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\TopicFR{Théorie des nombres}
\TopicEN{Number theory}

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\title{Rational points on a certain genus 2 curve}

\author{\lastname{Nguyen} \firstname{Xuan Tho}}
\address{Hanoi University of Science and Technology}
\email{tho.nguyenxuan1@hust.edu.vn}

\thanks{This paper is funded by the Ministry of Education and Training of Vietnam under the project B2022-CTT-03, 2022-2023.}

%~\keywords{rational points, elliptic curve Chabauty, fake-2 Selmer sets}
\subjclass{14G05, 14H99}

\begin{abstract}
We give a correct proof to the fact that all rational points on the curve
\[
y^2=(x^2+1)(x^2+3)(x^2+7)
\]
are $\pm \infty$ and $(\pm 1,\,\pm 8)$. This corrects previous works of Cohen~\cite{Cohen} and Duquesne~\cite{S,S1}.
\end{abstract}

\begin{document}
\maketitle

\section{Introduction}

The goal of this paper is to prove the following theorems

\begin{theo}\label{T1}
All rational points on the curve
\begin{equation}\label{E1}
\mathcal{C}_1\colon y^2=(x^2+1)(x^2+3)(x^2+7)
\end{equation}
are $\pm \infty$ and $(\pm 1,\,\pm 8)$.
\end{theo}

The available proofs of Theorems~\ref{T1} in~\cite{Cohen} and~\cite{S} are unfortunately incorrect. The curve~\eqref{E1} appeared in the work of Flynn and Wetherell~\cite{Flynn}. However, Flynn and Wetherell also pointed out that their method failed in determining all rational points on the curve~\eqref{E1}. Duquesne later gave an incorrect proof of Theorem~\ref{T1} in his thesis~\cite[pp.~64--69]{S}. {Duquesne also reported Theorem~\ref{T1} in~\cite[Theorem~4]{S1} and refered to his thesis for the proof}. Furthermore, Duquesne's proof was reproduced in Cohen's book~\cite[Theorem~13.3.10, pp.~459--462]{Cohen}. Duquesne argued as the following: since $\Q(i)$ is a unique factorization domain and
\[
(x+i)(x^2+3)(x-i)(x^2+7)=y^2,
\]
it follows that
\[
\left\{
\begin{aligned}
\alpha y_1^2&=(x+i)(x^2+3),\\
\alpha y_2^2&=(x-i)(x^2+7),
\end{aligned}
\right.
\]
where $\alpha$ can be taken as a divisor of the resultant of $(x+i)(x^2+3)$ and $(x-i)(x^2+7)$. The problem is that polynomials $(x+i)(x^2+3)$ and $(x-i)(x^2+7)$ have odd (three) degrees and so the denominator of $x$ needs to be considered. Specifically, write $x = X/Z$, where $X,Z\in \Z$ and $\gcd(X,Z) = 1$, so~\eqref{E1} takes the form
\[
Y^2 = (X^2 + Z^2)(X^2 + 3Z^2)(X^2 + 7Z^2).
\]
Certainly,
\begin{equation}\label{E2}
\left\{
\begin{aligned}
\beta Y_1^2&= (X^2 + 3Z^2)(X + iZ),\\
\beta Y_2^2&= (X^2 + 7Z^2)(X-iZ),
\end{aligned}
\right.
\end{equation}
where $\beta\in \Q(i)$. Now
\[
\mathrm{Resultant}_{X}((X^2 + 3Z^2)(X + iZ), (X^2 + 7Z^2)(X -iZ)) = -2^7\cdot 3\cdot i\cdot Z^9,
\]
so that $\beta$ may be taken as a squarefree divisor of $2\cdot 3 \cdot Z$ in $\Z[i]$. Certainly $\gcd(\beta,Z) = 1$, otherwise $\gcd(X,Z)>1$, so $\beta| 2\cdot 3$. Just as for $\alpha$. However, the system~\eqref{E2} on dividing by $Z^3$ corresponds to
\[
\left\{
\begin{aligned}
\dfrac{\beta}{Z}\cdot y_1^2 &= (x^2 + 3)(x + i),\\[0.5em]
\dfrac{\beta}{Z}\cdot y_2^2& = (x^2 + 7)(x -i),
\end{aligned}
\right.
\]
so that the original $\alpha$ has to depend on (the square-free part of) $Z$.


