\documentclass[11pt,reqno]{amsart}
\usepackage{jaswal}
\AddToHook{env/proposition/begin}{\crefalias{theorem}{proposition}}
\AddToHook{env/lemma/begin}{\crefalias{theorem}{lemma}}
\AddToHook{env/remark/begin}{\crefalias{theorem}{remark}}

\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\F}{\mathbb{F}}
\newcommand{\PP}{\mathbb{P}}
\newcommand{\tors}{\mathrm{tors}}
\newcommand{\rank}{\operatorname{rank}}
\newcommand{\ZZ}[1]{\Z/#1\Z}
\newcommand{\Tf}{\ZZ{2}\times\ZZ{4}}

\begin{document}

\title[Torsion $\Z/2\Z\times\Z/4\Z$ and rank at least six]{Infinitely many elliptic curves with torsion
$\boldsymbol{\Z/2\Z\times\Z/4\Z}$ and rank at least six}
\author{Manvir Jaswal}
\address{Toronto, Canada}
\email{manvirjaswal@icloud.com}
\urladdr{https://manvirjaswal.github.io}
\date{October 27, 2026}
\subjclass[2020]{Primary 11G05; Secondary 14J27, 11G35}
\keywords{Elliptic curves, Mordell--Weil rank, torsion subgroup, specialization, elliptic surfaces, Diophantine triples}
\hypersetup{pdftitle={Infinitely many elliptic curves with torsion \unichar{"2124}/2\unichar{"2124} \unichar{"00D7} \unichar{"2124}/4\unichar{"2124} and rank at least six},pdfauthor={Manvir Jaswal}}

\begin{abstract}
We show that there are infinitely many elliptic curves over $\Q$ with torsion subgroup $\Tf$ and rank at least $6$,
answering the case $\Tf$ of Problem~2.6 in Dujella's list of open problems. The largest rank previously known to occur
infinitely often with this torsion was $5$, due to Eroshkin. The curves are fibers of a one-parameter family of
Dujella and Peral, which we show has rank exactly $3$ over the rational function field. The family is a
base change of an elliptic K3 surface, and quadratic twists of that surface give it three further independent
points over the function field of an elliptic curve of rank one. Congruences then carry the independence of the six points, and
the torsion $\Tf$, from one rational point of that curve to infinitely many. The result is unconditional. It contradicts the prediction of
Park, Poonen, Voight and Wood that only finitely many curves with this torsion have rank greater than $5$, although
the curves form a sparse family. A second family, over an elliptic curve of rank two, gives another proof.
\end{abstract}

\maketitle
\contents

\section{Introduction}\label{sec:intro}

By Mazur's theorem \cite{Mazur}, the torsion subgroup of an elliptic curve over $\Q$ is one of fifteen groups. For each
such group $T$ one asks how large the rank of a curve with $E(\Q)_\tors\cong T$ can be, and which ranks occur
infinitely often. Dujella maintains tables of the records \cite{DujTors,DujGeneric}: for each $T$, the largest known
rank of a single curve, the largest known rank of a curve over $\Q(t)$, and the largest known rank that occurs for
infinitely many curves, a lower bound for $b_T=\limsup\{\rank E(\Q): E(\Q)_\tors\cong T\}$ in the notation of
\cite{PPVW}.

The elliptic curves over $\Q$ whose torsion subgroup contains $\Tf$ are, up to isomorphism, those of the form
\[
E_{a,b}\colon\ y^2=x(x+a^2)(x+b^2),\qquad a,b\in\Q,\ ab(a^2-b^2)\neq0
\]
(see \cite[Theorem~5.3]{BekkerZarhin}, and \cite[\S 6.7]{DP21} for torsion exactly $\Tf$): the points $(0,0)$, $(-a^2,0)$, $(-b^2,0)$ have
order $2$, and $(ab,ab(a+b))$ has order $4$, with double $(0,0)$. Over $\Q(t)$, Lecacheux \cite[Th\'eor\`eme~9]{Lecacheux}, Elkies
\cite{Elkies07} and Eroshkin (see \cite{DP14}) constructed families of rank at least $3$, and Dujella and Peral
\cite{DP14} one of rank exactly $4$, still the largest rank over $\Q(t)$ in Dujella's table. Elkies obtained rank at
least $4$ for an infinite family by a quadratic base change over an elliptic curve of positive rank \cite[Lecture~III, p.~13]{Elkies07}.
As of October 15, 2026 the largest known rank of a curve over $\Q$ with this torsion is $9$, for curves of Dujella and
Peral \cite{DP14,DP21} and of Klagsbrun (see \cite{DujZ2Z4}), and the largest rank known to occur infinitely often
is $5$, which Dujella's table \cite{DujGeneric} credits to Eroshkin (2009, personal communication). Dujella's list of open problems
\cite[Problem~2.6]{Dopen} asks:
\begin{quote}
``Are there infinitely many elliptic curves over $\Q$ with torsion group $\Z/2\Z\times\Z/4\Z$ (respectively $\Z/6\Z$)
and rank $\ge 6$?''
\end{quote}
\cref{thm:main} answers the first part of this question. The case $\Z/6\Z$ remains open.

\begin{theorem}\label{thm:main}
There are infinitely many elliptic curves $E$ over $\Q$ with $E(\Q)_\tors\cong\Tf$ and $\rank E(\Q)\ge 6$. Among them
there are infinitely many distinct $j$-invariants.
\end{theorem}

We obtain these curves as fibers of a one-parameter family $X$ of Dujella and Peral \cite{DP21} over $\Q(m)$, of
rank $3$, which is the base change by $n=m^2$ of an elliptic K3 surface $Y$ over $\Q(n)$. Over a field obtained
from $\Q(m)$ by adjoining square roots, the rank of $X$ is the sum of the ranks over $\Q(m)$ of the quadratic twists
of $X$ by the corresponding square classes. So a quadratic twist $Y^q$ of $Y$ of positive rank, with $q$ not a square in $\Q(m)$, gives new points of $X$
over the function field of any cover of the $m$-line on which $q$ becomes a square. For these points to give
infinitely many curves over $\Q$, this cover must have infinitely many rational points, and so, by Faltings'
theorem \cite{Faltings}, genus at most one. The
new ingredient is a curve $C$ of genus one and rank one over whose function field $X$ acquires three further
independent points. To find it, we used averages of traces of Frobenius to decide which twists of $Y$ to look at
(see below). They led us to the twists by $5(2n+243)$ and by $5n(3n+242)$, which become trivial
over the curve $C$ given by $u^2=5(3m^2+242)$ and $v^2=5(2m^2+243)$ (since $n=m^2$, $5n(3n+242)$ and
$5(3n+242)$ have the same square class). Over $\Q(C)$ three square classes of $\Q(m)$
become trivial, those of $5(3m^2+242)$, $5(2m^2+243)$ and their product. The twist of $X$ by each of them has rank
one over $\Q(m)$ (\cref{prop:rank6}), and these three twists give the three new points.

To pass to curves over $\Q$, we specialize at one rational point $c_0$ of $C$, where the six points stay independent
(\cref{prop:indep}): reduction modulo six primes gives six homomorphisms to $\ZZ3$, and their values on the six
points form a matrix that is invertible over $\F_3$. These reductions depend only on the point of $C$
modulo the six primes, so the same argument applies at every rational point $c$ of $C$ congruent to $c_0$
modulo them. The torsion is handled in the same way: a curve $E_{a,b}$ has torsion larger than $\Tf$ only if $ab$ or
$-ab$ is a square (\cref{lem:tors}), which a congruence modulo $13^2$ rules out, since $13$ divides $ab$ exactly once
at $c_0$. As $C$ has positive rank, infinitely many of its rational points satisfy all these congruences
(\cref{prop:cong}), and the fibers at these points form an explicit infinite set of curves as in \cref{thm:main},
with infinitely many $j$-invariants because $j$ is a nonconstant function on $C$.
For the stronger statement of \cref{thm:precise}(c) below, which concerns all but finitely many points of $C(\Q)$,
we use instead Silverman's specialization theorem \cite[Theorem~C]{Silverman83} for the rank and, for the torsion,
Faltings' theorem, applied to two double covers of $C$. They leave a finite set of
exceptions that we have not determined (\cref{sec:proof}).

Families over an elliptic curve of positive rank go back to N\'eron: his curves of rank at least $11$ are the fibers
of such a family (see \cite[Lecture~I]{Elkies07}). With prescribed torsion they were used by Lecacheux and by
Rabarison (see \cite[\S 5.4]{DP15}) and by Elkies \cite[Lecture~III]{Elkies07}. In this way Dujella and Peral
obtained infinitely many curves of rank at least $3$ with torsion $\Z/8\Z$ and with torsion $\Z/2\Z\times\Z/6\Z$
\cite[\S\S 5.4, 6.6]{DP15}. Dujella, Kazalicki and Peral gave further such families for $\Z/2\Z\times\Z/6\Z$
\cite[\S 9]{DKP21}. For $\Z/4\Z$, Dujella and Peral made explicit a remark of Elkies \cite[Lecture~II, p.~9]{Elkies07}: pairs of quadratic sections of his elliptic K3 surface with torsion $\Z/4\Z$ and Mordell--Weil rank $4$ give several infinite families of rank at least $6$ parametrized by elliptic curves of positive rank \cite[\S 4]{DP21}. Elkies obtained the torsion $\Tf$, as here, by a quadratic base change of a curve with torsion
$\Z/4\Z$ \cite[Lecture~II]{Elkies07}. With no torsion prescribed, Elkies announced in \cite[Lectures~II and~III]{Elkies07},
and proved in \cite{Elkies26}, that pairs of quadratic base changes of an elliptic K3 surface of Mordell--Weil rank
$17$ give a family of rank at least $19$ over an elliptic curve of positive rank, and hence infinitely many curves
of rank at least $19$. The author's preprint \cite{Jaswal} describes a construction by the same mechanism for the
torsion $\Z/2\Z\times\Z/6\Z$: quadratic twists of positive rank, found with averages of traces of Frobenius, become
trivial over the function field of a curve of genus one and rank one, and Silverman's theorem gives the
specialization; for \cref{thm:main} we use congruences instead.

\cref{thm:main} bears on the heuristic of Park, Poonen, Voight and Wood \cite{PPVW}, which suggests that ranks
of elliptic curves over $\Q$ are bounded. Adapted to prescribed torsion as in \cite[\S 8.3]{PPVW}, the argument of \cite[Theorem~7.3.3]{PPVW} predicts that if the number of
curves with torsion $T$ and height at most $H$ is of order $H^{1/d}$, as Harron and Snowden prove
\cite[Theorem~1.1 and \S 1.2]{HarronSnowden}, then $H^{1/d-(r-1)/24+o(1)}$ of them have rank at least $r$ when $1\le r<1+24/d$, at most $H^{o(1)}$
when $r=1+24/d$, and only finitely many when $r>1+24/d$. This suggests that $b_T\le 1+\lfloor 24/d\rfloor$. For $T=\Tf$ one has $d=6$, and the table of \cite[\S 8.3]{PPVW} lists the rank bound
$5$ next to the known lower bound $5$. Dujella and Peral noted that the two bounds coincide for this group and for
$\Z/6\Z$, $\Z/8\Z$ and $\Z/2\Z\times\Z/6\Z$ \cite{DP19,DP21}, and Dujella \cite{Duj23} that these lower bounds are then best
possible if the heuristic is correct.

\cref{thm:main} gives $b_T\ge 6$ for $T=\Tf$, so for this group the known lower bound $5$ is not best
possible, and $b_T$ exceeds the bound $5$ that the heuristic suggests. Our curves
form a sparse family, however: they are parametrized by the rational points of $C$, so
by N\'eron's count of points of bounded height \cite[Ch.~III, Th\'eor\`eme~4 and Corollaire~2]{Neron52} only
$(\log H)^{1/2+o(1)}$ of them have height at most $H$ (\cref{rem:height}). This count is within the model's bound of $H^{o(1)}$ curves of rank at least $5$, but the model
predicts only finitely many curves of rank at least $6$: what fails is the table's prediction for the family of all
curves with torsion $\Tf$. Park, Poonen, Voight and Wood note that special families of elliptic curves may behave
differently for arithmetic reasons \cite[Remark~1.1.2]{PPVW}, and, as Elkies pointed out (see
\cite[Remark~1.1.3]{PPVW}), their model would have to be refined to predict the rank distribution in parametric
families of high generic rank such as ours. Dujella, Kazalicki and Peral had suggested, from root numbers in
families of rank $3$, that the heuristic needs some adjustment for $\Z/8\Z$ and $\Z/2\Z\times\Z/6\Z$ \cite{DKP21};
for $\Z/2\Z\times\Z/6\Z$, where the table of \cite[\S 8.3]{PPVW} lists the rank bound $3$, the author's preprint \cite{Jaswal} gives a family of rank at
least $4$.

\cref{thm:main} follows from the more precise \cref{thm:precise}. Let
\begin{equation}\label{eq:ab}
a=7(m-99)(m+1)(m-11)(m+9),\qquad b=-7(m+99)(m-1)(m+11)(m-9),
\end{equation}
and let
\begin{equation}\label{eq:X}
X\colon\ y^2=x(x+a^2)(x+b^2)
\end{equation}
over $\Q(m)$, the family of Dujella and Peral with $\tau=1/7$ in other coordinates (see below). Its fiber $X_{m_0}$ at
$m_0\in\Q$ is the curve $E_{a(m_0),b(m_0)}$, an elliptic curve with torsion
containing $\Tf$ unless $m_0\in\{0,\pm1,\pm9,\pm11,\pm99\}$ (\cref{sec:family}). Let $C$ be the affine curve
\begin{equation}\label{eq:C}
C\colon\quad u^2=5(3m^2+242),\qquad v^2=5(2m^2+243)
\end{equation}
in the coordinates $(m,u,v)$. Three points $P_1,P_2,P_3$ of $X$ are defined over $\Q(m)$, and three more,
$P_4,P_5,P_6$, over the function field $\Q(C)$ (\cref{lem:sections} and \cref{prop:points}).

