref{lem:1.4:frey}(2) gives type $\mathrm I_1^*$, $f_2=3$, $v_2(\Delta_{\min})=8$. If $z$ is even, then $r\ge7$, $f_2=1$, and $v_2(\Delta_{\min})=2r-8=10v_2(z)-4\equiv1\pmod5$, so 2 divides $N_5$. At $\ell=3,5,7$, $v_\ell(\Delta_{\min})=2+10v_\ell(xyz)$, and Lemma~\ref{lem:1.4:level} gives $N_5$ (\code{1.4.6a}; the 2-adic data are also checked with PARI on all residue classes of $\varepsilon x$ and $z$ modulo $2^6$ in \code{1.4.6c}). \end{proof} \paragraph{Step 2: irreducibility and level lowering.} The trace sets $T_q=T_q(21,5,4)$ are (\code{1.4.6b}) \[ \begin{array}{c|c@{\qquad}c|c@{\qquad}c|c} q&T_q&q&T_q&q&T_q\\\hline 11&\{1,4\}&41&\F_5^\times&151&\{0,2,3\}\\ 13&\F_5^\times&61&\{2,3\}&181&\{2,3\}\\ 17&\F_5^\times&101&\F_5^\times&211&\F_5^\times\\ 31&\{0,1,4\}&131&\F_5^\times&& \end{array} \] \begin{lem}\label{lem:1.4.6-irr} $E[5]$ is irreducible. There are a newform $f\in S_2(\Gamma_0(N_5))$, with $N_5\in\{210,840\}$ as in Lemma~\ref{lem:1.4.6-frey}, and a prime ideal $\lambda\mid5$ of $\mathcal O_{K_f}$ such that $a_q(f)\bmod\lambda\in T_q$ for the eleven primes $q$ of the table. \end{lem} \begin{proof} As $31\equiv1\pmod5$, $T_{31}\cap\{2,3\}=\emptyset$ and $f_2(E)\le5$, Lemma~\ref{lem:1.4:sieve}(2) applies; then Lemma~\ref{lem:1.4:sieve}(3) gives $f$ and $\lambda$, since no $q$ of the table divides $10\cdot21\cdot5\cdot4$. \end{proof} \paragraph{Step 3: the sieve.} \begin{lem}\label{lem:1.4.6-sieve} No newform of level 210 or 840 satisfies the conditions of Lemma~\ref{lem:1.4.6-irr}. More precisely, $S_2(\Gamma_0(210))^{\mathrm{new}}$ consists of five rational newforms, two excluded at $q=11$, two at $q=31$ and one at $q=61$; $S_2(\Gamma_0(840))^{\mathrm{new}}$ has dimension 12 and 11 Galois orbits, ten rational and one with coefficient field $\Q(\sqrt2)$, in which 5 is inert; of these 11 pairs (orbit, $\lambda$), five are excluded at $q=11$, one at $q=31$, two at $q=41$ and three at $q=61$. \end{lem} \begin{proof} This is a finite computation (\code{1.4.6c}); the dimensions and the Hecke polynomials are confirmed by \code{1.4qb} and \code{1.4qc}. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.4.6}] A solution other than $(0,0,0)$ gives a primitive one, and then, by Lemmas~\ref{lem:1.4.6-frey} and~\ref{lem:1.4.6-irr}, a newform of level 210 or 840 that Lemma~\ref{lem:1.4.6-sieve} excludes. \end{proof} \subsection{\texorpdfstring{$17x^5+6y^5+3z^5=0$ \ ($H=832$)}{17x^5+6y^5+3z^5=0 (H=832)}}\label{sec:1.4.7} \begin{thm}\label{thm:1.4.7} The only integer solution of $17x^5+6y^5+3z^5=0$ is $(x,y,z)=(0,0,0)$. \end{thm} \paragraph{Setup.} Modulo 3 the equation gives $3\mid x$, and with $x=3X$ it becomes \[ 1377X^5+2y^5+z^5=0,\qquad 1377=3^4\cdot17 . \] We apply the common framework of Section~\ref{sec:1.4} (Lemmas~\ref{lem:1.4:terms}--\ref{lem:1.4:sieve}) to this equation: a primitive solution gives a Frey curve $E$ with $E[5]$ irreducible and a newform of level $102$ or $1632$ (Steps 1--2), and the auxiliary primes 41, 11 and 131 exclude all of them (Step 3). \paragraph{Step 1: the Frey curve.