%~Mouliné par MaN_auto v.0.35.3 (13ef1c44) 2025-08-29 13:17:08
\documentclass[CRMATH,Unicode,XML,biblatex]{cedram}

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

\addbibresource{CRMATH_Nguyen-Xuan_20250373.bib}

\usepackage{amssymb}

\newcommand{\Z}{{\mathbb Z}}
\newcommand{\Q}{{\mathbb Q}}

\DeclareMathOperator{\rank}{rank}

%%%---------------------------------------------------------------------------

%%% DELIMITERS

\usepackage{mathtools}
\DeclarePairedDelimiter{\parens}{\lparen}{\rparen}

%%%---------------------------------------------------------------------------

\graphicspath{{./figures/}}

\newcommand*{\mk}{\mkern -1mu}
\newcommand*{\Mk}{\mkern -2mu}
\newcommand*{\mK}{\mkern 1mu}
\newcommand*{\MK}{\mkern 2mu}

\hypersetup{urlcolor=purple, linkcolor=blue, citecolor=red}

\newcommand*{\relabel}{\renewcommand{\labelenumi}{(\theenumi)}}
\newcommand*{\romanenumi}{\renewcommand*{\theenumi}{\roman{enumi}}\relabel}
\newcommand*{\Romanenumi}{\renewcommand*{\theenumi}{\Roman{enumi}}\relabel}
\newcommand*{\alphenumi}{\renewcommand*{\theenumi}{\alph{enumi}}\relabel}
\newcommand*{\Alphenumi}{\renewcommand*{\theenumi}{\Alph{enumi}}\relabel}

%%%---------------------------------------------------------------------------

\makeatletter
\def\@setafterauthor{%
  \vglue3mm%
\begingroup\hsize=13cm\advance\hsize\abstractmarginL\raggedright
\noindent
   \leftskip\abstractmarginL
  \normalfont\Small
  \@afterauthor\par
\endgroup
\vskip2pt plus 3pt minus 1pt
}
\makeatother

%%%---------------------------------------------------------------------------

\title{Note on a symmetric Diophantine equation}
\alttitle{Note sur une équation diophantienne symétrique}

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

\thanks{The author is supported by the Vietnam National Foundation for Science and Technology Development (NAFOSTED) (grant number 101.04-2023.21) and by Vietnam Institute for Advanced Study in Mathematics (VIASM) from April 2025 to May 2025. The author gratefully acknowledges the Institute for their support and funding}
\CDRGrant[Vietnam National Foundation for Science and Technology Development (NAFOSTED)]{101.04-2023.21}

\keywords{\kwd{Arithmetic geometry} \kwd{Diophantine equations} \kwd{rational points}}
\altkeywords{\kwd{Géométrie arithmétique} \kwd{équations diophantiennes} \kwd{points rationnels}}

\subjclass{14G05, 14G12}

\begin{abstract}
Using an elementary argument, we show that for all rational numbers $\alpha$ such that neither $\alpha$ nor $3\alpha$ is a rational square, the equation
\[
x^4-4\alpha x^2-4\alpha y^2+y^4=-6\alpha^2
\]
has no rational solutions. This answers Hindes' two questions and generalizes his theorem (Theorem~1.1) in ``Rational points on certain families of symmetric equations'', \textit{Int.\ J.\ Number Theory} \textbf{11} (2015), no.\ 6, pp.\ 1821--1838.
\end{abstract}

\begin{altabstract}
En utilisant un argument élémentaire, nous montrons que pour tous les nombres rationnels $\alpha$ tels que ni $\alpha$ ni $3\alpha$ n'est un carré rationnel, l'équation
\[
x^4-4\alpha x^2-4\alpha y^2+y^4=-6\alpha^2
\]
n'a pas de solutions rationnelles. Ceci répond aux deux questions de Hindes et généralise son théorème (Theorem~1.1) dans ``Rational points on certain families of symmetric equations'', \textit{Int.\ J.\ Number Theory} \textbf{11} (2015), no.\ 6, pp.\ 1821--1838.
\end{altabstract}

\begin{document}

\maketitle

\section{Introduction}\label{Sec1}

Let $a,b\in \Q$. Consider the symmetric quartic equation
\begin{equation}\label{EQ1}
x^4+ax^2+ay^2+y^4=b
\end{equation}
with rational numbers $x,y$. When $a=0$, equation~\eqref{EQ1} reduces to the Fermat quartic
\begin{equation}\label{EQ2}
x^4+y^4=b.
\end{equation}
Dem'janenko~\cite{Dem} studied rational solutions of~\eqref{EQ2} using the elliptic curve $y^2=x(x^2+b)$. Silverman~\cite{Silverman} extended Dem'janenko's result to number fields. Hindes~\cite{Hindes} further extended Dem'janenko's method to study rational solutions of a family of equations~\eqref{EQ1} with $a\neq 0$. For each pair of rational numbers $(a,b)$ with $b(a^2+4b)(a^2+2b)\neq 0$, let us define the curves
\begin{equation}\label{EQ3}
\begin{cases}
F=F_{(a,b)}\colon x^4+ax^2+a y^2+y^4=b,\\
E=E_{(a, b)}\colon y^2=x\parens[\big]{x^2-4 a x-(16 b+4 a^2)}.
\end{cases}
\end{equation}
There are two maps $\phi_1, \phi_2 \colon F_{(a, b)} \to E_{(a, b)}$ given by
\[
\begin{cases}
\phi_1(x, y)=\parens[\big]{-4 x^2, x(8 y^2+4 a)},\\ \phi_2(x, y)=\parens[\big]{-4 y^2, y (8 x^2+4 a)}.
\end{cases}
\]
For each squarefree positive integer $\alpha$, let $F^{(\alpha)}=F_{(\alpha \cdot a, \alpha^2 \cdot b)}$ and $E^{(\alpha)}=E_{(\alpha \cdot a, \alpha^2 \cdot b)}$. Hindes~\cite[Theorem~1.1]{Hindes} proved the following theorem.