\needspace{8\baselineskip}
\begin{theorem}\label{thm:precise}
\begin{enumerate}[(a)]
\item The smooth projective model $\overline C$ of $C$ has genus one and infinitely many rational points. It is
isomorphic over $\Q$ to the elliptic curve
\[
E_C\colon y^2=x^3+x^2-137620x-10504000,
\]
of conductor $9240$ and rank one, and $C(\Q)=\overline C(\Q)$.
\item The six points $P_1,\dots,P_6\in X(\Q(C))$ of \cref{prop:points} are linearly independent, and
$\rank X(\Q(C))=6$.
\item At every $c\in C(\Q)$ except the sixteen points $(\pm1,\pm35,\pm35)$ and $(\pm99,\pm385,\pm315)$ the fiber
$X_{m(c)}$ is an elliptic curve with torsion subgroup containing $\Tf$. For all but finitely many $c\in C(\Q)$ the
torsion subgroup of $X_{m(c)}(\Q)$ is $\Tf$ and the rank of $X_{m(c)}(\Q)$ is at least $6$.
\end{enumerate}
\end{theorem}

For example, $c_0=(7401/2399,\ 88235/2399,\ 86835/2399)\in C(\Q)$ gives, after scaling, the curve $E_{a_0,b_0}$ with
$(a_0,b_0)=(7037226227200,\ 3329722383911)$, which has torsion $\Tf$ and rank exactly $6$
(\cref{prop:indep} and \cref{sec:proof}).

The family $X$ is the one-parameter family that Dujella and Peral obtained in \cite[\S 7.2]{DP21}
from their model with two parameters $\alpha$ and $\tau$ by making $\tau^2+\alpha^2+2$ a square through a parameter $t$
and fixing $\tau=1/7$, written in other coordinates
(\cref{lem:dp}). The model goes back
to Dujella's study of the curves induced by Diophantine triples \cite[\S 4]{Duj07}, where it is used at $\alpha=2$, and
appears as \cite[(5)]{DP14} and \cite[(13)]{DP21}. Dujella and Peral chose $\tau=1/7$, motivated by experiments and by the
splitting of one coefficient of their model into seven linear factors. On this family they found a curve of rank $9$,
at $t=900/781$ in their coordinate (see \cref{lem:dp}), and seven curves of rank $8$ \cite[\S 7.2]{DP21}; one of
these, at $t=245/138$, they had found before \cite[\S 3.2]{DP19}. They do not determine the rank of the
family over $\Q(t)$; we show that it is exactly $3$ (\cref{prop:rank3}).

\subsection*{\texorpdfstring{How the curve $C$ was found}{How the curve C was found}} The polynomials $a$ and $b$ satisfy $a(-m)=-b(m)$ and $b(-m)=-a(m)$, so the
coefficients of \eqref{eq:X} are polynomials in $n=m^2$, and $X$ is the base change to $\Q(m)=\Q(n)(\sqrt n)$ of an
elliptic surface $Y$ over the $n$-line, an elliptic K3 surface with a point of order $4$ (\cref{lem:Y}). For a twist
$Y^q$ of $Y$ and a prime $p$, let $A_p(Y^q)$ be the average
over $n_0\in\F_p$ of the traces of Frobenius of the fibers of $Y^q$ modulo $p$ (with trace $0$ at the singular fibers).
Nagao \cite{Nagao} conjectured that $-\frac1x\sum_{p\le x}A_p\log p$ tends to the rank over $\Q(n)$, and Rosen and
Silverman showed that Tate's conjecture for the surface implies it if, in addition, the $L$-function $L_2$ of the surface has no zeros on the line $\operatorname{Re}s=2$ \cite[Theorem~1.3]{RosenSilverman}. As a heuristic guide in the search
for twists of positive rank, we computed
plain averages of $-A_p(Y^q)$ over primes $p$ for the twists by many linear polynomials in $n$ and by their products
with $n$ (\cref{sec:computations}). For $Y$ and $Y^n$ they are $1.00$ and $1.99$, in line with $\rank X(\Q(m))=3$, and
only three other twists of this kind have averages near $1$: the twists by
$q_2=5(2n+243)$, by $nq_1$ with $q_1=5(3n+242)$, and by $n(4n+45)$, with averages $1.04$, $1.04$ and $1.06$. The first
two are exchanged by the automorphism $n\mapsto 9801/n$ of $Y$, and the twist by $q_1q_2/25=(3n+242)(2n+243)$ has
average $1.00$. The third leads to no better base: over the curves of genus one on which $4n+45$ and one of $q_1$, $q_2$ become
squares, $X$ has rank only $5$, and the curve on which all three become squares has genus $5$
(\cref{sec:conic}).

The curve $C$ is the cover of the $m$-line on which $q_1$ and $q_2$ become squares; over its function field the
twists of $Y$ by $q_2$, $nq_1$ and $q_1q_2$ become trivial, and each contributes a point (\cref{lem:twists}). Over
the conic on which only $q_1$ becomes a square, $X$ has rank at least $4$, and Dujella and Peral's curve of rank $9$
is a fiber of $X$ over this conic (\cref{prop:conic}). The twist by $h=99n^2-9998n+970299$ has average $1.93$; the
polynomial $h$ comes from the image, under the symmetry $(\alpha,\tau)\mapsto(\alpha,1/\tau)$ of their model, of
Dujella and Peral's condition that $\alpha^2\tau^2+2\alpha^2+1$ be a square (\cref{sec:conic}). Over the curve of
genus one on which $h$ becomes a square, the twist of $X$ by $h$, of rank $2$, becomes trivial, and $X$ has rank
exactly $5$ (\cref{prop:G}). So the choice of the base curve is what lifts the rank to $6$.

\subsection*{A second family} The family of \cref{sec:second} comes from the model of Dujella and Peral on the line
$\alpha=33/19$, a value at which a quadratic twist of the model acquires two independent sections, found by an exact computation. Its base
is a curve of genus one and rank two, over whose function field the family acquires six independent points: one on
the family itself; two on its quadratic twist by $29\bigl(\tau^2-(161/627)^2\bigr)$, together with their images
under the involution $\tau\mapsto1/\tau$ of the base line; and one on its twist by
$-\bigl(\tau^2-(161/627)^2\bigr)\bigl(\tau^2-(627/161)^2\bigr)$, whose branch points are exchanged by the involution
(\cref{thm:second}). This family gives a second proof of \cref{thm:main}.

\subsection*{Formalization} All the numbered results of this paper are formalized in Lean~4 with \mbox{Mathlib}~\cite{Lean}. The formal proofs follow the arguments of \crefrange{sec:family}{sec:second} and prove the computations they use; where they differ from the paper, the documentation says how. Fourteen of the results, including \cref{thm:main,thm:precise,thm:second}, take as hypotheses theorems of other authors in arithmetic and algebraic geometry, stated in general form and listed with their sources in the formalization.

\section{The family of Dujella and Peral}\label{sec:family}

The curve $X$ is the family of Dujella and Peral with $\tau=1/7$, which a symmetry of their model carries to the line
$\alpha=7$ used here (\cref{lem:dp} and the paragraph after it). It has rank exactly $3$ over $\Q(m)$ (\cref{prop:rank3}).

Dujella and Peral \cite[\S 6.7]{DP21} consider the curves $y^2=(\xi x+1)(1-x/\xi)(\zeta x+1)$ induced by the
Diophantine triples $\{\xi,-1/\xi,\zeta\}$ with $\xi=(\alpha\tau+1)/(\tau-\alpha)$ and
$\zeta=4\alpha\tau/\bigl((\alpha\tau+1)(\tau-\alpha)\bigr)$. After a change of coordinates these are
\begin{equation}\label{eq:DP}
y^2=x\bigl(x+(A_+B_+)^2\bigr)\bigl(x+(A_-B_-)^2\bigr),\qquad A_\pm=\tau\pm\alpha,\ B_\pm=1\pm\alpha\tau,
\end{equation}
which, multiplied out, is their model \cite[(13)]{DP21} and \cite[(5)]{DP14}:
\begin{multline*}
y^2=x^3+2(\alpha^2+\tau^2+4\alpha^2\tau^2+\alpha^4\tau^2+\alpha^2\tau^4)x^2\\
+(\tau+\alpha)^2(\alpha\tau-1)^2(\tau-\alpha)^2(\alpha\tau+1)^2x.
\end{multline*}
Asking that $x=(\tau+\alpha)^2(\alpha\tau-1)(\alpha\tau+1)$, respectively $x=(\tau+\alpha)(\alpha\tau-1)^2(\tau-\alpha)$,
be the abscissa of a point leads to their conditions, because at these divisors of the coefficient of $x$ the right side
of \eqref{eq:DP} is a square times $\tau^2+\alpha^2+2$, respectively $\alpha^2\tau^2+2\alpha^2+1$:
\[
\tau^2+\alpha^2+2=\square\quad\text{\cite[(14)]{DP21}},\qquad\qquad \alpha^2\tau^2+2\alpha^2+1=\square\quad\text{\cite[(16)]{DP21}}.
\]
At $\alpha=7$ the second condition reads $49\tau^2+99=\square$, and $\tau=(99-m^2)/(14m)$ gives
$49\tau^2+99=\bigl((m^2+99)/(2m)\bigr)^2$. Then
\[
\begin{aligned}
14m\,A_+&=-(m-99)(m+1), & 14m\,B_+&=-7(m-11)(m+9),\\
14m\,A_-&=-(m+99)(m-1), & 14m\,B_-&=7(m+11)(m-9),
\end{aligned}
\]
so that $a=(14m)^2A_+B_+$ and $b=(14m)^2A_-B_-$ in \eqref{eq:ab}, and $X$ is \eqref{eq:DP} at $\alpha=7$, $\tau=(99-m^2)/(14m)$
after the substitution $x\mapsto(14m)^4x$. One has
\[
a-b=14(m^4-2m^2+9801),\qquad a+b=-1400\,m(m^2-99),
\]
so $ab(a^2-b^2)$ vanishes at $m_0\in\Q$ only for $m_0\in\{0,\pm1,\pm9,\pm11,\pm99\}$. The family of
\cite[\S 7.2]{DP21} uses the first condition instead: writing
\[
\tau^2+\alpha^2+2=\bigl((\tau^2+1)/\tau+(\alpha-1/\tau)t\bigr)^2
\]
gives $\alpha=(t^2-2(\tau^2+1)t+1)/(\tau(t^2-1))$, and they take $\tau=1/7$, so that
\[
\alpha_t:=(49t^2-100t+49)/(7(t^2-1)).
\]

\needspace{4\baselineskip}
\begin{lemma}\label{lem:dp}
Let $m=(t+1)/(t-1)$. Then
\[
a=\frac{112\,t(5t-4)(5t-6)(49t-50)}{(t-1)^4},\qquad b=\frac{112\,(4t-5)(6t-5)(50t-49)}{(t-1)^4},
\]
and $A_+B_+(\alpha_t,\tfrac17)=4a'/(343(t^2-1)^2)$, $A_-B_-(\alpha_t,\tfrac17)=-4b'/(343(t^2-1)^2)$, where
$a'=t(5t-4)(5t-6)(49t-50)$ and $b'=(4t-5)(6t-5)(50t-49)$. Hence $a=\rho\cdot A_+B_+(\alpha_t,\tfrac17)$ and
$b=-\rho\cdot A_-B_-(\alpha_t,\tfrac17)$ with $\rho=(98(t+1)/(t-1))^2$, and $(x,y)\mapsto(\rho^2x,\rho^3y)$ is an
isomorphism over $\Q(t)=\Q(m)$ from the curve \eqref{eq:DP} at $(\alpha_t,1/7)$, the family of \cite[\S 7.2]{DP21}, to
$X$.
\end{lemma}

\begin{proof}
These are identities in $\Q(t)$. A curve $y^2=x(x+r^2)(x+s^2)$ depends only on $r^2$ and $s^2$, and scaling both by
$\rho^2$ is the substitution $x\mapsto\rho^2x$.
\end{proof}

\cref{lem:dp} reflects a symmetry of \eqref{eq:DP}. Write $\sigma_1=\tau^2+\alpha^2+2$ and $\sigma_2={\alpha^2\tau^2+2\alpha^2+1}$ for the
expressions in the two conditions. The map $\psi(\alpha,\tau)=(1/\tau,\alpha)$ satisfies $(A_+B_+)\circ\psi=A_+B_+/\tau^2$,
$(A_-B_-)\circ\psi=-A_-B_-/\tau^2$ and $\sigma_2\circ\psi=\sigma_1/\tau^2$, so it preserves \eqref{eq:DP} up to $x\mapsto\tau^4x$ and maps
the points where the first condition holds to points where the second holds. It maps the line $\tau=1/7$ onto the
line $\alpha=7$, and the substitution $m=(t+1)/(t-1)$ of \cref{lem:dp} gives $(99-m^2)/(14m)=\alpha_t$. Dujella and
Peral had a further reason for $\tau=1/7$ \cite[\S 7.2]{DP21}: at $\alpha=(t^2-2(\tau^2+1)t+1)/(\tau(t^2-1))$ the
coefficient of $x$ in their model has two quadratic factors in $t$, the numerators of $A_+$ and $A_-$, which both have
discriminant $8\tau^2(\tau^2+1)$, a square at $\tau=1/7$. Under $\psi$ this corresponds to the fact that
$2(\alpha^2+1)=100$ is a square at $\alpha=7$. Their curve of
rank~$9$ at $t=900/781$ is the fiber of $X$ at $m=1681/119$, and their curves of rank~$8$ at
$t=201/170,\,245/138,\,-800/459,\,1610/1417,\,1955/1754,\,2254/2215,\,2301/1670$ are the fibers at
$m=371/31,\,383/107,\,341/1259,\,3027/193,\,3709/201,\,4469/39,\,3971/631$.