For a specific example, consider the curve
\[
y^2 = (x^2 + 1)(x^2 + 15)(x^2 + 18),
\]
where the resultant of $(x+i)(x^2 +15)$ and $(x-i)(x^2 +18)$ equals $-2^2 \cdot 3^2 \cdot 7 \cdot 17i$. The curve has a rational point with $x = 3/5$, at which point
\begin{align*}
(x+i)(x^2+15) &= (1-i)^{15}(4+i)\cdot \dfrac{3}{125}\\
&\equiv (1-i)(4+i)\cdot 3\cdot 5 \pmod{\Z[i]^2}.
\end{align*}
The resultant technique will work when applied to polynomials of even degree.

In the next sections, we will prove Theorem~\ref{T1}. The main tools are the elliptic curve Chabauty method in combination with Bruin and Stoll's fake-2 Selmer set~\cite{Stoll1}, which have been implemented in MAGMA~\cite{MAGMA}. See Stoll~\cite{Stoll2,Stoll3} for concrete examples. The MAGMA codes for the computation of the 2-fake Selmer sets for each curve $\mathcal{C}_1$ and $\mathcal{C}_2$ are given. {The MAGMA codes for the elliptic curve Chabauty routine are available at \url{https://www.overleaf.com/read/mgsfpypvvbrb}}


\section{A proof of Theorem~\ref{T1}} 

\begin{proof}[Step 1]
We compute the fake-2 Selmer set $\Sel_{\mathrm{fake}}^{(2)}(\mathcal{C}_1)$:\\
{\bf MAGMA codes:}{
\begin{lstlisting}
P<x> := PolynomialRing(Rationals());
C1 := HyperellipticCurve((x^2+1)*(x^2+3)*(x^2+7));
Sel, mSel := TwoCoverDescent(C1);
#Sel;
A<th> := Domain(mSel);
Sel eq {mSel(x0 - th): x0 in
{th+1,1,-1}};
\end{lstlisting}
}
{\bf Output:}
\begin{lstlisting}
3
true
\end{lstlisting}

The last line shows that for every rational point $(x,y)\in \mathcal{C}(\Q)$, there exists $a\in \Q$ such that
\begin{align}\label{C1}
x-\alpha \in a\Q(\alpha)^2\quad \forall \alpha &\in\{i,\sqrt{-3},\sqrt{-7}\},\vphantom{\dfrac{x-\alpha}{1-\alpha}}
\\[0.5em]
\text{or}\mkern 30mu\label{C2}
\dfrac{x-\alpha}{1-\alpha}\in a\Q(\alpha)^2\quad\forall \alpha&\in \{i,\sqrt{-3},\sqrt{-7}\},
\\[0.5em]
\text{or}\quad\label{C3}
\dfrac{x-\alpha}{-1-\alpha}\in a\Q(\alpha)^2\quad\forall \alpha&\in \{i,\sqrt{-3},\sqrt{-7}\}.
\end{align}
\let\qed\relax
\end{proof}

\begin{proof}[Step 2]
We use the elliptic curve Chabauty method:

\begin{proof}[Case~(\ref{C1})]
Then
\[
\mathcal{E}_1\colon z^2= (x-i)(x-\sqrt{-7})(x^2+3),
\]
where $z\in K=\Q(i,\sqrt{-7})$. The curve $\mathcal{E}_1$ has rank $2$. The elliptic curve Chabauty routine in MAGMA~\cite{MAGMA} works at $p=5$ and shows that there are no points $(x,z)$ in $\mathcal{E}_1(K)$ with the rational $x$-coordinate.
\let\qed\relax
\end{proof}

\begin{proof}[Case~(\ref{C2})] Then
\[
\mathcal{E}_2\colon z^2=2(1-i)(1-\sqrt{-3})(x-i)(x-\sqrt{-3})(x^2+7),
\]
where $z\in K=\Q(i,\sqrt{-3})$. The curve $\mathcal{E}_2$ has rank $2$. The elliptic curve Chabauty routine works at $p=5$ {with the auxiliary prime $13$,} and shows that every point $(x,z)\in \mathcal{E}_2(K)$ with $x\in \Q$ satisfies $x=\pm 1$. Hence in~\eqref{E1}{,} we have $x=\pm 1$ and $y=\pm 8$.
\let\qed\relax
\end{proof}

\begin{proof}[Case~(\ref{C3})]
By taking the complex conjugate and mapping $x$ to $-x$, we have Case~\eqref{C2}. Hence $x=\pm1$ {and $y=\pm 8$.}
\let\qed\relax
\end{proof}
\let\qed\relax
\end{proof}

\subsection*{Acknowledgment}

The author would like to thanks Professor Andrew Bremner for his help during the preparation of this paper.

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