\documentclass[11pt]{article} \usepackage[margin=1in]{geometry} \usepackage{amsmath,amssymb,amsthm,mathtools} \usepackage{enumitem} \usepackage{booktabs} \usepackage{array} \usepackage{xcolor} \usepackage[colorlinks=true,linkcolor=blue!60!black,citecolor=blue!60!black,urlcolor=blue!60!black]{hyperref} \usepackage[expansion=false]{microtype} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{corollary}[theorem]{Corollary} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{problem}[theorem]{Problem} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\C}{\mathbb{C}} \newcommand{\R}{\mathbb{R}} \newcommand{\F}{\mathbb{F}} \newcommand{\hh}{\widehat h} \newcommand{\lmax}{\lambda_{\max}} \newcommand{\lmin}{\lambda_{\min}} \newcommand{\status}[1]{\textup{\textsf{[#1]}}} \DeclareMathOperator{\tr}{tr} \DeclareMathOperator{\disc}{disc} \DeclareMathOperator{\lc}{lc} \DeclareMathOperator{\rank}{rank} \DeclareMathOperator{\diag}{diag} \title{A linear-in-rank bound for integral arithmetic progressions\\ on elliptic curves} \author{Jake Crim} \date{October 2026} \begin{document} \maketitle \begin{abstract} Every known bound on the length $N$ of an $x$-progression on an elliptic curve of rank $r$ has the shape $N\le C^{1+r}$. We prove a bound that is linear in the rank for \emph{integral} progressions, with an error depending only on the height of the model. Let $y^2=x^3+a_2x^2+a_4x+a_6$ be an integral model and $H_f=1+|a_2|+|a_4|+|a_6|$. For every $\varepsilon>0$, every integral $x$-progression of length $N$ satisfies \[ N\le\bigl(\tfrac32+\varepsilon\bigr)s+C_\varepsilon\sqrt{1+\log H_f}, \] where $s\le r$ is the rank of the span of the progression. In particular $N\le(\tfrac32+o(1))s$ whenever $\log H_f=o(N^2)$. The proof has two halves. \begin{itemize} \item The \emph{local theory} consists of exact point counts, the root-class lemma, pinned values, and a forbidden range of primes reaching $N^2/49$ by a shift-and-square argument. It forces $\log|b|\gg N^2$, where $b$ is the common difference. \item The \emph{spectral cap} is proved in an exact form. Splitting the Néron--Tate pairing into local kernels of positive type and summing the regularised diagonal terms over all places gives an identity in which the bad-prime corrections cancel. Combined with a uniform archimedean separation lemma, this yields $\lmax(G)\le\tfrac34\log X+\tfrac14\psi(N-1)+\tfrac54\log H_f+2$, where $X=\max|x_i|$. \end{itemize} Over function fields we prove an exact colouring formula for the Gram matrix of a progression of integral sections, which gives $N\le\min(6\chi,\tfrac32 s)$. We also classify the $3$-term Gram types on rational elliptic surfaces and show that all of them occur, and we show that all sign data are realised on elliptic $K3$ surfaces. The main bound is vacuous for every known example (all have $N\le14$); its interest is that it is linear rather than exponential in the rank. \end{abstract} \medskip \noindent\textbf{2020 Mathematics Subject Classification.} 11G05, 11G50, 11B25, 14J27. \noindent\textbf{Keywords.} Elliptic curves, arithmetic progressions, canonical height, local heights, Mordell--Weil lattice, Bremner's conjecture. \tableofcontents \section{Introduction}\label{sec:intro} Let $E/\Q$ be an elliptic curve given by a Weierstrass equation. An \emph{$x$-progression} of length $N$ is a sequence of points $P_0,\dots,P_{N-1}\in E(\Q)$ whose $x$-coordinates form a non-constant arithmetic progression. Bremner \cite{Bremner} observed that such points tend to be independent in the Mordell--Weil group, and conjectured that long progressions force large rank. This was motivated in part by \cite{BST}, where it is shown that the curves $y^2=x(x^2-n^2)$ of rank one have no non-trivial integral progressions. Garcia-Fritz and Pasten \cite{GFP19} proved this for families with fixed $j$-invariant, in the form $1+r\ge c(j_0)\log N$. Combined with the uniform Mordell--Lang theorem of Gao--Ge--Kühne \cite{GGK}, their method gives $N\le c^{1+r}$ \cite{GFP26b}; Harrison, Mudgal and Schmidt \cite{HMS} give another proof with an effective absolute constant. All of these bounds are exponential in the rank. In this paper we prove a bound that is \emph{linear} in the rank for \emph{integral} progressions, at the cost of a mild dependence on the height of the model. Write the model as \begin{equation}\label{eq:Wintro} y^2=x^3+a_2x^2+a_4x+a_6,\qquad a_k\in\Z,\qquad H_f=1+|a_2|+|a_4|+|a_6|, \end{equation} and let $s$ be the rank of the subgroup generated by the progression, so $s\le r=\rank E(\Q)$. \begin{theorem}\label{thm:main} For every $\varepsilon>0$ there is an effective constant $C_\varepsilon$ such that every integral $x$-progression of length $N$ on~\eqref{eq:Wintro} satisfies \[ N\le\bigl(\tfrac32+\varepsilon\bigr)s+C_\varepsilon\sqrt{1+\log H_f}. \] In particular $N\le(\tfrac32+o(1))s$ along any sequence of progressions with $\log H_f=o(N^2)$. \end{theorem} The result also holds for every integral cubic model (Remark~\ref{rem:models}). It is Corollaries~\ref{cor:32} and~\ref{cor:general} below. The proof has two halves. \emph{Local theory} (Sections~\ref{sec:local}--\ref{sec:denominator}). A progression of length $N$ reduces to too many points modulo small good primes. Combining exact point counts with a root-class lemma and Hasse's bound, and then a shift-and-square argument based on Weil's bound, we show that every prime $5\le p\le((N-77)/7)^2$ divides $2b\,\lc(f)\disc(f)$ (Theorem~\ref{thm:forbidden}). Hence $\log|b|\gg N^2$, so the points of a long progression have large height. \emph{Spectral cap} (Sections~\ref{sec:trace}--\ref{sec:kernels}). Let $G$ be the Gram matrix of the Néron--Tate pairing on the progression. Since $\rank G=s$, the trace inequality $\tr G\le s\,\lmax(G)$ bounds $N$ as soon as $\lmax(G)$ is controlled. We split the pairing into local kernels of positive type, using an antisymmetrised positivity lemma (Lemma~\ref{lem:posdef}), and prove an exact identity for the regularised diagonal in which all bad-prime corrections cancel (Theorem~\ref{thm:exact}). With a uniform archimedean separation lemma this gives \[ \lmax(G)\le\tfrac34\log X+\tfrac14\psi(N-1)+\tfrac54\log H_f+2,\qquad X=\max_i|x_i| \] (Theorem~\ref{thm:cap}). Comparing with $\tr G\approx\tfrac12N\log X$ gives Theorem~\ref{thm:main}. \emph{Function fields} (Section~\ref{sec:ff}). Over $\C(t)$ the corresponding statements are exact: the Gram matrix of a progression of integral sections is determined by sign data at the zeros of the common difference (Theorem~\ref{thm:ff}). We classify the $3$-term types on rational elliptic surfaces and realise all sign data on elliptic $K3$ surfaces (Theorems~\ref{thm:types} and~\ref{thm:k3}). Section~\ref{sec:prior} discusses the relation of the method to earlier work of Baker--Petsche, Hindry--Silverman and Helfgott--Venkatesh. Section~\ref{sec:GFP} compares the consequences with those of \cite{GFP19}, and Section~\ref{sec:status} lists open problems. Statements marked \emph{computer-assisted} rely on computations described in Appendix~\ref{app:comp}. \subsection*{Conventions} \begin{itemize}[leftmargin=*] \item $h(x)=\log\max(|u|,|v|)$ for $x=u/v$ in lowest terms. \item $\hh$ is the canonical height in Silverman's normalisation, so $\hh\approx\tfrac12h(x)$. \item Our pairing is $\langle P,Q\rangle=\tfrac14\bigl(\hh(P+Q)-\hh(P-Q)\bigr)$, so $\langle P,P\rangle=\hh(P)$. For points $P_i$ we write $G_{ij}=\langle P_i,P_j\rangle$ and $\hh_i=\hh(P_i)$. \item The \emph{span rank} $s$ is the rank of the subgroup generated by the $P_i$; it equals $\rank G$. \item $\psi(n)=\log\operatorname{lcm}(1,\dots,n)=\sum_p m_p(n)\log p$, with $m_p(n)=\max_{1\le k\le n}v_p(k)$. Also $\vartheta(x)=\sum_{p\le x}\log p$. \end{itemize} \section{Setting}\label{sec:setting} For the local theory (Sections~\ref{sec:local}--\ref{sec:denominator}) we work in an integral model \begin{equation}\label{eq:model} C:\ y^2=f(x),\qquad f\in\Z[x],\quad \deg f=m\in\{3,4\},\quad \disc f\neq0, \end{equation} with points $P_i=(x_i,y_i)$, $x_i=a+ib$ $(0\le i4$, which is impossible. \item \emph{$p=7$.} Hasse forces $z_7\ge 7-2\sqrt7>1.7$, so $z_7=3$ and $a_7=-4$. The constraint $3\le 14-N$ requires $N\le11$. \item \emph{$p=5$.} Hasse forces $z_5\ge1$, so $z_5\in\{1,3\}$ and $a_5\in\{-4,-2\}$. The constraint $z_5\le10-N$ excludes $z_5=3$ for $N\ge8$ and excludes everything for $N\ge10$. \item \emph{$p=3$.} Every $z_3\in\{0,1,3\}$ is allowed by Hasse. The constraint $z_3\le 6-N$ leaves $\{0,1,3\}$ for $N=3$, $\{0,1\}$ for $N=4,5$, and $\{0\}$ for $N\ge6$. \end{itemize} \end{proof} \section{The forbidden range}\label{sec:forbidden} We use Weil's bound in the following form \cite[Thm.