The torsion points of $X$ over $\Q(m)$ include $(0,0)$, $(-a^2,0)$, $(-b^2,0)$ and the point $(ab,\,ab(a+b))$ of
order~$4$, with double $(0,0)$. The next lemma gives three points of infinite order.

\needspace{4\baselineskip}
\begin{lemma}\label{lem:sections}
The points $P_1,P_2,P_3$ with the following coordinates lie in $X(\Q(m))$:
\begin{align*}
x(P_1)&=-49(m-1)^2(m-11)^2(m+9)^2(m+99)^2,\\
x(P_1)+a^2&=-19208\,m(m^2-99)(m-11)^2(m+9)^2,\\
x(P_1)+b^2&=392\,m(m^2-99)(m-1)^2(m+99)^2,\\
y(P_1)&=-19208\,m(m^2-99)(m-1)^2(m-11)^2(m+9)^2(m+99)^2;\displaybreak[0]\\[4pt]
x(P_2)&=49(m-99)(m-1)(m+1)(m+99)(m-9)^2(m+11)^2,\\
x(P_2)+a^2&=98(m-99)(m+1)(m^2-99)(m^2+99)^2,\\
x(P_2)+b^2&=98(m-1)(m+99)(m^2-99)(m-9)^2(m+11)^2,\\
y(P_2)&=-686\,(m-99)(m-1)(m+1)(m+99)(m^2-99)\\&\qquad\cdot(m^2+99)(m-9)^2(m+11)^2;\displaybreak[0]\\[4pt]
x(P_3)&=-49(m-11)(m-9)(m+1)(m+99)(m+9)^2(m+11)^2,\\
x(P_3)+a^2&=-3920\,m(m-11)(m+1)(m+9)^2(2m-33)^2,\\
x(P_3)+b^2&=3920\,m(m-9)(m+99)(m-6)^2(m+11)^2,\\
y(P_3)&=27440\,m(m-6)(m-9)(m-11)(m+1)(m+99)\\&\qquad\cdot(2m-33)(m+9)^2(m+11)^2.
\end{align*}
\end{lemma}

\begin{proof}
In each case the product $x(x+a^2)(x+b^2)$ of the displayed factorizations is $y^2$; for instance
$(-49)(-19208)\cdot392=19208^2$.
\end{proof}

In the notation above, at $\alpha=7$, $x(P_1)=-(14m)^4\bigl((\tau-\alpha)(1+\alpha\tau)\bigr)^2$ and
$x(P_2)=(14m)^4(\tau+\alpha)(\alpha\tau-1)^2(\tau-\alpha)$. So $P_1$ is the point that every curve of the model
carries, the image of the point $(0,1)$ of $y^2=(\xi x+1)(1-x/\xi)(\zeta x+1)$ \cite[\S 6.7]{DP21} (the point $P$ of
\cite[\S 4]{Duj07}), and $P_2$ is the point whose
existence is the second condition of Dujella and Peral. The point $P_3$ comes from the splitting of $B_+$ and $B_-$ on the
line $\alpha=7$, where $\alpha^2+1=2\cdot5^2$: one has $x(P_3)=(14m)^2B_+B_-\,(m+1)(m+9)(m+11)(m+99)$, which involves
the linear factors $m+9$ of $14mB_+$ and $m+11$ of $14mB_-$.

\begin{proposition}\label{prop:rank3}
$\rank X(\Q(m))=3$, and $P_1,P_2,P_3$ are linearly independent.
\end{proposition}

\begin{proof}
\emph{Lower bound.} At $m=37/5$ the fiber is $E_{a,b}$ with $a=198746352/625$ and $b=87707648/625$. For a
curve $y^2=x(x-e_2)(x-e_3)$ with rational $e_2,e_3$ let
\[
\kappa\colon E(\Q)\to(\Q^\times/\Q^{\times2})^2,\qquad P\mapsto\bigl(x(P),\,x(P)-e_2\bigr),
\]
with the usual values at the points of order $2$; it is the $2$-descent homomorphism, with kernel $2E(\Q)$
\cite[Proposition~X.1.4]{SilvermanAEC}. Here $e_2=-a^2$, $e_3=-b^2$, and in squarefree representatives
\[
\begin{aligned}
\kappa(P_1)&=(-1,\ 5\cdot7\cdot37\cdot79), & \kappa(P_2)&=(-2\cdot3\cdot19\cdot229,\ 3\cdot79\cdot229),\\
\kappa(P_3)&=(-2\cdot3\cdot19,\ 3\cdot7\cdot37), & \kappa(-a^2,0)&=(-1,\ 5\cdot7\cdot37\cdot79\cdot20233),
\end{aligned}
\]
\[
\kappa(0,0)=(1,1),\qquad \kappa(ab,ab(a+b))=(3\cdot19\cdot23\cdot41\cdot229,\ 3\cdot5\cdot37\cdot41\cdot79\cdot229).
\]
These five nontrivial vectors are independent: $20233$ occurs only in $\kappa(-a^2,0)$, $23$ only in
$\kappa(ab,ab(a+b))$, then $229$ only in $\kappa(P_2)$, and $2$ only in $\kappa(P_3)$. The fiber has torsion
$\Tf=\langle(-a^2,0),(ab,ab(a+b))\rangle$: by Mazur's theorem \cite{Mazur} a larger
torsion subgroup would contain a point of order $8$, which would need $ab$ or $-ab$ to be a square \cite[Main Theorem~1]{Ono}, and $ab$ has squarefree part
$3\cdot19\cdot23\cdot41\cdot229$ (the first entry of $\kappa(ab,ab(a+b))$) while $-ab<0$. So $\kappa$ maps the torsion
subgroup onto $\langle\kappa(-a^2,0),\kappa(ab,ab(a+b))\rangle$. Suppose that $\sum n_iP_i$ is a torsion point with
$n_i\in\Z$ not all zero. A point whose double is a torsion point is itself one, so we may halve the $n_i$ until one of
them is odd, and then the image of $\sum n_iP_i$ gives a relation among the five vectors. So the specialized
points, and hence $P_1,P_2,P_3$, are independent (specialization at a smooth fiber is a homomorphism).

\emph{Upper bound.} In \cite[\S 2]{DP14}, Dujella and Peral proved that their curve over $\Q(t)$ with torsion
$\Z/2\Z\times\Z/4\Z$ has rank exactly $4$, with an earlier form of the criterion of Gusi\'c and Tadi\'c. We use \cite[Theorem~1.1]{GusicTadic}: for a nonconstant
curve $y^2=(x-e_1)(x-e_2)(x-e_3)$ with $e_i\in\Z[m]$ and $m_0\in\Q$ with nonsingular fiber, if $g(m_0)$ is not a square
for any nonconstant squarefree divisor $g\in\Z[m]$ of $(e_1-e_2)(e_1-e_3)$, of $(e_2-e_1)(e_2-e_3)$ or of
$(e_3-e_1)(e_3-e_2)$, then specialization at $m_0$ is injective. The proof uses that the group of points of the curve
over $\Q(m)$ is finitely generated \cite[\S 1]{GusicTadic}, which holds here by the theorem of Lang and N\'eron
\cite{LangNeron}. For $X$, whose $j$-invariant is not constant, $(e_1,e_2,e_3)=(0,-a^2,-b^2)$, and these
products are $a^2b^2$, $a^2(a^2-b^2)$ and $-b^2(a^2-b^2)$. Their nonconstant squarefree divisors are, up to sign,
products of distinct factors among the eight linear factors of $ab$ and the three polynomials $m$, $m^2-99$ and
$m^4-2m^2+9801$, times a
squarefree divisor of the content, which is $7^4$ for $a^2b^2$ and $2^4\cdot5^2\cdot7^4$ for the other two products.
At $m=14/3$ the eleven factors $m-99$, $m+1$, $m-11$, $m+9$,
$m+99$, $m-1$, $m+11$, $m-9$, $m$, $m^2-99$, $m^4-2m^2+9801$ have squarefree parts $-3\cdot283$, $3\cdot17$, $-3\cdot19$, $3\cdot41$, $3\cdot311$, $3\cdot11$, $3\cdot47$, $-3\cdot13$, $2\cdot3\cdot7$, $-5\cdot139$, $73\cdot11353$. These and $-1$, $2$, $5$,
$7$ are independent in $\Q^\times/\Q^{\times2}$, so no such divisor takes a square value at $14/3$. The fiber at
$14/3$ is isomorphic to $E_{3747769,\,2090231}$, where $3747769=17\cdot19\cdot41\cdot283$ and
$2090231=11\cdot13\cdot47\cdot311$. It has full rational $2$-torsion and a $2$-Selmer group of dimension $5$
(computed with \texttt{mwrank}~\cite{CremonaBook}), so its rank is at most $3$; hence
$\rank X(\Q(m))\le3$.
\end{proof}

\section{A K3 surface and three of its twists}\label{sec:twists}

The family $X$ is a base change of an elliptic K3 surface $Y$, and quadratic twists of $Y$ are the source of its
further points: a point on such a twist becomes a point of $X$ over any extension of $\Q(m)$ in which the twisting
polynomial is a square. \cref{lem:twists} gives points on three twists, and over the function field of the curve $C$
of \cref{sec:C} they become the points $P_4$, $P_5$ and $P_6$ of \cref{prop:points}. We follow Dujella and Peral, who raised the rank of a curve with torsion $\Z/4\Z$ over a rational function field to $6$ in this way \cite[\S 4]{DP24}: there each new point amounts to a square condition, and two such conditions with a common factor hold together over a curve of genus~$0$.

Write $n=m^2$. Since $a(-m)=-b(m)$ and $b(-m)=-a(m)$, the polynomials
\[
\begin{aligned}
S&:=a^2+b^2=98\,(n^4+9996n^3-1960394n^2+97970796n+96059601),\\
w&:=ab=-49(n-1)(n-81)(n-121)(n-9801)
\end{aligned}
\]
lie in $\Q[n]$, and $X$ is the base change to $\Q(m)$ of
\begin{equation}\label{eq:Y}
Y\colon\ y^2=x(x^2+Sx+w^2)
\end{equation}
over $\Q(n)$. For $q\in\Q(n)^\times$ write $Y^q\colon y^2=x(x^2+qSx+q^2w^2)$ for the quadratic twist of $Y$ by $q$;
a point $(x,y)\in Y^q(K)$, for a field $K\supseteq\Q(n)$ containing a square root $\sqrt q$ of $q$, gives the point
$(x/q,\,y/(q\sqrt q))\in Y(K)$.

\begin{lemma}\label{lem:Y}
\begin{enumerate}[(a)]
\item $S-2w=(a-b)^2=196(n^2-2n+9801)^2$ and $S+2w=(a+b)^2=1960000\,n(n-99)^2$. The point $(-w,\,w(a-b))\in Y(\Q(n))$
has order $4$, with double $(0,0)$.
\item The discriminant $\Delta=16w^4(S^2-4w^2)$ of \eqref{eq:Y} is a constant times
\[
(n-1)^4(n-81)^4(n-121)^4(n-9801)^4\,n\,(n-99)^2(n^2-2n+9801)^2.
\] The singular fibers of $Y$ are of type $I_4$ at
$n=1,81,121,9801$, of type $I_2$ at $n=99$ and at the two roots of $n^2-2n+9801$, and of type $I_1$ at $n=0$ and
$n=\infty$; their Euler numbers sum to $24$, and $Y$ is a relatively minimal elliptic K3 surface.
\item $S(9801/n)=(99/n)^4S(n)$ and $w(9801/n)=(99/n)^4w(n)$, so $n\mapsto9801/n$ lifts to an automorphism of $Y$.
\end{enumerate}
\end{lemma}

\begin{proof}
Parts (a) and (c) are polynomial identities, and $x=-w$ gives $x(x^2+Sx+w^2)=w^2(S-2w)$; doubling $(-w,w(a-b))$ gives $x=0$.
(b) The coefficients $S$ and $w^2$ of $x^2$ and $x$ have degrees $4$ and $8$ in $n$, so the
substitution $n=1/s$, $x=x'/s^4$, $y=y'/s^6$ gives the equation $y'^2=x'\bigl(x'^2+s^4S(1/s)x'+s^8w(1/s)^2\bigr)$ with
coefficients in $\Q[s]$ and discriminant $s^{24}\Delta(1/s)$. Since $\Delta$ has degree $23$, this discriminant vanishes
to order $1$ at $s=0$, so the fiber at $\infty$ is of type $I_1$ and contributes $1$ to the Euler number. At a zero of $w$, where $S=(a-b)^2\neq0$, the fiber is multiplicative of type $I_4$, and at the zeros
of $S^2-4w^2=(a-b)^2(a+b)^2$ of multiplicity $2$ and $1$, where the invariant $c_4=16(S^2-3w^2)$ of \eqref{eq:Y} is $16w^2\neq0$, it is of type $I_2$ and $I_1$
respectively \cite[Table~(IV.3.1)]{Miranda}. A relatively minimal elliptic surface over $\PP^1$ with a section and Euler number $24$ is a
K3 surface \cite[(III.4.2)--(III.4.6)]{Miranda}.
\end{proof}

The twists we use were found with averages of traces of Frobenius, a heuristic guide to the rank: the average
of a twist $Y^q$ is the plain average of $-A_p(Y^q)$ over the primes $300<p<3000$, with $A_p$ as in \cref{sec:intro}.
We computed it for the twists of $Y$ by the linear polynomials in $n$ of
\cref{sec:computations} and by their products with $n$. Apart from $Y$ and $Y^n$, whose averages are $1.00$ and
$1.99$, only three of these twists have averages above $0.45$, all three close to $1$: the twists by $q_2$, by
$nq_1$ and by $n(4n+45)$, where
\[
q_1=5(3n+242),\qquad q_2=5(2n+243),\qquad q_{12}=q_1q_2/25=(3n+242)(2n+243).
\]
The automorphism of \cref{lem:Y}(c) exchanges the twists by $q_2$ and $nq_1$, up to squares. Over $\Q(C)$, where
$q_1$ and $q_2$ become squares, these two twists become trivial together with the twist by $q_{12}$, whose class is
the product of theirs over $\Q(m)$; as $q_{12}$ is quadratic in $n$ and not divisible by $n$, this twist is not among those scanned, and its
average is $1.00$. The twist by $n(4n+45)$ leads to no better base: over the curves of genus one on which $4m^2+45$
and one of $q_1$, $q_2$ become squares, $X$ has rank only $5$ (\cref{sec:conic}). We found the points below by
computing generators of the twists by $q_2$, $nq_1$ and $q_{12}$ at the integers $2\le n_0\le160$ at which these
twists have rank one, and interpolating in $n$ the values of $-x/(qw)-qw/x$, where $x$ is the $x$-coordinate of the
generator on $Y^q$, for $q=q_2$, $nq_1$, $q_{12}$; these values do not change when $(0,0)$ is added to the generator,
and the interpolating functions have degree $4$.