} \begin{lem}\label{lem:1.4.7-frey} Let $(x,y,z)$ be a primitive solution. Then $x=3X$, $(X,y,z)$ is a primitive solution of $1377X^5+2y^5+z^5=0$, $Xyz\ne0$, the terms $1377X^5$, $2y^5$, $z^5$ are pairwise coprime, and $X$, $z$ are odd. Let $\varepsilon=\pm1$ with $\varepsilon X\equiv3\pmod4$, and let $E=E_{u,v}$ with $(u,v,w)=\varepsilon(1377X^5,2y^5,z^5)$. Then $E$ is semistable at every odd prime, with multiplicative reduction at 3 and 17, $v_3(\Delta_{\min})=8+10v_3(X)$, $v_{17}(\Delta_{\min})=2+10v_{17}(X)$, and $5\nmid N_5$. If $y$ is odd, then $f_2(E)=5$ and $N_5=1632=2^5\cdot3\cdot17$; if $y$ is even, then $f_2(E)=1$, $v_2(\Delta_{\min})=10v_2(y)-6$ and $N_5=102=2\cdot3\cdot17$. \end{lem} \begin{proof} Modulo 3, $17x^5+6y^5+3z^5\equiv2x^5$, so $3\mid x$; dividing $17\cdot3^5X^5+6y^5+3z^5=0$ by 3 gives the new equation. A common prime factor of $X,y,z$ would divide $x,y,z$, so $(X,y,z)$ is primitive. The coefficients $3^4\cdot17$, $2$, $1$ are pairwise coprime with exponents at most 4, and only one of them is $\pm1$. By Lemma~\ref{lem:1.4:terms}, $Xyz\ne0$, the terms are pairwise coprime, and since $2y^5$ is even, $X$ and $z$ are odd. As $1377\equiv1\pmod4$, $u\equiv\varepsilon X\equiv-1\pmod4$. Here $r=1+5v_2(y)$; Lemma~\ref{lem:1.4:frey}(2) gives $f_2=5$ if $y$ is odd, and $f_2=1$, $v_2(\Delta_{\min})=2r-8=10v_2(y)-6\equiv4\pmod5$ if $y$ is even. Lemma~\ref{lem:1.4:level} gives the odd primes and $N_5$; $5\nmid1377\cdot2$ (\code{1.4.7a}; the 2-adic data are also checked with PARI in \code{1.4.7c}). \end{proof} \paragraph{Step 2: irreducibility and level lowering.} The trace sets $T_q=T_q(1377,2,1)$ are (\code{1.4.7b}) \[ \begin{array}{c|ccccc} q&11&31&41&131&241\\\hline T_q&\{2,3\}&\{0,1,4\}&\{0\}&\{0,2,3\}&\{0,2,3\} \end{array} \] For example, modulo 41 the curve has 25 points, none with a zero coordinate, and the traces at these points are $\pm10$. \begin{lem}\label{lem:1.4.7-irr} $E[5]$ is irreducible. There are a newform $f\in S_2(\Gamma_0(N_5))$, with $N_5\in\{102,1632\}$ as in Lemma~\ref{lem:1.4.7-frey}, and a prime ideal $\lambda\mid5$ of $\mathcal O_{K_f}$ such that $a_q(f)\bmod\lambda\in T_q$ for the five primes $q$ of the table. \end{lem} \begin{proof} As $31\equiv1\pmod5$, $T_{31}\cap\{2,3\}=\emptyset$ and $f_2(E)\le5$, Lemma~\ref{lem:1.4:sieve}(2) applies (so does $q=41$); Lemma~\ref{lem:1.4:sieve}(3) gives $f$ and $\lambda$, since no $q$ of the table divides $10\cdot1377\cdot2$. \end{proof} \paragraph{Step 3: the sieve.} \begin{lem}\label{lem:1.4.7-sieve} No newform of level 102 or 1632 satisfies the conditions of Lemma~\ref{lem:1.4.7-irr}. More precisely, $S_2(\Gamma_0(102))^{\mathrm{new}}$ consists of three rational newforms, one excluded at $q=41$ and two at $q=11$. $S_2(\Gamma_0(1632))^{\mathrm{new}}$ has dimension 32 and 20 Galois orbits: 12 rational, four quadratic and four cubic, in whose coefficient fields 5 is inert, so there are 20 pairs (orbit, $\lambda$). The condition at $q=41$ leaves 8 pairs and the condition at $q=11$ leaves two rational newforms, with $(a_{41},a_{11})=(-5,3)$ and $(-5,-3)$; their coefficients $a_{131}=9$ and $-9$ are $4$ and $1$ modulo 5, not in $T_{131}$. \end{lem} \begin{proof} This is a finite computation (\code{1.4.7c}); the dimensions and the Hecke polynomials are confirmed by \code{1.4qb} and \code{1.4qc}. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.4.7}] A solution other than $(0,0,0)$ gives a primitive one, and then, by Lemmas~\ref{lem:1.4.7-frey} and~\ref{lem:1.4.7-irr}, a newform of level 102 or 1632 that Lemma~\ref{lem:1.4.7-sieve} excludes. \end{proof} \subsection{\texorpdfstring{$17x^5+9y^5+5z^5=0$ \ ($H=992$)}{17x^5+9y^5+5z^5=0 (H=992)}}\label{sec:1.4.8} \begin{thm}\label{thm:1.4.8} The only integer solution of $17x^5+9y^5+5z^5=0$ is $(x,y,z)=(0,0,0)$. \end{thm} \paragraph{Setup.} We use the common framework of Section~\ref{sec:1.4} (Lemmas~\ref{lem:1.4:terms}--\ref{lem:1.4:sieve}). Here the Frey curve is semistable everywhere, the only level is $510$ (Steps 1--2), and the auxiliary primes 41 and 11 exclude all newforms of that level (Step 3). \paragraph{Step 1: the Frey curve.} \begin{lem}\label{lem:1.4.8-frey} Let $(x,y,z)$ be a primitive solution. Then $xyz\ne0$, the terms $17x^5$, $9y^5$, $5z^5$ are pairwise coprime, and exactly one of $x,y,z$ is even. Let $t_j$ be the even term, $t_i$ the first and $t_k$ the second of the two odd terms in the order $17x^5,9y^5,5z^5$, $\varepsilon=\pm1$ with $\varepsilon t_i\equiv-1\pmod4$, and $E=E_{u,v}$ with $(u,v,w)=\varepsilon(t_i,t_j,t_k)$. Then $E$ is semistable, with multiplicative reduction at $2,3,5,17$, \[ \begin{gathered} v_2(\Delta_{\min})=10e-8\quad(e=v_2(xyz)\ge1),\\ v_3(\Delta_{\min})=4+10v_3(y),\qquad v_5(\Delta_{\min})=2+10v_5(z),\qquad v_{17}(\Delta_{\min})=2+10v_{17}(x), \end{gathered} \] and $N_5=510=2\cdot3\cdot5\cdot17$. \end{lem} \begin{proof} The coefficients $17$, $9=3^2$, $5$ are pairwise coprime, with exponents at most 2, and none is $\pm1$. By Lemma~\ref{lem:1.4:terms}, $xyz\ne0$, the terms are pairwise coprime and exactly one is even; as the coefficients are odd, exactly one variable is even, and $r=5e$ with $e=v_2(xyz)\ge1$. By Lemma~\ref{lem:1.4:frey}(2), $f_2=1$ and $v_2(\Delta_{\min})=2r-8=10e-8\equiv2\pmod5$, so 2 divides $N_5$. The odd primes and $N_5$ are given by Lemma~\ref{lem:1.4:level} (\code{1.4.8a}; the 2-adic data are also checked with PARI for each choice of the even variable in \code{1.4.8c}). \end{proof} \paragraph{Step 2: irreducibility and level lowering.} The trace sets $T_q=T_q(17,9,5)$ are (\code{1.4.8b}) \[ \begin{array}{c|cccc} q&11&31&41&71\\\hline T_q&\{2,3\}&\{2,3\}&\{0\}&\{2,3\} \end{array} \] Modulo 41 the curve has 25 points, none with a zero coordinate, and the traces at these points are $\pm10$. \begin{lem}\label{lem:1.4.8-irr} $E[5]$ is irreducible. There are a newform $f\in S_2(\Gamma_0(510))$ and a prime ideal $\lambda\mid5$ of $\mathcal O_{K_f}$ such that $a_q(f)\bmod\lambda\in T_q$ for $q=11,31,41,71$. \end{lem} \begin{proof} As $41\equiv1\pmod5$, $T_{41}\cap\{2,3\}=\emptyset$ and $f_2(E)=1$, Lemma~\ref{lem:1.4:sieve}(2) applies; Lemma~\ref{lem:1.4:sieve}(3) gives $f$ and $\lambda$. \end{proof} \paragraph{Step 3: the sieve.