~11.23]{IK}, \cite[Thm.~2C$'$]{Schmidt}, \cite[Thm.~5.41]{LN}. The hypothesis in \cite{IK} reads ``$f$ is not a $d$-th power''. It has to be understood over $\overline{\F}_p$: for a nonsquare $c\in\F_p^\times$ the constant $f=c$ is not a square in $\F_p[X]$, yet $\bigl|\sum_x\chi(c)\bigr|=p$. \begin{lemma}[Weil]\label{lem:weil} Let $p$ be odd and let $\chi$ be the quadratic character of $\F_p$. Let $G\in\F_p[X]$ be a polynomial that is not a constant multiple of a square, with $d$ distinct roots in $\overline{\F}_p$. Then $\bigl|\sum_{x\in\F_p}\chi(G(x))\bigr|\le(d-1)\sqrt p$. \end{lemma} \begin{lemma}[Shift and square]\label{lem:shift} Let $p>m$, $p\notin S$ and $N\le p$. Then for every integer $H$ with $m+1\le Hm=|R|$, so this is impossible. Lemma~\ref{lem:weil} therefore bounds each of the $H(H-1)$ off-diagonal terms by $(2m-1)\sqrt p$. \end{proof} \begin{corollary}\label{cor:7sqrtp} Let $p\notin S$ and $N\le p$. \begin{enumerate}[label=\textup{(\alph*)}] \item $N\le(2m+1)\sqrt p\,(1+o(1))$ as $p\to\infty$. For cubics this reads $N\le7\sqrt p(1+o(1))$. \item If $m=3$ and $p\ge144$, then $N\le 7\sqrt p+77$. \end{enumerate} \end{corollary} \begin{proof} Put $u=\sqrt p$ and $H=\lfloor u\rfloor$. If $N\le H$ there is nothing to prove, so assume $H3$, and $H\ge u-1$, so \[ N\le u+\frac{6(u-1)u^2}{(u-4)^2}=7u+\frac{6u(7u-16)}{(u-4)^2}. \] The last term is decreasing in $u$ for $u>4$, and equals $76.5$ at $u=12$. \end{proof} \begin{theorem}[Forbidden range, cubic case]\label{thm:forbidden} Let $m=3$ and $N\ge12$. Each prime $p$ in the following ranges lies in $S$: \begin{enumerate}[label=\textup{(\roman*)}] \item $5\le p\le N$; \item $NN$ and $p\notin S$. Then the $N$ residues $x_i\bmod p$ are distinct, and $z_p\le3$. Proposition~\ref{prop:counts}(a) and the affine form of Hasse's bound \cite[Thm.~11.25]{IK} give $2N-z_p\le\#C^{\mathrm{aff}}(\F_p)\le p+2\sqrt p$. So any prime with $p+2\sqrt p<2N-3$ must lie in $S$. (iii) is the contrapositive of Corollary~\ref{cor:7sqrtp}(b). For the consequence: if $p\le2N-2\sqrt{2N}-5$, then $p+2\sqrt p\le 2N-5<2N-3$. So the ranges (i)--(iii) cover every prime in $[5,Y_N]$. \end{proof} \begin{corollary}[Height of the progression]\label{cor:height} Let $m=3$ and $N\ge12$. Then \[ \log|2b\,\lc(f)\disc(f)|\ \ge\ \vartheta(Y_N)-\log3 . \] In particular $\log|2b\,\lc(f)\disc(f)|\ge(1-o(1))N^2/49$. By \cite[(3.16)]{RS}, $\vartheta(x)\ge x(1-1/\log x)$ for $x\ge41$. \end{corollary} \begin{proof} The primes in $S$ are distinct prime divisors of $2b\lc(f)\disc(f)$. By Theorem~\ref{thm:forbidden}, together with $2\in S$, they include every prime $p\le Y_N$ except possibly $3$. \end{proof} \begin{table}[ht] \centering \begin{tabular}{rrrrr} \toprule $N$ & point counting reaches & shift-and-square reaches & $N^2/49$ & $Y_N$ (explicit)\\ \midrule 50 & 79 & -- & 51 & 75\\ 130 & 223 & 229 & 345 & 222\\ 145 & 251 & 293 & 429 & 250\\ 200 & 359 & 619 & 816 & 355\\ 400 & 739 & 2819 & 3265 & 2129\\ 1000 & 1907 & 19213 & 20408 & 17386\\ \bottomrule \end{tabular} \caption{The largest prime forced into $S$ by Theorem~\ref{thm:forbidden}(ii), and by inequality~\eqref{eq:shift} optimised exactly over $H$. Shift-and-square first beats point counting at $N=130$. See Appendix~\ref{app:comp}.} \label{tab:forbidden} \end{table} \begin{remark}[Reading of Theorem~\ref{thm:forbidden}]\label{rem:honest} The common difference $b$ can absorb the forced primes. Theorem~\ref{thm:forbidden} is therefore a statement about the \emph{height of the progression} (or of the model), not about the curve alone. Point counting alone, range~(ii), forces only $\vartheta(2N)\sim2N$. The shift-and-square range improves this by a factor of order $N$. That improvement is what makes $\psi(N-1)$ negligible in Section~\ref{sec:linear} and yields the constant $3/2$; with range~(ii) alone the same argument gives $7/4$. The $\sqrt p$ scale is the natural limit of one-variable methods: square-root cancellation is all that Lemma~\ref{lem:weil} provides. Burgess's bound \cite[Thm.~12.6]{IK} treats $\chi(n)$ with a linear argument, via \cite[Cor.~11.24]{IK}, and no analogue for $\chi(g(t))$ with $\deg g\ge2$ is known. A random model for runs of squares predicts runs of length about $3\log_2p$, so the true forbidden range is probably exponential in $N$. Nothing below uses this. \end{remark} \begin{remark}[Quartics] For $m=4$ the same arguments give: every prime $11\le p\le N$ lies in $S$; and every prime $p\ge N$ with $N>9\sqrt p(1+o(1))$ lies in $S$. For the first claim, a smooth genus-one quartic model has at most $p+1+2\sqrt p$ points, while Proposition~\ref{prop:counts}(a) gives $2p-4$ affine points. The second claim is Corollary~\ref{cor:7sqrtp}(a). \end{remark} \section{Primes dividing the common difference}\label{sec:pb} \begin{theorem}[Damped mechanism]\label{thm:damped} In the model~\eqref{eq:model}, let $p\ge3$ with $v_p(b)=1$, $p\nmid\lc(f)$, and $p\mid y_0$, $p\mid y_1$. Then: \begin{enumerate}[label=\textup{(\alph*)}] \item $f(a)\equiv f'(a)\equiv0\pmod p$, so $p\mid\disc f$; \item every $P_k$ reduces to the singular point $(a\bmod p,0)$ of $C\bmod p$; \item if $m=3$ and the model is minimal at $p$, then $E$ has bad reduction at $p$. \end{enumerate} \end{theorem} \begin{proof} All $x_k$ are congruent to $a$ mod $p$, and $p\mid y_0$ gives $f(a)\equiv0$. From $p\mid y_0,y_1$ we get $p^2\mid f(x_0)$ and $p^2\mid f(x_1)$. Since $x_1=x_0+b$ with $p^2\mid b^2$, \[ f(x_1)\equiv f(x_0)+bf'(x_0)\pmod{p^2}, \] so $p^2\mid bf'(x_0)$. As $v_p(b)=1$, this gives $p\mid f'(x_0)$, hence $f'(a)\equiv0$. This proves~(a). For~(b): $y_k^2=f(x_k)\equiv f(a)\equiv0$, so every $P_k$ reduces to $(a,0)$, which is singular by~(a). For~(c) (with $m=3$): $\Delta=16\disc f$, and for a model minimal at $p$, bad reduction is equivalent to $p\mid\Delta$. \end{proof} \begin{definition} Let $p$ divide the common difference of the progression in a model minimal at $p$. The prime $p$ is a \emph{smooth zero} if every $P_i$ reduces to a nonsingular point mod $p$, and a \emph{damped zero} otherwise. \end{definition} \begin{theorem}[Smooth-prime formula]\label{thm:smoothformula} Let $p$ be odd and let $y^2=f(x)$ (with $f$ monic) be minimal at $p$. Let $P_i,P_j\in E(\Z_p)$ satisfy the following: \begin{itemize} \item $x_i\equiv x_j\pmod p$; \item both reduce to nonsingular points; \item $y_i\equiv\varepsilon y_j\not\equiv0\pmod p$ with $\varepsilon=\pm1$. \end{itemize} Then the local pairing $\lambda_p(i,j):=\tfrac14\bigl(\lambda_p(P_i+P_j)-\lambda_p(P_i-P_j)\bigr)$ is \[ \lambda_p(i,j)=-\varepsilon\cdot\tfrac14v_p(x_i-x_j)\log p . \] Here $\lambda_p$ is the local height of \S\ref{subsec:localheights}. \end{theorem} \begin{proof} Take $\varepsilon=+1$. By the addition law, \[ x(P_i-P_j)=\Bigl(\frac{y_i+y_j}{x_i-x_j}\Bigr)^2-a_2-x_i-x_j . \] Here $y_i+y_j\equiv2y_j$ is a unit and $v_p(x_i-x_j)\ge1$, so $v_p(x(P_i-P_j))=-2v_p(x_i-x_j)$. Hence $P_i-P_j\in E_1$ with level $v_p(x_i-x_j)$. Reduction is a homomorphism $E_0(\Q_p)\to\widetilde E_{ns}(\F_p)$. The reductions of $P_i$ and $P_j$ coincide and are not $2$-torsion (since $\bar y\ne0$), so $P_i+P_j$ reduces to $2\bar P\ne\bar O$ and has level $0$. By (L2) of \S\ref{subsec:localheights}, $\lambda_p(P_i+P_j)-\lambda_p(P_i-P_j)=-v_p(x_i-x_j)\log p$. The case $\varepsilon=-1$ is the same with the roles of $P_i\pm P_j$ exchanged. \end{proof} At a smooth zero all $P_i$ reduce to $\pm\bar P$ for a single $\bar P$, since all the $x_i$ agree mod $p$. When $\bar P\notin E[2]$, Theorem~\ref{thm:smoothformula} shows that the $p$-adic part of the Gram matrix is a rank-one sign pattern $-\varepsilon\varepsilon^T$, weighted by $v_p(x_i-x_j)$. \begin{proposition}[All zeros damped on the nine $8$-term solutions]\label{prop:nine}\textup{(computer-assisted)} Consider the nine $8$-term solutions listed by their $y$-values $Y_k$ $(0\le k\le7)$ in Table~\ref{tab:nine}. They were found by a search over cubics with $0\le y_0,\dots,y_3\le300$; the completeness of that search is not used here. Write $Y_k^2=c_3k^3+c_2k^2+c_1k+c_0$, and use the progression model \[ Y^2=X^3+c_2X^2+c_1c_3X+c_0c_3^2,\qquad X_k=c_3k, \] which has common difference $b=c_3$. Then: \begin{itemize} \item every prime dividing the common difference $d_{\min}$ of the minimal model is damped (23 of 23), of Kodaira types $\mathrm{IV},\mathrm I_3,\mathrm I_4,\mathrm I_7,\mathrm I_8,\mathrm I_{16}, \mathrm I_1^*$; \item each of the 18 odd damped primes satisfies $v_p(b)=v_p(d_{\min})=1$ and $p\mid Y_k$ for all $k$, so Theorem~\ref{thm:damped} explains all of them. \end{itemize} \end{proposition} \begin{table}[ht] \centering\small \begin{tabular}{ll} \toprule $(Y_0,\dots,Y_7)$ & primes of $d_{\min}$ (all damped), Kodaira type\\ \midrule $(1,7,7,5,1,1,7,13)$ & none ($d_{\min}=1$; this is $5077a1$)\\ $(167,223,223,197,167,167,223,323)$ & $5,7,13$: IV\\ $(3,49,53,45,29,7,21,53)$ & $3$: $\mathrm I_7$; $11$: $\mathrm I_3$\\ $(41,127,139,125,97,71,83,139)$ & $2,7$: IV; $13$: $\mathrm I_3$\\ $(82,121,128,121,110,107,124,163)$ & $2$: $\mathrm I_{16}$; $7$: IV\\ $(31,52,47,34,35,64,107,158)$ & $2$: $\mathrm I_8$; $5,7$: $\mathrm I_4$\\ $(139,23,31,115,173,209,217,181)$ & $2$: $\mathrm I_1^*$; $7,11$: $\mathrm I_4$\\ $(169,157,119,67,55,131,233,349)$ & $2$: $\mathrm I_1^*$; $7,11$: $\mathrm I_4$\\ $(217,251,205,109,17,187,379,595)$ & $3$: IV; $5,7,11$: $\mathrm I_4$\\ \bottomrule \end{tabular} \caption{Proposition~\ref{prop:nine}.