\needspace{4\baselineskip}
\begin{lemma}\label{lem:twists}
With $N=n^2-2n+9801$, the points $(x_i,y_i)$ with the following coordinates lie on $Y^{q_2}$, $Y^{nq_1}$ and $Y^{q_{12}}$ over $\Q(n)$:
\[
\begin{aligned}
x_4&=49(n-1)(n-81)(n-121)^2(2n+243)^2 &&\text{on }Y^{q_2},\\
x_5&=49(n-9801)(n-121)(n-81)^2(3n+242)^2 &&\text{on }Y^{nq_1},\\
x_6&=49(n-9801)(n-1)(n-121)^2(2n+243)^2 &&\text{on }Y^{q_{12}},
\end{aligned}
\]
and
\[
\begin{aligned}
y_4&=-686\,(n-1)(n-36)(n-81)(n-121)^2(2n+243)^2N,\\
y_5&=-686\,(n-9801)(n-121)(4n-1089)(n-81)^2(3n+242)^2N,\\
y_6&=-1715\,(n-9801)(n-1)(n-99)(n-121)^2(2n+243)^2N.
\end{aligned}
\]
Up to sign, $(x_5,y_5)$ is the image of $(x_4,y_4)$ under the automorphism of \cref{lem:Y}(c).
\end{lemma}

\begin{proof}
Expanding gives
\[
\begin{aligned}
x_4^2+q_2Sx_4+q_2^2w^2&=9604\,(n-1)(n-81)(n-36)^2(n-121)^2(2n+243)^2N^2,\\
x_5^2+nq_1Sx_5+n^2q_1^2w^2&=9604\,(n-9801)(n-121)(n-81)^2(3n+242)^2(4n-1089)^2N^2,\\
x_6^2+q_{12}Sx_6+q_{12}^2w^2&=60025\,(n-9801)(n-1)(n-99)^2(n-121)^2(2n+243)^2N^2,
\end{aligned}
\]
so that $x_i(x_i^2+qSx_i+q^2w^2)=y_i^2$ for $q=q_2$, $nq_1$, $q_{12}$ and $i=4$, $5$, $6$ respectively. The last assertion holds since
\[
q_2(9801/n)=(9/n)^2\,nq_1(n)\qquad\text{and}\qquad x_4(9801/n)=(9/n)^2(99/n)^4\,x_5(n).
\]
\end{proof}

\section{\texorpdfstring{The curve $C$ and six points}{The curve C and six points}}\label{sec:C}

On $C$ one has $u^2=q_1(m^2)$ and $v^2=q_2(m^2)$, so the twists of \cref{lem:twists} become trivial over
$\Q(C)=\Q(m,u,v)$: $q_2=v^2$, $nq_1=(mu)^2$ and $q_{12}=(uv/5)^2$.

\begin{lemma}\label{lem:C}
Let $\overline C\subset\PP^3$ be the closure of $C$, defined by $u^2=15m^2+1210z^2$ and $v^2=10m^2+1215z^2$.
\begin{enumerate}[(a)]
\item $\overline C$ is a smooth curve of genus one, and $(m,u,v)=(1,35,35)$ is a rational point.
\item $\overline C$ is isomorphic over $\Q$ to $E_C\colon y^2=x^3+x^2-137620x-10504000$, which has conductor
$9240=2^3\cdot3\cdot5\cdot7\cdot11$ (LMFDB label 9240.bf4 \cite{LMFDB}), torsion subgroup
$\Tf$, generated by $(-190,2970)$ and
$(-80,0)$, and rank $1$; the point $(-322,726)$ has infinite order. In particular $\overline C(\Q)$ is infinite, and
$C(\Q)=\overline C(\Q)$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) The pencil $\lambda_1(u^2-15m^2-1210z^2)+\lambda_2(v^2-10m^2-1215z^2)$ consists of diagonal quadrics with
determinant $25\lambda_1\lambda_2(3\lambda_1+2\lambda_2)(242\lambda_1+243\lambda_2)$, which has four distinct roots in
$\PP^1$. So the binary quartic that \cite[\S 2]{CFS} attaches to the pair of quadrics has nonzero discriminant, and by
\cite[Theorem~2.8(ii)]{CFS} the base locus $\overline C$ is a smooth curve of genus one.
(b) The lines $u=35+k(m-1)$ through $(m,u)=(1,35)$ meet the conic $u^2=15m^2+1210$ again at
$m=(k^2-70k+15)/(k^2-15)$, $u=-5(7k^2-6k+105)/(k^2-15)$, and then $v^2=5(2m^2+243)$ becomes
\[
\eta^2=g(k):=49k^4-56k^3+514k^2-840k+11025,\qquad \eta=v(k^2-15)/5.
\]
This gives a birational map from $\overline C$ to the smooth projective model of $\eta^2=g(k)$. The discriminant of $g$
is $2^{16}\cdot3^{12}\cdot5^{2}\cdot7^{4}\cdot11^{4}\neq0$, and its leading coefficient is the square $49$, so this curve has rational
points at infinity and is isomorphic over $\Q$ to its Jacobian. The invariants of $g$ in the sense of
\cite[\S 2]{Cremona} are $I(g)=6605776$ and $J(g)=18071642752$. By \cite[Theorem~2.8(iii)]{CFS} the Jacobian is
$y^2=x^3-27c_4x-54c_6$ with $c_4=2^4I(g)$ and $c_6=2^5J(g)$, and the substitution $(x,y)\mapsto(4x,8y)$ turns this into
$y^2=x^3-27I(g)x-27J(g)$, with minimal model $E_C$. On $E_C$,
\[
3\cdot(-322,726)=\Bigl(-\frac{1901193922}{2521^2},\ \frac{31980410728794}{2521^3}\Bigr)
\]
is not $2521$-integral, whereas a torsion point of an integral Weierstrass equation has $p$-integral coordinates at
every odd prime $p$ \cite[Theorem~VII.3.4]{SilvermanAEC}; so $(-322,726)$ has infinite order. The curve $E_C$ has
full rational $2$-torsion and a $2$-Selmer group of dimension $3$ (computed with \texttt{mwrank}
\cite{CremonaBook}), so its rank is at most $1$, hence $1$. Finally, a rational point of $\overline C$ with $z=0$ would need
$u^2=15m^2$ and $v^2=10m^2$ with $m\neq0$, and $15$ and $10$ are not squares.
\end{proof}

\begin{proposition}\label{prop:points}
With $n=m^2$, $q_1=u^2$, $q_2=v^2$ and $x_i,y_i$ as in \cref{lem:twists}, the points
\[
P_4=\Bigl(\frac{x_4}{q_2},\ \frac{y_4}{q_2v}\Bigr),\qquad P_5=\Bigl(\frac{x_5}{nq_1},\ \frac{y_5}{nq_1\,mu}\Bigr),
\qquad P_6=\Bigl(\frac{x_6}{q_{12}},\ \frac{y_6}{q_{12}\,uv/5}\Bigr),
\]
together with $P_1,P_2,P_3$ of \cref{lem:sections}, lie in $X(\Q(C))$. Explicitly,
\[
\begin{aligned}
x(P_4)&=\tfrac{49}{5}(m^2-1)(m^2-81)(m^2-121)^2(2m^2+243),\\
x(P_5)&=\tfrac{49}{5}\,m^{-2}(m^2-9801)(m^2-121)(m^2-81)^2(3m^2+242),\\
x(P_6)&=49\,(m^2-9801)(m^2-1)(m^2-121)^2(2m^2+243)/(3m^2+242).
\end{aligned}
\]
\end{proposition}

\begin{proof}
This follows from \cref{lem:twists} and the map $(x,y)\mapsto(x/q,\,y/(q\sqrt q))$ defined before \cref{lem:Y},
applied with $\sqrt{q_2}=v$, $\sqrt{nq_1}=mu$ and $\sqrt{q_{12}}=uv/5$.
\end{proof}

\needspace{4\baselineskip}
\begin{remark}\label{rem:eigen}
The six points lie in four different eigenspaces of $X(\Q(C))\otimes\Q$. The extension $\Q(C)/\Q(m)$ is Galois with group
$(\ZZ2)^2$, generated by the sign changes of $u$ and $v$, and $X(\Q(C))\otimes\Q$ is the sum of the eigenspaces of its
four characters, $X(\Q(m))\otimes\Q$ and the parts on which only $u\mapsto-u$, only $v\mapsto-v$, or both act by $-1$. The points
$P_1,P_2,P_3$ lie in the first, $P_5$ in the second, $P_4$ in the third and $P_6$ in the fourth.
\cref{prop:indep} proves their independence directly.
\end{remark}

\section{\texorpdfstring{Independence and the rank over $\Q(C)$}{Independence and the rank over Q(C)}}\label{sec:indep}

\begin{proposition}\label{prop:indep}
The points $P_1,\dots,P_6$ of \cref{prop:points} are linearly independent in $X(\Q(C))$.
\end{proposition}

\begin{proof}
We specialize at one rational point of $C$ and show that the six specialized points are independent by reducing
them modulo six primes. Each reduction gives a homomorphism to $\ZZ3$ that kills the torsion, whose order is prime
to $3$, and the six homomorphisms, evaluated at the six points, give a matrix that is invertible over $\F_3$.

Let $c_0=(m_0,u_0,v_0)=(7401/2399,\ 88235/2399,\ 86835/2399)\in C(\Q)$, a point found by a search of the points
of $C(\Q)$ of small height (\cref{sec:computations}). The fiber
$X_{m_0}$ is smooth, all coordinates
of the $P_i$ are regular at $c_0$ (their denominators are products of powers of $5$, $m$, $u$ and $v$), and
specialization at $c_0$ is a group homomorphism $X(\Q(C))\to X_{m_0}(\Q)$, since
the Weierstrass model is smooth over the local ring of $\overline C$ at $c_0$. So it suffices that the points
$P_i(c_0)$ are independent. Here
\[
a(m_0)=8689566945346560000/2399^4,\qquad b(m_0)=4111541199653302800/2399^4,
\]
and the scaling $(x,y)\mapsto(\nu^2x,\nu^3y)$ with $\nu=2399^4/1234800$, which makes $a$ and $b$ coprime integers,
maps $X_{m_0}$ to $E_0:=E_{a_0,b_0}$, where
\[
\begin{aligned}
a_0&=7037226227200=2^9\cdot5^2\cdot13\cdot47\cdot59\cdot101\cdot151,\\
b_0&=3329722383911=7^2\cdot11\cdot17\cdot31\cdot41\cdot43\cdot61\cdot109.
\end{aligned}
\]
Let $R_i$ be the image of $P_i(c_0)$, with the $y$-coordinates of \cref{lem:sections} and \cref{prop:points} and
with $u_0,v_0>0$. Then
\[
\begin{aligned}
x(R_1)&=-9134182662998624112839888896/625,\\
x(R_2)&=-20409047029579692378895000,\\
x(R_3)&=-19944153128696423007969280,\\
x(R_4)&=-80466963140208120656325184314880/5755201,\\
x(R_5)&=6804553116957406970610499537448960/6086089,\\
x(R_6)&=-224941667455133109155372655848800/6355441.
\end{aligned}
\]
The torsion subgroup of $E_0(\Q)$ contains $\Tf$, so by Mazur's theorem \cite{Mazur} its order is $8$ or $16$, prime to $3$,
so it lies in $3E_0(\Q)$. For a prime $p$ of good reduction at which $3$
divides $\#\widetilde E_0(\F_p)$ and the $3$-primary part of $\widetilde E_0(\F_p)$ is cyclic,
$P\mapsto(\#\widetilde E_0(\F_p)/3)\,\widetilde P$ is a homomorphism $E_0(\Q)\to\widetilde E_0(\F_p)[3]\cong\ZZ3$ with
kernel containing $3E_0(\Q)$, so it kills the torsion subgroup. The primes
\[
p=23,\ 37,\ 53,\ 83,\ 89,\ 113,\qquad \#\widetilde E_0(\F_p)=24,\ 48,\ 48,\ 96,\ 96,\ 96,
\]
are of this kind: they are odd and do not divide $a_0b_0(a_0^2-b_0^2)$, and since $9\nmid\#\widetilde E_0(\F_p)$ the
$3$-primary part is cyclic. With respect to a generator of $\widetilde E_0(\F_p)[3]$ (so up to sign), the images of
$R_1,\dots,R_6$ at the $j$-th of these primes are the entries of the $j$-th row of
\[
M=\begin{pmatrix}
1&0&1&1&1&0\\ 1&1&0&0&1&1\\ 2&0&1&0&0&2\\ 2&2&1&1&0&1\\ 2&0&2&2&1&0\\ 2&0&2&1&1&0
\end{pmatrix},\qquad \det M\equiv1\pmod 3.
\]
We computed $M$ in PARI by reducing the $R_i$ modulo $p$ (the denominators of the $x(R_i)$ are prime to the six primes),
multiplying them by $\#\widetilde E_0(\F_p)/3$ and comparing the results with the points of order $3$ of
$\widetilde E_0(\F_p)$, which have $x$-coordinate $9$, $11$, $40$, $75$, $79$ and $4$ at $p=23$, $37$, $53$, $83$,
$89$ and $113$. Up to the signs of its rows and columns, which change only the sign of $\det M$, the matrix $M$ is
determined by the printed $x(R_i)$.
If $\sum_in_iR_i$ were a torsion point for some nonzero $(n_i)\in\Z^6$, we could divide the $n_i$ by their greatest common divisor and assume $\gcd(n_i)=1$. Then
$(n_i\bmod3)$ is a nonzero column vector with $M\,(n_i\bmod3)=0$, which is impossible. So $R_1,\dots,R_6$ are independent, and so are $P_1,\dots,P_6$.
\end{proof}