} \begin{lem}\label{lem:1.4.8-sieve} No newform of level 510 satisfies the conditions of Lemma~\ref{lem:1.4.8-irr}. More precisely, $S_2(\Gamma_0(510))^{\mathrm{new}}$ has dimension 9 and 8 Galois orbits, seven rational and one with coefficient field $\Q(\sqrt6)$, in which 5 splits; this gives 9 pairs (orbit, $\lambda$). The condition at $q=41$ leaves three pairs: two rational newforms, with $a_{11}=0$ and $4$, and one prime ideal $\lambda\mid5$ of $\Q(\sqrt6)$, at which $a_{41}=2\sqrt6+2\equiv0$ and $a_{11}=0$. None of $0,4,0$ lies in $T_{11}$. \end{lem} \begin{proof} This is a finite computation (\code{1.4.8c}); the dimension and the Hecke polynomials are confirmed by \code{1.4qb} and \code{1.4qc}. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.4.8}] A solution other than $(0,0,0)$ gives a primitive one, and then, by Lemmas~\ref{lem:1.4.8-frey} and~\ref{lem:1.4.8-irr}, a newform of level 510 that Lemma~\ref{lem:1.4.8-sieve} excludes. \end{proof} \subsection{\texorpdfstring{$14x^5+13y^5+2z^5=0$ \ ($H=928$)}{14x^5+13y^5+2z^5=0 (H=928)}}\label{sec:1.4.9} \begin{thm}\label{thm:1.4.9} The only integer solution of $14x^5+13y^5+2z^5=0$ is $(x,y,z)=(0,0,0)$. \end{thm} \paragraph{Setup.} Modulo 2 the equation gives $2\mid y$, and with $y=2Y$ it becomes \[ 7x^5+208Y^5+z^5=0,\qquad 208=2^4\cdot13 . \] We apply the common framework of Section~\ref{sec:1.4} (Lemmas~\ref{lem:1.4:terms}--\ref{lem:1.4:sieve}) to this equation. The Frey curve has good or multiplicative reduction at 2 with $5\mid v_2(\Delta_{\min})$, so the prime 2 disappears from the level, which is $91$ (Step 1). Irreducibility follows from Mazur's theorem (Step 2), and the single auxiliary prime 11 excludes all newforms of level 91 (Step 3). \paragraph{Step 1: the Frey curve.} \begin{lem}\label{lem:1.4.9-frey} Let $(x,y,z)$ be a primitive solution. Then $y=2Y$, $(x,Y,z)$ is a primitive solution of $7x^5+208Y^5+z^5=0$, $xYz\ne0$, the terms $7x^5$, $208Y^5$, $z^5$ are pairwise coprime, and $x$, $z$ are odd. Let $\varepsilon=\pm1$ with $\varepsilon x\equiv1\pmod4$, and let $E=E_{u,v}$ with $(u,v,w)=\varepsilon(7x^5,208Y^5,z^5)$. Then $E$ is semistable, with multiplicative reduction at 7 and 13, $v_7(\Delta_{\min})=2+10v_7(x)$, $v_{13}(\Delta_{\min})=2+10v_{13}(Y)$; at 2 it has good reduction if $Y$ is odd, and multiplicative reduction with $v_2(\Delta_{\min})=10v_2(Y)$ if $Y$ is even; at 5 it has good reduction, or multiplicative reduction with $5\mid v_5(\Delta_{\min})$. Its level is $N_5=91=7\cdot13$. \end{lem} \begin{proof} Modulo 2, $14x^5+13y^5+2z^5\equiv y^5$, so $y=2Y$, and dividing by 2 gives the new equation; a common prime factor of $x,Y,z$ would divide $x,y,z$. The coefficients $7$, $2^4\cdot13$, $1$ are pairwise coprime with exponents at most 4, and only one is $\pm1$. By Lemma~\ref{lem:1.4:terms}, $xYz\ne0$, the