\documentclass[11pt,reqno]{amsart} \usepackage[a4paper,margin=27mm]{geometry} \usepackage{amsmath,amssymb,amsthm} \usepackage{booktabs,longtable,array} \usepackage{xcolor} \usepackage[colorlinks=true,linkcolor=blue!50!black,citecolor=blue!50!black,urlcolor=blue!50!black]{hyperref} \usepackage{enumitem} \usepackage{pdflscape} % ---- TODO macro: red, inline, with a margin flag ---- \newcommand{\todo}[1]{\textcolor{red}{\textbf{[TODO: #1]}}\marginpar{\textcolor{red}{\footnotesize\textbf{TODO}}}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\C}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \newcommand{\NS}{\operatorname{NS}} \newcommand{\contr}{\operatorname{contr}} \newcommand{\Gal}{\operatorname{Gal}} \newcommand{\Jac}{\operatorname{Jac}} \newcommand{\Pic}{\operatorname{Pic}} \newcommand{\tors}{\mathrm{tors}} \newcommand{\MW}{\mathrm{MW}} \newcommand{\sol}[6]{(#1,#2,#3\,|\,#4,#5,#6)} \newcommand{\F}{\mathbb{F}} \newcommand{\R}{\mathbb{R}} \newcommand{\Fr}{\operatorname{Frob}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\Km}{\operatorname{Km}} \newcommand{\Sym}{\operatorname{Sym}} \newcommand{\Cl}{\operatorname{Cl}} \newcommand{\apphead}[1]{\par\medskip\noindent\textbf{#1}\par\nobreak\smallskip\noindent\ignorespaces} \theoremstyle{plain} \newtheorem{theorem}{Theorem}[section] \newtheorem{proposition}[theorem]{Proposition} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{question}[theorem]{Question} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{observation}[theorem]{Observation} \numberwithin{equation}{section} \setlength{\emergencystretch}{2em} \allowdisplaybreaks \title[Brudno's families and Bremner's K3 surface]{Brudno's families of equal sums of sixth powers\\ and Bremner's K3 surface} \author{Alon Amit} \address{Los Altos, CA, USA} \email{alon.amit@gmail.com} \author{Claude 5.1} \thanks{Claude 5.1 is a large language model developed by Anthropic; see \S\ref{sec:ai} for its role in this paper.} \date{\today} \begin{document} \begin{abstract} Brudno (1970, 1976) and Brudno--Kaplansky (1974) found an elliptic family and a degree-four polynomial family of solutions of $x^6+y^6+z^6=u^6+v^6+w^6$. We explain both in terms of the K3 surface $V\subset\PP^5$, the complete intersection of three quadrics studied by Bremner (1981) and Kuwata (2007). The Brudno--Kaplansky curve is Cremona's curve $480\mathrm{f}4$ and is the fibre at $\lambda=\infty$ of the elliptic pencil on $V$ that Bremner calls the Brudno pencil; the curve $z^2=x^4+38x^2y^2+y^4$ of 1970 is $480\mathrm{f}1$, and the classical recursions of Lebesgue and Desboves that Brudno iterated are the multiplication maps $[2]$ and $[-3]$ on it. Brudno's 1976 quartic is the unique rational curve of degree four on $V$ (Bremner's Theorem~4.4) and a section of the Brudno pencil; its parameter is the fibre coordinate $\lambda$, and the involution Brudno noticed is a linear symmetry of $V$. Using the Mordell--Weil lattice of the pencil we prove a closed formula for the degree of a section, $\deg C_S=2\langle S-\gamma/4,\,S-\gamma/4\rangle$, enumerate all sections of degree $\le12$ (four of them new, of degrees $7,10,11,11$), and enumerate, by a lattice computation, all smooth rational curves of degree $\le 12$ on $V$ over $\Q(i)$ and over $\Q$, reproducing Bremner's theorems and testing his conjecture. A census of the $74$ primitive solutions with all terms $\le 1343$ identifies which of them lie on $V$, on known curves, on sections, and on fibres of jumping rank, and records some phenomena off $V$. In the second part we compute equations and parametrisations of the $48$ new curves that are not sections and show that seven of the eight census solutions on no known curve lie on them, leaving a single unexplained point; we determine the geometry over $\C$ of the sextic fourfold $M_6$, of the threefold $W=M_6\cap M_2$ (which has exactly $710$ nodes and defect $95$) and of $V$ (transcendental lattice $U(2)\oplus\langle12\rangle$, zeta function through the newform $320\mathrm f2$), and discuss in what sense the geometry drives the arithmetic; and we show that Bremner's conjecture follows, with non-explicit finite sets, from the theorems of Sterk and Bright--Logan--van~Luijk, describe how to make it explicit (Borcherds--Shimada), and run the first step of Bremner's own cone procedure, which identifies the degree-$8$ pencil involutions that the proof requires. \end{abstract} \maketitle \tableofcontents %====================================================================== \section{Introduction}\label{sec:intro} We study the Diophantine equation \begin{equation}\label{eq:sextic} x^6+y^6+z^6=u^6+v^6+w^6 , \end{equation} whose first non-trivial solution $3^6+19^6+22^6=10^6+15^6+23^6$ was found by Subba Rao in 1934 \cite{SubbaRao1934}. Lander, Parkin and Selfridge \cite{LPS1967} listed the ten primitive solutions with common sum below $2.5\cdot10^{14}$ and noticed that all but one of them also satisfy \begin{equation}\label{eq:k2} x^2+y^2+z^2=u^2+v^2+w^2 . \end{equation} Simcha Brudno then wrote three short papers on \eqref{eq:sextic} (after an earlier numerical note \cite{Brudno1969}, which we do not use): in 1970 \cite{Brudno1970} he produced infinitely many solutions from a trigonometric identity, which led him to the genus-one quartic $C\colon z^2=x^4+38x^2y^2+y^4$ and to the nineteenth-century recursions of Lebesgue and Desboves on it; with Kaplansky \cite{BK1974} he reduced the system \eqref{eq:sextic}, \eqref{eq:k2}, $v=y-z$, $w=y+z$ to the rational points of a cubic curve $E_1$ of rank one, and recorded Tate's remark that the obvious point $P$ generates $E_1(\Q)$ modulo torsion; and in 1976 \cite{Brudno1976} he exhibited, without any indication of how it was found, the two-parameter polynomial solution of degree four {\small \begin{equation}\label{eq:brudno-quartic} \begin{aligned} x &= 2m^4 + 4m^3n - 5m^2n^2 - 12mn^3 - 9n^4, &\quad u &= m^4 - 3m^3n - 14m^2n^2 - 15mn^3 - 9n^4,\\ y &= 3m^4 + 9m^3n + 18m^2n^2 + 21mn^3 + 9n^4, &\quad v &= 3m^4 + 8m^3n + 9m^2n^2,\\ z &= -m^4 - 10m^3n - 17m^2n^2 - 12mn^3, &\quad w &= 2m^4 + 12m^3n + 19m^2n^2 + 18mn^3 + 9n^4, \end{aligned} \end{equation}} which satisfies \eqref{eq:sextic}, \eqref{eq:k2} and the linear relation $3x+y+z=3u+v+w$ identically. The question that motivated this paper is: where does \eqref{eq:brudno-quartic} come from? The answer is in Bremner's 1981 paper \cite{Bremner1981}, although Bremner never says so. Bremner observed that the known solutions of \eqref{eq:sextic}, \eqref{eq:k2}, after permuting and changing signs of the unknowns, satisfy the three quadratic equations \begin{equation}\label{eq:V} V\colon\quad Q(x,u)=Q(w,z),\quad Q(y,v)=Q(u,x),\quad Q(z,w)=Q(v,y),\qquad Q(s,t)=s^2+st-t^2, \end{equation} which define a K3 surface $V\subset\PP^5$; cubing the three equations and adding gives \eqref{eq:sextic}, adding them gives \eqref{eq:k2}. On $V$ the Brudno--Kaplansky substitution becomes an elliptic pencil $E\to\PP^1_\lambda$ (the \emph{Brudno pencil}), $E_1$ being its fibre at $\lambda=\infty$, and Brudno's quartic is the unique curve of degree four on $V$ of arithmetic genus zero over $\Q$ up to the symmetries of $V$ (Bremner's Theorem 4.4); Bremner even writes it down (p.