\needspace{4\baselineskip}
\begin{proposition}\label{prop:rank6}
For $q\in\Q(m)^\times$ let $X^q\colon y^2=x(x+qa^2)(x+qb^2)$ be the quadratic twist of $X$ by $q$. With $n=m^2$,
each of the twists $X^{q_1}$, $X^{q_2}$ and $X^{q_{12}}$ has rank $1$ over $\Q(m)$. Hence $\rank X(\Q(C))=6$, and
$P_1,\dots,P_6$ generate a subgroup of finite index in $X(\Q(C))$.
\end{proposition}

\begin{proof}
Since $\Q(C)=\Q(m)(u,v)$ with $u^2=q_1$, $v^2=q_2$ and $q_1q_2=25q_{12}$, the decomposition of
\cref{rem:eigen} gives
\[
\rank X(\Q(C))=\rank X(\Q(m))+\rank X^{q_1}(\Q(m))+\rank X^{q_2}(\Q(m))+\rank X^{q_{12}}(\Q(m)),
\]
because the part of $X(\Q(C))\otimes\Q$ on which $u\mapsto-u$ acts by $-1$ and $v\mapsto-v$ trivially is the image of
$X^{q_1}(\Q(m))\otimes\Q$ under $(x,y)\mapsto(x/q_1,\,y/(q_1u))$, and similarly for the other two parts. As
$\rank X(\Q(m))=3$ (\cref{prop:rank3}) and $\rank X(\Q(C))\ge6$ (\cref{prop:indep}), the ranks of the three twists
add up to at least $3$, so it suffices to show that each of them has rank at most $1$ over $\Q(m)$: then each has
rank exactly $1$, and $\rank X(\Q(C))=6$.

We apply \cite[Theorem~1.1]{GusicTadic} as in \cref{prop:rank3} to $X^q$, with
$(e_1,e_2,e_3)=(0,-qa^2,-qb^2)$, for $q=q_1,q_2,q_{12}\in\Z[m]$; the curve $X^q$ has the same nonconstant $j$-invariant
as $X$. The three products are now $q^2a^2b^2$, $q^2a^2(a^2-b^2)$ and $-q^2b^2(a^2-b^2)$. Their irreducible factors
are the eleven of \cref{prop:rank3} and those of $q$, which are among $3m^2+242$ and $2m^2+243$, and their contents
involve only the primes $2$, $5$ and $7$. For $X^{q_1}$ we take $m=14/3$, where the eleven factors have the squarefree
parts listed in \cref{prop:rank3} and $3m^2+242$ has squarefree part $2\cdot3\cdot461$. For $X^{q_2}$ and
$X^{q_{12}}$ we take $m=50/3$, where the eleven factors, in the same order, have squarefree parts $-3\cdot13\cdot19$, $3\cdot53$, $3\cdot17$, $3\cdot7\cdot11$, $3\cdot347$, $3\cdot47$, $3\cdot83$, $3\cdot23$, $2\cdot3$, $1609$ and $113\cdot241\cdot257$, and those
of $2m^2+243$ and $3m^2+242$ are $7187$ and $2\cdot3\cdot1613$. In each case these classes and $-1$, $2$, $5$, $7$ are independent in
$\Q^\times/\Q^{\times2}$, so no nonconstant squarefree divisor of the three products takes a square value, which is the
hypothesis of \cite[Theorem~1.1]{GusicTadic}. The fibers of $X^{q_1}$ at $14/3$ and of $X^{q_2}$ and $X^{q_{12}}$
at $50/3$ have full rational $2$-torsion and $2$-Selmer groups of dimension $3$ (computed with
\texttt{mwrank}~\cite{CremonaBook}); as the $2$-torsion accounts for $2$ of these dimensions, their ranks are at
most $1$, and that theorem gives $\rank X^q(\Q(m))\le1$.

Finally, $X(\Q(C))$ is finitely generated \cite{LangNeron}, so its subgroup generated by the six independent points,
which has rank $6$, has finite index.
\end{proof}

\section{Proof of the theorems}\label{sec:proof}

We prove \cref{thm:precise} first, with the theorems of Silverman and Faltings, and then \cref{thm:main},
by congruences that carry the computation at $c_0$ to infinitely many points of $C(\Q)$ (\cref{prop:cong}).
The torsion of the fibers is controlled by the following criterion, whose proof follows that of
\cite[Main Theorem~1]{Ono}.

\begin{lemma}\label{lem:tors}
Let $a,b\in\Q$ with $ab(a^2-b^2)\neq0$. If neither $ab$ nor $-ab$ is a square, then $E_{a,b}(\Q)_\tors\cong\Tf$.
\end{lemma}

\begin{proof}
By Mazur's theorem \cite{Mazur} the only larger possibility is $\Z/2\Z\times\Z/8\Z$, which needs a point of order $8$, that is, a half of a point of order $4$.
On $E_{a,b}$ the points of order $4$ are the halves of the points of order $2$. The map $\kappa$ of
\cref{prop:rank3} (with $e_1=0$) has first component $x$ and kills $2E_{a,b}(\Q)$. At $(-a^2,0)$ and $(-b^2,0)$
its first component is $-a^2$ or $-b^2$, which is negative and so not a square, so these points have no rational
halves. The halves of $(0,0)$ are the points $(ab,\pm ab(a+b))$ and $(-ab,\pm ab(a-b))$, at which the first component
of $\kappa$ is $ab$ or $-ab$, and a half of such a point exists only if this component is a square. So $\Z/2\Z\times\Z/8\Z$ requires $ab$ or $-ab$ to be a square.
\end{proof}

\begin{proof}[Proof of \cref{thm:precise}]
Part (a) is \cref{lem:C}, and part (b) is \cref{prop:indep,prop:rank6}.

(c) By \cref{sec:family}, the fiber at $c\in C(\Q)$ is an elliptic curve with torsion containing $\Tf$ unless
$m(c)\in\{0,\pm1,\pm9,\pm11,\pm99\}$, and of these values only $\pm1$ and $\pm99$ occur on $C$, at the sixteen points
listed ($m(c)=\infty$ does not occur, by \cref{lem:C}).

The $j$-invariant of $X$ is $256(a^4-a^2b^2+b^4)^3/\bigl(a^4b^4(a^2-b^2)^2\bigr)$, a quotient of polynomials in $m$
of degrees $48$ and $46$ that are coprime because $a$ and $b$ are; so it has degree $48$ in $m$. As $m$ is a
nonconstant function on $C$, $X/\Q(C)$ is not isotrivial, and its $\Q(C)/\Q$-trace is zero. Silverman's
specialization theorem \cite[Theorem~C]{Silverman83} says that for a family of abelian varieties over a smooth
projective curve, all defined over a number field, whose generic fiber has trace zero, specialization is injective at
all but finitely many rational points of the curve with smooth fiber. We apply it to the Kodaira--N\'eron model of $X$ over
$\overline C$, whose sections are the points of $X(\Q(C))$ and whose fiber at $c$ is $X_{m(c)}$ when that curve is
smooth. So the specialization map $X(\Q(C))\to X_{m(c)}(\Q)$ is injective for all but finitely many
$c\in C(\Q)$ with smooth fiber, and at those $c$, $\rank X_{m(c)}(\Q)\ge6$ by \cref{thm:precise}(b).

It remains to show that the torsion is exactly $\Tf$ for all but finitely many $c$. By \cref{lem:tors}, a
larger torsion subgroup requires $ab(m(c))$ or $-ab(m(c))$ to be a nonzero square. On $\overline C$ the function
\[
ab=-49(m^2-1)(m^2-81)(m^2-121)(m^2-9801)
\]
has $32$ simple zeros, the four points above each of its eight rational roots, and poles of the even order $8$ at
the four points above $m=\infty$. Neither these roots nor $\infty$ is a branch point of $m\colon\overline C\to\PP^1$,
whose branch points are the roots of $(3m^2+242)(2m^2+243)$. So each of the double covers $r^2=\pm ab$
of $\overline C$ is branched at exactly $32$ points and has genus $17$, by the Riemann--Hurwitz formula. By Faltings'
theorem \cite[Satz~7]{Faltings} each has finitely many rational points, so $\pm ab(m(c))$ is a square for only finitely
many $c\in C(\Q)$.
\end{proof}

We have not determined the finite set of points $c\in C(\Q)$ at which the specialization map is not injective or the
torsion subgroup is larger than $\Tf$: the constants in Silverman's theorem can be computed in principle but are too
large to be practical \cite[\S 1]{GusicTadic}, and Faltings' theorem is not effective. At $c_0$ the six points stay independent (\cref{prop:indep}) and the
torsion subgroup is $\Tf$: since $13\,\|\,a_0b_0$ and $-a_0b_0<0$, neither $a_0b_0$ nor $-a_0b_0$ is a square
(\cref{lem:tors}).
The rank of $E_{a_0,b_0}(\Q)$ is exactly $6$: \texttt{mwrank}~\cite{CremonaBook} gives
this curve a $2$-Selmer group of dimension $8$, and its $2$-torsion subgroup is $(\ZZ2)^2$, so the rank is at most $6$.

The computation at $c_0$ carries over, by congruences, to an infinite set of rational points of~$C$.

\begin{proposition}\label{prop:cong}
Let $c_0=(m_0,u_0,v_0)$ be the point of \cref{prop:indep}, and let $\mathcal S$ be the set of points
$(m,u,v)\in C(\Q)$ such that $\operatorname{ord}_{13}(m-m_0)\ge2$ and, for each of the six primes $p=23,37,53,83,89,113$ of
\cref{prop:indep}, the coordinates $m$, $u$, $v$ are $p$-integral and congruent to $m_0$, $u_0$, $v_0$
modulo $p$. Then $\mathcal S$ is infinite. For every $c=(m,u,v)\in\mathcal S$ the fiber $X_m=X_{m(c)}$ is an elliptic curve
with $X_m(\Q)_\tors\cong\Tf$, and the specializations $P_1(c),\dots,P_6(c)$ of the points of \cref{lem:sections}
and \cref{prop:points} are linearly independent in $X_m(\Q)$, so that $\rank X_m(\Q)\ge6$.
\end{proposition}

\begin{proof}
\emph{Torsion.} At $m_0=7401/2399$ the eight linear factors $m\pm1$, $m\pm9$, $m\pm11$, $m\pm99$ of
$ab=-49(m^2-1)(m^2-81)(m^2-121)(m^2-9801)$ are $13$-adic units except $m-99$, since $m_0\equiv8\pmod{13}$, and
$\operatorname{ord}_{13}(m_0-99)=1$, since $m_0-99=-2^2\cdot3\cdot5^2\cdot13\cdot59/2399$. If $\operatorname{ord}_{13}(m-m_0)\ge2$, then
$\operatorname{ord}_{13}(m-99)=1$, and the other seven factors are congruent modulo $13^2$ to their values at $m_0$, so they are
$13$-adic units and $\operatorname{ord}_{13}(ab(m))=1$. In particular $ab(m)\neq0$, and neither $ab(m)$ nor $-ab(m)$ is a square.

\emph{Good reduction.} Let $p$ be one of the six primes. The numbers $m_0$, $3m_0^2+242$, $a(m_0)$, $b(m_0)$,
$a(m_0)-b(m_0)$ and $a(m_0)+b(m_0)$ are $p$-adic units, and $u_0$, $v_0$ are $p$-integral. Since $a,b\in\Z[m]$ and
$m\equiv m_0\pmod p$, the number $a(m)b(m)(a(m)^2-b(m)^2)$ is a $p$-adic unit. So $X_m$ is an elliptic curve, the
model $y^2=x(x+a(m)^2)(x+b(m)^2)$ has good reduction at $p$, and its reduction is that of $X_{m_0}$. By the first part
and \cref{lem:tors}, $X_m(\Q)_\tors\cong\Tf$.

\emph{The six points.} The coordinates of $P_1$, $P_2$, $P_3$ are polynomials in $m$ with integer coefficients, and
those of $P_4$, $P_5$, $P_6$ lie in $\Z[1/5,1/m,1/(3m^2+242)][m,u,v]$. For the $x$-coordinates this can be read off
\cref{prop:points}. For the $y$-coordinates, $u^2=q_1$ and $v^2=q_2$ give $y(P_4)=y_4v/q_2^2$,
$y(P_5)=y_5u/(m^3q_1^2)$ and $y(P_6)=y_6uv/(5q_{12}^2)$, and $y_4$, $y_5$, $y_6$ of \cref{lem:twists} are
divisible by $(2n+243)^2$, $(3n+242)^2$ and $(2n+243)^2$ respectively. So for $c\in\mathcal S$ the points $P_i(c)$
have $p$-integral coordinates, congruent modulo $p$ to those of $P_i(c_0)$, and they have the same reductions as the
$P_i(c_0)$.