}\label{tab:nine} \end{table} \begin{remark}[Which model to test in]\label{rem:model} The hypothesis $p\mid y_k$ of Theorem~\ref{thm:damped} is not invariant under changes of coordinates $y\mapsto y+sx+t$. It must therefore be tested in the progression model, not in a minimal model reached by such a change. In the progression model it holds for every odd damped prime in Table~\ref{tab:nine}. \end{remark} \begin{proposition}[Smooth zeros exist]\label{prop:smooth}\textup{(computer-assisted)} Smooth zeros at good primes occur on progressions of length $7$ and $8$, and hence of every length $\le8$. \begin{enumerate}[label=\textup{(\alph*)}] \item \textbf{Length $7$.} For $r,s\in\Z$ put \[ A=|r^2-2rs-s^2|,\quad B=|r^2+2rs-s^2|,\quad C=r^2+s^2,\quad c=\frac{B^2-A^2}{12}, \] and $q(k)=A^2+c\,k(k-4)(k-5)$. Then \[ q(0)=q(4)=q(5)=A^2,\qquad q(1)=q(2)=q(6)=B^2,\qquad q(3)=C^2 . \] The last identity holds because $A^2+B^2=2C^2$. Take $r=343$, $s=1$; all of $A,B,C$ are then even, and halving them gives \[ Y=(58481,59167,59167,58825,58481,58481,59167). \] This gives a $7$-term integral progression on a curve of rank exactly $7$. The prime $7$ is a smooth zero: it has good reduction, $v_7(d_{\min})=1$, and the sign vector is $\varepsilon=(+,+,+,-,+,+,+)$. The other primes of $d_{\min}=34314=2\cdot3\cdot7\cdot19\cdot43$ are damped; $19$ and $43$ have type IV. \item \textbf{Length $8$.} With $s=1$, requiring $q(7)=(7B^2-5A^2)/2$ to be a square gives $W^2=r^4+24r^3+2r^2-24r+1$. This is a genus-one curve with Jacobian $E'=[0,2,0,-580,1144]$, of rank $1$ with generator $G=(-18,80)$. The map is $r=(12x-24\pm y)/(284-2x)$. The point $146G$ gives an $8$-term progression with $Y_k$ of about $8300$ digits. In a model minimal at $7$ (by Tate's algorithm), $7$ has good reduction, $v_7(b)=1$, all $Y_k$ are $7$-adic units, and $Y\bmod7$ takes values $\pm1$. So $7$ is a smooth zero. \end{enumerate} \end{proposition} \begin{problem} Does a $7$-term $x$-progression with no collinear triple have a smooth zero at a good prime? The family in Proposition~\ref{prop:smooth} has many collinear triples. \end{problem} \section{Denominators of non-integral progressions}\label{sec:denominator} \begin{proposition}[Denominator rule]\label{prop:denom} Let $y^2=x^3+a_2x^2+a_4x+a_6$ be integral, and let $x_1+x_3=2x_2$ with $x_i=x(P_i)$. Fix a prime $p$ and put $e_i=\max(0,-v_p(x_i)/2)$ and $e=\max_ie_i$. Assume $e>0$. \begin{enumerate}[label=\textup{(\alph*)}] \item Let $p$ be odd. Then $e$ is attained at least twice. The pattern $e_1=e_3>e_2$ forces $p\equiv1\pmod4$. The pattern $e_1=e_2>e_3$ (or $e_2=e_3>e_1$) forces $p\equiv\pm1\pmod8$. In particular, for $p\equiv3\pmod8$ all three $e_i$ are equal. \item Let $p=2$ and $a_2=0$. Then $e_1=e_2=e_3$. \end{enumerate} \end{proposition} \begin{proof} Let $e_i>0$. In the formal group, $x=z^{-2}-a_2-a_4z^2-\cdots$ with $v_p(z)=e_i$. Hence $u_i:=p^{2e_i}x_i=w_i^2(1-a_2z^2-\cdots)$, where $w_i=p^{e_i}/z$ is a unit. So $u_i$ is a unit congruent to a nonzero square mod $p$. When $a_2=0$ the correction factor is $1+O(z^4)$, so $u_i\equiv1\pmod8$ if $p=2$. (a) If $e$ were attained only at $i$, then $x_i$ would have strictly smaller valuation than the other two terms of $x_1+x_3-2x_2=0$. This is impossible. \begin{itemize} \item If $e_1=e_3=e>e_2$, then $u_1+u_3=2p^{2e}x_2\equiv0\pmod p$. So $-1$ is a square mod $p$. \item If $e_1=e_2=e>e_3$, then $u_1\equiv2u_2\pmod p$. So $2$ is a square mod $p$. \end{itemize} (b) Use $v_2(2)=1$ and $u_i\equiv1\pmod8$. \begin{itemize} \item $e_1=e_3=e>e_2$: then $v_2(u_1+u_3)=1$, while $v_2(2^{2e+1}x_2)\ge3$. \item $e_1=e_2=e>e_3$: then $v_2(u_1+2^{2e}x_3)=0$, while $v_2(2u_2)=1$. \item A unique maximum: the same valuation argument as in~(a). \end{itemize} All three cases are contradictions. \end{proof} \section{The trace inequality}\label{sec:trace} \begin{lemma}\label{lem:trace} Let $G$ be real symmetric positive semidefinite of rank $s$. Then $\tr G\le s\,\lmax(G)$. In particular, if $M$ has unit diagonal, then $N=\tr M\le s\,\lmax(M)$. \end{lemma} \begin{proof} $\tr G$ is the sum of the $s$ nonzero eigenvalues, each of which is at most $\lmax(G)$. \end{proof} The Néron--Tate pairing is positive definite on $E(\Q)\otimes\R$, so $\rank G=s$. The quantity that matters is therefore $\lmax(G)$, not the mean squared cosine. Arguments through $N^2=(\tr M)^2\le s\tr(M^2)$ (Alon's lemma) lose everything once the typical cosine stays above $s^{-1/2}$. \section{Local kernels of positive type}\label{sec:kernels} \subsection{An antisymmetrised positivity lemma} \begin{lemma}\label{lem:posdef} Let $\mathcal G$ be a compact abelian group, and let $F:\mathcal G\to\R$ be continuous and even with absolutely convergent Fourier series $F=\sum_\chi\widehat F(\chi)\chi$. Assume $\widehat F(\chi)\ge0$ for every character $\chi$ with $\chi^2\ne1$. Then for all $g_1,\dots,g_n\in\mathcal G$ the matrix \[ S_{ij}=F(g_i-g_j)-F(g_i+g_j)\qquad(1\le i,j\le n) \] is positive semidefinite. Its diagonal is $S_{ii}=F(0)-F(2g_i)$. Adding a constant to $F$ does not change $S$. \end{lemma} \begin{proof} For $c\in\R^n$ put $a_\chi=\sum_ic_i\chi(g_i)$. Then \[ \sum_{i,j}c_ic_jS_{ij}=\sum_\chi\widehat F(\chi)\bigl(|a_\chi|^2-a_\chi^2\bigr). \] Since $F$ is real and even, $\widehat F(\bar\chi)=\widehat F(\chi)\in\R$, and $a_{\bar\chi}=\bar a_\chi$. Two kinds of terms appear: \begin{itemize} \item a real character ($\chi=\bar\chi$) has $a_\chi\in\R$, so its term vanishes; \item a pair $\{\chi,\bar\chi\}$ with $\chi^2\ne1$ contributes $\widehat F(\chi)\bigl(2|a_\chi|^2-2\operatorname{Re}a_\chi^2\bigr)=4\widehat F(\chi)(\operatorname{Im}a_\chi)^2\ge0$. \end{itemize} A constant only changes the coefficient of the trivial character, which is real. \end{proof} \subsection{Local heights}\label{subsec:localheights} We use the Néron functions $\lambda_v:E(\Q_v)\smallsetminus\{O\}\to\R$ in Silverman's normalisation \cite[Ch.~VI]{SilATAEC}. They are model-independent, even and continuous. We need the following facts. \begin{enumerate}[label=(L\arabic*),leftmargin=*] \item $\hh(P)=\sum_v\lambda_v(P)$ for $P\in E(\Q)\smallsetminus\{O\}$ \cite[Thm.~VI.2.1]{SilATAEC}. \item Let $p$ be a prime and use a model minimal at $p$. Let $E_0\supseteq E_1\supseteq E_2\supseteq\cdots$ be the usual filtration of $E(\Q_p)$, where $E_k=\{R:v_p(x(R))\le-2k\}\cup\{O\}$ for $k\ge1$. Define the level $\ell_p(R)=\max\{k:R\in E_k\}$ (so $\ell_p(R)=0$ if $R\notin E_1$) and $\gamma_p=\tfrac1{12}v_p(\Delta_{\min})\log p$. \begin{itemize} \item For $R\in E_0\smallsetminus\{O\}$: $\lambda_p(R)=\gamma_p+\ell_p(R)\log p$ \cite[Thm.~VI.4.1]{SilATAEC}. \item For $R\notin E_0$: $\lambda_p(R)=\gamma_p+c_p(\pi R)\log p$, where $\pi:E(\Q_p)\to\Phi_p$ is the projection to the component group, $c_p(0)=0$, and $c_p$ is even and depends only on the component. Its values are listed in Table~\ref{tab:cp} \cite[Thm.