~574) as the image of a line under a fibrewise involution of the Brudno pencil. In one sentence: Brudno's 1970/1974 family is a fibre of the Brudno pencil, his 1976 family is a section of it, and the two are therefore the same object seen twice. The paper has two parts. Sections~\ref{sec:brudno} and~\ref{sec:bremner} summarise, briefly, what Brudno, Brudno--Kaplansky, Bremner, Delorme, Choudhry and Kuwata proved; we add the identification of $E_1$ with Cremona's curve $480\mathrm{f}4$ and of the Lebesgue and Desboves recursions with $[2]$ and $[-3]$ on $\Jac(C)=480\mathrm{f}1$. Sections~\ref{sec:quartic}--\ref{sec:census} contain the new results: \begin{enumerate}[label=(\roman*)] \item the dictionary between Brudno's quartic and Bremner's degree-four curve, the fact that it is the section $g_1+g_3+2T$ of the Brudno pencil with $\lambda=(3m+2n)/(2m+3n)$, that the hyperplane $3x+y+z=3u+v+w$ is its linear span and cuts $V$ residually in two lines and a conic, and that Brudno's involution $(m:n)\mapsto(-3(m+n):m+3n)$ is a linear symmetry of $V$ (Section~\ref{sec:quartic}); also the identification of the published parametric solutions of Bremner, Delorme and Choudhry with one another and with sections of the pencil; \item the Mordell--Weil lattice of the Brudno pencil over $\Q(\lambda)$: a proof that Bremner's three generators generate, and the degree formula $\deg C_S=2\langle S-\gamma/4,S-\gamma/4\rangle$ for the curve $C_S\subset\PP^5$ traced by a section $S$, with $\gamma=-g_1+g_2+g_3$; the complete list of sections of degree $\le12$, which contains four curves not in the literature, of degrees $7,10,11,11$ (Section~\ref{sec:sections}); \item a complete enumeration of the smooth rational curves of degree $\le 12$ on $V$, over $\Q(i)$ and over $\Q$, by a rigorous induction in the N\'eron--Severi lattice; Bremner's Theorems 4.2--4.4 are recovered, $52$ new $\Q$-rational curves of degrees $7$--$12$ appear up to symmetry, and Bremner's conjecture that all such curves come from finitely many by finitely many automorphisms is tested through degree $12$: it fails if one only uses the involutions of the Brudno pencil, and holds if one adds the involutions of the $14$ rational elliptic pencils of degree $\le 6$ (Section~\ref{sec:lattice}); \item a census of the $74$ primitive solutions of \eqref{eq:sextic} with all terms $\le1343$: $59$ satisfy \eqref{eq:k2}, $41$ lie on $V$, $33$ of those on a known curve or on a section, and the remaining $8$ lie on fibres of the Brudno pencil whose rank jumps; Delorme's and Guy's question whether every solution of \eqref{eq:sextic}, \eqref{eq:k2} lies on $V$ is answered by the smallest counterexample $\#14$, and a $(3,6)$-multigrade solution is pointed out (Section~\ref{sec:census}). \end{enumerate} The second part (Sections~\ref{sec:newcurves}--\ref{sec:bremner-conj}) was added after the first version of the paper and answers the questions it left open: \begin{enumerate}[label=(\roman*),resume] \item equations, ideals and birational parametrisations of all $48$ non-section curves of degree $\le12$, and the incidence of the $41$ census solutions on $V$ with the $62$ curves of degree $\le12$: seven of the eight solutions on no known curve and no section lie on the new curves, and exactly one solution on $V$, $\#49$, lies on no $\Q$-rational smooth rational curve of degree $\le12$ (Section~\ref{sec:newcurves}); \item the geometry over $\C$ of the sextic fourfold $M_6$, of the threefold $W=M_6\cap M_2$ and of $V$: fundamental groups, Hodge numbers, the $710$ nodes of $W$ and its defect $95$, the N\'eron--Severi and transcendental lattices of $V$ and its zeta function, point counts over $\F_p$ verifying all of it, and a discussion of the sense in which the geometry governs the arithmetic over finite fields and over $\Q$ (Section~\ref{sec:geometry}); \item Bremner's conjecture: it follows for every K3 surface over every field of characteristic $\ne2$, with non-explicit finite sets, from the theorems of Sterk and of Bright, Logan and van Luijk together with a Dirichlet-domain argument; the explicit version for $V$ is a computation by Borcherds' method (Shimada, \textsc{Oscar}), for which we provide the embedding $\NS(V)\hookrightarrow\mathrm{II}_{1,25}$; and the first step of Bremner's cone procedure, run on our data, shows that the pencils of degree $\le6$ cannot complete the proof and exhibits the degree-$8$ pencil involutions that do the missing work (Section~\ref{sec:bremner-conj}). \end{enumerate} Section~\ref{sec:open} lists open problems. Appendix~\ref{app:misprints} collects misprints in \cite{Bremner1981,Kuwata2007,Delorme1992,Choudhry1994,Choudhry2000} found while checking every formula of those papers; Appendix~\ref{app:comp} describes the computations; Appendices~\ref{app:table}--\ref{app:param} contain the tables and the parametrisations of the new curves of degree $\le9$. Known results are summarised briefly, with references to the precise statements; all new results are given in full. Every numerical statement below was verified by exact computation (Appendix~\ref{app:comp}); statements that we could not verify, or bibliographic data we could not check, are marked in red as \textcolor{red}{\textbf{[TODO: \dots]}}. \subsection{Statement on the use of AI}\label{sec:ai} This paper was written by Claude, a large language model developed by Anthropic, in the course of a conversation with the first-named author, Alon Amit, in October 2026; the model configured for the conversation was Claude 5.1 (model identifier \texttt{claude-fable-5-1}). Since the reader of a mathematics paper is entitled to know who did what, we describe the division of labour. The first-named author set the problem and directed the work. He asked where Brudno's two families come from, what the map from the Brudno--Kaplansky curve to integer solutions is, whether the polynomial solutions can be shown to be finite in number, and which elliptic curve explains the second family; he supplied the papers of Brudno, Brudno--Kaplansky, Bremner, Kuwata, Delorme and Choudhry, chose among the approaches that were proposed to him at each stage (in particular to work on $M_6$ only, to lead with a literature search, to enumerate curves through degree $12$, and later to compute the non-section curves, the geometry over $\C$ and the status of Bremner's conjecture), reviewed the intermediate results, and decided what to pursue. He did not write the text or the code. Claude read the sources, searched the literature, formulated the statements, found the arguments and wrote the proofs, designed and ran the computations (Python/SymPy, PARI/GP, Sage/Singular and C; parts of the longer computations were delegated to parallel instances of the same model, whose results were then re-verified independently by the main instance, as recorded in Appendix~\ref{app:comp}), and wrote the paper, the accompanying Sage notebook and the interactive applet. The pronoun ``we'' in the text refers to the two authors. What this means for the reader. Every numerical and algebraic statement was checked by exact computation, and the checks are described in Appendix~\ref{app:comp}; where a statement rests on a hypothesis (such as (H), $\rho(V_{\bar\Q})=19$) or is only numerical (such as Observation~\ref{obs:geo-W}) this is said in place. The proofs, the summaries of the literature and the choice of what to cite are the model's, and have been reviewed by the first-named author but not refereed by anyone else. Theorem and section numbers quoted from the literature were taken from the sources as the model read them---in a few cases through reviews or summaries rather than the original text, which is noted where it happens---and bibliographic data that could not be verified are marked \textcolor{red}{\textbf{[TODO]}}. A language model can misread a source or produce an argument that looks right and is not, and it can do so fluently; the authors have tried to keep such errors out by insisting on exact computation wherever a statement could be computed, but the reader should keep the possibility in mind, especially for the non-computational parts (Sections~\ref{sec:geo-answer}, \ref{sec:bremner-conj}). Responsibility for the decision to circulate the paper, and for its final form, rests with the first-named author. \subsection*{Conventions}\label{sec:conventions} Throughout, $V\subset\PP^5$ is the surface \eqref{eq:V} with coordinates $(x,y,z,u,v,w)$, exactly as on p.~548 of \cite{Bremner1981} (Bremner's $V$; Kuwata's $K_B$; Delorme's system (II); Choudhry's system (2)--(4)). Bremner's lines $\Gamma_1,\dots,\Gamma_{32}$ and conics $Q_1,\dots,Q_{24}$ are those of his Tables 1 and 2 (with the misprint corrections of Appendix~\ref{app:misprints}); $\Gamma_1,\dots,\Gamma_8$ and $Q_1,\dots,Q_{12}$ are the ones defined over $\Q$. $H$ denotes the hyperplane class. The Brudno pencil is Bremner's equation (5), \begin{equation}\label{eq:pencil} y=a+(\lambda-1)b,\quad z=(\lambda-1)a-b,\quad v=-(\lambda-1)a-\lambda b,\quad w=\lambda a-(\lambda-1)b, \end{equation} under which \eqref{eq:V} reduces to the genus-one quartic $E_\lambda\subset\PP^3_{(x:u:a:b)}$, \begin{equation}\label{eq:Elambda} x^2-u^2=(\lambda^2-1)(a^2+b^2),\qquad xu=\lambda a^2-(3\lambda^2-5\lambda+3)ab-\lambda b^2 . \end{equation} With Bremner's abbreviations \[ A=\lambda^2+1,\quad B=\lambda^2-1,\quad P=2\lambda^2-2\lambda+1,\quad Q=\lambda^2-2\lambda+2,\quad U=3\lambda^2-5\lambda+3, \] the Weierstrass model is Bremner's $(14')$, \begin{equation}\label{eq:W} \mathcal E\colon\quad y^2=x^3-2(5PQ-2B^2)\,x^2+25P^2Q^2\,x\qquad\text{over }\Q(\lambda), \end{equation} reached from $E_\lambda$ by the map of Bremner's Appendix I (recalled in \S\ref{sec:zero}). The torsion point \[ T=(5PQ,\;10BPQ) \] has order $4$, and Bremner's three generators of infinite order over $\Q(\lambda)$ (his Remark on p.