\emph{Independence.} The scaling $\nu=2399^4/(2^4\cdot3^2\cdot5^2\cdot7^3)$ of \cref{prop:indep} is a
$p$-adic unit, so it induces an isomorphism from the reduction of $X_{m_0}$ to that of $E_0$, which carries the
reductions of the $P_i(c_0)$ to those of the $R_i$. Hence the six homomorphisms $X_m(\Q)\to\ZZ3$ defined as in the
proof of \cref{prop:indep}, one for each prime, kill the torsion subgroup and take on
$P_1(c),\dots,P_6(c)$ the values given by the matrix $M$. Since $\det M\not\equiv0\pmod3$, the points
$P_1(c),\dots,P_6(c)$ are independent, as there.

\emph{$\mathcal S$ is infinite.} Take $c_0$ as the origin of the group law on $\overline C$. By \cref{lem:C}(b),
$\overline C$ is then isomorphic to $E_C$, so $\overline C(\Q)$ contains a point $c'$ of infinite order. For $p=13$
let $\Omega_p\subset\overline C(\Q_p)$ be the set of affine points with $m\in\Z_{13}$ and $\operatorname{ord}_{13}(m-m_0)\ge2$, and for
each of the six other primes the set of affine points with $m,u,v\in\Z_p$ congruent to $m_0,u_0,v_0$ modulo $p$. Each
$\Omega_p$ is an open neighborhood of $c_0$, so $\Omega=\prod_p\Omega_p$ is an open neighborhood of the identity in
the compact group $\Gamma=\prod_p\overline C(\Q_p)$, the products running over the seven primes. For the diagonal
embedding of $\overline C(\Q)$ in $\Gamma$ one has $\mathcal S=\overline C(\Q)\cap\Omega$. Let $\Omega'$ be an open
neighborhood of the identity with $\Omega'-\Omega'\subset\Omega$. Finitely many translates of $\Omega'$ cover $\Gamma$,
so one of them contains $kc'$ for infinitely many $k\ge1$, say for $k_1<k_2<\cdots$. Then the points $(k_j-k_1)c'$,
$j\ge2$, lie in $\Omega'-\Omega'\subset\Omega$, hence in $\mathcal S$, and they are distinct because $c'$ has infinite
order.
\end{proof}

\begin{proof}[Proof of \cref{thm:main}]
By \cref{prop:cong} the infinite set $\mathcal S$ consists of points $c$ at which $X_{m(c)}$ has torsion
subgroup $\Tf$ and rank at least $6$. The map $c\mapsto j(X_{m(c)})$ is a nonconstant rational
function on $\overline C$ of degree at most $48\cdot4=192$, since $m$ has degree $4$ on $\overline C$; so each value is
taken at most $192$ times, and infinitely many $j$-invariants occur.
\end{proof}

\begin{remark}\label{rem:height}
Here, as in \cite[\S 5.1]{PPVW}, the height of an elliptic curve over $\Q$ is $\max(|4A^3|,|27B^2|)$ for its model
$y^2=x^3+Ax+B$ with $A,B\in\Z$ and no prime $p$ with $p^4\mid A$ and $p^6\mid B$. For $m=x_1/x_0$ in lowest terms the
invariants $c_4$, $c_6$ of the model $E_{x_0^4a(m),\,x_0^4b(m)}$ of $X_m$ are binary forms in $(x_0,x_1)$ of degrees $16$ and $24$
with no common zero, since every singular fiber of \eqref{eq:X}, including the one at $m=\infty$, is nodal; so the
height of $X_m$ lies between two positive constant multiples of $\max(|x_0|,|x_1|)^{48}$. Then
\cite[Ch.~III, Th\'eor\`eme~4 and Corollaire~2]{Neron52}, applied to the function $m$ on $\overline C$, whose Jacobian
has rank one, together with the bound $192$ gives $(\log H)^{1/2+o(1)}$ curves $X_{m(c)}$, $c\in C(\Q)$, of height at
most $H$. In the same way, for the family $F$ of \cref{sec:second} over the curve $G_2$, whose Jacobian has
rank two, the forms of degrees $4$ and $6$ in its parameter $W$ and the bound $96$ proved there give
$(\log H)^{1+o(1)}$.
\end{remark}

\section{\texorpdfstring{Other bases for $X$}{Other bases for X}}\label{sec:conic}

We compare $C$ with other bases for $X$, none of which is needed for \cref{thm:main}: over a conic
through Dujella and Peral's curve of rank $9$, $X$ has rank at least $4$ (\cref{prop:conic}), and over a
curve of genus one that comes from a symmetry of their model, it has rank exactly $5$ (\cref{prop:G}).

The curve $C$ is the fiber product over the $m$-line of the conics $u^2=5(3m^2+242)$ and $v^2=5(2m^2+243)$, which
the involution $m\mapsto99/m$ of the family (the involution $n\mapsto9801/n$ of \cref{lem:Y}(c)) exchanges up to
squares. Over each conic alone two of the twists of \cref{sec:twists} stay nontrivial and one becomes trivial.

\begin{proposition}\label{prop:conic}
Let $D\colon u^2=5(3m^2+242)$, parametrized by $m=(k^2-70k+15)/(k^2-15)$, $u=-5(7k^2-6k+105)/(k^2-15)$.
\begin{enumerate}[(a)]
\item The points $P_1,P_2,P_3,P_5$ lie in $X(\Q(D))=X(\Q(k))$ and are independent, so $X$ over $\Q(k)$ has rank at
least $4$.
\item Its fiber at $k=25/11$ is the fiber of $X$ at $m=1681/119$, the curve of rank $9$ of Dujella and Peral
\cite[\S 7.2]{DP21} at $t=900/781$, with minimal model
\[
\begin{aligned}
y^2+xy=x^3&-1443705842368492991301675445569878391286260\,x\\
&+518524534126954116322153225511662398609137586525877978241619600.
\end{aligned}
\]
\end{enumerate}
\end{proposition}

\begin{proof}
(a) $P_5$ involves only $m$ and $\sqrt{nq_1}=mu$, which lie in $\Q(k)$. At $k_0=35/41$ one has $(m,u)=(7401/2399,\
88235/2399)$, the first two coordinates of $c_0$, so the four points specialize to four of the six independent points of
\cref{prop:indep}. (b) $m(25/11)=1681/119=(t+1)/(t-1)$ at
$t=900/781$, and the minimal model is computed from \eqref{eq:X}; it is the model printed in \cite[\S 7.2]{DP21}.
\end{proof}

In the coordinate $t$ of \cref{lem:dp}, $D$ is the conic $u'^2=1225t^2-2390t+1225$, with $u'=u(t-1)$, and at
$t=900/781$ one has $1225t^2-2390t+1225=(7715/781)^2$. So Dujella and Peral's curve of rank $9$ is a fiber of a family
of rank at least $4$ over $\Q(k)$.

Another base of genus one comes from a symmetry of the model. The map $(\alpha,\tau)\mapsto(\alpha,1/\tau)$
preserves \eqref{eq:DP} up to $x\mapsto\tau^4x$ and carries $\alpha^2\tau^2+2\alpha^2+1$ to
$(2\alpha^2\tau^2+\alpha^2+\tau^2)/\tau^2$, so it turns the second condition of Dujella and Peral into the
condition that $2\alpha^2\tau^2+\alpha^2+\tau^2$ be a square. At $\alpha=7$ and $\tau=(99-m^2)/(14m)$,
\[
2\alpha^2\tau^2+\alpha^2+\tau^2=99\tau^2+49=h(m)/(14m)^2,\qquad h(m)=99m^4-9998m^2+970299.
\]
Let $G\colon \eta^2=h(m)$. The twist of $Y$ by $h$, a quadratic polynomial in $n$, is the one with average $1.93$ in the
search described in the introduction.

\begin{proposition}\label{prop:G}
\begin{enumerate}[(a)]
\item $G$ is a curve of genus one with the rational point $(m,\eta)=(57,31836)$; its Jacobian is
$y^2=x^3+x^2-26097400x+38851416548$, of conductor $683760$, with torsion $\Tf$ and rank $2$.
\item The quadratic twist $X^h\colon y^2=x(x+ha^2)(x+hb^2)$ has rank $2$ over $\Q(m)$, and $\rank X(\Q(G))=5$.
\end{enumerate}
\end{proposition}

\begin{proof}
(a) $h$ is squarefree of degree $4$ (its discriminant is $2^{18}\cdot3^{8}\cdot5^{4}\cdot7^{8}\cdot11^{4}\cdot37^{2}$) and $h(57)=31836^2$; the
Jacobian is computed from the invariants of $h$ as in \cref{lem:C}. With
\texttt{mwrank}~\cite{CremonaBook} one finds a $2$-Selmer group of dimension $4$ (with full rational $2$-torsion), so
the rank is at most $2$. The points $(-5194,185220)$ and $(-4249,270270)$ are independent, so the rank is $2$. Indeed, the
cubic has the roots $-5734$, $1666$ and $4067$, and the $2$-descent map $P\mapsto(x(P)+5734,\ x(P)-1666)$ of \cref{prop:rank3} sends the two points to
$(3\cdot5,\,-5\cdot7)$ and $(3\cdot5\cdot11,\,-5\cdot7)$; it sends the generators $(-784,242550)$ and $(1666,0)$ of the torsion subgroup to $(2\cdot11,\,-2)$ and $(2\cdot37,\,-2\cdot37)$. These four vectors are independent: $37$ occurs only in the last, then $2$ only in the third and $11$ only in
the second, and the first is nontrivial.

(b) Since $\Q(G)=\Q(m)(\sqrt h)$, $\rank X(\Q(G))=\rank X(\Q(m))+\rank X^h(\Q(m))=3+\rank X^h(\Q(m))$
(\cref{prop:rank3}). \emph{Upper bound.} For $X^h$, $(e_1,e_2,e_3)=(0,-ha^2,-hb^2)$; the nonconstant
squarefree divisors in $\Z[m]$ of $h^2a^2b^2$, $h^2a^2(a^2-b^2)$ and $-h^2b^2(a^2-b^2)$ are now built from the factors
of \cref{prop:rank3} and $h$. At $m=-80/3$ these twelve factors have squarefree parts $-3\cdot13\cdot29$, $-3\cdot7\cdot11$, $-3\cdot113$, $-3\cdot53$, $3\cdot7\cdot31$, $-3\cdot83$, $-3\cdot47$, $-3\cdot107$, $-3\cdot5$, $7\cdot787$, $593\cdot1433$, $3\cdot1291\cdot2083$. With $-1$,
$2$, $5$, $7$ they are independent in $\Q^\times/\Q^{\times2}$, so no such divisor takes a square value at $-80/3$. The
fiber of
$X^h$ at $-80/3$ has full rational $2$-torsion and a $2$-Selmer group of dimension $4$, so its rank is at most $2$, and
by \cite[Theorem~1.1]{GusicTadic} $\rank X^h(\Q(m))\le2$. \emph{Lower bound.} The substitution
$m\mapsto m'=(\eta-98m)/(99-m^2)$ satisfies $\tau(m')=1/\tau(m)$. Put $\delta=(99-m^2)/(14m')$, so that $a(m')=a(m)/\delta^2$ and
$b(m')=b(m)/\delta^2$, and write $x(m')$ and $y(m')$ for the coordinates $x$ and $y$ with $m$ replaced by $m'$. Then
the map $(x,y)\mapsto(\delta^4x(m'),\delta^6y(m'))$ carries points of $X$ over $\Q(m')$ to points of $X$ over $\Q(G)$. Let
$P_2'$ and $P_3'$ be the images of $P_2$ and $P_3$. At $m=291/53$, $\eta=2447004/2809$, scaling $x$ by
$(7890481/7056)^2$ takes the fiber to $E=E_{379291648,\,212565977}$, with torsion $\Tf$ by \cref{lem:tors} ($43\,\|\,ab$ and $-ab<0$), and
takes $P_1,P_2,P_3,P_2',P_3'$ to points with $x$-coordinates
\[
\begin{array}{lll}
-85985831567294464, & -58444996670344024, & -2248161523739262976/49,\\
-62331445860559616, & -48899358183829504. &
\end{array}
\]
We use the primes $p=37,41,61,107,113$, where $\#\widetilde E(\F_p)=48,48,72,96,120$; at $p=61$ the group is
$\Z/36\Z\times\ZZ2$, with cyclic $3$-primary part. Computed as in \cref{prop:indep}, the images of these points at
the $j$-th of these primes in $\widetilde E(\F_p)/3\widetilde E(\F_p)\cong\ZZ3$ are the entries of the
$j$-th of the rows
\[
(2,1,1,0,1),\quad(1,0,1,2,0),\quad(0,2,0,0,2),\quad(0,0,0,2,1),\quad(2,2,0,0,0)
\]
of a matrix of rank $5$ over $\F_3$, each row defined up to sign by the choice of generator. So the five points are independent, and $\rank X(\Q(G))\ge5$.
The upper bound gives $\rank X(\Q(G))\le5$, so the five independent points
$P_1,P_2,P_3,P_2',P_3'$ span a subgroup of finite index, and the image $P_1'$ of $P_1$ is linearly dependent on them.
\end{proof}

The fiber at $m=291/53$, isomorphic to $E_{379291648,\,212565977}$, has rank $6$, one more than $\rank X(\Q(G))$, and its
Tate--Shafarevich group has no element of order $2$. Indeed, its $2$-Selmer group has dimension $8$ (computed with
\texttt{mwrank}), and its points with the $x$-coordinates of $P_1$, $P_2$, $P_3$,
$P_3'$ listed above and with $x$-coordinates $585506734092845056$ and $1093412290152185253846016/17732521$ have images
under the map $\kappa$ of \cref{prop:rank3} which, with those of the torsion, span a space of dimension $8$.
This fiber shows what $C$ adds: over $G$ the rank $6$ occurs at single fibers such as this one, whereas
over $C$ it is the rank of $X(\Q(C))$ (\cref{thm:precise}(b)).