terms are pairwise coprime, and since $208Y^5$ is even, $x$ and $z$ are odd. As $7\equiv-1\pmod4$, $u=7\varepsilon x^5\equiv-\varepsilon x\equiv-1\pmod4$. Here $r=v_2(208Y^5)=4+5v_2(Y)$. If $Y$ is odd, then $r=4$ and $E$ has good reduction at 2 by Lemma~\ref{lem:1.4:frey}(2). If $Y$ is even, then $r\ge9$, the reduction is multiplicative and $v_2(\Delta_{\min})=2r-8=10v_2(Y)$; since $r\equiv4\pmod5$, 2 does not divide $N_5$ (Lemma~\ref{lem:1.4:level}). The rest is Lemma~\ref{lem:1.4:level}, with $5\nmid7\cdot208$ (\code{1.4.9a}; the 2-adic data are also checked with PARI in \code{1.4.9b}). \end{proof} \paragraph{Step 2: irreducibility and level lowering.} \begin{lem}\label{lem:1.4.9-irr} $E[5]$ is irreducible, and there are a newform $f\in S_2(\Gamma_0(91))$ and a prime ideal $\lambda\mid5$ of $\mathcal O_{K_f}$ such that $a_{11}(f)\bmod\lambda\in T_{11}(7,208,1)$. \end{lem} \begin{proof} By Lemma~\ref{lem:1.4.9-frey}, $E$ has full rational 2-torsion, is semistable at every odd prime, has $f_2(E)\le1$, and has good reduction at 5 or multiplicative reduction with $5\mid v_5(\Delta_{\min})$. Lemma~\ref{lem:1.4:flat} gives irreducibility, and Lemma~\ref{lem:1.4:sieve}(3) gives $f$ and $\lambda$ ($11\nmid10\cdot7\cdot208$). \end{proof} \begin{lem}\label{lem:1.4.9-eleven} Every point of $7a^5+208b^5+c^5=0$ over $\F_{11}$ has $a=0$ and $bc\ne0$. Hence $11\mid x$, $11\nmid Yz$, and $T_{11}(7,208,1)=\{2,3\}$. \end{lem} \begin{proof} The nonzero fifth powers modulo 11 are $\pm1$, and $208\equiv-1\pmod{11}$. If $abc\ne0$, then $7a^5-b^5+c^5\equiv7\pm1\pm1$ or $-7\pm1\pm1$, which is one of $\pm5,\pm7,\pm9$ and is not $0$ modulo 11. If $b=0$ and $ac\ne0$, or $c=0$ and $ab\ne0$, the two remaining terms add up to $\pm7\pm1\not\equiv0$, and two vanishing coordinates force the third to vanish. So $a=0$, and then $b^5=c^5$ with $bc\ne0$: the five points $(0:1:c)$ with $c\in\{1,3,4,5,9\}$. All of them have a zero coordinate, so $T_{11}=\{\pm12\bmod5\}=\{2,3\}$ (\code{1.4.9a}). \end{proof} \paragraph{Step 3: the newforms of level 91.} \begin{lem}\label{lem:1.4.9-sieve} $S_2(\Gamma_0(91))^{\mathrm{new}}$ has dimension 7 and four Galois orbits, of dimensions $1,1,2,3$. The minimal polynomials of their eigenvalues $a_{11}$ are \[ T,\qquad T+6,\qquad T^2-18,\qquad T^3-2T^2-6T+8 . \] None of them has $2$ or $3$ as a root modulo 5. \end{lem} \begin{proof} The newforms are computed with PARI/GP (\code{1.4.9b}; the dimension and the Hecke polynomial are confirmed by \code{1.4qb} and \code{1.4qc}). The values of the four polynomials at $T=2$ and $T=3$ are $(2,3)$, $(3,4)$, $(1,1)$ and $(1,4)$ modulo 5. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.4.9}] Suppose that there is a solution other than $(0,0,0)$, and take a primitive one. By Lemmas~\ref{lem:1.4.9-irr} and~\ref{lem:1.4.9-eleven}, there is a newform $f$ of level 91 with $a_{11}(f)\equiv2$ or $3\pmod\lambda$ for a prime ideal $\lambda\mid5$. The minimal polynomial of $a_{11}(f)$ has integer coefficients and vanishes at $a_{11}(f)$, so its reduction modulo 5 vanishes at $2$ or $3$, contrary to Lemma~\ref{lem:1.4.9-sieve}. \end{proof} \subsection{\texorpdfstring{$4x^5+4y^5+11z^5=0$ \ ($H=608$)}{4x^5+4y^5+11z^5=0 (H=608)}}\label{sec:1.4.10} \begin{thm}\label{thm:1.4.10} The integer solutions of $4x^5+4y^5+11z^5=0$ are exactly $(x,y,z)=(n,-n,0)$, $n\in\Z$. \end{thm} \paragraph{Setup.