~556, with the misprint $(P,3B,Q)\to(P,3BP,Q)$ corrected and transported to \eqref{eq:W}) are \begin{equation}\label{eq:gens} \begin{gathered} g_1=(5P^2,\,15BP^2),\qquad g_2=\bigl(5Q(2\lambda-1)^2,\;10Q(\lambda-1)(2\lambda-1)(3\lambda^2-\lambda+1)\bigr),\\ g_3=\bigl(5P(\lambda-2)^2,\;10P(\lambda-1)(\lambda-2)(\lambda^2-\lambda+3)\bigr). \end{gathered} \end{equation} A section is written $S=n_1g_1+n_2g_2+n_3g_3+kT$ and abbreviated $(n_1,n_2,n_3,k)$ or, when only the lattice part matters, $(n_1,n_2,n_3)$. The zero section is the line $\Gamma_5$ (see Remark~\ref{rem:zero}). For a section $S$, $C_S\subset V$ denotes the smooth rational curve it traces in $\PP^5$; its degree is $H\cdot C_S$. The $74$ primitive solutions of \eqref{eq:sextic} with all six terms $\le1343$ are numbered $\#1,\dots,\#74$ by increasing common sum (Appendix~\ref{app:table}); we write a solution as $\sol{a}{b}{c}{x}{y}{z}$, the side containing the larger maximal term first. ``$k=2$'' means that \eqref{eq:k2} holds; ``on $V$'' means that some signed permutation of the six terms lies on $V$. The symmetry group of $V$ is the group of the $96$ signed permutations preserving \eqref{eq:V}, which acts on $\PP^5$ through a group of order $48$ (Bremner's Appendix II); ``up to symmetry'' always refers to this group. %====================================================================== \section{Brudno's results, briefly}\label{sec:brudno} \subsection{The 1970 family and the quartic $C$}\label{sec:1970} Brudno \cite{Brudno1970} starts from the identity \[ (\sin\alpha\sin\beta)^6+(\cos\alpha-\cos\beta\sin\alpha)^6+(\cos\alpha+\cos\beta\sin\alpha)^6 = 1+\cos^6\alpha+(\cos\beta\sin\alpha)^6, \] which ``has only one positive rational solution'' in $\cos^2\beta$, namely $\cos^2\beta=(1+9\cos^2\alpha)^{-1}$. Rationalising with $\cos\alpha=2xy/(x^2+y^2)$, $\sin\alpha=(x^2-y^2)/(x^2+y^2)$, $\cos\beta=(x^2+y^2)/z$, $\sin\beta=6xy/z$ reduces the problem to the genus-one quartic \begin{equation}\label{eq:C} C\colon\quad z^2=x^4+38x^2y^2+y^4 \end{equation} (his equation (6)), and clearing denominators gives the family \begin{equation}\label{eq:1970family} \begin{gathered} A=6xy(x^2-y^2),\quad B=2xyz-(x^4-y^4),\quad C=2xyz+(x^4-y^4),\\ D=(x^2+y^2)z,\quad E=2xyz,\quad F=x^4-y^4, \end{gathered} \end{equation} with $A^6+B^6+C^6=D^6+E^6+F^6$ and $A^2+B^2+C^2=D^2+E^2+F^2$ identically modulo \eqref{eq:C}. Note $B=E-F$, $C=E+F$. The seed $(x,y,z)=(1,2,13)$ gives $36^6+37^6+67^6=15^6+52^6+65^6$, the second solution of \cite{LPS1967} ($\#2$ below); the point $P_0=(2:1:13)$ of $C$ gives the same solution. Brudno then iterates two classical recursions on $C$: Lebesgue's \cite{Lebesgue1853} (quoted via Dickson \cite{Dickson}) \begin{equation}\label{eq:Leb} x_2=x_1^4-y_1^4,\qquad y_2=2x_1y_1z_1, \end{equation} and Desboves's \cite{Desboves1879} \begin{equation}\label{eq:Des} x_2=x_1(4y_1^4z_1^2-g^2),\qquad y_2=y_1(4x_1^4z_1^2-g^2),\qquad g=x_1^4-y_1^4 . \end{equation} From the seed, \eqref{eq:Leb} gives the solution \[ 1351769^6+11601720^6+15873751^6=7260991^6+8612760^6+16171009^6 \] and \eqref{eq:Des} a $17$-digit solution; the ``irrational'' parameter choice $x_1:y_1=17\pm7\sqrt5$ gives $902^6+10591^6+10947^6=2618^6+9689^6+11493^6$. These are the solutions $4P$, $6P$, $3P$ of the next paragraph. \subsection{Brudno--Kaplansky: the curve $E_1$}\label{sec:BK} Brudno and Kaplansky \cite{BK1974} substitute (their (4)) \[ y=pa+(q-p)b,\quad z=(q-p)a-pb,\quad v=(q-p)a+qb,\quad w=qa-(q-p)b, \] under which \eqref{eq:sextic} and \eqref{eq:k2} reduce to $x^2-u^2=(q^2-p^2)(a^2+b^2)$ and $xu=pqa^2-(3p^2-5pq+3q^2)ab-pqb^2$ (``somewhat remarkably, after division by $3$ the expression (7) is exactly the square of \dots''). This is \eqref{eq:pencil}--\eqref{eq:Elambda} with $(p,q)=(1,\lambda)$ and $v$ replaced by $-v$. They pursue only $(p,q)=(0,1)$, i.e.\ $v=y-z$, $w=y+z$ (the shape of \eqref{eq:1970family}), where the system becomes $x^2-u^2=a^2+b^2$, $xu=3ab$ and, with $T=1/(x-1)$, $W=(2T+1)x/a$, the cubic \begin{equation}\label{eq:E1} E_1\colon\quad W^2=(2T+1)(10T^2+2T+1). \end{equation} On $E_1$, $R=(0,1)$ has order $4$ and generates the torsion (a check with Lutz's criterion \cite{Lutz1937}; in textbook form the Nagell--Lutz theorem \cite{Nagell1935,Lutz1937}, \cite[VIII.7]{Silverman} \todo{Lutz's ``Theorem III, p.~244'' cited by Brudno--Kaplansky was not consulted}), $P=(-1/4,3/4)$ has infinite order and corresponds to the trivial solution $(3,1,2;2,3,1)$, the eight points $\pm S+jR$ give the same solution, Brudno's 1970 solutions ``pair off with $2P,3P,4P$ and $6P$'', and $5P$ is computed (with transcription errors corrected in \cite{Brudno1976}). A note added in proof says: ``John Tate showed us how to prove that $G^*$ is generated by the image of $P$. The solution of Eqs.\ (1)--(3) is thus complete.'' No argument is recorded. Inverting the substitutions, a point $(T,W)$ of $E_1$ gives the solution \begin{equation}\label{eq:E1map} x=(T+1)W,\quad y=(2T+1)(T+1),\quad z=-TW,\quad u=3T(2T+1),\quad v=y-z,\quad w=y+z, \end{equation} which satisfies \eqref{eq:sextic} and \eqref{eq:k2} identically modulo \eqref{eq:E1} (verified by computation). \begin{proposition}\label{prop:480f4} $E_1$ is the curve $480\mathrm{f}4$ of Cremona's tables \cite{Cremona}, with minimal model $y^2=x^3-x^2+15x+225$ reached by $(T,W)\mapsto(20T+5,20W)$. Under this map $P\mapsto(0,15)$, $R\mapsto(5,20)$ and the $2$-torsion point $(-1/2,0)\mapsto(-5,0)$. Cremona's table lists $(0,15)$ as the generator of $E_1(\Q)$ modulo torsion and $(5,20)$ as the torsion generator; $E_1(\Q)\cong\Z\times\Z/4$, the rank being certified by $2$-descent, and $\hat h(P)=0.516278469943469\ldots$ is the regulator. Thus Tate's unrecorded argument is replaced by the tables: $P$ generates $E_1(\Q)/\tors$. \end{proposition} \begin{proof} Verified with PARI/GP and against Cremona's files \texttt{allcurves} and \texttt{allgens}: conductor $480=2^5\cdot3\cdot5$, minimal discriminant $-2^{12}\cdot3^2\cdot5^4$, $j=2^6\cdot11^3/(3^2\cdot5^4)$, Kodaira types $\mathrm{I}_3^*,\mathrm{I}_2,\mathrm{I}_4$ at $2,3,5$; \texttt{ellrank} returns lower and upper bound $1$; \texttt{ellsaturation} of $\{P\}$ at primes $\le200$ returns $\{P\}$; Cremona's listed generator is $(0,15)=P$ and his regulator equals $\hat h(P)$. \todo{The LMFDB page of $480\mathrm{f}4$ could not be loaded in our environment; the LMFDB label may differ from Cremona's.} \end{proof} Table~\ref{tab:multiples} lists the multiples $nP$, $n\le7$, in $(T,W)$-coordinates together with the corresponding solutions \eqref{eq:E1map}; $n=2,3,4,5,6$ are exactly Brudno's solutions of \S\ref{sec:1970} and the corrected $5P$ of \cite{Brudno1976}; $n=7$ is new. \begin{table}[ht]\centering\footnotesize\setlength{\tabcolsep}{4pt} \caption{Multiples of $P=(-1/4,3/4)$ on $E_1$ and the solutions \eqref{eq:E1map} (absolute values, primitive).}\label{tab:multiples} \begin{tabular}{@{}cl@{}}\toprule $n$ & $nP=(T,W)$; solution\\\midrule 1 & $(-1/4,\;3/4)$; \quad $\sol{3}{2}{1}{2}{1}{3}$ (trivial)\\[2pt] 2 & $(-3/16,\;-25/32)$; \quad $\sol{65}{52}{15}{36}{67}{37}$\\[2pt] 3 & $(119/4,\;2937/4)$; \quad $\sol{10947}{902}{10591}{2618}{11493}{9689}$\\[2pt] 4 & $(-2479/8000,\;114231/160000)$; \quad $\sol{16171009}{8612760}{7260991}{11601720}{1351769}{15873751}$\\[2pt] 5 & $(-111961/917764,\;-363593823/439608956)$;\\ & $\sol{165809277507}{151561337462}{23038103009}{63175337782}{174599440471}{128523234453}$\\[2pt] 6 & $(111355677/15335056,\;2756217479575/30026039648)$;\\ & $(14313884517104705,\,2420573921978212,\,12581285648588145\,|$\\ & $\quad 6382739481279084,\,15001859570566357,\,10160711726609933)$\\[2pt] 7 & $(-84669865201/231320597764,\;36510434863885803/55627746029688956)$;\\ & $(10138323873382807040643,\,4138890740067511917838,\,5853434897467259943361\,|$\\ & $\quad 7168855857422411252878,\,1714544157399748025523,\,9992325637534771861199)$\\ \bottomrule \end{tabular} \end{table} \begin{proposition}[the 1970 quartic and the two recursions]\label{prop:isogeny} Let $C_c\colon z^2=x^4+cx^2y^2+y^4$ with origin $O=(1:0:1)$, and let $C=C_{38}$. \begin{enumerate}[label=\textup{(\alph*)}] \item $\Jac(C)$ is the curve $480\mathrm{f}1$ (minimal model $y^2=x^3-x^2-30x+72$, torsion $(\Z/2)^2$, rank $1$), the optimal curve of the isogeny class of $E_1=480\mathrm{f}4$; the class $480\mathrm{f}$ consists of $480\mathrm{f}1,\dots,480\mathrm{f}4$ with $480\mathrm{f}1$ $2$-isogenous to each of the other three. In particular $\Jac(C)$ is $2$-isogenous to, but not isomorphic to, $E_1$. \item The map $\varphi(x,y,z)=(x^2/y^2,\,xz/y^3)$ is a $2$-isogeny from $C_c$ onto $E'_c\colon w^2=t^3+ct^2+t$ \textup(for $c=38$, $E'_{38}\cong480\mathrm{f}3$\textup), with $\varphi(O)=O$ and $\varphi\circ\iota=-\varphi$ for $\iota(x,y,z)=(x,-y,z)$, so that $\iota=[-1]$. \item Lebesgue's map \eqref{eq:Leb}, completed by $z_2=x_1^8+2cx_1^6y_1^2+6x_1^4y_1^4+2cx_1^2y_1^6+y_1^8$, is an endomorphism of $C_c$ and equals $[2]$. Desboves's map \eqref{eq:Des}, after reduction modulo the equation of $C_c$, reads $x_2=x_1(x_1^8-6x_1^4y_1^4-4cx_1^2y_1^6-3y_1^8)$, $y_2=y_1(3x_1^8+4cx_1^6y_1^2+6x_1^4y_1^4-y_1^8)$ and, completed by $z_2=z_1S(x_1,y_1)$ with the form $S$ of degree $16$ determined by the equation of $C_c$ \textup(up to sign\textup), equals $[-3]$; the other choice of sign of $z_2$ gives $[3]$ followed by the translation by the $2$-torsion point $(1:0:-1)$. \item The seed $P_0=(2:1:13)$ generates $C(\Q)$ modulo torsion, and the composite of the family map \eqref{eq:1970family} with Brudno--Kaplansky's parametrisation sends $C(\Q)$ onto the index-two subgroup $2\Z P$ of $E_1(\Q)/\tors$. Consequently, starting from any rational point of $C$, the recursions \eqref{eq:Leb} and \eqref{eq:Des} only reach the solutions $2^a3^b\cdot 2P$ \textup($a,b\ge0$\textup) of $E_1$, never $3P$ or $5P$; Brudno's $3P$ came from the $\Q(\sqrt5)$-point $x:y=17\pm7\sqrt5$ of $C$. \end{enumerate} \end{proposition} \begin{proof} (a) PARI's \texttt{ellfromeqn} turns \eqref{eq:C} into $Y^2=X^3+38X^2-4X-152=(X+38)(X-2)(X+2)$, whose minimal model is $[0,-1,0,-30,72]=480\mathrm{f}1$; the isogeny matrix of the class $480\mathrm{f}$ is read off Cremona's file \texttt{allisog}, and \texttt{ellisisom} confirms that $480\mathrm{f}1\not\cong480\mathrm{f}4$. (b) Direct substitution; $E'_{38}$ has minimal model $[0,-1,0,-480,4212]=480\mathrm{f}3$. (c) The identities $\varphi\circ\mathrm{Leb}=[2]\circ\varphi$ and $t(\varphi\circ\mathrm{Des})=t([3]\circ\varphi)$ hold as identities of rational functions of $x,y,z,c$ modulo the equation of $C_c$ (checked with sympy; the doubling and tripling formulas on $E'_c$ were written out explicitly). A morphism $f\colon C_c\to C_c$ with $\varphi\circ f=[n]\circ\varphi$ differs from $[n]$ by a constant in $\ker\varphi$, which vanishes if $f(O)=O$; Lebesgue's map fixes $O$, so it is $[2]$. For Desboves's map the two lifts $z_2=\pm z_1S$ give $\varphi\circ\mathrm{Des}=\pm[3]\circ\varphi$, each identity holding on a closed subset and the two covering $C_c$, so one of them holds identically; at $P_0=(2,1,13)$, $c=38$, one finds $\varphi(P_0)=(4,26)$, a point of infinite order, and $\varphi(\mathrm{Des}P_0)=-[3]\varphi(P_0)$ for the lift $z_2=+z_1S$; since $\mathrm{Des}(O)=(-1:0:1)=O$ in the weighted projective plane, that lift is $[-3]$. The other lift differs by $(x,y,z)\mapsto(x,y,-z)$, which is the reflection $R\mapsto(1:0:-1)-R$, whence the last statement. (d) $\hat h(\varphi(P_0))=2.06511387977388\ldots$ equals the regulator of $480\mathrm{f}3$ in Cremona's table, so $\varphi(P_0)$ generates $E'(\Q)/\tors$; as $\deg\varphi=2$, $P_0$ generates $C(\Q)/\tors$. The family map \eqref{eq:1970family} is a morphism from $C$ onto the Brudno--Kaplansky quartic curve in $\PP^3$ (the solutions \eqref{eq:1970family} have $B=E-F$, $C=E+F$ and satisfy \eqref{eq:k2}), which Brudno--Kaplansky's substitutions identify with $E_1$; so $\psi\colon C\to E_1$ is a morphism of genus-one curves, $\psi=h+c$ with $h$ a homomorphism. By \cite{BK1974} the seed, its Lebesgue image $2P_0$ and its Desboves image $-3P_0$ give the solutions of $2P$, $4P$, $6P$, and a solution determines its point on $E_1$ up to $\pm$ and the torsion (a signed permutation preserving the Brudno--Kaplansky locus induces an automorphism of the space quartic $E_1\subset\PP^3$ preserving its hyperplane class, hence of the form $R\mapsto\pm R+t$ with $4t=0$). Writing $h(P_0)+c\equiv\varepsilon_1 2P$, $2h(P_0)+c\equiv\varepsilon_2 4P$, $-3h(P_0)+c\equiv\varepsilon_3 6P$ modulo torsion ($\varepsilon_i=\pm1$), subtraction gives $h(P_0)\equiv(2\varepsilon_2-\varepsilon_1)2P$ and $-4(2\varepsilon_2-\varepsilon_1)=3\varepsilon_3-\varepsilon_1$, which forces $\varepsilon_2=\varepsilon_1=-\varepsilon_3$, $h(P_0)\equiv\pm2P$ and $c$ torsion. Hence $\psi(C(\Q))\equiv2\Z P$ modulo torsion. The family map is $2:1$ onto its image (the fibre over the image of $(1:2:13)$ is $\{(1:2:13),(2:1:-13)\}$, verified by computation), so $h$ has degree $2$ and is the $2$-isogeny $480\mathrm{f}1\to480\mathrm{f}4$, which therefore sends the generator $P_0$ to $\pm2P$. \end{proof} \begin{remark} Brudno's attributions of \eqref{eq:Leb} to Lebesgue and of \eqref{eq:Des} to Desboves are quoted from his paper; we did not locate the formulas in the original memoirs \todo{(the Numdam summaries of \cite{Lebesgue1853} and \cite{Desboves1879} were consulted, not the full texts; the attribution most plausibly comes from Dickson \cite{Dickson}, Ch.~XXII, p.~637)}. \end{remark} \subsection{The 1976 quartic}\label{sec:1976} Brudno \cite{Brudno1976} states \eqref{eq:brudno-quartic} without derivation (``The principal aim of this paper is to exhibit the following explicit solution to (1) in terms of parameters $m,n$''), and records: the identity $3x+y+z=3u+v+w$ (his (5)); the substitution $m':n'=-3(m+n):(m+3n)$ (his (6)), which yields $x':y':z':u':v':w'=u:v:w:x:y:z$ (his (7); the scale factor is $36$), ``so any particular primitive solution \dots arises from two distinct ratios $m:n$''; the content of the six forms, which has no prime factor $>5$; the geometric remark that $(x,y,z)$ and $(u,v,w)$ are lattice points on a common sphere, a common surface $X^6+Y^6+Z^6=b$ and a double cone of axis $3:1:1$; and the coverage statement: ``Of the ten smallest primitive solutions to (1), listed in [LPS], all but the sixth satisfy (2). Only the second satisfies (3), while (4) gives rise to all except the seventh (and, of course, the sixth).'' We add two facts, verified by computation: the six forms \eqref{eq:brudno-quartic} satisfy no identity $\sum x^k=\sum u^k$ for $k=1,3,4,5$; and their $6\times5$ coefficient matrix has rank $5$, so $3x+y+z=3u+v+w$ spans the space of linear relations and the image of $\PP^1$ is a rational normal quartic curve spanning the hyperplane $H_{\mathrm{Br}}\colon 3x+y+z=3u+v+w$. The only hint about the origin of \eqref{eq:brudno-quartic} is the acknowledgement of ``Denis Bartley without whose help the computations could not have been performed''. %====================================================================== \section{Bremner's surface, and later work}\label{sec:bremner} \subsection{Bremner 1981}\label{sec:bremner81} Bremner \cite{Bremner1981} introduces $V$ as follows (pp.~544--545): the Brudno--Kaplansky substitution \eqref{eq:pencil} is the restriction of $(y,z,v,w)$ to one ruling of the rank-four quadric $Q(z,w)=Q(v,y)$; at Swinnerton-Dyer's suggestion the LPS table was tested against this quadric, with the ``startling observation'' that all known solutions except the sixth satisfy it \emph{and} its two images under the cyclic permutation of $(x,y,z)$ and $(u,v,w)$, i.e.\ \eqref{eq:V}, after relabelling and sign changes. Cubing and adding gives \eqref{eq:sextic}; the mechanism is the identity \begin{equation}\label{eq:cube} Q(s,t)^3-Q(t,s)^3=2(s^6-t^6),\qquad Q(s,t)-Q(t,s)=2(s^2-t^2), \end{equation} which Bremner does not write out. His main results are: \begin{itemize} \item $V$ is a smooth complete intersection of three quadrics, hence a K3 surface, of degree $H^2=8$ (\S2). \item $V$ contains exactly $32$ lines and $24$ conics over $\Q(i)$ (Tables 1--2; Theorem 4.2), $8$ lines and $12$ conics being defined over $\Q$, and no irreducible cubic over $\Q(i)$ (Theorem 4.3). The lines and conics span $\NS(V_{\Q(i)})$, which has rank $19$ with an explicit $\Z$-basis of $19$ divisors (Theorem 3.3); $\NS(V_\Q)$ has rank $13$ with basis $\Lambda_1,\dots,\Lambda_{13}=\Gamma_1,\Gamma_2,\Gamma_3,\Gamma_5,\Gamma_6,\Gamma_7,\Gamma_8,\Gamma_9+\Gamma_{10},\Gamma_{19}+\Gamma_{20},\Gamma_{27}+\Gamma_{28},Q_1,Q_2,Q_5$ (Corollary, p.~563). We recomputed the Gram matrices: discriminant $48$ over $\Q(i)$ and $768$ over $\Q$. \item The Brudno pencil $E$ has Weierstrass model \eqref{eq:W}; by a $2$-descent over $\Q(i)(\lambda)$, $\mathcal E(\Q(i)(\lambda))\cong\Z/4\times\Z^3$ (Theorem 3.2), and over $\Q(\lambda)$ the rank is still $3$ with the generators \eqref{eq:gens} (Remark, p.