Since $G$ has the rational point $(57,31836)$ and its Jacobian has rank $2$, the set $G(\Q)$ is infinite.
Silverman's theorem and the argument of \cref{sec:proof}, now with double covers
$r^2=\pm ab$ of $G$ of genus $9$, then give infinitely many curves with torsion $\Tf$ and rank at least $5$ from $G(\Q)$; this recovers the
bound credited to Eroshkin in \cite{DujGeneric}. Over $G$
one twist of rank $2$ becomes trivial, whereas over $C$ three twists of rank $1$ do (\cref{prop:rank6}).

The third twist found by the search of \cref{sec:intro}, by $n(4n+45)$, gives over $\Q(m)$ the twist of $X$ by
$4m^2+45$, which has rank $1$, but the twists of $X$ by $(4m^2+45)q_1$, $(4m^2+45)q_2$ and $(4m^2+45)q_{12}$ have
rank $0$ over $\Q(m)$. So $X$ has rank $5$ over each of the two curves of genus one on which $4m^2+45$ and one of
$q_1$, $q_2$ become squares, and rank $7$ over the curve of genus $5$ (by the Riemann--Hurwitz formula) on which all
three become squares, which has only finitely many rational points by Faltings' theorem. These ranks follow as in
\cref{prop:rank6}. The twist of $Y$ by $n(4n+45)$ has the point with $x$-coordinate
$49(n-121)(n-1)n^2(11n-1971)^2$, which has infinite order since its specialization at $m=42$ does. The upper bounds
come from the criterion of Gusi\'c and Tadi\'c at $m=42$ for the twist by $4m^2+45$, at $m=-58/3$ for the twists by
$(4m^2+45)q_1$ and $(4m^2+45)q_2$, and at $m=-46/7$ for the twist by $(4m^2+45)q_{12}$, where the fibers have $2$-Selmer
groups of dimensions $3$, $2$, $2$ and $2$.
At $m=42$ the eleven factors of \cref{prop:rank3}, in the same order, have squarefree parts $-3\cdot19$, $43$, $31$, $3\cdot17$, $3\cdot47$, $41$, $53$, $3\cdot11$, $2\cdot3\cdot7$, $5\cdot37$ and $346441$, and $4m^2+45$ has squarefree part $3\cdot263$. At $m=-58/3$ they have squarefree parts $-3\cdot5\cdot71$, $-3\cdot5\cdot11$, $-3\cdot7\cdot13$, $-3\cdot31$, $3\cdot239$, $-3\cdot61$, $-3$, $-3\cdot5\cdot17$, $-2\cdot3\cdot29$, $2473$ and $97\cdot4969$, and those of $4m^2+45$, $3m^2+242$ and $2m^2+243$ are $83\cdot167$, $2\cdot3\cdot5\cdot409$ and $5\cdot1783$. At $m=-46/7$ they are $-7\cdot739$, $-3\cdot7\cdot13$, $-3\cdot7\cdot41$, $7\cdot17$, $7\cdot647$, $-7\cdot53$, $7\cdot31$, $-7\cdot109$, $-2\cdot7\cdot23$, $-5\cdot547$ and $1153\cdot24113$, and those of $4m^2+45$, $3m^2+242$ and $2m^2+243$ are $47\cdot227$, $2\cdot9103$ and $16139$. In each case these classes and $-1$,
$2$, $5$, $7$ are independent in $\Q^\times/\Q^{\times2}$, and the contents of the three products again involve only
the primes $2$, $5$ and $7$, so no nonconstant squarefree divisor of the three products takes a square value.

\section{A second family}\label{sec:second}

We give a second proof of \cref{thm:main}, independent of the first, with a family over a curve of genus one
and rank two. Since the base has rank two rather than one, the family gives $(\log H)^{1+o(1)}$ curves of height at most
$H$, against $(\log H)^{1/2+o(1)}$ for $C$ (\cref{rem:height}).

Let $\mu=19/33$, $\lambda=\mu^2=361/1089$, $\beta=(161/1089)^2$ and $\gamma=\lambda^2/\beta=(361/161)^2$, and let
\begin{equation}\label{eq:R}
F\colon\ y^2=x(x^2+Ux+V^2),\qquad U=2(1+\lambda)^2W+2(W+\lambda)^2,\quad V=-(W-1)(W-\lambda^2),
\end{equation}
over $\Q(W)$. Then $U-2V=4(W+\lambda)^2$ and
$U+2V=4(1+\lambda)^2W$, so $(-V,\,2V(W+\lambda))$ is a point of order $4$
with double $(0,0)$, and the other points of order $2$ are defined over $\Q(s_1)$ with $W=s_1^2$. Over $\Q(s_1)$, $F$ is
the curve $E_{(s_1-1)(s_1-\lambda),\,(s_1+1)(s_1+\lambda)}$, and since
$A_+B_+=\alpha^3(s_1+1)(s_1+\lambda)$ and $A_-B_-=-\alpha^3(s_1-1)(s_1-\lambda)$ at $(\alpha,\tau)=(33/19,\,33s_1/19)$, it is the
model \eqref{eq:DP} of Dujella and Peral on the line $\alpha=33/19$, with no square condition imposed.

These parameters come from an exact computation with $\lambda$ and $\beta$ left free, in which we prescribe the shape of a
section, as Dujella and Peral do to find new points \cite[\S 4]{DP24}. In the model $F^q$ used below, the twist of $F$ by a
constant multiple of $W-\beta$ has a section whose $x$-coordinate is a constant times $(W-1)(W-\lambda^2)$ when
$(\lambda,\beta)$ lies on the curve
\[
\lambda^4-16\lambda^3+(16\beta+30)\lambda^2+(160\beta-16)\lambda+64\beta^2+16\beta+1=0,
\]
and one whose $x$-coordinate is a constant times $(W-1)(W-\beta)$ when it lies on the curve
\[
16\lambda^4+32\lambda^3+(8-8\beta)\lambda^2-(24\beta+8)\lambda+\beta^2-18\beta+1=0,
\]
apart from degenerate cases: $\lambda=-1$, where $F$ is singular, and the points $(1,0)$ of the first curve and $(1,1)$ of the
second, where the constant multiple is $0$. Other than $(-1,1)$ and $(1/9,1/81)$, where $\beta=\lambda^2$, the two curves meet
at only one rational point of the $(\lambda,\beta)$-plane, $\bigl(361/1089,(161/1089)^2\bigr)$. There both multiples of $W-\beta$
are $29(W-\beta)$ up to squares, so the twist by $29(W-\beta)$ has the two points $Q_2$ and $Q_3$ below. The involution $W\mapsto\lambda^2/W$ gives two more, $Q_4$ and $Q_5$, and the
twist by $-(W-\beta)(W-\gamma)$, whose branch points $W=\beta$ and $W=\gamma=\lambda^2/\beta$ are exchanged by the involution, carries
one more, $Q_6$; the remaining point, $Q_1$, lies on $F$ itself.

Let $G_2$ be the
curve
\[
G_2\colon\quad W=s_1^2,\qquad 29(W-\beta)=s_2^2,\qquad -29\beta(W-\gamma)=s_3^2,
\]
and put $q_A=29(W-\beta)$, $q_B=-29\beta(W-\gamma)$ and $q_H=-(W-\beta)(W-\gamma)$, so that $q_Aq_B/q_H=(4669/1089)^2$.

\emph{The points.} For $q\in\Q(W)^\times$ let $F^q\colon y^2=x(x^2+qUx+q^2V^2)$. The points $Q_1\in F(\Q(W))$,
$Q_2,Q_3\in F^{q_A}(\Q(W))$ and $Q_6\in F^{q_H}(\Q(W))$ with
\[
\begin{aligned}
x(Q_1)&=\mu^{-2}(W-\lambda^2)^2, & x(Q_2)&=\bigl(\tfrac{1015}{1089}\bigr)^2(W-1)(W-\lambda^2),\\
x(Q_3)&=\bigl(\tfrac{928}{1089}\bigr)^2(W-1)(W-\beta), & x(Q_6)&=\bigl(\tfrac{240}{289}\bigr)^2(W-\beta)^2(W-\gamma)^2
\end{aligned}
\]
satisfy
\[
\begin{aligned}
x^2+Ux+V^2&=\tfrac{2102500}{130321}\,(W+\lambda)^2(W-\lambda^2)^2 &&\text{at }x=x(Q_1),\\
x^2+q_AUx+q_A^2V^2&=841\,(W-1)(W-\lambda^2)\,f_2^2 &&\text{at }x=x(Q_2),\\
x^2+q_AUx+q_A^2V^2&=841\,(W-1)(W-\beta)^3\bigl(W-\tfrac{625}{1089}\bigr)^2 &&\text{at }x=x(Q_3),\\
x^2+q_HUx+q_H^2V^2&=\tfrac{671898241}{6975757441}\,(W-\beta)^2(W-\gamma)^2f_6^2 &&\text{at }x=x(Q_6),
\end{aligned}
\]
with $f_2=W^2-\tfrac{648517}{1185921}W+\tfrac{3467044}{1291467969}$ and
$f_6=W^2-\tfrac{14138244652530322}{796818233864961}W+\lambda^2$, so that in each case $x(x^2+qUx+q^2V^2)$ is a square in
$\Q[W]$ ($671898241=161^4$ and $6975757441=289^4$). Since $U(\lambda^2/W)=(\lambda/W)^2U(W)$, $V(\lambda^2/W)=(\lambda/W)^2V(W)$ and $q_A(\lambda^2/W)=q_B/W$, the
map
\[
\iota\colon(x,y)\mapsto\bigl((W^2/\lambda)^2x(\lambda^2/W),\ (W^2/\lambda)^3y(\lambda^2/W)\bigr)
\]
is an isomorphism of groups $F^{q_A}(\Q(W))\to F^{Wq_B}(\Q(W))$, the composite of the field automorphism
$W\mapsto\lambda^2/W$ with an isomorphism of curves. Put $Q_4=\iota(Q_2)$ and $Q_5=\iota(Q_3)$. On $G_2$ one has
$q_A=s_2^2$, $Wq_B=(s_1s_3)^2$ and $q_H=(1089s_2s_3/4669)^2$, so, as in \cref{prop:points}, the six points
give points of $F$ over $\Q(G_2)$, which we denote by the same names.

\begin{theorem}\label{thm:second}
\begin{enumerate}[(a)]
\item $G_2$ is a curve of genus one with infinitely many rational points. Its Jacobian has conductor
$182747931661680$, torsion subgroup $\Tf$ and rank~$2$.
\item The points $Q_1,\dots,Q_6$ defined above lie in $F(\Q(G_2))$ and are linearly independent, so
$\rank F(\Q(G_2))\ge6$.
\item At the sixteen rational points of $G_2$ with $W\in\{1,\lambda^2\}$ the fiber of $F$ is singular. At every other
rational point it is an elliptic curve whose torsion
subgroup contains $\Tf$, and for all but finitely many rational points of $G_2$ its torsion subgroup is exactly $\Tf$
and its rank is at least $6$. Infinitely many of these curves are pairwise non-isomorphic.
\item At the point of $G_2$ with $s_1=23306521/20091929$ described below, the fiber is isomorphic to $E_{a,b}$ with
$(a,b)=(460019200,\ 11180308689)$; its torsion subgroup is exactly $\Tf$, and the six points specialize to independent
points.
\end{enumerate}
\end{theorem}

\begin{proof}[Proof of \cref{thm:second}]
(a) Parametrizing the conic $29(s_1^2-\beta)=s_2^2$ through $(s_1,s_2)=(161/1089,0)$ by
$s_1=\tfrac{161}{1089}\,(\theta^2+29)/(\theta^2-29)$, $s_2=\theta(s_1-\tfrac{161}{1089})$ turns the third equation into
\[
\eta^2=1326538900\,\theta^4-77611154441\,\theta^2+1115619214900,\qquad \eta=\tfrac{1185921}{58}\,s_3(\theta^2-29),
\]
which is squarefree (its discriminant is nonzero), so $G_2$ has genus one.
Its Jacobian has the minimal model
\[
y^2=x^3+x^2-7927479575830823114160\,x+271658305399029819040875379804308,
\]
of conductor $182747931661680=2^4\cdot3\cdot5\cdot7\cdot11\cdot17\cdot19\cdot23\cdot29\cdot197\cdot233$ and torsion $\Tf$. It has
full rational $2$-torsion and a $2$-Selmer group of dimension $4$ (computed with \texttt{mwrank}
\cite{CremonaBook}), so its rank is at most $2$. Its points $(50796866898,\ 198589582959768)$ and
$(389256165869/4,\ 164296202311285275/8)$ are independent, so the rank is $2$: the cubic has the roots
$-102809641014$, $51068871386$ and $51740769627$, and the $2$-descent map
$P\mapsto(x(P)+102809641014,\ x(P)-51068871386)$ of \cref{prop:rank3} sends the two points to
$(2,\,-2)$ and $(197,\,7\cdot23\cdot197)$, and the generators $(41550472818,\ 3741958510639272)$ and $(51068871386,\ 0)$ of the torsion subgroup to
$(2\cdot197\cdot233,\,-2\cdot197\cdot233)$ and $(29\cdot197\cdot233,\,-29\cdot197\cdot233)$. These four
vectors are independent: $29$ occurs only in the last, $7$ only in the second, then $233$ only in the third, and $2$ in
the first. As a check, twice the point
$(50796866898,\ 198589582959768)$ has $x$-coordinate with denominator $7^2\cdot13^2$. Since $G_2$ has the rational
point $\theta=10/3$ and its Jacobian has rank $2$, the set $G_2(\Q)$ is infinite.