} Modulo 4 the equation gives $2\mid z$, and with $z=2w$ it becomes \[ x^5+y^5+88w^5=0,\qquad 88=2^3\cdot11 . \] The solutions with a zero coordinate are elementary (Step 1). For the others we apply the common framework of Section~\ref{sec:1.4} (Lemmas~\ref{lem:1.4:terms}--\ref{lem:1.4:sieve}): the Frey curve gives a newform of level $22$ or $88$, irreducibility follows from Mazur's theorem (Step 2), and the single auxiliary prime 3 excludes all newforms of these levels (Step 3). \paragraph{Step 1: reduction and the solutions with a zero coordinate.} \begin{lem}\label{lem:1.4.10-basic} If $(x,y,z)$ is a primitive solution, then $z=2w$ and $(x,y,w)$ is a primitive solution of $x^5+y^5+88w^5=0$. If moreover $xyz=0$, then $(x,y,z)=\pm(1,-1,0)$. Hence the integer solutions with $xyz=0$ are the $(n,-n,0)$, $n\in\Z$. \end{lem} \begin{proof} Modulo 4, $11z^5\equiv0$, so $z$ is even; dividing $4x^5+4y^5+11\cdot2^5w^5=0$ by 4 gives the new equation, and a common prime factor of $x,y,w$ would divide $x,y,z$. The coefficients $1,1,2^3\cdot11$ are pairwise coprime with exponents at most 3. By Lemma~\ref{lem:1.4:terms}(1), $x=0$ would force $|1|=|88|=1$, and so would $y=0$; hence a zero coordinate can only be $w=0$, and then $(x,y)=\pm(1,-1)$. Every nonzero solution is a multiple of a primitive one, and $(n,-n,0)$ is a solution for every $n$. \end{proof} \begin{lem}\label{lem:1.4.10-frey} Let $(x,y,z)$ be a primitive solution with $xyz\ne0$, and $z=2w$. Then the terms $x^5$, $y^5$, $88w^5$ are pairwise coprime, and $x$, $y$ are odd. Let $\varepsilon=\pm1$ with $\varepsilon x\equiv3\pmod4$, and let $E=E_{u,v}$ with $(u,v,w')=\varepsilon(x^5,88w^5,y^5)$. Then $E$ is semistable at every odd prime, with multiplicative reduction at 11 and $v_{11}(\Delta_{\min})=2+10v_{11}(w)$, and it has good reduction at 5 or multiplicative reduction with $5\mid v_5(\Delta_{\min})$. If $w$ is odd, then $E$ has type $\mathrm{III}^*$ at 2, $f_2(E)=3$ and $N_5=88=2^3\cdot11$; if $w$ is even, then $f_2(E)=1$, $v_2(\Delta_{\min})=10v_2(w)-2$ and $N_5=22=2\cdot11$. \end{lem} \begin{proof} By Lemma~\ref{lem:1.4:terms}(2) applied to $x^5+y^5+88w^5=0$, the terms are pairwise coprime, and $88w^5$ is the even one. Here $r=v_2(88w^5)=3+5v_2(w)$. If $w$ is odd, then $r=3$ and Lemma~\ref{lem:1.4:frey}(2) gives type $\mathrm{III}^*$, $f_2=3$, $v_2(\Delta_{\min})=10$. If $w$ is even, then $r\ge8$, $f_2=1$ and $v_2(\Delta_{\min})=2r-8=10v_2(w)-2\equiv3\pmod5$. The rest is Lemma~\ref{lem:1.4:level}, with $5\nmid88$ (\code{1.4.10a}; the 2-adic data are also checked with PARI in \code{1.4.10b}). \end{proof} \paragraph{Step 2: irreducibility, level lowering and the prime 3.