~556: ``we did verify (for fun \dots)'' that they generate; no proof is given). The reducible fibres are two $\mathrm I_2$ (at $\lambda=\pm1$: $Q_5+Q_{10}$ and $Q_6+Q_{12}$) and four $\mathrm I_4$ (at the roots of $PQ$, i.e.\ $\lambda=1\pm i,\,(1\pm i)/2$, each a $4$-cycle of lines); there are also four $\mathrm I_1$ fibres at the roots of $5PQ-B^2$ (Bremner lists only the reducible fibres). \item Up to the symmetries of $V$ the irreducible curves of arithmetic genus $0$ on $V$ over $\Q$ of degree $4,5,6$ are: one of degree $4$, two of degree $5$, one of degree $6$ (Theorem 4.4, with the classes written in terms of the lines and conics). \item For a genus-one pencil $F$ and curves $C_1,C_2$ with $C_i\cdot F=1$, $R\mapsto C_1+C_2-R$ (group law on the fibre) is an involution of $V$ (\S5); the four explicit parametrisations of \S6.2 (degrees $4,5,5,6$) are images of $\Gamma_1,\Gamma_8,\Gamma_1,Q_1$ under the involutions of the Brudno pencil with $C_1+C_2=\Gamma_3+Q_1,\ Q_1+Q_2,\ \Gamma_3+\Gamma_4,\ \Gamma_6+\Gamma_7$. \item Conjecture (\S7): every curve of arithmetic genus $0$ on $V$ over $\Q$ is obtained from one of a finite set of curves by repeated application of automorphisms from a finite set, the degree increasing at each application of a non-linear automorphism; the analogous statement is proved for a K3 surface in the quintic fourfold in \cite{Bremner1981fifth}. \end{itemize} Bremner leaves open the rank of $\NS(V_{\C})$ (``plausibly'' $19$) and does not mention the hyperplane $3x+y+z=3u+v+w$, Brudno's involution, Kodaira types, heights, or the identification of his explicit curves with Brudno's. \subsection{Kuwata 2007} Kuwata \cite{Kuwata2007} shows that $V$ is the minimal desingularisation of a Kummer quartic surface with $16$ nodes arising from a quadratic line complex, namely the Kummer surface of the Jacobian of the genus-two curve \[ \mathcal C\colon\quad y^2=(x^2+1)(x^2+2x+5)(x^2-2x+5), \] whose reduced automorphism group is $S_3$; $\Jac(\mathcal C)$ is isogenous to $E\times E$ with $E\colon y^2=x(x^2+4x+20)$ of conductor $320$ and without complex multiplication (Theorem 5.1), so that $\rho(V_{\bar\Q})=19$ (Corollary 5.2), answering Bremner's question. He also lists the $32$ lines with their images on the two Kummer quartics, describes Bremner's symmetry group of order $48$ on the Kummer model, and tabulates the nine LPS solutions on $V$. In Kuwata's framework Bremner's three quadrics are the three $\Q$-rational members of rank four of the net (the vertices of three conjugate pairs of lines of the discriminant sextic), and the three ``Brudno-type'' pencils $E,\delta E,\delta^2E$ ($\delta$ the cyclic symmetry $(x,y,z,u,v,w)\mapsto(y,z,x,v,w,u)$) are their rulings (verified by computation; not in \cite{Kuwata2007}). \subsection{Delorme 1992} Delorme \cite{Delorme1992} rediscovers \eqref{eq:V} (his system (II)) by an elementary route: for a point of $V$ the three $2\times2$ ``link matrices'' $M_i$ carrying $(x_i,y_i)$ to $(x_{i+1},y_{i+1})$ have determinant $-1$ and $\operatorname{tr}(M_3M_2M_1)=0$; link matrices are parametrised by a conic $\cong\PP^1$ (Theorems 1--3). In geometric language, which Delorme does not use, $V$ is thereby birational over $\Q$ to the hypersurface $X_1X_2X_3=Y_1Y_2Y_3$ of tridegree $(2,2,2)$ in $(\PP^1)^3$, $X_i=s_i^2+(s_i+t_i)^2$, $Y_i=t_i^2+(s_i+t_i)^2$, whose three projections are genus-one pencils. Every ``solving triple'' yields two solutions (the two eigenvectors), which he calls conjugate; nine parametric solutions are produced, which he names $S(5;1)$, $S(4;4;a)$, $S(4;4;b)$, $S(5;9)$, $S(9;5)$, $S(8;8;a)$, $S(8;8;b)$, $S(7;11)$, $S(11;7)$ by their degrees and the degrees of their conjugates. He notes that $S(4;4;a)$ is Brudno's quartic (up to exchanging $x_1,x_2$) and that $S(4;4;b)$ is the same curve reparametrised by $(s,t)\mapsto(2s+3t,-3s-2t)$; that the seventh LPS solution lies on $V$ but on none of his curves (Remark 2); that no solution of \eqref{eq:sextic}, \eqref{eq:k2} off $V$ was known (Remark 3, ``It is still an unsolved problem to find a counterexample''); and that it ``would be interesting'' to interpret his results in Bremner's framework (Remark 4). In the $(\PP^1)^3$ model the Brudno--Kaplansky curve is the fibre of the second projection over $(s_2:t_2)=(2:-3)$ (his Remark 1, in our reading). \subsection{Choudhry 1994 and 2000} Choudhry \cite{Choudhry1994} gives an elementary involution of $V$ (move along the line through a point of $V$ cut out by the two $3$-planes of the quadrics $Q(x,u)=Q(w,z)$ and $Q(z,w)=Q(v,y)$ through it, and take the residual intersection with the third quadric) and applies it to Brudno's quartic to obtain a new rational curve of degree $8$ (his (12), p.~840); applied to a trivial conic it returns Bremner's sextic. Choudhry \cite{Choudhry2000} answers the question, attributed to Guy \cite{Guy}, whether every solution of \eqref{eq:sextic}, \eqref{eq:k2} lies on $V$: he produces infinitely many solutions of the three equations $\sum x^r=\sum u^r$, $r=1,2,6$, from an elliptic curve of rank one, and tabulates $26$ of them; none lies on $V$ under any signed permutation (verified by computation, for all $26$). His smallest is $83^6+211^6+300^6=124^6+185^6+303^6$ with $83+211-300=-124-185+303$. %====================================================================== \section{Brudno's quartic on $V$}\label{sec:quartic} \subsection{The zero section}\label{sec:zero} Bremner's Appendix I maps $E_\lambda\subset\PP^3$ to the model (14) by $X:Y:Z=\tau:2B\omega/(\lambda A^2\rho):\rho$, with $\rho,\sigma,\tau,\pi,\omega$ explicit forms, and (14) to \eqref{eq:W} by $x=5PQ\,X/Z$, $y=5PQ\,Y/Z$. His Table 6 states that the line $\Gamma_6$ (the point $(x:u:a:b)=(\lambda:1:1:0)$ of $E_\lambda$) maps to the origin $(0:1:0)$ and $\Gamma_5$ (the point $(-\lambda:-1:1:0)$) to the $2$-torsion point $(0:0:1)$. \begin{remark}\label{rem:zero} With the map of Appendix I as printed, the point of $E_\lambda$ that goes to $(0:1:0)$ is $\Gamma_5$, not $\Gamma_6$; $\Gamma_6$ goes to the $2$-torsion point $(0,0)$. (The forms $(\tau,2B\omega/(\lambda A^2\rho),\rho)$ have base points on $E_\lambda$: the plane $\rho=0$ cuts $E_\lambda$ in $2\Gamma_5+\Gamma_6+p$ and $\tau=0$ in $3\Gamma_6+p$; the limits at the base points were computed exactly from a second-order local parametrisation.) The two rows of Table 6 are thus interchanged relative to Bremner's own formula; the other rows agree with the formula, except for sign or conjugation discrepancies in the $Y$-coordinates of four entries (Appendix~\ref{app:misprints}, (B11)), which do not affect the rank computation. We take $\Gamma_5$ as zero section throughout. If one insists on $\Gamma_6$ as origin, every torsion coordinate $k$ below becomes $k+2\pmod 4$; the lattice part $(n_1,n_2,n_3)$, the heights, the degree formula and all curve lists are unchanged. \end{remark} With this origin the eight rational lines and eight of the rational conics are the sections \begin{equation}\label{eq:lines-sections} \begin{gathered} \Gamma_5=0,\qquad \Gamma_4=T,\qquad \Gamma_6=2T,\qquad \Gamma_3=3T,\\ \Gamma_8=-g_1,\qquad \Gamma_1=-g_1+T,\qquad \Gamma_7=-g_1+2T,\qquad \Gamma_2=-g_1+3T,\\ Q_9=g_2,\qquad Q_1=g_2+T,\qquad Q_{11}=g_2+2T,\qquad Q_3=g_2+3T,\\ Q_7=g_3,\qquad Q_4=g_3+T,\qquad Q_8=g_3+2T,\qquad Q_2=g_3+3T, \end{gathered} \end{equation} (verified by computation) while $Q_5,Q_{10},Q_6,Q_{12}$ are fibre components. In particular $g_1,g_2,g_3$ are, up to torsion, the classes of the line $\Gamma_8$ and of the conics $Q_9$ and $Q_7$. \subsection{The dictionary} \begin{theorem}\label{thm:quartic} Let $(x,y,z,u,v,w)$ be Brudno's forms \eqref{eq:brudno-quartic}. \begin{enumerate}[label=\textup{(\alph*)}] \item The point $(v,-u,w,-x,y,-z)$ lies on $V$ for all $(m,n)$. Evaluated at $(m,n)=(s,1)$, the six forms $v,-u,w,-x,y,-z$ coincide exactly with the six polynomials of Bremner's degree-four parametrisation on p.