~5.2]{Sil88}, \cite[\S2]{CPS}. \end{itemize} \item Write $E(\C)=\C/\Lambda$ for the lattice $\Lambda$ of the model~\eqref{eq:W}, with $x=\wp(z)-a_2/3$ and $2y=\wp'(z)$, so $dx/2y=dz$. Then $\Delta(\Lambda)=\Delta_W$, and \[ \lambda_\infty(z)=-\log\bigl|e^{-\frac12z\eta(z)}\sigma(z)\Delta(\Lambda)^{1/12}\bigr|, \] where $\eta$ is the $\R$-linear extension of the quasi-period map \cite[\S VI.3]{SilATAEC}. Since $\sigma(z)=z+O(z^5)$, \[ \lambda_\infty(z)+\log|z|\ \longrightarrow\ -\tfrac1{12}\log|\Delta_W|\qquad(z\to0). \] \end{enumerate} \begin{table}[ht] \centering \begin{tabular}{llll} \toprule type & $\Phi_p(\overline{\F}_p)$ & $-c_p$ on non-zero classes & $\widehat{c_p}(\chi)$, $\chi^2\ne1$\\ \midrule $\mathrm I_m$ (split) & $\Z/m$ & $i(m-i)/(2m)$ on class $i$ & $1/(4m\sin^2(\pi k/m))$ for $\chi=e(k\cdot/m)$\\ $\mathrm I_m$ (non-split) & $\Phi(\Q_p)\subseteq\Z/2$ & as above & (no such $\chi$)\\ $\mathrm{II},\mathrm{II}^*$ & $0$ & -- & --\\ $\mathrm{III},\mathrm{III}^*$ & $\Z/2$ & $\tfrac14$, $\tfrac34$ & (no such $\chi$)\\ $\mathrm{IV}$, $\mathrm{IV}^*$ & $\Z/3$ & $\tfrac13$, $\tfrac23$ (constant) & $\tfrac19$, $\tfrac29$\\ $\mathrm I_m^*$, $m$ odd & $\Z/4$ & $\tfrac12$ (order $2$), $\tfrac{m+4}8$ (order $4$) & $\tfrac18$\\ $\mathrm I_m^*$, $m$ even & $(\Z/2)^2$ & $\tfrac12$, $\tfrac{m+4}8$, $\tfrac{m+4}8$ & (no such $\chi$)\\ \bottomrule \end{tabular} \caption{Component corrections in Silverman's normalisation: half the values of \cite[Table~2]{CPS}, since $\lambda^{\rm CPS}=2\lambda+\tfrac16\log|\Delta|$ \cite[(11)]{CPS}; and their Fourier coefficients on characters of order $>2$.}\label{tab:cp} \end{table} \subsection{Non-archimedean kernels} For a prime $p$ and an integer $K\ge0$ define $F_{p,K}:E(\Q_p)\to\R$ by \[ F_{p,K}(R)=c_p(\pi R)\log p+\log p\cdot\min\bigl(\ell_p(R),K\bigr) =c_p(\pi R)\log p+\log p\sum_{k=1}^K\mathbf 1_{E_k}(R), \] with $\ell_p(O)=\infty$, so that $F_{p,K}(O)=K\log p$. \begin{lemma}\label{lem:padic} $F_{p,K}$ satisfies the hypotheses of Lemma~\ref{lem:posdef} on $\mathcal G=E(\Q_p)$. Moreover, for $R\ne O$, \begin{equation}\label{eq:padicexact} \lambda_p(R)-F_{p,K}(R)=\gamma_p+\log p\cdot\bigl(\ell_p(R)-K\bigr)^+ . \end{equation} \end{lemma} \begin{proof} $E(\Q_p)$ is a compact abelian group, and $F_{p,K}$ factors through the finite quotient $E(\Q_p)/E_{K+1}$, so its Fourier series is a finite sum. $F_{p,K}$ is even because $c_p$ is even and each $E_k$ is a subgroup. The indicator of a finite-index subgroup $H$ has Fourier coefficients $[\mathcal G:H]^{-1}\ge0$ on $H^\perp$ and $0$ elsewhere. The function $c_p\circ\pi$ has coefficient $\widehat{c_p}(\chi)$ on characters trivial on $E_0$, and $0$ elsewhere. A character $\chi$ of $\Phi_p(\Q_p)$ with $\chi^2\ne1$ is the restriction of characters of $\Phi_p(\overline{\F}_p)$, all of order $>2$. So Proposition~\ref{prop:cp} below gives $\widehat F_{p,K}(\chi)\ge0$ whenever $\chi^2\ne1$. Identity~\eqref{eq:padicexact} follows directly from (L2). \end{proof} \begin{proposition}[Bad-prime positivity]\label{prop:cp} For every Kodaira type, $\widehat{c_p}(\chi)\ge0$ for every character $\chi$ of $\Phi_p(\overline{\F}_p)$ of order $>2$. Here $\widehat c(\chi)=|\Phi|^{-1}\sum_gc(g)\bar\chi(g)$. \end{proposition} \begin{proof} If $\Phi\subseteq(\Z/2)^2$, there is nothing to prove. The remaining cases are as follows. \begin{itemize} \item \emph{$\mathrm{IV}$ and $\mathrm{IV}^*$} ($\Phi=\Z/3$). Evenness makes $c$ constant, equal to $-\alpha<0$, on the two non-zero classes. So $\widehat c(\chi)=-\tfrac\alpha3(\omega+\bar\omega)=\tfrac\alpha3>0$. \item \emph{$\mathrm I_m^*$ with $m$ odd} ($\Phi=\Z/4$). Let $\chi(1)=i$. Evenness gives $c(1)=c(3)$, so $\widehat c(\chi)=\tfrac14\bigl(-ic(1)-c(2)+ic(3)\bigr)=-\tfrac14c(2)=\tfrac18$. \item \emph{$\mathrm I_m$} ($\Phi=\Z/m$, $c(i)=-i(m-i)/(2m)$). Let $\zeta=e^{-2\pi ik/m}\ne1$. Using $\sum_{i0 . \] This was also checked numerically for $m<200$ (Appendix~\ref{app:comp}). \end{itemize} \end{proof} \subsection{The archimedean kernel} Let $A$ be the covolume of $\Lambda$, and let $\Lambda^*=\{\xi\in\C:\operatorname{Re}(\bar\xi\omega)\in\Z \ \forall\omega\in\Lambda\}$ be the dual lattice. Characters of $\C/\Lambda$ are $e_\xi(z)=e^{2\pi i\operatorname{Re}(\bar\xi z)}$ with $\xi\in\Lambda^*$. Let $\omega_1$ be a shortest nonzero period. \begin{lemma}[$\lambda_\infty$ is a Green function]\label{lem:green} In the sense of distributions on $\C$, \[ \Delta\lambda_\infty=\frac{2\pi}A-2\pi\sum_{\omega\in\Lambda}\delta_\omega . \] Consequently $\widehat{\lambda_\infty}(\xi)=1/(2\pi A|\xi|^2)>0$ for $\xi\in\Lambda^*\smallsetminus\{0\}$. \end{lemma} \begin{proof} The function $-\log|\sigma|$ has Laplacian $-2\pi\sum_\omega\delta_\omega$, since $\sigma$ is entire with simple zeros exactly on $\Lambda$. The term $\log|\Delta|^{1/12}$ is constant. For the remaining term, write $\eta(z)=\alpha z+\beta\bar z$. The Legendre relation $\eta(\omega_1)\omega_2-\eta(\omega_2)\omega_1=2\pi i$, with the orientation $\operatorname{Im}(\bar\omega_1\omega_2)=A$, gives $\beta\cdot2iA=2\pi i$, so $\beta=\pi/A$. Hence \[ \operatorname{Re}\bigl(\tfrac12z\eta(z)\bigr)=\tfrac12\operatorname{Re}(\alpha z^2)+\tfrac\pi{2A}|z|^2 \] has Laplacian $2\pi/A$. This gives the formula for $\Delta\lambda_\infty$. Now let $g=\sum_{\xi\ne0}e_\xi/(2\pi A|\xi|^2)$. By Poisson summation, $\sum_{\xi\in\Lambda^*}e_\xi=A\sum_\omega\delta_\omega$, so $\Delta g=2\pi/A-2\pi\sum_\omega\delta_\omega$. Thus $\lambda_\infty-g$ is a $\Lambda$-periodic harmonic distribution, hence constant by Weyl's lemma. \end{proof} Let $\nu_\rho$ be the uniform probability measure on the disc $\{|w|\le\rho\}$. Put $\mu=\nu_\rho*\nu_\rho$, which is radial, supported in $\{|w|\le2\rho\}$, and has second moment $M_2=\int|w|^2d\mu$. Define the smoothed kernel \[ F_\infty(z)=\int\lambda_\infty(z+w)\,d\mu(w). \] \begin{lemma}[Smoothing]\label{lem:smooth} Assume $2\rho<|\omega_1|$. \begin{enumerate}[label=\textup{(\alph*)}] \item $F_\infty$ satisfies the hypotheses of Lemma~\ref{lem:posdef} on $\C/\Lambda$. \item If $d(z,\Lambda)>2\rho$, then $F_\infty(z)=\lambda_\infty(z)+\tfrac{\pi}{2A}M_2$. \item $F_\infty(0)=-\log\rho+\tfrac14-\tfrac1{12}\log|\Delta_W|+\tfrac\pi{2A}M_2$. \end{enumerate} \end{lemma} \begin{proof} (a) $\widehat F_\infty(\xi)=\widehat{\lambda_\infty}(\xi)|\widehat{\nu_\rho}(\xi)|^2\ge0$. The disc transform is $\widehat{\nu_\rho}(\xi)=2J_1(2\pi\rho|\xi|)/(2\pi\rho|\xi|)=O(|\xi|^{-3/2})$, so $\widehat F_\infty(\xi)=O(|\xi|^{-5})$. This is summable over $\Lambda^*$, which makes $F_\infty$ continuous with an absolutely convergent Fourier series. Evenness is inherited from $\lambda_\infty$. (b) By Lemma~\ref{lem:green}, $w\mapsto\lambda_\infty(z+w)-\tfrac\pi{2A}|w|^2$ is harmonic on $|w|\le2\rho$. Its average over each circle $|w|=t$ therefore equals its value at $0$. Since $\mu$ is radial, $\int\lambda_\infty(z+w)\,d\mu=\lambda_\infty(z)+\tfrac\pi{2A}M_2$. (c) On $|w|<|\omega_1|$, write $\lambda_\infty(w)=-\log|w|+k(w)$. Here $k$ is smooth with $\Delta k=2\pi/A$, and $k(0)=-\tfrac1{12}\log|\Delta_W|$ by (L3). As in~(b), $\int k\,d\mu=k(0)+\tfrac\pi{2A}M_2$. For the logarithmic part, the potential of $\nu_\rho$ is \[ U(z)=\int-\log|z-w|\,d\nu_\rho(w)=-\log\rho+\tfrac12(1-|z|^2/\rho^2)\qquad(|z|\le\rho). \] Since $\nu_\rho$ is symmetric, $\int-\log|w|\,d\mu=\iint-\log|w_1+w_2|\,d\nu_\rho d\nu_\rho=\int U\,d\nu_\rho=-\log\rho+\tfrac14$. \end{proof} \subsection{The general cap and the exact diagonal} \begin{theorem}[Spectral cap, general form]\label{thm:general} Let $P_1,\dots,P_n\in E(\Q)$ with $P_i\pm P_j\ne O$ for $i\ne j$ and $2P_i\ne O$. For each place $v$ let $F_v$ satisfy the hypotheses of Lemma~\ref{lem:posdef} on $E(\Q_v)$, with $F_v(P_i\pm P_j)=\lambda_v(P_i\pm P_j)-\kappa_v$ ($i\ne j$) for constants $\kappa_v$. Assume $F_v=0$ and $\kappa_v=\gamma_v$ for all but finitely many $v$. Put \[ S^v_{ij}=F_v(P_i-P_j)-F_v(P_i+P_j),\qquad D_i=\sum_vS^v_{ii}, \qquad B_i=\sum_v\bigl[\lambda_v(2P_i)+F_v(0)-F_v(2P_i)\bigr]. \] Then \[ G\ \preceq\ \diag\bigl(\hh(P_i)+\tfrac14D_i\bigr)=\tfrac14\diag(B_i), \qquad\text{so}\qquad \lmax(G)\le\tfrac14\max_iB_i . \] \end{theorem} \begin{proof} For $i\ne j$, (L1) gives $4G_{ij}=\hh(P_i+P_j)-\hh(P_i-P_j)=\sum_v\bigl(\lambda_v(P_i+P_j)-\lambda_v(P_i-P_j)\bigr)=-\sum_vS^v_{ij}$; the constants $\kappa_v$ cancel. Hence \[ 4G=\diag(4\hh_i+D_i)-\sum_vS^v, \] and each $S^v\succeq0$ by Lemma~\ref{lem:posdef}. Finally, $4\hh_i=\hh(2P_i)=\sum_v\lambda_v(2P_i)$ by (L1), so $4\hh_i+D_i=B_i$. \end{proof} \begin{theorem}[Exact regularised diagonal]\label{thm:exact} In Theorem~\ref{thm:general}, take the following kernels: \begin{itemize} \item at each prime $p$, $F_p=F_{p,K_p}$ with $K_p\ge\max_{i\ne j}\ell_p(P_i\pm P_j)$; \item at the archimedean place, $F_\infty$ as in Lemma~\ref{lem:smooth}, with $2\rho<\min(|\omega_1|,\delta)$, where \[ \delta=\min\bigl\{d(P_i\pm P_j,\Lambda)\ (i\ne j),\ d(2P_i,\Lambda)\bigr\}. \] \end{itemize} Then the hypotheses of Theorem~\ref{thm:general} hold, and \begin{equation}\label{eq:exact} B_i=-\log\rho+\tfrac14+\tfrac1{12}\log\Bigl|\frac{\Delta_{\min}}{\Delta_W}\Bigr| +\sum_p\max\bigl(K_p,\ell_p(2P_i)\bigr)\log p . \end{equation} \end{theorem} \begin{proof} The hypotheses hold by Lemmas~\ref{lem:padic} and~\ref{lem:smooth}(a),(b), with $\kappa_p=\gamma_p$ and $\kappa_\infty=-\tfrac\pi{2A}M_2$. For