} \begin{lem}\label{lem:1.4.10-irr} $E[5]$ is irreducible, and there are a newform $f\in S_2(\Gamma_0(N_5))$, $N_5\in\{22,88\}$ as in Lemma~\ref{lem:1.4.10-frey}, and a prime ideal $\lambda\mid5$ of $\mathcal O_{K_f}$ such that $a_3(f)\bmod\lambda\in T_3(1,1,88)$. \end{lem} \begin{proof} $E$ has full rational 2-torsion, is semistable at every odd prime, has $f_2(E)\le3$, and has good reduction at 5 or multiplicative reduction with $5\mid v_5(\Delta_{\min})$ (Lemma~\ref{lem:1.4.10-frey}). Lemma~\ref{lem:1.4:flat} gives irreducibility, and Lemma~\ref{lem:1.4:sieve}(3) gives $f$ and $\lambda$ ($3\nmid10\cdot88$). \end{proof} \begin{lem}\label{lem:1.4.10-three} $T_3(1,1,88)=\{0,1,4\}$. \end{lem} \begin{proof} Over $\F_3$ we have $a^5=a$ and $88\equiv1$, so the curve is the line $a+b+c=0$, with the four points $(1:1:1)$, $(1:2:0)$, $(1:0:2)$, $(0:1:2)$. At $(1:1:1)$ the curve $E_{1,1}$ is $Y^2=X(X-1)(X+1)=X^3-X$, which has four points over $\F_3$, so its trace is 0. The other three points have a zero coordinate and contribute $\pm(3+1)\equiv\pm4$. Hence $T_3=\{0,4,1\}$ (\code{1.4.10a}). \end{proof} \paragraph{Step 3: the newforms of levels 22 and 88.} \begin{lem}\label{lem:1.4.10-sieve} $S_2(\Gamma_0(22))^{\mathrm{new}}=0$. $S_2(\Gamma_0(88))^{\mathrm{new}}$ has dimension 3 and two Galois orbits, of dimensions 1 and 2, and the minimal polynomials of their eigenvalues $a_3$ are $T+3$ and $T^2-T-4$. Neither has a root in $\{0,1,4\}$ modulo 5. \end{lem} \begin{proof} The newforms are computed with PARI/GP (\code{1.4.10b}; the dimensions and the Hecke polynomial are confirmed by \code{1.4qb} and \code{1.4qc}). The values of $T+3$ at $0,1,4$ are $3,4,2$ modulo 5, and those of $T^2-T-4$ are $1,1,3$; in fact $T^2-T-4$ is irreducible modulo 5, its discriminant 17 being a nonsquare. \end{proof} \begin{proof}[Proof of Theorem~\ref{thm:1.4.10}] By Lemma~\ref{lem:1.4.10-basic}, it suffices to show that there is no primitive solution with $xyz\ne0$. For such a solution, Lemma~\ref{lem:1.4.10-irr} gives a newform of level 22 or 88; there is none of level 22, and for one of level 88, $a_3(f)\bmod\lambda\in\{0,1,4\}$ by Lemma~\ref{lem:1.4.10-three}, so the reduction modulo 5 of the minimal polynomial of $a_3(f)$ would have a root in $\{0,1,4\}$, contrary to Lemma~\ref{lem:1.4.10-sieve}. \end{proof} \section*{Reproducibility} \addcontentsline{toc}{section}{Reproducibility} The separate reproducibility archive contains the computational material for the thirty-eight proofs in this paper and its shared dependencies, together with the LaTeX source and BibTeX database. Its top-level README gives the build instructions, software requirements and current certificate map, and distinguishes historical computational outputs from the editorial and packaging checks performed for this edition. Generated large witness files are not needed to typeset the paper and can be reconstructed using the documented commands. \bibliographystyle{amsalpha} \bibliography{references} \end{document}