~574 of \cite{Bremner1981}. Brudno's curve lies on $V$ under exactly $96$ signed permutations, one orbit of the symmetry group. \item In this placement Brudno's curve is a section of the Brudno pencil: the fibre coordinate of the point with parameter $(m:n)$ is \[ \lambda=\frac{3m+2n}{2m+3n}, \] and as an element of $\mathcal E(\Q(\lambda))$ the curve is \[ S=g_1+g_3+2T=(1,0,1,2). \] The sixteen sections $(\pm1,0,1,k)$, $(\pm1,1,0,k)$, $k\in\Z/4$, form one orbit under the stabiliser of the pencil in the symmetry group of $V$, and all of them are symmetric images of Brudno's curve; no other section is. \item Brudno's curve is the unique irreducible curve of degree $4$ and arithmetic genus $0$ on $V$ over $\Q$ up to symmetry \textup(Bremner's Theorem 4.4\textup); its class in Bremner's notation is $\Gamma_5-\Gamma_6+Q_1+Q_2$ up to symmetry, and it is a smooth rational curve \textup($D^2=-2$\textup). \item The hyperplane $H_{\mathrm{Br}}\colon 3x+y+z=3u+v+w$, which in the coordinates of $V$ reads $x-3y+z+3u-v+w=0$, is the linear span of the curve, and \[ V\cap H_{\mathrm{Br}}=C_{\mathrm{Br}}+\Gamma_2+\Gamma_8+Q_9 , \] i.e.\ the residual intersection consists of two rational lines and a rational conic; in classes, $C_{\mathrm{Br}}\sim H-\Gamma_2-\Gamma_8-Q_9$. \item Brudno's involution $(m:n)\mapsto(-3(m+n):m+3n)$ is the restriction to the curve of the symmetry \[ \sigma\colon(x,y,z,u,v,w)\mapsto(v,u,-w,y,x,-z) \] of $V$, which interchanges the first and third quadric of \eqref{eq:V} \textup(up to sign\textup) and preserves the second. It does not preserve $\lambda$ \textup(it sends $\lambda=(3m+2n)/(2m+3n)$ to $(7m+3n)/(3(m-n))$\textup) and interchanges the Brudno pencil with one of its two symmetric copies; Brudno's curve is a section of both and a bisection of the third \textup(``tridegree'' $(1,1,2)$\textup). \end{enumerate} \end{theorem} \begin{proof} (a) Direct substitution into \eqref{eq:V}; the search over all $2^6\cdot6!$ signed permutations found exactly $96$ placements, and Bremner's polynomials are those of $(v,-u,w,-x,y,-z)$ at $n=1$. (b) The fibre coordinate of a point of $V$ is the common root $\lambda$ of $\lambda^2(v+y+z)-\lambda(2v+3z)+(2v-y+z)$ and $\lambda^2(w+y-z)-\lambda(2w+3y)+(2w+y+z)$ (solve \eqref{eq:pencil} for $\lambda$); on Brudno's curve the gcd of these two polynomials is $(2m+3n)\lambda-(3m+2n)$. Since $\lambda$ has degree one in $(m:n)$ the curve is a section, and its Mordell--Weil coordinates were found by reparametrising by $\lambda$, mapping to \eqref{eq:W} and decomposing exactly in the basis $(g_1,g_2,g_3,T)$ with the height pairing of \S\ref{sec:lattice-MW}; the stabiliser action is described in \S\ref{sec:action}, and the statement that no further section is a symmetric image is the exact geometric test described there. (c) Bremner's Theorem 4.4 and its Remark; the class and $D^2=-2$ were recomputed from the intersection numbers of the parametrised curve with the $56$ lines and conics (Section~\ref{sec:lattice}). (d) The rank of the coefficient matrix is $5$ (\S\ref{sec:1976}). The hyperplane $x-3y+z+3u-v+w=0$ contains exactly the lines $\Gamma_2,\Gamma_8$ and the conic $Q_9$ among the eight rational lines and twelve rational conics (check that the linear form lies in the span of their defining forms). The residual intersection $H|_V-C_{\mathrm{Br}}$ is an effective divisor of degree $8-4=4$ containing $\Gamma_2+\Gamma_8+Q_9$, which has degree $4$; hence equality. As a check, $H-\Gamma_2-\Gamma_8-Q_9$ equals the class of Bremner's curve (i) computed in Section~\ref{sec:lattice}, $-\Gamma_2+\Gamma_5+\Gamma_7-\Gamma_8+(\Gamma_{27}+\Gamma_{28})+Q_2$, using $H\sim\Gamma_1+\Gamma_5+Q_1+Q_2+Q_5$ and Bremner's relation $Q_9\sim\Gamma_1-\Gamma_7-(\Gamma_{27}+\Gamma_{28})+Q_1+Q_5$. (e) Brudno's substitution sends $(x,y,z,u,v,w)$ to $36\,(u,v,w,x,y,z)$; transported through the dictionary of (a) this is $\sigma$, and $\sigma$ maps the three quadrics $F_1,F_2,F_3$ of \eqref{eq:V} to $-F_3,-F_2,-F_1$ (direct computation). The Brudno pencil is the ruling of the third quadric (Bremner, p.~545), so $\sigma$ carries it to the ruling of the first. The tridegree is computed in \S\ref{sec:action}. \end{proof} \begin{remark} Bremner's $\lambda(s)=(3s+2)/(2s+3)$ for his curve (i) is the formula of (b) with $(m,n)=(s,1)$. He identifies the curve as the image of the line $\Gamma_1$ under the involution of the Brudno pencil with $C_1+C_2=\Gamma_3+Q_1$ (up to symmetry); in the coordinates \eqref{eq:lines-sections} this reads $R\mapsto 3T+(g_2+T)-R$, so $\Gamma_1=-g_1+T\mapsto g_1+g_2+3T$, which is in the orbit of (b). Thus Brudno's quartic is, in the language of Mordell--Weil lattices, simply a conic minus a line: $g_1+g_3=[Q_7]-[\Gamma_8]$ up to torsion. \end{remark} \subsection{The published parametrisations}\label{sec:known} The literature contains fifteen explicit parametric solutions of \eqref{eq:sextic} lying on $V$: Brudno's quartic, Bremner's four curves (i)--(iv), Delorme's nine, and Choudhry's octic (Choudhry's sextic is Bremner's (iv), as he says). We placed each of them on $V$ (each in exactly $96$ signed permutations), computed its class in $\NS(V)$ from its intersection numbers with the $56$ lines and conics (Section~\ref{sec:lattice}) and, when it is a section of a Brudno-type pencil, its Mordell--Weil coordinates (Section~\ref{sec:sections}). \begin{theorem}\label{thm:known} All fifteen curves are smooth rational curves on $V$ \textup($D^2=-2$, birationally parametrised\textup). Up to the symmetries of $V$ they are ten distinct curves: \begin{enumerate}[label=\textup{(\arabic*)}] \item Brudno's quartic $=$ Bremner \textup{(i)} $=$ Delorme $S(4;4;a)=S(4;4;b)$; section $(1,0,1)$, tridegree $(1,1,2)$; \item Bremner \textup{(ii)} $=$ Delorme $S(5;1)$ \textup(degree $5$\textup); section of all three Brudno-type pencils, coordinates $(1,1,1)$ or $(0,1,1)$, tridegree $(1,1,1)$; \item Bremner \textup{(iii)} $=$ Delorme $S(5;9)$ \textup(degree $5$\textup); section $(1,0,0)$ or $(-2,0,0)$, tridegree $(1,3,3)$; \item Bremner \textup{(iv)} \textup(degree $6$; $=$ Choudhry's sextic\textup); section $(-1,-1,0)$ or $(-1,0,-1)$, tridegree $(1,5,4)$; \item Delorme $S(7;11)$; tridegree $(3,3,3)$, a $3$-section of all three pencils; \item Delorme $S(8;8;a)$ and \textup{(7)} $S(8;8;b)$; tridegree $(2,3,3)$, distinct curves exchanged by Bremner's involution with $C_1+C_2=Q_1+Q_2$ \textup(up to symmetry\textup); \setcounter{enumi}{7} \item Choudhry's octic; section $(0,-1,0)$, i.e.\ $-g_2$, tridegree $(1,6,5)$: it is the image of the conic $Q_1=g_2+T$ under the fibrewise negation $R\mapsto-R$ \textup(Bremner's involution with $C_1+C_2=\Gamma_3+\Gamma_4$\textup), and also the image of Brudno's quartic under the involution with $C_1+C_2=\Gamma_6+\Gamma_7$; \item Delorme $S(9;5)$; section $(-1,1,1)$ or $(2,1,1)$, tridegree $(1,3,3)$; it is the image of Bremner's \textup{(iii)} under the involution with $C_1+C_2=Q_1+Q_2$; \item Delorme $S(11;7)$; tridegree $(3,3,3)$; $S(7;11)$ and $S(11;7)$ are exchanged by the involution with $C_1+C_2=Q_1+Q_2$. \end{enumerate} In particular the curves of Delorme's conjugate pairs $S(5;9)/S(9;5)$, $S(8;8;a)/S(8;8;b)$, $S(7;11)/S(11;7)$ are related by Bremner's involution $R\mapsto Q_1+Q_2-R$ of the Brudno pencil, up to symmetry. \end{theorem} \begin{proof} Computation; see Sections~\ref{sec:sections} and~\ref{sec:lattice}. Two smooth rational curves with classes in the same symmetry orbit are symmetric images of each other by Lemma~\ref{lem:irr}, so the identifications follow from the classes. The images under the involutions were computed on the lattice (\S\ref{sec:involutions}). \end{proof} The exceptions in Delorme's Remark~2 and in Brudno's coverage statement are now explained: the seventh LPS solution $\#7=\sol{51}{113}{136}{40}{125}{129}$ lies on $V$ but on none of the fifteen curves and on no section of any Brudno-type pencil (Section~\ref{sec:census}). %====================================================================== \section{The Mordell--Weil lattice of the Brudno pencil and the degree of a section}\label{sec:sections} \subsection{Heights and generators}\label{sec:lattice-MW} The discriminant of \eqref{eq:W} is $-160000\,B^2P^4Q^4(5PQ-B^2)$, of degree $24$ in $\lambda$, so \eqref{eq:W} is a minimal model of the K3 surface (Euler characteristic $\chi=2$) with no singular fibre at $\infty$: fibres $\mathrm I_4$ at the four roots of $PQ$, $\mathrm I_2$ at $\lambda=\pm1$, $\mathrm I_1$ at the four roots of $5PQ-B^2=9\lambda^4-30\lambda^3+47\lambda^2-30\lambda+9$. We use Shioda's height pairing \cite{Shioda1990}: for sections $S,S'$, \begin{equation}\label{eq:height} \langle S,S\rangle=4+2\,(S\cdot O)-\sum_v\contr_v(S),\qquad \langle S,S'\rangle=2+(S\cdot O)+(S'\cdot O)-(S\cdot S')-\sum_v\contr_v(S,S'), \end{equation} with the usual local contributions ($\contr_v(S)=i(n-i)/n$, $\contr_v(S,S')=i(n-j)/n$ for $i\le j$, when $S,S'$ meet the components $i,j$ of an $\mathrm I_n$ fibre, the components being numbered cyclically from the identity component). Component indices were read off the valuations of the $x$-coordinate on the model \eqref{eq:W}; the orientation is fixed by requiring $T$ to meet component $+1$ at all four $\mathrm I_4$ fibres. One finds $S\cdot O=0$ for $S\in\{T,g_1,g_2,g_3\}$, the component data $T:(1,1;1,1)$, $g_1:(2,0;1,1)$, $g_2:(0,1;1,0)$, $g_3:(3,0;1,0)$ (at the $P$-fibres, the $Q$-fibres, $\lambda=1$, $\lambda=-1$ respectively), and the Gram matrix \begin{equation}\label{eq:gram} \bigl(\langle g_i,g_j\rangle\bigr)=\begin{pmatrix}1&-\tfrac12&-\tfrac12\\-\tfrac12&2&\tfrac12\\-\tfrac12&\tfrac12&2\end{pmatrix},\qquad \det=3,\qquad \langle S,S\rangle=n_1^2+2n_2^2+2n_3^2-n_1n_2-n_1n_3+n_2n_3 . \end{equation} (All of this was checked exactly in function-field arithmetic, and cross-checked against $\langle S,S\rangle=(4+2\,(4S\cdot O))/16$, since $4S$ meets the identity component everywhere, and against the homomorphism property of the component map.) \begin{lemma}\label{lem:generate} Assume Bremner's Theorems 3.2 and 3.3, and $\rho(V_{\bar\Q})=19$ \textup(hypothesis \textup{(H)} of Section~\ref{sec:lattice}, Kuwata's theorem\textup). Then $g_1,g_2,g_3$ generate $\mathcal E(\Q(\lambda))$ modulo torsion, and $\mathcal E(\Q(\lambda))=\langle g_1,g_2,g_3\rangle\oplus\langle T\rangle\cong\Z^3\times\Z/4$. \end{lemma} \begin{proof} Under (H), $\NS(V_{\bar\Q})=\NS(V_{\Q(i)})$ has discriminant $48$ (Theorem 3.3, recomputed), and by the Shioda--Tate sequence every section of $\mathcal E$ over $\bar\Q(\lambda)$ is defined over $\Q(i)$, so the geometric Mordell--Weil lattice is $\MW(\Q(i)(\lambda))$, of rank $3$ with torsion $\Z/4$ (Theorem 3.2). Shioda's determinant formula $|\det\NS|=|\det\MW|\cdot\prod_v n_v/|\mathcal E_\tors|^2$ (with $n_v=4,4,4,4,2,2$ for the reducible fibres) gives $\det\MW(\Q(i)(\lambda))=48\cdot4^2/(4^4\cdot2^2)=3/4$. Since $\det\langle g_1,g_2,g_3\rangle=3$, the sublattice $\langle g_1,g_2,g_3\rangle$ has index $2$ in $\MW(\Q(i)(\lambda))$. Now $\mathcal E(\Q(\lambda))$ contains $g_1,g_2,g_3$ and the whole torsion subgroup $\langle T\rangle$ of $\mathcal E(\Q(i)(\lambda))$, and it is a proper subgroup of $\mathcal E(\Q(i)(\lambda))$ because the line $\Gamma_9$ is a section defined over $\Q(i)$ but not over $\Q$. Hence $\mathcal E(\Q(\lambda))/\tors$ is a proper sublattice of $\MW(\Q(i)(\lambda))$ containing a sublattice of index $2$, so it equals $\langle g_1,g_2,g_3\rangle$. \end{proof} \begin{remark} The lattice spanned by the lines and conics is primitive of rank $19$ with discriminant $48$, matching the Picard lattice $U(2)\oplus E_8^2\oplus\langle-12\rangle$ recorded in \cite{ALT}. The lemma turns Bremner's ``verified for fun'' into a proof (conditional on (H)). The heights are $\langle g_1,g_1\rangle=1$, $\langle g_2,g_2\rangle=\langle g_3,g_3\rangle=2$. \end{remark} \begin{remark}[specialisation at $\lambda=\infty$] The fibres of \eqref{eq:W} at $\lambda=0$ and $\lambda=\infty$ (interchanged by the symmetry $\lambda\mapsto1/\lambda$) are both $y^2=x^3-16x^2+100x$, which is $E_1=480\mathrm{f}4$ ($c_4=-704$, $c_6=-198656$, $\Delta=-23040000$). Although Bremner's map has $\lambda$ in a denominator, $\omega\equiv0\bmod\lambda$ and the specialisation at $\lambda=0$ is a genuine isomorphism $E_0\to\mathcal E_0$. The specialisation homomorphism $\mathcal E(\Q(\lambda))\to\mathcal E_0(\Q)$ has kernel of rank $2$ ($g_2\mapsto3T$, $g_3-g_1\mapsto2T$) and image all of $\mathcal E_0(\Q)=\langle g_1\rangle\times\langle T\rangle$, with $g_1\mapsto P$ (height $0.516278\ldots$, the regulator of $480\mathrm{f}4$). Mapping back to $V$, $g_1,2g_1,\dots,6g_1$ give exactly the solutions $P,\dots,6P$ of Table~\ref{tab:multiples}. So Brudno--Kaplansky's generator is the specialisation of Bremner's $g_1$. \end{remark} \subsection{The degree formula} Let $\gamma=-g_1+g_2+g_3$; then $\langle\gamma,\gamma\rangle=8$ and $\langle\gamma,S\rangle=-2n_1+3n_2+3n_3$. \begin{theorem}[degree of a section]\label{thm:degree} For every section $S=n_1g_1+n_2g_2+n_3g_3+kT$ of the Brudno pencil, the curve $C_S\subset\PP^5$ has degree \begin{equation}\label{eq:degree} \begin{aligned} \deg C_S&=2\langle S,S\rangle-\langle\gamma,S\rangle+1=2\langle S-\gamma/4,\;S-\gamma/4\rangle\\ &=2n_1^2+4n_2^2+4n_3^2-2n_1n_2-2n_1n_3+2n_2n_3+2n_1-3n_2-3n_3+1 . \end{aligned} \end{equation} In particular the degree does not depend on the torsion part $k$, $\deg C_S\equiv n_2+n_3+1\pmod2$, $\deg C_S=1$ exactly for the eight lines $S\in\{kT,-g_1+kT\}$, and there is no section of degree $3$ or $12$. \end{theorem} \begin{proof} We use the following facts, all verified exactly against the $56\times56$ intersection matrix of the lines and conics (which span $\NS(V_{\Q(i)})\otimes\Q$). \begin{enumerate}[label=\textup{(\arabic*)}] \item $H\sim\Gamma_1+\Gamma_5+Q_1+Q_2+Q_5$ (Bremner, Appendix II: $\Pi-\Gamma_1-\Gamma_5\sim Q_1+Q_2+Q_5$, $\Pi$ his notation for the hyperplane class). Here $\Gamma_5=O$ and $Q_5$ is the identity component of the $\mathrm I_2$ fibre at $\lambda=1$ ($Q_5\cdot\Gamma_5=1$). \item By \eqref{eq:lines-sections}, $[\Gamma_1]=-g_1+T$, $[Q_1]=g_2+T$, $[Q_2]=g_3+3T$, so $[\Gamma_1]+[Q_1]+[Q_2]\equiv\gamma$ modulo torsion; $\Gamma_1\cdot O=0$, $Q_1\cdot O=Q_2\cdot O=1$. The fibre components met by $(\Gamma_1,Q_1,Q_2)$ are $(3,1,2)$ at each $P$-fibre, $(1,2,3)$ at each $Q$-fibre, $(0,0,0)$ at $\lambda=1$ and $(0,1,1)$ at $\lambda=-1$. \item For a section $S$, by (1), \[ \deg C_S=H\cdot S=S\cdot O+\sum_{C\in\{\Gamma_1,Q_1,Q_2\}}C\cdot S+S\cdot Q_5 . \] By \eqref{eq:height}, $C\cdot S=2+C\cdot O+S\cdot O-\langle C,S\rangle-\sum_v\contr_v(C,S)$, and $S\cdot Q_5=[\,i_1(S)=0\,]$, where $i_1(S)$ is the component met at $\lambda=1$. Summing, and using that torsion is orthogonal to everything, \[ \deg C_S=4(S\cdot O)+8-\langle\gamma,S\rangle-\sum_v\sum_C\contr_v(C,S)+[\,i_1(S)=0\,]. \] By \eqref{eq:height} again, $4(S\cdot O)=2\langle S,S\rangle-8+2\sum_v\contr_v(S)$, so \[ \deg C_S=2\langle S,S\rangle-\langle\gamma,S\rangle+\sum_v\Bigl(2\contr_v(S)-\sum_C\contr_v(C,S)\Bigr)+[\,i_1(S)=0\,]. \] \item The fibre terms are evaluated with (2). At an $\mathrm I_4$ fibre the three curves meet the components $\{1,2,3\}$, and for $S$ on component $i=1,2,3$ one has $2\cdot\tfrac34-(\tfrac34+\tfrac12+\tfrac14)=2\cdot1-(\tfrac12+1+\tfrac12)=2\cdot\tfrac34-(\tfrac14+\tfrac12+\tfrac34)=0$, while for $i=0$ all terms vanish. At $\lambda=-1$ exactly two of the curves meet component $1$, so the term is $2\cdot\tfrac12-2\cdot\tfrac12=0$ if $S$ meets component $1$ and $0$ otherwise. At $\lambda=1$ no $C$ meets component $1$, so the term $2\contr_1(S)+[\,i_1(S)=0\,]$ equals $1$ whether $i_1(S)=0$ or $1$. Hence $\deg C_S=2\langle S,S\rangle-\langle\gamma,S\rangle+1$, and the remaining expressions follow from \eqref{eq:gram} and $\langle\gamma,\gamma\rangle=8$. \end{enumerate} The last assertions: the quadratic form $2\langle S-\gamma/4,S-\gamma/4\rangle$ is positive definite, its minimum value $1$ is attained exactly at $(0,0,0)$ and $(-1,0,0)$, the parity statement is read off \eqref{eq:degree}, and the values $3$ and $12$ are not represented (enumeration of the finitely many lattice points with $\deg\le12$). \end{proof} Conceptually, $\gamma$ is the Mordell--Weil class of the restriction of $H$ to the generic fibre: $E_\lambda\subset\PP^3$ is a space quartic, a plane section of it is a divisor of degree $4$ whose sum in the group law (with origin $\Gamma_5$) is $\gamma$ modulo torsion; the formula is then the standard expression of $H\cdot S$ through the height pairing. The theorem was discovered empirically on all $1372$ sections with $|n_i|\le3$ and all $2916$ sections with $|n_i|\le4$ (degrees up to $289$), on each of which the six binary forms were computed and checked on $V$; no torsion correction was ever needed. \subsection{The action of the symmetries on sections}\label{sec:action} Exactly $32$ of the $96$ signed permutations ($16$ projectively) preserve the Brudno pencil; the other $64$ carry it to $\delta E$ or $\delta^2E$, the rulings of the other two quadrics of \eqref{eq:V}. The stabiliser acts on $\PP^1_\lambda$ through the Klein four-group $\{\lambda,\,1/\lambda,\,(2\lambda-3)/(3\lambda-2),\,(3\lambda-2)/(2\lambda-3)\}$; the elements acting trivially on $\lambda$ act on every fibre as translations by torsion (e.g.