all but finitely many $p$ every level involved is $0$; take $K_p=0$ there, and then $F_p=0$. At a prime $p$, \eqref{eq:padicexact} applied to $2P_i$ gives \[ \lambda_p(2P_i)+F_p(0)-F_p(2P_i)=\gamma_p+\max\bigl(K_p,\ell_p(2P_i)\bigr)\log p . \] At $\infty$, Lemma~\ref{lem:smooth}(b),(c) gives \[ \lambda_\infty(2P_i)+F_\infty(0)-F_\infty(2P_i)=-\log\rho+\tfrac14-\tfrac1{12}\log|\Delta_W| . \] Sum over all places and use $\sum_p\gamma_p=\tfrac1{12}\log|\Delta_{\min}|$. \end{proof} \begin{remark}\label{rem:cancel} Three features of~\eqref{eq:exact} are worth noting. \begin{itemize} \item The component corrections $c_p$ have cancelled. Bad primes enter only through positivity (Proposition~\ref{prop:cp}) and through levels. In particular no bound for the local height differences $\alpha_p$ of \cite{CPS} is needed. \item The archimedean smoothing is \emph{exact} on the off-diagonal entries, because $\lambda_\infty$ has constant Laplacian away from $\Lambda$. Smoothing is therefore two-sided automatically: it holds for test vectors of every sign pattern. \item The only analytic input is the separation $\delta$. \end{itemize} \end{remark} \subsection{Archimedean separation} \begin{proposition}[Distance lemma]\label{prop:arch} For the model~\eqref{eq:W}, define \[ \kappa_\infty:=\max\Bigl(\sup_{Q\in E(\C)\smallsetminus\{O\}}\bigl[-\log d(Q,\Lambda)-\tfrac12\log\max(1,|x(Q)|)\bigr],\ -\log|\omega_1|\Bigr). \] Then \[ \kappa_\infty\le 2+\tfrac1{12}\bigl(\log|\Delta_W|+\log^+|j|\bigr)+\tfrac12\log(1+|a_2|/3)\le 3.61+\tfrac43\log H_f . \] \end{proposition} \begin{proof} Write $\Lambda=\omega_1\Lambda_\tau$ with $\Lambda_\tau=\Z+\tau\Z$, where $\tau$ lies in the standard fundamental domain: $|\tau|\ge1$, $|\operatorname{Re}\tau|\le\tfrac12$, and $y=\operatorname{Im}\tau\ge\sqrt3/2$. \emph{Step 1: the normalised lattice.} We claim that for all $z\notin\Lambda_\tau$, \[ -\log d(z,\Lambda_\tau)-\tfrac12\log\max(1,|\wp_\tau(z)|)\le c_0:=1.68 . \] The nonzero points of $\Lambda_\tau$ have mutual distances $\ge1$. Discs of radius $\tfrac12$ around them are therefore disjoint, so $\#\{\omega\ne0:|\omega|\le R\}\le4(R+\tfrac12)^2$. Partial summation then gives $\sum_{\omega\ne0}|\omega|^{-3}\le19$. For $|w|\le\tfrac14$ and $\omega\ne0$, \[ |(w-\omega)^{-2}-\omega^{-2}|\le4|w|/|\omega|^3,\qquad\text{so}\qquad|\wp_\tau(w)-w^{-2}|\le76|w|. \] Put $r_1=152^{-1/3}\approx0.187$. If $|w|1$. Now let $d=d(z,\Lambda_\tau)$. \begin{itemize} \item If $d0$. Hence \[ \Bigl|\sum c_nq^n\Bigr|\le j(i\sqrt3/2)-e^{\pi\sqrt3}-744<1335, \] so $|j|\ge e^{2\pi y}-2079$. Therefore $2\pi y\le\log^+|j|+\log2080$, and \[ -\log|\omega_1|=\tfrac1{12}\bigl(\log|\Delta_W|-\log|\Delta(\Lambda_\tau)|\bigr) \le\tfrac1{12}\bigl(\log|\Delta_W|+\log^+|j|\bigr)-1.19 . \] This bounds both terms in the definition of $\kappa_\infty$. \emph{Step 4: in terms of $H_f$.} We have $|\disc f|\le54H_f^4$, so $\log|\Delta_W|\le\log864+4\log H_f$. Also $|c_4|=|16a_2^2-48a_4|\le64H_f^2$ and $|\Delta_W|\ge1$, so $\log^+|j|\le3\log(64H_f^2)$. \end{proof} \subsection{Integral progressions} \begin{theorem}[Spectral cap for integral progressions]\label{thm:cap} Let $W$ be as in~\eqref{eq:W}, and let $P_0,\dots,P_{N-1}\in E(\Z)$ with $x_i=a+ib$, $b\ne0$, $N\ge3$. Let $I\subseteq\{0,\dots,N-1\}$ be any set of indices with $y_i\ne0$ for $i\in I$, and let $G_I=(\langle P_i,P_j\rangle)_{i,j\in I}$. Then, with $L=\log X$, \begin{align*} \lmax(G_I)\ &\le\ \tfrac14\Bigl(3L+\psi(N-1)+\kappa_\infty+\tfrac72\log H_f+3.58\Bigr)\\ &\le\ \tfrac34L+\tfrac14\psi(N-1)+\tfrac54\log H_f+2 . \end{align*} \end{theorem} \begin{proof} Let $u\in\Z$ be such that $\Delta_W=u^{12}\Delta_{\min}$. Write $u_p=v_p(u)$, so that $\sum_pu_p\log p=\tfrac1{12}\log|\Delta_W/\Delta_{\min}|$. Write $\ell_p^W(R)=\max(0,-\tfrac12v_p(x(R)))$, with $x$ the coordinate of $W$. \emph{Step 1: comparing levels.} The model $W$ is obtained from a minimal model by $x=u^2x'+r$ with $r\in\Z$ \cite[VII.1.3]{SilAEC}. If $R\in E_k$ with $k>u_p$, then $v_p(x(R))=2u_p-2k<0$. Hence $\ell_p(R)\le\ell_p^W(R)+u_p$ for every $R$. \emph{Step 2: levels of $P_i\pm P_j$.} Put $D=x_i-x_j=(i-j)b$, and let $u_\pm=x(P_i\pm P_j)$. A direct computation (also verified symbolically, Appendix~\ref{app:comp}) gives \[ (u_++u_-)D^2=2\bigl[(x_i+x_j)(x_ix_j+a_4)+2a_2x_ix_j+2a_6\bigr], \qquad u_+u_-D^2=(x_ix_j-a_4)^2-4a_6(x_i+x_j)-4a_2a_6 . \] Both right-hand sides are integers, so $v_p(u_++u_-)$ and $v_p(u_+u_-)$ are at least $-2v_p(D)$. If $v_p(u_+)$ and $v_p(u_-)$ are both negative, the product bound gives $\ell_p^W(u_+)+\ell_p^W(u_-)\le v_p(D)$. If exactly one is negative, the sum bound gives the same inequality. Hence $\ell_p^W(P_i+P_j)+\ell_p^W(P_i-P_j)\le v_p(D)\le v_p(b)+m_p(N-1)$. \emph{Step 3: levels of $2P_i$.} We have $x(2P)=\bigl(x^4-2a_4x^2-8a_6x+a_4^2-4a_2a_6\bigr)/(4y^2)$, so $\ell_p^W(2P_i)\le v_p(2y_i)$. \emph{Step 4: choice of $K_p$ and summation.} Take $K_p=u_p+v_p(b)+m_p(N-1)$. This is admissible by Steps 1--2. By Step~3, \[ \max\bigl(K_p,\ell_p(2P_i)\bigr)\le u_p+v_p(b)+m_p(N-1)+v_p(2y_i). \] Summing over $p$, the term $\sum u_p\log p$ cancels the term $\tfrac1{12}\log|\Delta_{\min}/\Delta_W|$ in~\eqref{eq:exact}. This leaves \begin{equation}\label{eq:Bi} B_i\le-\log\rho+\tfrac14+\log|b|+\psi(N-1)+\log|2y_i|. \end{equation} \emph{Step 5: archimedean separation.} Since $X\ge(N-1)|b|/2\ge|b|\ge1$, we have $X^3/b^2\ge X\ge1$, and $y_i^2=f(x_i)\le H_fX^3$. We bound $|x(Q)|$ for each point that enters $\delta$. \begin{itemize} \item \emph{$Q=P_i\pm P_j$.} By the addition law, $|x(P_i\pm P_j)|\le4H_fX^3/b^2+|a_2|+2X\le7H_fX^3/b^2$, using $H_f+2X\le3H_fX^3/b^2$. \item \emph{$Q=2P_i$.} Every root of $f$ has absolute value at most $H_f$. If $|x_i|\ge2H_f$, then $|f(x_i)|\ge|x_i|^3/8$, and the numerator of $x(2P_i)$ is at most $8H_f^2|x_i|^4$; so $|x(2P_i)|\le16H_f^2X$. If $|x_i|<2H_f$, then $|y_i|\ge1$, so $|x(2P_i)|\le2H_f^2(2H_f)^4=32H_f^6$. In either case $|x(2P_i)|\le48H_f^6X^3/b^2$. \end{itemize} Let $M=48H_f^6X^3/b^2\ge1$, so that $\max(1,|x(Q)|)\le M$ for every $Q$ entering $\delta$. Proposition~\ref{prop:arch} gives $-\log\delta\le\tfrac12\log M+\kappa_\infty$ and $-\log|\omega_1|\le\kappa_\infty$. Since~\eqref{eq:Bi} holds for every admissible $\rho<\tfrac12\min(\delta,|\omega_1|)$, we may let $\rho$ increase to this value: \[ -\log\bigl(\tfrac12\min(\delta,|\omega_1|)\bigr)\ \le\ \log2+\kappa_\infty+\tfrac12\log M =\log2+\kappa_\infty+\tfrac12\log48+3\log H_f+\tfrac32L-\log|b| . \] Combine this with~\eqref{eq:Bi} and $\log|2y_i|\le\log2+\tfrac12\log H_f+\tfrac32L$. The $\log|b|$ terms cancel, and we get \[ B_i\le3L+\psi(N-1)+\kappa_\infty+\tfrac72\log H_f+(2\log2+\tfrac12\log48+\tfrac14). \] The constant in parentheses is less than $3.58$. Now apply Theorem~\ref{thm:general} to the points $P_i$, $i\in I$; the hypotheses $P_i\pm P_j\ne O$ and $2P_i\ne O$ hold because the $x_i$ are distinct and $y_i\ne0$. The final form follows from Proposition~\ref{prop:arch}. \end{proof} \begin{corollary}[Height form]\label{cor:hform} Let $C^-(E)=\sup_{P\in E(\Q)\smallsetminus\{O\}}\bigl(\tfrac12h(x(P))-\hh(P)\bigr)$. Silverman \cite[Thm.~1.1]{Sil90} gives $C^-(E)\le\tfrac1{24}h(j)+\mu(E)+0.973$, where $\mu(E)=\tfrac1{12}h(\Delta)+\tfrac1{12}h_\infty(j)+\tfrac12h_\infty(b_2/12)+\tfrac12\log2^*$. For the model~\eqref{eq:W} we have $b_2=4a_2$, $h(\Delta)\le\log864+4\log H_f$ and $h(j),h_\infty(j)\le\log\max(|c_4|^3,|\Delta_W|)\le12.48+6\log H_f$. Hence \begin{equation}\label{eq:Cminus} C^-(E)\le3.45+\tfrac{19}{12}\log H_f . \end{equation} In the setting of Theorem~\ref{thm:cap}, \[ \lmax(G_I)\le\tfrac32\max_i\hh(P_i)+\tfrac14\psi(N-1)+\tfrac32C^-(E)+\tfrac54\log H_f+2 . \] \end{corollary} \begin{proof} Take $i^*$ with $|x_{i^*}|=X$. Then $\tfrac12L=\tfrac12h(x_{i^*})\le\hh(P_{i^*})+C^-(E)$. \end{proof} \begin{remark}[Numerical check]\label{rem:capcheck} Theorem~\ref{thm:exact}, with the actual levels and the actual $\delta$, was tested on $104$ integral progressions of length $\ge4$. These were all such progressions on the curves $y^2=x^3+a_2x^2+a_4x+a_6$ with $a_2\in\{0,\pm1,2\}$, $|a_4|,|a_6|\le60$ and $-60\le x<200$. The ratio $\lmax(G)/\tfrac14\max_iB_i$ never exceeded $0.819$. The ratio of $\lmax(G)$ to the explicit bound of Theorem~\ref{thm:cap} never exceeded $0.27$. On the $8$-term progression $x=-12,-8,\dots,16$ on $y^2=x^3-112x+400$ (which is $5077a1$), and on its rescalings $x\mapsto u^2x$ for $u=2,3,5,6$ (non-minimal models), the exact ratio was the same in every model: $0.626$. This is consistent with the model independence of~\eqref{eq:exact}. See Appendix~\ref{app:comp}. \end{remark} \begin{remark}[The archimedean diagonal place by place]\label{rem:Dinf} The archimedean contribution to the diagonal has the heuristic size $D_i^\infty\approx\tfrac32\log X-\log|b|-\tfrac12\log|x_i|$. Lemma~\ref{lem:smooth} gives it exactly: \[ D_i^\infty=S^\infty_{ii}=F_\infty(0)-F_\infty(2P_i) =-\log\rho+\tfrac14-\tfrac1{12}\log|\Delta_W|-\lambda_\infty(2P_i). \] To recover the heuristic size: \begin{itemize} \item Insert $-\log\rho\le\tfrac32L-\log|b|+\kappa_\infty+\tfrac12\log48+3\log H_f+\log2$ from Step~5 of Theorem~\ref{thm:cap}. \item If $|x_i|\ge8H_f$, the duplication formula gives $|x(2P_i)|\ge|x_i|/6$. \item A lower bound $\lambda_\infty(Q)\ge\tfrac12\log\max(1,|x(Q)|)-\tfrac1{12}\log|\Delta_W|-O(1+\log H_f)$ is the archimedean half of the explicit height-difference bounds \cite[Thm.