(b) The field $\Q(G_2)$ is $\Q(W)(\sqrt W,\sqrt{q_A},\sqrt{q_B})$, and $W$, $q_A$, $q_B$ are independent modulo
squares in $\Q(W)$, since their divisors modulo $2$ are supported on $\{0,\infty\}$, $\{\beta,\infty\}$ and
$\{\gamma,\infty\}$. So $\Q(G_2)/\Q(W)$ is Galois with group $(\ZZ2)^3$, and $F(\Q(G_2))\otimes\Q$ is the sum of the
eigenspaces of the eight characters. The eigenspace of the character attached to a square class $q$ is
$F^q(\Q(W))\otimes\Q$. The points lie in the eigenspaces of the four distinct classes $1$, $q_A$, $Wq_B$ and $q_Aq_B\equiv
q_H$. So they are independent as soon as $Q_1$ and $Q_6$ have infinite order and $Q_2,Q_3$ are independent in
$F^{q_A}(\Q(W))$ (then so are $Q_4=\iota(Q_2)$ and $Q_5=\iota(Q_3)$). We specialize at $W=4$, where $F$, $F^{q_A}$ and $F^{q_H}$ are
smooth, and write $F^{q_A}_4$ and $F^{q_H}_4$ for their fibers there. The curve $F^{q_A}_4$ has torsion subgroup $(\ZZ2)^2$, since its reductions at the good primes $7$ and $13$
have $8$ and $12$ points and its cubic has the three rational roots $0$,
$e_2=-881981589323167/156267624249$ and $e_3=-451693636824103/1406408618241$. With $\kappa(P)=(x(P),x(P)-e_2)$ as in
\cref{prop:rank3},
\[
\begin{aligned}
\kappa(Q_2)&=(3\cdot23\cdot79\cdot2539,\ 3\cdot29\cdot2539), & \kappa(Q_3)&=(3\cdot2017\cdot2339,\ 3\cdot29\cdot2017),\\
\kappa(0,0)&=(1,\ 29\cdot2017\cdot2339), & \kappa(e_2,0)&=(-29\cdot2017\cdot2339,\ 29\cdot53\cdot89),
\end{aligned}
\]
which are independent ($23$ and $53$ each occur once, and then $3$); so, as in \cref{prop:rank3}, $Q_2(4)$
and $Q_3(4)$ are independent modulo torsion, and so are $Q_2$ and $Q_3$. The point $Q_1(4)$, with
\[
x(Q_1(4))=21283118169769/466219936809,
\]
is not $19$-integral on the integral model $y^2=x^3+1089^2Ux^2+1089^4V^2x$ (at $W=4$), where its $x$-coordinate has
denominator $3^2\cdot11^2\cdot19^2$; so it has infinite order \cite[Theorem~VII.3.4]{SilvermanAEC}. Likewise
$Q_6(4)$, with
\[
x(Q_6(4))=11230014732441718248505600/974374006659413003857521,
\]
is not $7$-integral on the integral model of $F^{q_H}_4$ obtained by scaling $x$ by $190933281^2$.

(c) The discriminant of \eqref{eq:R} is $16V^4(U^2-4V^2)$, and $U^2-4V^2=16(1+\lambda)^2W(W+\lambda)^2$. At a rational
point of $G_2$, $W=s_1^2$ with $s_1=\tfrac{161}{1089}(\theta^2+29)/(\theta^2-29)\neq0$, so the fiber is singular exactly when
$W\in\{1,\lambda^2\}$, that is, at the sixteen points with $\theta\in\{\pm10/3,\pm87/10,\pm25/4,\pm116/25\}$. At the
other rational points the fiber is smooth, with
torsion containing $\Tf$. The $j$-invariant of $F$ is a nonconstant function of $W$, of degree $12$, so as in
\cref{sec:proof} Silverman's theorem \cite[Theorem~C]{Silverman83} shows that specialization is injective at all
but finitely many points of $G_2(\Q)$, which is infinite by (a); there the rank is at least $6$ by (b). Since $W$ has
degree $8$ on $G_2$, each $j$-invariant occurs at most $96$ times.

It remains to show that the torsion is exactly $\Tf$ at all but finitely many rational points. Over $\Q(s_1)$, $F$ is
the curve $E_{(s_1-1)(s_1-\lambda),(s_1+1)(s_1+\lambda)}$, for which $ab=-V$. So, by \cref{lem:tors}, whose proof follows
\cite[Main Theorem~1]{Ono}, torsion $\Z/2\Z\times\Z/8\Z$ at a rational point $P$ requires $V$ or $-V$ to be a nonzero
rational square at $W(P)$. The map $W\colon G_2\to\PP^1$ of degree $8$ is branched only over $0$, $\beta$, $\gamma$ and $\infty$, so $V$ has
$16$ simple zeros on $G_2$, the points over $W=1$ and $W=\lambda^2$, and poles of order $4$ at the four points over
$\infty$, each with ramification index $2$. So the double covers $r^2=\pm V$ of $G_2$ are branched at exactly $16$ points
and have genus $9$, by the Riemann--Hurwitz formula. By Faltings' theorem \cite[Satz~7]{Faltings} each has finitely
many rational points, so $\pm V(W)$ is a nonzero square at only finitely many rational points of $G_2$.

(d) The quartic of (a) has the rational point $\theta=857/140$, with $s_1=23306521/20091929$,
$s_2=1120373240/180827361$ and $s_3=11142156700/7293370227$. There the six points are defined over $\Q$. Scaling $x$ by
$(3633170498469369/1046900)^2$ takes the fiber to $E=E_{a,b}$ with
\[
(a,b)=(460019200,\ 11180308689)=(2^9\cdot5^2\cdot83\cdot433,\ 3^3\cdot11\cdot17\cdot31\cdot61\cdot1171),
\]
where $Q_1,\dots,Q_6$ have, in this order by rows, the $x$-coordinates
\[
\begin{array}{ll}
499284308691438490000/9, & 2215757908015311217546474200/19037652529,\\
339558475170049273921536/3258025, & 291785582846327357358433632/113273449,\\
-4563185528671931392, & 3381199714987226818560000/83521.
\end{array}
\]
The curve has torsion exactly $\Tf$, since $83\,\|\,ab$ and $-ab<0$ (\cref{lem:tors}), and squarefree conductor
\[
2\cdot3\cdot5\cdot7\cdot11\cdot13\cdot17\cdot23\cdot31\cdot53\cdot61\cdot83\cdot89\cdot137\cdot241\cdot401\cdot433
\cdot457\cdot1171\cdot2089.
\] At the primes $p=41,47,59,67,79,97$ one has $\#\widetilde E(\F_p)=48,48,72,72,96,96$, and at $p=59$ and $67$ the
group is $\Z/36\Z\times\ZZ2$, with cyclic $3$-primary part. The images of the six
points at the $j$-th of these primes in $\widetilde E(\F_p)/3\widetilde E(\F_p)$ are the entries of the $j$-th of the rows
\[
(1,1,1,1,1,2),\ (1,0,1,2,1,0),\ (0,2,2,0,0,0),\ (2,2,1,0,0,1),\ (2,1,0,1,1,1),\ (1,2,1,1,0,2)
\]
of a matrix with determinant $2$ modulo $3$. Each row is defined up to sign by the choice of generator, and each
column up to sign by that of the $y$-coordinate of its point, which is not fixed here. So the six points are
independent, as in \cref{prop:indep}.
\end{proof}

The curve $E_{460019200,\,11180308689}$ of (d) has rank $7$, and its Tate--Shafarevich group has no element of
order $2$. Indeed, its $2$-Selmer group has dimension $9$ (computed with \texttt{mwrank}), and the points $Q_1$, $Q_3$, $Q_5$ of (d), together
with its points with $x$-coordinates
\[
\begin{array}{ll}
-3478949377821720000, & -1520527880418835200,\\
-3984141305499992809392000/32041, & 180588656033172910080/169,
\end{array}
\]
have images under the map $\kappa$ of \cref{prop:rank3} which, with those of the torsion, span a space
of dimension $9$. The six points of (d) would not do here: with the torsion, their images under $\kappa$ span a space of
dimension $5$ only, even though the six points are independent. This curve bears on the
question, raised at the end of this section, whether rank $7$ occurs infinitely often with torsion $\Tf$.

\begin{remark}
Without Faltings' theorem one still obtains infinitely many of these curves with torsion exactly $\Tf$ and rank at
least $6$, by a $3$-adic analogue of the $13$-adic condition of \cref{prop:cong}. At $\theta=857/140$ one
has $\operatorname{ord}_3(V(W))=-1$. The same holds at every rational point of $G_2$ with
$\operatorname{ord}_3(\theta-857/140)\ge6$, since $V(W)$ is a constant times a product of eight linear polynomials in $\theta$ with integer
coefficients, divided by $(\theta^2-29)^4$, and each of these nine polynomials has $3$-adic valuation at most $5$ at
$857/140$. So $\pm V(W)$ is not a square in $\Q_3$ at these points, and by the proof of (c) the torsion there is
exactly $\Tf$. Since $G_2$
has a rational point and its Jacobian has positive rank, this $3$-adic neighborhood of $\theta=857/140$ contains
infinitely many rational points of $G_2$: if $P$ is a rational point of infinite order on the Jacobian and $N\ge1$ is
such that $NP$ lies in the kernel of reduction modulo $3$, a pro-$3$ group, then the translates of the point with
$\theta=857/140$ by $3^kNP$, for $k=1,2,\dots$, are distinct and tend to it $3$-adically. By Silverman's theorem, as in
the proof of (c), the rank is at least $6$ at all but finitely many of them.
\end{remark}

The case $\Z/6\Z$ of \cite[Problem~2.6]{Dopen} remains open. We do not know whether rank $7$ occurs infinitely often
with torsion $\Tf$.

\appendix
\section{Computations}\label{sec:computations}

We used PARI/GP 2.17.3 \cite{PARI} and SageMath 10.9 \cite{Sage}, which includes Cremona's
\texttt{mwrank}~\cite{CremonaBook}. The divisor tests of Gusi\'c and Tadi\'c were done in PARI and again with
python-flint \cite{FLINT}. Apart from the one noted below, the upper bounds for ranks from $2$-Selmer groups were computed with \texttt{mwrank} and again with PARI's \texttt{ellrank}, with the same results. The
independence of the six points at $c_0$, proved in PARI with the matrix $M$ of \cref{prop:indep}, was proved again
in SageMath by a $2$-descent, with the program listed under \cref{prop:indep} in \cite{Record}. There the images
under the map $\kappa$ of \cref{prop:rank3} of $P_1(c_0),\dots,P_6(c_0)$ and of the two generators of the torsion of
$X_{m_0}$ span a space of dimension $6$ only. The two relations come from two combinations of these points and of torsion points that lie
in $2X_{m_0}(\Q)$, and replacing two of the six points by halves of these combinations, that is, by points whose
doubles are these combinations, gives points whose images, with those of the torsion, span a space of dimension $8$.
The proof of \cref{thm:main} needs no upper bound for a rank: its computational inputs are polynomial identities and
factorizations (including the degree of $j$), the point $(-322,726)$ of infinite order on $E_C$, and the valuations
and the matrix $M$ at $c_0$.

The point $c_0$, at $k=35/41$ in the parameter of the proof of \cref{lem:C}, came from a search of the points of
$C(\Q)$ of height at most $10^5$ in $k$; its fiber is one of the two smooth fibers, up to isomorphism, found there,
both of rank $6$ \cite{Record}; for the other, the upper bound $6$ was computed with \texttt{ellrank} only.

The averages of \cref{sec:intro} are plain averages over the $368$ primes $p$ with $300<p<3000$. The twists
there are those by $s(n-r_0)$ and by $sn(n-r_0)$, where $r_0$ runs over the rationals whose
numerator and denominator are at most $1000$ in absolute value and $s$ over the squarefree integers with $|s|\le 100$
(so no $s$ is divisible by $p$). Apart from $r_0=0$ and $s=1$, where the twists are $Y$ and $Y^n$, exactly three averages exceed
$0.45$, those of the three twists named there. Among the eighteen twists by $q$ and by $nq$ with $q$
one of $4n+45$, $q_1$, $q_2$, $(4n+45)q_1$, $(4n+45)q_2$, $q_1q_2/25$, $h$, $(4n+45)h$ and $q_1h$, the twists by
$q_1q_2/25$ and by $h$ have averages $1.00$ and $1.93$, the three twists above reappear, and the other thirteen lie
between $-0.09$ and $0.07$.

The programs and their outputs are deposited in
\cite{Record}, and the Lean formalization in \cite{Lean}.

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\section*{Declarations}\phantomsection\label{sec:declarations}

\subsection*{Use of AI}
This paper, its results and its Lean formalization were produced by an AI research system that the author designed, built and directs, running on Anthropic's Claude Opus~5.5 in Claude Code and OpenAI's GPT-6 Astra in Codex. The author chooses the problems and prioritizes the work. Separate AI agents checked each result, and a second program confirmed the computations the proofs use, with any exceptions listed in the public deposit. The system also searched the literature and checked the citations against reliable sources. The author read the paper, reran the computations from the deposit and takes full responsibility for all of the paper's contents.

\subsection*{Data availability}
The programs and data used in this paper, with instructions for rerunning them, are deposited at Zenodo~\cite{Record}.

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