\ $(x,y,z,u,v,w)\mapsto(-x,y,z,-u,v,w)$ is $+2T$ and $\mapsto(-x,-z,y,u,w,-v)$ is $+T$), so $S,S+T,S+2T,S+3T$ are always symmetric images of one another. On the lattice part the stabiliser acts through \begin{gather*} \sigma_1\colon(n_1,n_2,n_3)\mapsto(n_1,n_3,n_2)\qquad[\lambda\mapsto1/\lambda],\\ \sigma_2\colon(n_1,n_2,n_3)\mapsto(-n_1+n_2+n_3-1,\,n_2,\,n_3)\qquad[\lambda\mapsto(3\lambda-2)/(2\lambda-3)], \end{gather*} both of which fix $\gamma/4=(-\tfrac14,\tfrac14,\tfrac14)$, as they must by Theorem~\ref{thm:degree}. (All of this was verified on random sections; the action is affine, $S\mapsto A_gS+t_g$.) A symmetry outside the stabiliser can map a section of $E$ to a section of $E$ only if the curve is also a section of $\delta E$ or $\delta^2E$; this was decided by an exact geometric test (reparametrise and compare), and no identification beyond those listed in Table~\ref{tab:sections} occurs. For a curve $C$ we call $(C\cdot E,\,C\cdot\delta E,\,C\cdot\delta^2E)$ its \emph{tridegree}; for a section of $E$ the second and third entries are the degrees of $\lambda\circ\delta^{-1}$ and $\lambda\circ\delta^{-2}$ on the curve. \subsection{All sections of degree $\le12$} Since the form $\langle S,S\rangle-\tfrac12\langle\gamma,S\rangle$ is positive definite, $\deg C_S\le12$ forces $|n_i|\le3$ (in fact all solutions have $|n_i|\le2$). Enumerating gives $36$ lattice classes, i.e.\ $144$ sections with torsion, in $12$ orbits under the stabiliser (Table~\ref{tab:sections}). Completeness of the list is conditional only on Lemma~\ref{lem:generate}. \begin{table}[ht]\centering\footnotesize\setlength{\tabcolsep}{4pt} \caption{Sections of the Brudno pencil of degree $\le12$, as $S=n_1g_1+n_2g_2+n_3g_3+kT$ (zero section $\Gamma_5$; the degree does not depend on $k$). One row per orbit of the stabiliser of the pencil; the orbit consists of the listed lattice classes with all four values of $k$.}\label{tab:sections} \begin{tabular}{@{}cllcl>{\raggedright\arraybackslash}p{4.6cm}@{}}\toprule orbit & $(n_1,n_2,n_3)$ & $\langle S,S\rangle$ & $\deg$ & tridegree & identification\\\midrule 0 & $(0,0,0),(-1,0,0)$ & $0,1$ & 1 & $(1,1,1)$ & lines $\Gamma_1,\dots,\Gamma_8$\\ 1 & $(0,1,0),(0,0,1)$ & $2,2$ & 2 & $(1,0,1)$ & conics $Q_1,Q_2,Q_3,Q_4,Q_7,Q_8,Q_9,Q_{11}$\\ 2 & $(1,0,1),(1,1,0),(-1,0,1),(-1,1,0)$ & $2,2,4,4$ & 4 & $(1,1,2)$ & Brudno's quartic ($=$ Bremner (i), $S(4;4;a)$, $S(4;4;b)$)\\ 3 & $(0,1,1),(1,1,1)$ & $5,4$ & 5 & $(1,1,1)$ & Bremner (ii) $=$ Delorme $S(5;1)$\\ 4 & $(1,0,0),(-2,0,0)$ & $1,4$ & 5 & $(1,3,3)$ & Bremner (iii) $=$ Delorme $S(5;9)$\\ 5 & $(-1,-1,0),(-1,0,-1)$ & $2,2$ & 6 & $(1,5,4)$ & Bremner (iv)\\ 6 & $(0,-1,1),(0,1,-1),(-1,-1,1),(-1,1,-1)$ & $3,3,4,4$ & 7 & $(1,4,4)$ & \textbf{new}; carries $\#1,2,3,22,27,33,42,54,72$\\ 7 & $(0,-1,0),(0,0,-1),(-2,-1,0),(-2,0,-1)$ & $2,2,4,4$ & 8 & $(1,6,5)$ & Choudhry's octic ($=-g_2$)\\ 8 & $(-1,1,1),(2,1,1)$ & $8,5$ & 9 & $(1,3,3)$ & Delorme $S(9;5)$\\ 9 & $(2,0,1),(2,1,0),(-2,0,1),(-2,1,0)$ & $4,4,8,8$ & 10 & $(1,4,5)$ & \textbf{new}; carries $\#2,27$\\ 10 & $(0,0,2),(0,2,0),(1,0,2),(1,2,0)$ & $8,8,7,7$ & 11 & $(1,4,4)$ & \textbf{new}; carries $\#2,4,8$\\ 11 & $(1,-1,1),(1,1,-1),(-2,-1,1),(-2,1,-1)$ & $4,4,7,7$ & 11 & $(1,6,6)$ & \textbf{new}; carries $\#2,20$\\ \bottomrule \end{tabular} \end{table} \begin{theorem}\label{thm:sections} Assume Lemma~\ref{lem:generate}. Up to the symmetries of $V$, the rational curves on $V$ that are sections of the Brudno pencil and have degree $\le12$ are exactly the twelve of Table~\ref{tab:sections}. Eight of them are the known curves \textup(the lines, the conics, and the six curves of Theorem~\ref{thm:known} that are sections\textup); the four others, of degrees $7,10,11,11$, are new. \end{theorem} The parameter of every section curve is $\lambda$ itself: writing $\lambda=m/n$, the six coordinates are binary forms $f_i(m,n)$ of degree $\deg C_S$ (content removed, sign normalised). For the degree-$7$ curve, the representative $S=-g_2+g_3=(0,-1,1,0)$ gives \input{curve7.tex} with $x^6+y^6+z^6=u^6+v^6+w^6$ and $x^2+y^2+z^2=u^2+v^2+w^2$ identically. It passes through nine of the $74$ solutions of Appendix~\ref{app:table}: $(m,n)=(1,3)$ gives Subba Rao's $\#1$, $(3,2)$ gives $\#2$, $(1,4)$ gives $\#3$, $(-1,3)$ gives $\#22$, $(3,7)$ gives $\#27=\sol{177}{281}{508}{148}{299}{507}$, $(2,1)$ gives $\#33$, $(3,1)$ gives $\#42=\sol{353}{375}{770}{158}{501}{763}$, $(7,3)$ gives $\#54$ and $(4,1)$ gives $\#72$. After Brudno's quartic ($19$ of $74$) it is the most ``productive'' known curve in this range; $\#27$ and $\#42$ lie on no previously published curve. The degree-$10$ curve, representative $S=-2g_1+g_3=(-2,0,1,0)$, is \input{curve10.tex} The two curves of degree $11$ are given in Appendix~\ref{app:classes}; the full list of the $144$ sections with their forms is in the accompanying data (Appendix~\ref{app:comp}). \begin{remark} The new curves are not far from the old ones in the lattice: the degree-$7$ orbit is $\iota_{\Gamma_3+Q_1}(C_{\mathrm{Br}})$, the degree-$10$ orbit is $\iota_{\Gamma_3+Q_1}(\text{Bremner (iii)})$, and the two degree-$11$ orbits are $\iota_{\Gamma_3+Q_1}(C_{\mathrm{Br}})$ and $\iota_{\Gamma_3+Q_1}(\text{Bremner (iv)})$ up to symmetry, where $\iota_{C_1+C_2}$ is Bremner's involution $R\mapsto C_1+C_2-R$ (\S\ref{sec:involutions}). The negatives of the quartic sections $(1,0,1)+kT$, $(1,1,0)+kT$ are sextics (orbit $5$), but the negatives of $(-1,0,1)+kT$, $(-1,1,0)+kT$ form the orbit $6$: the degree-$7$ curve is ``minus Brudno's quartic'' for the other half of its torsion translates. On the fibre $\lambda=3$, which carries seven of the $74$ solutions, $\mathcal E_3(\Q)\cong\Z^2\times\Z/4$ with basis $g_1,g_2$ (the specialisation has $g_3=g_1+T$) and $\#1=-g_1+g_2+2T$, $\#42=g_1-g_2+2T=-\#1$: Subba Rao's solution and $\#42$ are negatives of each other. \end{remark} %====================================================================== \section{Smooth rational curves of degree $\le12$ on $V$}\label{sec:lattice} \subsection{Two lemmas} Throughout this section we assume \begin{itemize} \item[(H)] $\rho(V_{\bar\Q})=19$ (Kuwata, Corollary 5.2; also \cite{ALT}). \end{itemize} Since $\NS(V_{\Q(i)})=\NS(V_{\bar\Q})^{\Gal}$ is a primitive sublattice of rank $19$ (Bremner, Theorem 3.3), (H) gives $\NS(V_{\bar\Q})=\NS(V_{\Q(i)})=\Z\langle\Gamma_1,\dots,Q_{24}\rangle$. The lattice computations below do not use (H); the statements about curves over $\bar\Q$ do. \begin{lemma}\label{lem:irr} Let $X$ be a K3 surface over an algebraically closed field of characteristic $0$, $H$ ample, and $D\in\NS(X)$ with $D^2=-2$ and $D\cdot H>0$. Then $D$ is effective, and $D$ is the class of an irreducible curve if and only if $D\cdot C\ge0$ for every irreducible curve $C$ with $C^2=-2$ and $C\cdot H0$, $C_i$ irreducible). From $D^2=\sum n_i(D\cdot C_i)=-2<0$ some component $C=C_i$ has $D\cdot C<0$. Then $C^2<0$ (an irreducible curve with $C^2\ge0$ is nef), hence $C^2=-2$ and $p_a(C)=0$, so $C$ is a smooth rational curve; and $C\cdot H\le D\cdot H$, with equality only if $D-C=0$ ($D-C$ is effective of degree $0$ and $H$ is ample), which would make $D$ irreducible. So $C$ is an irreducible $(-2)$-curve of degree $<\deg D$ with $D\cdot C<0$. Finally, if $\Gamma\in|D|$ is irreducible and $\Gamma'\in|D|$ is another member, either $\Gamma\not\subset\Gamma'$ and $\Gamma'\cdot\Gamma\ge0$ contradicts $\Gamma'\cdot\Gamma=\Gamma^2=-2$, or $\Gamma'=\Gamma+(\text{effective}\sim0)=\Gamma$. \end{proof} \begin{lemma}\label{lem:galois} Under \textup{(H)}, every irreducible $(-2)$-curve $\Gamma$ on $V_{\bar\Q}$ is defined over $\Q(i)$, and $\Gamma$ is defined over $\Q$ if and only if its class is fixed by complex conjugation. The irreducible $(-2)$-curves not defined over $\Q$ come in conjugate pairs with conjugate classes. \end{lemma} \begin{proof} The class $[\Gamma]\in\NS(V_{\bar\Q})=\NS(V_{\Q(i)})$ is fixed by $\Gal(\bar\Q/\Q(i))$; for $\tau$ in that group, $\Gamma^\tau$ is an irreducible curve with the same class, so $\Gamma^\tau=\Gamma$ (otherwise $\Gamma\cdot\Gamma^\tau=-2<0$), and $\Gamma$ is defined over $\Q(i)$. Likewise, if $[\Gamma]$ is fixed by conjugation then $\bar\Gamma=\Gamma$ and $\Gamma$ is defined over $\Q$; the converse is clear. \end{proof} \subsection{The enumeration} Lemma~\ref{lem:irr} gives an induction on the degree: the irreducible $(-2)$-curves of degree $d$ are exactly the $(-2)$-classes $D$ of degree $d$ with $D\cdot C\ge0$ for every irreducible $(-2)$-curve $C$ of degree $