~5.5]{Sil90} (after the normalisation change \cite[(11)]{CPS}), \cite[\S\S7--10]{CPS}. \end{itemize} Together these give \[ D_i^\infty\le\tfrac32L-\log|b|-\tfrac12\log|x_i|+O(1+\log H_f), \] Theorem~\ref{thm:cap} does not use this per-place form. It sums $\lambda_v(2P_i)$ over all places into $4\hh(P_i)$, so no archimedean height-difference bound enters the cap itself. \end{remark} \begin{remark}[Other integral cubic models]\label{rem:models} Every integral cubic model reduces to~\eqref{eq:W} without changing the points, the span, or the fact that the $x$-coordinates form a progression. \begin{itemize} \item For $y^2=cx^3+a_2x^2+a_4x+a_6$, the substitution $(X,Y)=(cx,cy)$ gives $Y^2=X^3+a_2X^2+ca_4X+c^2a_6$. The new progression has difference $cb$, and $H\le c^2H_f$. \item For a general Weierstrass equation, the substitution $(X,Y)=(4x,4(2y+a_1x+a_3))$ gives $Y^2=X^3+b_2X^2+8b_4X+16b_6$. The new progression has difference $4b$, and the new height is polynomial in the old coefficients. \end{itemize} So Theorem~\ref{thm:cap} and Section~\ref{sec:linear} apply to all integral cubic models, with $\log H_f$ replaced by $O(1+\log H)$. Quartic models $y^2=g(x)$ are different: the map to the Jacobian does not carry progressions to progressions, so they are left open (Section~\ref{sec:status}). \end{remark} \section{The linear bound}\label{sec:linear} \begin{lemma}\label{lem:spacing} For an integer progression $x_i=a+ib$ $(0\le iN$, then $t$ lies outside $[0,N-1]$, so $|i-t|\ge M-(N-1)$ for every $i$, and $(M-N+1)/M>1/N$. \end{itemize} \end{proof} \begin{theorem}[Linear bound, explicit form]\label{thm:linear} Let $W$ be as in~\eqref{eq:W}, and let $P_0,\dots,P_{N-1}\in E(\Z)$ with $x_i=a+ib$, $b\ne0$, $N\ge12$. Let $s$ be the span rank. Then \begin{enumerate}[label=\textup{(\alph*)}] \item $L\ \ge\ \log|b|\ \ge\ \vartheta(Y_N)-\log6-\log|\disc f|$; \item $(N-4)\bigl(\tfrac12L-\tfrac12\log(2N)-C^-(E)\bigr)\ \le\ s\bigl(\tfrac34L+\tfrac14\psi(N-1)+\tfrac54\log H_f+2\bigr)$. \end{enumerate} \end{theorem} \begin{proof} (a) Corollary~\ref{cor:height} with $\lc(f)=1$ gives the second inequality. The first is $X\ge|b|$. (b) Let $I$ be the set of indices $i\ne i_0$ (with $i_0$ from Lemma~\ref{lem:spacing}) such that $y_i\ne0$. At most three $x_i$ are roots of $f$, so $|I|\ge N-4$. For $i\in I$, \[ \hh(P_i)\ge\tfrac12\log\max(1,|x_i|)-C^-(E)\ge\tfrac12L-\tfrac12\log2N-C^-(E). \] The principal submatrix $G_I$ has rank $\le s$. Lemma~\ref{lem:trace} and Theorem~\ref{thm:cap} give \[ \textstyle\sum_{i\in I}\hh(P_i)=\tr G_I\le s\,\lmax(G_I). \] If the bracket on the left of~(b) is negative, (b) holds trivially. \end{proof} \begin{corollary}\label{cor:32} If $N\to\infty$ along a sequence of integral progressions on models with $\log H_f=o(N^2)$, then \[ N\le\bigl(\tfrac32+o(1)\bigr)s\le\bigl(\tfrac32+o(1)\bigr)r . \] \end{corollary} \begin{proof} We have $|\disc f|\le54H_f^4$, and $\vartheta(Y_N)=(1+o(1))N^2/49$. So~(a) gives $L\ge(1-o(1))N^2/49$. Every other term in~(b) is $o(L)$: $\psi(N-1)\le1.04N$ \cite[Lemma~4.3]{HS}, \cite{RS}, $\log2N$, $\log H_f$, and $C^-(E)$ by~\eqref{eq:Cminus}. Dividing~(b) by $L$ gives $N-4\le s\,(\tfrac32+o(1))$. \end{proof} \begin{corollary}[General form]\label{cor:general} For every $\varepsilon>0$ there is an effective $C_\varepsilon$ such that every integral $x$-progression on a model~\eqref{eq:W} satisfies \[ N\le\bigl(\tfrac32+\varepsilon\bigr)s+C_\varepsilon\sqrt{1+\log H_f}. \] \end{corollary} \begin{proof} Choose $K=K(\varepsilon)$ and $N_1=N_1(\varepsilon)$ large. Suppose $N\ge N_1$ and $N^2\ge K(1+\log H_f)$. Then by Theorem~\ref{thm:linear}(a), $L\ge N^2/60$ once $K$ and $N_1$ are large. Moreover, by~\eqref{eq:Cminus}, $C^-(E)+\tfrac54\log H_f+2\le6(1+\log H_f)\le6N^2/K\le360L/K$, and $\psi(N-1)\le 63L/N$. So the ratio of the bracket on the right of~(b) to the bracket on the left is at most $\tfrac32+\varepsilon$, and $N\le(\tfrac32+\varepsilon)s+4$. Otherwise $N<\max\bigl(N_1,\sqrt{K(1+\log H_f)}\bigr)$. \end{proof} \begin{table}[ht] \centering \begin{tabular}{lrrrrrrrrr} \toprule $N$ & 20 & 50 & 100 & 150 & 200 & 500 & 1000 & 5000 & $\infty$\\ \midrule bound on $(N-4)/s$ & 3.10 & 2.03 & 1.89 & 1.78 & 1.69 & 1.56 & 1.53 & 1.505 & $3/2$\\ \bottomrule \end{tabular} \caption{The ratio $\bigl(\tfrac34L+\tfrac14\psi(N-1)\bigr)/\bigl(\tfrac12L-\tfrac12\log2N\bigr)$ with $L$ the forced lower bound (curve constants omitted, i.e.\ $H_f$ bounded). The forced primes are those of Table~\ref{tab:forbidden}, optimised exactly. With point counting alone the ratio would tend to $7/4$.} \label{tab:ratio} \end{table} \begin{remark} The theorem is vacuous for every known example: all have $N\le8$ on cubics, and $N\le14$ on quartics (MacLeod \cite{MacL}). It becomes meaningful around $N\approx50$, provided $\log H_f$ is small compared with $N^2$. The constant $\tfrac32$ is the limit of the method. The diagonal~\eqref{eq:exact} contains the separation term ($\approx\tfrac32L-\log|b|$) and the denominator of $x(2P_i)$ ($\approx\tfrac32L$), while the trace only sees $\hh\approx\tfrac12L$. The cosine bound $\lmax(M)\le\tfrac32$ is attained in the function-field example~\ref{ex:A2A1}. \end{remark} \begin{corollary}[Families]\label{cor:families} For fixed $\varepsilon>0$: \begin{enumerate}[label=\textup{(\alph*)}] \item \emph{Quadratic twists.} Fix $y^2=f(x)$ as in~\eqref{eq:W}, and for squarefree $D$ consider the twist model $y^2=x^3+Da_2x^2+D^2a_4x+D^3a_6$. It has $H\le|D|^3H_f$. So its longest integral progression satisfies \[ \beta_{\rm int}\le(\tfrac32+\varepsilon)\,r(E^{(D)})+O_{\varepsilon,E}\bigl(\sqrt{\log|D|}\bigr)\ \ll_E\ \omega(D)+\sqrt{\log|D|}, \] using $r(E^{(D)})\ll_E1+\omega(D)$ from a $2$-descent. \item \emph{Mordell and congruent-number curves.} On $y^2=x^3+n$ and on $y^2=x^3-n^2x$, integral progressions satisfy $\beta_{\rm int}\le(\tfrac32+\varepsilon)r+O_\varepsilon(\sqrt{\log|n|})$. \end{enumerate} \end{corollary} \section{Function fields: the exact version}\label{sec:ff} Let $\mathcal S\to\mathbb P^1$ be an elliptic surface over $\C$ with zero section $O$, arithmetic genus $\chi$, and minimal Weierstrass equation $y^2=x^3+A_2x^2+A_4x+A_6$ with $\deg A_k\le k\chi$. Assume all singular fibres are irreducible. A section $P$ is \emph{integral} ($P\cdot O=0$) exactly when $x(P),y(P)$ are polynomials of degree $\le2\chi$ and $\le3\chi$ respectively. For such sections, Shioda's formula \cite{Shioda}, \cite[Ch.~6]{SchuettShioda} reads \[ \langle P,Q\rangle=\chi-(P\cdot Q)\quad(P\ne Q),\qquad\langle P,P\rangle=2\chi . \] \begin{theorem}[Exact colouring formula]\label{thm:ff} Let $P_0,\dots,P_{N-1}$ be integral sections with $x_i=a+ib$, where $a,b\in\C[t]$ and $b\ne0$. Regard $b$ as a section of $\mathcal O(2\chi)$, with zeros $v$ of multiplicity $e_v$ (including $\infty$ with multiplicity $2\chi-\deg b$), so that $\sum_ve_v=2\chi$. Assume $y_i(v)\ne0$ for all $i$ and $v$. Fix a square root $\eta_v$ of the common value $f_v(a(v))$, and put $c_v(i)=y_i(v)/\eta_v\in\{\pm1\}$. Then \[ \langle P_i,P_j\rangle=-\tfrac12\sum_ve_v\,c_v(i)c_v(j)\quad(i\ne j),\qquad\langle P_i,P_i\rangle=2\chi . \] Equivalently, the cosine matrix is $M=\tfrac12(3I-C)$ with $C=\sum_vw_vc_vc_v^T$ and $w_v=e_v/2\chi$, so that $\sum_vw_v=1$. \end{theorem} \begin{proof} Two distinct integral sections can meet only over points where $x_i=x_j$. Since $x_i-x_j=(i-j)b$ and $i-j\in\C^\times$, these are exactly the zeros of $b$. Over a finite zero $v$, $y_i(v)^2=y_j(v)^2=f_v(a(v))\ne0$, so the sections meet iff $c_v(i)=c_v(j)$. At the common point $\partial_y(y^2-f)=2y\ne0$, so $(t,x)$ are local coordinates on $\mathcal S$. The sections are the graphs $x=x_i(t)$ and $x=x_j(t)$, and their intersection multiplicity is $\operatorname{ord}_v(x_i-x_j)=e_v$. At $\infty$, use the chart $s=1/t$, $X=s^{2\chi}x$, $Y=s^{3\chi}y$; the same computation gives $e_\infty=2\chi-\deg b$. Hence \[ P_i\cdot P_j=\sum_ve_v\,\mathbf 1[c_v(i)=c_v(j)]=\chi+\tfrac12\sum_ve_vc_v(i)c_v(j). \] Substitute this into Shioda's formula. \end{proof} \begin{corollary}\label{cor:ff} In the setting of Theorem~\ref{thm:ff}: \begin{enumerate}[label=\textup{(\alph*)}] \item $N\le3/\max_vw_v\le6\chi$, and $N\le\tfrac32s$. Here $s\le\rho-2\le10\chi-2$ by Shioda--Tate. \item If $\chi=1$ (rational elliptic surfaces), then $N\le6$. Equality forces two simple zeros with $c_1\perp c_2$, $s=4$, and a relation lattice of dimension $2$. \item A double zero of $b$ gives $C=cc^T$, so $N\le3$, and $N=3$ forces a dependence. \end{enumerate} \end{corollary} \begin{proof} $M\succeq0$ forces $\lmax(C)\le3$. Since $C\succeq w_vc_vc_v^T$, we have $\lmax(C)\ge w_vN$. Also $\max_v w_v\ge1/(2\chi)$, because there are at most $2\chi$ zeros. Next, $C\succeq0$ gives $\lmax(M)\le\tfrac32$, and Lemma~\ref{lem:trace} gives $N\le s\lmax(M)\le\tfrac32 s$. For (b) with $N=6$: we need $w_1=w_2=\tfrac12$ and $\lmax(C)=3$. The nonzero eigenvalues of $\tfrac12(c_1c_1^T+c_2c_2^T)$ are $\tfrac12(6\pm|c_1\cdot c_2|)$, so $c_1\cdot c_2=0$. Then $C$ has the eigenvalue $3$ twice, so $M$ has nullity $2$ and $s=4$. For (c): if $C=cc^T$, then $\lmax(C)=N\le3$, and $N=3$ gives $\lmin(M)=0$. \end{proof} \begin{example}[An $A_2\oplus A_1$ progression]\label{ex:A2A1}\textup{(computer-assisted)} On the rational elliptic surface \[ y^2=x^3+(22t^2-46t+13)x^2+(8t^4-24t^3+9t^2+4t-2)x+(t^3-2t^2-4t+1)^2, \] whose discriminant is squarefree of degree $12$ (so all fibres are $\mathrm I_1$), the sections \[ x=-t,0,t\quad\text{with}\quad y=t^3-6t^2+3t-1,\ \ t^3-2t^2-4t+1,\ \ t^3+2t^2-5t+1 \] are integral. Here $b=t$ has simple zeros at $0$ and $\infty$, with colourings $(-,+,+)$ and $(+,+,+)$. Theorem~\ref{thm:ff} gives the Gram matrix $\left(\begin{smallmatrix}2&0&0\\0&2&-1\\0&-1&2\end{smallmatrix}\right)$, of type $A_2\oplus A_1$, with $\lmax(M)=\tfrac32$. \end{example} \subsection{Realising sign data: rational and $K3$ surfaces} Throughout this subsection, a $3$-term progression is indexed by $i\in\{-1,0,1\}$, with $x_i=a+ib$. Replacing $P_i$ by $-P_i$ leaves the $x$-progression unchanged and flips $c_v(i)$ at every zero $v$ simultaneously. We call two sets of sign data \emph{equivalent} if they differ by such flips, by a global sign at a single zero (the choice of $\eta_v$), and by relabelling the zeros. \begin{lemma}[Interpolation criterion]\label{lem:interp} Let $a,b\in\C[t]$ with $\deg a\le2\chi$, and let $b$ be squarefree of degree exactly $2\chi$, with zeros $v_1,\dots,v_{2\chi}$. Let $y_{-1},y_0,y_1\in\C[t]$ have degree $\le3\chi$. Then there is a monic cubic $f=x^3+A_2x^2+A_4x+A_6$ with $\deg A_k\le k\chi$ and $y_i^2=f(a+ib)$ for $i=-1,0,1$ if and only if, at every zero $v$ of $b$, \begin{equation}\label{eq:interp} y_{-1}(v)^2=y_0(v)^2=y_1(v)^2\qquad\text{and}\qquad \bigl(y_1y_1'-2y_0y_0'+y_{-1}y_{-1}'\bigr)(v)=0 . \end{equation} If $y_0(v)\ne0$ and $y_i(v)=c_v(i)\eta_v$, the second condition is the linear condition $c_v(1)y_1'(v)-2c_v(0)y_0'(v)+c_v(-1)y_{-1}'(v)=0$. \end{lemma} \begin{proof} Lagrange interpolation at the three nodes $a-b,a,a+b$ shows that every monic cubic through the three points $(x_i,y_i^2)$ has the form \[ f(x)=(x-a+b)(x-a)(x-a-b)+y_0^2+(x-a)D_1+(x-a)^2D_2, \] where \[ D_1=\frac{y_1^2-y_{-1}^2}{2b},\qquad D_2=\frac{y_1^2-2y_0^2+y_{-1}^2}{2b^2}, \] and this $f$ is unique. So $f$ has polynomial coefficients if and only if $b\mid y_1^2-y_{-1}^2$ and $b^2\mid y_1^2-2y_0^2+y_{-1}^2$. Since $b$ is squarefree, these divisibilities are exactly the conditions~\eqref{eq:interp}. The first condition gives the value conditions, and the derivative of the second-difference numerator gives the derivative condition. The degree bounds are then automatic. The numerators have degree $\le6\chi$ and $\deg b=2\chi$, so $\deg D_2\le2\chi$ and $\deg D_1\le4\chi$. Expanding $f$ in powers of $x$ gives $\deg A_2\le2\chi$, $\deg A_4\le4\chi$ and $\deg A_6\le6\chi$. \end{proof} So the progressions with prescribed sign data at the zeros of $b$ form a linear family. There are $3$ linear conditions per zero, on $3(3\chi+1)$ coefficients. ``All fibres $\mathrm I_1$'' ($\Delta$ squarefree of degree $12\chi$, which in particular makes the Weierstrass model minimal and all fibres irreducible) is a Zariski-open condition on this family. Hence a single member with all fibres $\mathrm I_1$ shows that the generic member has all fibres $\mathrm I_1$ too. \begin{theorem}[$3$-term types on rational elliptic surfaces]\label{thm:types} Let $\chi=1$, and assume all singular fibres are irreducible. \begin{enumerate}[label=\textup{(\alph*)}] \item \emph{Classification.} Suppose $b$ has two simple zeros $v_1,v_2$. Up to equivalence, the sign data are classified by $d=c_{v_1}c_{v_2}\in\{\pm1\}^3/\pm1$ (coordinatewise product), which takes four values. The Gram matrices are as follows. \begin{center} \renewcommand{\arraystretch}{2.3} \begin{tabular}{lll} \toprule $d$ & Gram matrix of $(P_{-1},P_0,P_1)$ & type\\ \midrule $(+,+,+)$ & $\left(\begin{smallmatrix}2&-1&-1\\-1&2&-1\\-1&-1&2\end{smallmatrix}\right)$ & $A_2$; $P_{-1}+P_0+P_1=O$ (collinear)\\ $(-,+,+)$ & $\left(\begin{smallmatrix}2&0&0\\0&2&-1\\0&-1&2\end{smallmatrix}\right)$ & $A_1\oplus A_2$, end term orthogonal\\ $(+,+,-)$ & $\left(\begin{smallmatrix}2&-1&0\\-1&2&0\\0&0&2\end{smallmatrix}\right)$ & $A_2\oplus A_1$, end term orthogonal\\ $(+,-,+)$ & $\left(\begin{smallmatrix}2&0&-1\\0&2&0\\-1&0&2\end{smallmatrix}\right)$ & $A_2\oplus A_1$, middle term orthogonal\\ \bottomrule \end{tabular} \end{center} If $b$ has a double zero, the Gram matrix is of type $A_2$ and the three points are collinear. \item \emph{Realisation.} All five cases occur with all fibres $\mathrm I_1$. \end{enumerate} \end{theorem} \begin{proof} (a) Use the flips to make $c_{v_1}=(+,+,+)$. Then $c_{v_2}=d$, and $d$ is defined up to the global sign at $v_2$. Theorem~\ref{thm:ff} gives $\langle P_i,P_j\rangle=-\tfrac12\bigl(1+d(i)d(j)\bigr)$ for $i\ne j$, which produces the four matrices. For a double zero, $\langle P_i,P_j\rangle=-c(i)c(j)$. After flips the Gram matrix is $A_2$, which has rank $2$ with kernel $(1,1,1)$. So $P_{-1}+P_0+P_1$ has height $0$. The Mordell--Weil group of a rational elliptic surface with irreducible fibres is torsion-free: it is the $E_8$ lattice \cite[Ch.~8]{SchuettShioda}. So $P_{-1}+P_0+P_1=O$. In the $d=(+,+,+)$ case the same argument applies. (b) Lemma~\ref{lem:interp} reduces realisation to linear algebra. One explicit member of each family with $\Delta$ squarefree of degree $12$ is listed in Appendix~\ref{app:comp}, item~A.9. In each one, $y_0(v)\ne0$ at the zeros of $b$, and the Gram matrix was recomputed independently of Theorem~\ref{thm:ff}, from intersection numbers. For example, for $d=(+,-,+)$ take \[ a=2t^2+2t+3,\quad b=t^2-t,\quad \begin{cases} y_{-1}=-16t^3+35t^2-15t+2,\\ y_0=-t^3+t^2-8t+2,\\ y_1=2t^3+3t^2-t+2, \end{cases} \] with $f$ given by Lemma~\ref{lem:interp}. For the double zero, take a collinear triple on the line $y=mx+k$ and put $f=(mx+k)^2+\prod_i(x-x_i)$ with $b=t^2$. \end{proof} \begin{theorem}[Sign data on elliptic $K3$ surfaces]\label{thm:k3} Let $\chi=2$, let $b$ have four simple zeros, and assume all fibres are $\mathrm I_1$. \begin{enumerate}[label=\textup{(\alph*)}] \item Normalise $c_{v_1}=(+,+,+)$. The sign data at the other three zeros then form one of $20$ multisets of classes. All $20$ are realised, so every equivalence class of sign data occurs. Their Gram matrices, given by Theorem~\ref{thm:ff}, agree with direct intersection counts. \item The old Gram matrix $\left(\begin{smallmatrix}4&-1&1\\-1&4&0\\1&0&4\end{smallmatrix}\right)$, with a cosine $+\tfrac14$, is realised by the sign data \[\bigl((+,+,+),\,(-,+,+),\,(+,+,-),\,(+,+,-)\bigr).\] \item Sign data with $c_v(-1)c_v(1)$ constant across the four zeros are realised as well. They give $\langle P_{-1},P_1\rangle=\pm2$, that is, a cosine $\pm\tfrac12$. \end{enumerate} \end{theorem} \begin{proof} (a) Each of the other three zeros carries one of four classes, and the zeros are unordered, which gives $\binom{6}{3}=20$ multisets. This overcounts equivalence classes, because the choice of reference zero is not fixed, but it covers all of them. For realisation, see item~A.9 of the appendix. \begin{itemize} \item For each class, Lemma~\ref{lem:interp} yields a linear family, from which a member was drawn. \item Each member has $\Delta$ squarefree of degree $24$, and $y_0(v)\ne0$ at all four zeros. \item In each case the Gram matrix was computed from intersection numbers, and it equals the formula. \end{itemize} (b) is one of the $20$ cases. Explicitly, $b=t(t-1)(t+1)(t-2)$ and $a=-3t^3+2t^2+t-3$, with \begin{align*} y_{-1}&=\tfrac{13}2t^6-20t^5+51t^4-38t^3-\tfrac{101}2t^2+63t-3,\\ y_0&=t^6-2t^5+35t^4-32t^3-38t^2+30t-3,\\ y_1&=-3t^6-t^5+2t^4-t^3-3t-3, \end{align*} and $f$ is given by Lemma~\ref{lem:interp}. (c) These are the cases $c_{v_k}\in\{(+,+,+),(+,-,+)\}$ for $k=2,3,4$, and all of them are realised. \end{proof} \begin{remark}[Signs and obtuseness]\label{rem:k3} A positive cosine is not an invariant of a progression: replacing $P_i$ by $-P_i$ keeps the $x$-progression and changes signs in the Gram matrix. The invariant quantity is $\Pi=G_{-1,0}\,G_{0,1}\,G_{-1,1}$. When $\Pi\le0$, such flips make every off-diagonal entry $\le0$, so the triple is obtuse up to sign. For four simple zeros, $\Pi\le0$ always holds. To see this, write $u_v=c_v(-1)c_v(0)$ and $w_v=c_v(0)c_v(1)$. Then $\Pi=-\tfrac18S_uS_wS_{uw}$, where $S_u=\sum_vu_v$. If $S_u$ and $S_w$ are both nonzero, each of $u$ and $w$ has at most one entry differing from the majority, and a check of the cases gives $S_uS_wS_{uw}\ge0$. For example, the matrix in~(b) becomes $\left(\begin{smallmatrix}4&-1&-1\\-1&4&0\\-1&0&4\end{smallmatrix}\right)$ after flipping $P_1$. \end{remark} \section{Relation to earlier work}\label{sec:prior} The analytic ingredients of Section~\ref{sec:kernels} are not new. What appears to be new is how they are combined. \begin{itemize} \item \textbf{Positive-type Néron functions.} Baker and Petsche \cite[Prop.~2, (13)]{BP} prove that the archimedean Néron function has nonnegative Fourier coefficients $\widehat\lambda(\gamma_\omega)=b/(2\pi|\omega'|^2)$; compare Lemma~\ref{lem:green}. They smooth it by convolution with a heat kernel, where we use a disc measure (Lemma~\ref{lem:smooth}). For multiplicative reduction they use the Fourier positivity of the Bernoulli term \cite[Prop.~5]{BP}. Their Theorem~8 sums local discrepancies into a global lower bound for $\sum_{i\ne j}\hh(P_i-P_j)$. \item \textbf{Fourier positivity at bad places.} Hindry and Silverman \cite[\S3, \S4]{HS} use the positivity of $\widehat{B_2}$, through the Blanksby--Montgomery--Fejér argument, to bound weighted sums of local heights at places of multiplicative reduction. To handle additive reduction they pass to multiples of the point. \item \textbf{Average Green-function bounds.} Faltings and Elkies bound the average of the Arakelov--Green function over pairs of $n$ points from below by $-O((\log n)/n)$. Wilms \cite{Wilms} gives an alternative, effective proof. This is the Riemann-surface analogue of Baker--Petsche's archimedean discrepancy, and like it, it controls the quadratic form at the all-ones vector only. \item \textbf{Gram matrices and lattice symmetry.} Prakash \cite{Prakash} bounds Néron--Tate heights by spectral-gap criteria for automorphisms acting on the Mordell--Weil lattice. That work uses the height Gram matrix, but for a different purpose: effective Mordell rather than progressions. \item \textbf{Quasi-orthogonality.} Helfgott and Venkatesh \cite[\S3]{HV} prove that integral points are nearly orthogonal in the Mordell--Weil lattice, using local heights and a Mumford-type gap principle. \end{itemize} What the present argument adds: \begin{enumerate}[label=(\alph*)] \item \emph{Antisymmetrisation} (Lemma~\ref{lem:posdef}). The kernel $F(g_i-g_j)-F(g_i+g_j)$ is insensitive to real characters, and it yields a matrix inequality $G\preceq\tfrac14\diag(B_i)$ valid for every test vector. Baker--Petsche's kernel $\lambda(P_i-P_j)$ controls only the quadratic form at the all-ones vector, i.e.\ $\mathbf 1^TG\mathbf 1-\tr G$, and positivity of the full Fourier series is needed there. \item \emph{Additive reduction without base change.} Positivity at characters of order $>2$ holds for every Kodaira type (Proposition~\ref{prop:cp}), and the component terms cancel from the diagonal (Theorem~\ref{thm:exact}). Baker--Petsche work over a field where $E$ is semistable; Hindry--Silverman replace $P$ by a multiple. \item \emph{The exact diagonal identity}~\eqref{eq:exact}, together with its application to progressions via the local theory of Sections~\ref{sec:local}--\ref{sec:denominator}. \end{enumerate} We have not found (a) or (c) in the literature checked (\cite{BP,Pet,HS,HV,CPS,Sil90,Wilms,Prakash}). A wider search of matrix forms of the Mumford gap principle is still advisable before claiming novelty. \section{Comparison with the exponential bounds}\label{sec:GFP} Several results bound progression length exponentially in the rank: \begin{itemize} \item Garcia-Fritz and Pasten \cite{GFP19}: $1+r\ge c(j_0)\log\max\{\beta_x,\beta_y\}$ for curves with fixed $j$-invariant, with $c(j_0)\asymp1/\log\log H(j_0)$. \item Combined with Gao--Ge--Kühne \cite{GGK}, their method gives the uniform bound $n\le c(d)^{1+\rank\Gamma}$ \cite[Thm.~1.1]{GFP26b}. \item Harrison, Mudgal and Schmidt \cite{HMS} prove $|A|\le C^{1+r}$ with an effective absolute $C$. \item \cite{GFP26a} deduces uniform boundedness of progression lengths from uniform boundedness of ranks, via Dimitrov--Gao--Habegger \cite{DGH}. \end{itemize} Corollary~\ref{cor:general} concerns integral progressions only. It is linear in the span rank, but has a $\sqrt{\log H_f}$ dependence on the model. The two kinds of result are complementary, and the useful bound is the minimum of the two. \begin{center} \small \begin{tabular}{p{3.4cm}p{4.6cm}p{6.2cm}} \toprule result & statement & effect of Corollaries~\ref{cor:general} and~\ref{cor:families}\\ \midrule \cite[Thm.~1.1]{GFP19} & $1+r\ge c(j_0)\log\beta$ & $s\ge(\tfrac23-\varepsilon)N-O_\varepsilon(\sqrt{1+\log H_f})$: linear, and about the span itself (Bremner's ``tend to be independent'')\\ \cite[Cor.~1.3]{GFP19} (twists) & $\beta(E^{(D)})\le C(E)^{\omega(D)+1}$ & $\beta_{\rm int}(E^{(D)})\ll_E\omega(D)+\sqrt{\log|D|}$\\ \cite[Cors.~1.4, 1.5]{GFP19} & bounded ranks $\Rightarrow$ bounded $\beta$; Mohanty & not improved: the $\sqrt{\log H_f}$ term blocks uniformity; unconditionally $\beta_{\rm int}(y^2=x^3+n)\le(\tfrac32+\varepsilon)r+O(\sqrt{\log|n|})$\\ \cite[Thms.~1.6, 1.7]{GFP19} & bounded average moments & moments not improved (the average of $(\log n)^{k/2}$ grows). Using the bounded averages of $\kappa^{r}$ ($\kappa>1$) that \cite{GFP19} uses, for $B\gg\sqrt{\log x}$ the tail $\Pr(\beta_{\rm int}>B)$ decays exponentially in $B$; the bound $\beta\le e^{O(1+r)}$ gives only polynomial decay\\ \cite[Thm.~1.2]{GFP19} (torsion) & no long progressions of torsion & not touched\\ \bottomrule \end{tabular} \end{center} \section{Open problems}\label{sec:status} \begin{enumerate} \item Remove the $\sqrt{\log H_f}$ term in Theorem~\ref{thm:main}, for instance via a Lang-type lower bound for $\hh$ on progression points. This would make the bound uniform in $E$. \item Extend the spectral cap to rational points with mixed denominators. The exact identity~\eqref{eq:exact} holds for arbitrary points; what changes is the level bound of Step~2 of Theorem~\ref{thm:cap}. \item Treat quartic models, such as MacLeod's $14$-term examples \cite{MacL}. The local theory applies, but the spectral half needs the local heights of the Jacobian. \item Does a $7$-term $x$-progression with no collinear triple have a smooth zero at a good prime? \item Extend Theorem~\ref{thm:types} to $N\ge4$ terms on rational elliptic surfaces. For $N=6$, Corollary~\ref{cor:ff} forces $c_1\perp c_2$. \item Determine whether the antisymmetrised kernel of Lemma~\ref{lem:posdef} appears in matrix forms of the Mumford gap principle; see Section~\ref{sec:prior}. \end{enumerate} \appendix \section{Computations}\label{app:comp} All computations were run with PARI/GP~2.17.2 (through \texttt{cypari2}) and SymPy. \begin{enumerate}[label=A.\arabic*,leftmargin=*] \item \emph{Identities.} The sum and product formulae for $x(P\pm Q)$ in Step~2 of Theorem~\ref{thm:cap}, and the duplication numerator in Step~3, were verified symbolically for $y^2=x^3+a_2x^2+a_4x+a_6$. \item \emph{Fourier coefficients.} $\widehat c(k)=1/(4m\sin^2(\pi k/m))$ for type $\mathrm I_m$ was checked to $10^{-9}$ for all $m<200$ and all $k$. \item \emph{Archimedean constants.} $j(i\sqrt3/2)=2309.5777\ldots$, so the tail in Step~3 of Proposition~\ref{prop:arch} is $1334.8$. The factor $|\prod(1-q^n)|^{24}$ was at least $0.957$ on $2000$ random $\tau$. A random search over $\tau$ and $z$ gave an empirical value $0.62