\begin{theo}\label{TH1}
Let $F$ and $E$ be the symmetric quartic and elliptic curves defined in~\eqref{EQ3} corresponding to $(a, b)=(-4,-6)$. Then the following statements hold.
\begin{enumerate}
\item \label{TH1_1} If $p$ is a prime number such that $p \equiv 1 \bmod 24$, then $F^{(p)}$ is everywhere locally solvable.
\item \label{TH1_2} The global root number $W(E^{(p)})=-1$ for all positive, odd primes.
\item \label{TH1_3} If $\alpha>7 \cdot 10^{74}$ is square-free and $\rank\parens[\big]{E^{(\alpha)}(\mathbb{Q})} \leq 1$, then $F^{(\alpha)}(\mathbb{Q})=\varnothing$.
\item \label{TH1_4} If $p \not \equiv \pm 1 \bmod 16$, then $\rank(E^{(p)}) \leq 2$.
\item \label{TH1_5} Assuming the parity conjecture, if $p>3 \cdot 10^{74}$ and $p \equiv 25 \bmod 48$, then $F^{(p)}(\mathbb{Q})=$ $\varnothing$, and $F^{(p)}$ breaks the Hasse principle.
\end{enumerate}
\end{theo}

Hindes posed the following two questions about Theorem~\ref{TH1}.
\begin{enumerate}
\item Is part~\eqref{TH1_3} of Theorem~\ref{TH1} still true when $\alpha<7\cdot 10^{74}$ and $\alpha\neq 3$?
\item Is $F^{(577)}(\Q)$ empty?
\end{enumerate}
In this paper, we answer these two questions by proving the following theorem.

\begin{theo}\label{T1}
Let $\alpha$ be a rational number such that $\alpha$ and $3\alpha$ are not rational squares. Then the equation
\begin{equation}\label{E1}
x^4-4\alpha x^2-4\alpha y^2+y^4=-6\alpha^2
\end{equation}
has no rational solutions.
\end{theo}

Theorem~\ref{T1} provides a significant generalization of parts~\eqref{TH1_3} and~\eqref{TH1_5} of Theorem~\ref{TH1}. The conditions $\alpha>7\cdot 10^{74}$ and $\rank\parens[\big]{E^{\alpha}(\Q)}\leq 1$ in part~\eqref{TH1_3} are not necessary. The assumption of the parity conjecture and the condition $p>3\cdot 10^{74}$ in part~\eqref{TH1_5} are not required. In contrast to Hindes' approach, our proof of Theorem~\ref{TH1} is short and elementary. For other studies on the rational solutions to~\eqref{EQ2}, see Bremner and Morton~\cite{BM}, Serre~\cite{Serre}, Flynn and Wetherell~\cite{Flynn}, Cohen~\cite{Cohen}, and Bremner and Tho~\cite{BT}.

Combining part~\eqref{TH1_1} of Theorem~\ref{TH1} with Theorem~\ref{T1}, we obtain the following corollary.

\begin{coro}
Let $p$ be a prime number such that $p\equiv 1 \bmod 24$. Then:
\begin{enumerate}
\item $F^{(p)}$ is everywhere locally soluble;
\item $F^{(p)}(\Q)=\emptyset $.
\end{enumerate}
In particular, $F^{(p)}$ is a counterexample to the Hasse principle.
\end{coro}

\section{Proof of Theorem~\ref{T1}}
\begin{proof}

Assume that $(x,y)$ is a rational solution to~\eqref{E1}. Equation~\eqref{E1} can be written in the form
\begin{equation}\label{E2}
(x^2-2\alpha)^2+(y^2-2\alpha)^2=2\alpha^2.
\end{equation}
Since the equation $X^2+Y^2=2$ has a parameterization
\[
X=\dfrac{-t^2-2t-1}{t^2+1}, \quad Y=\dfrac{t^2-2t-1}{t^2+1},
\]
it follows from~\eqref{E2} that there exists a rational number $t$ such that
\[
\begin{cases}
x^2-2\alpha&=\dfrac{(-t^2-2t-1)\alpha}{t^2+1},\\
y^2-2\alpha&=\dfrac{(t^2-2t-1)\alpha}{t^2+1}.
\end{cases}
\]
Hence
\begin{equation}\label{E3}
\begin{cases}
x^2&=\dfrac{(t^2-2t+3)\alpha}{t^2+1},\\
y^2&=\dfrac{(3t^2-2t+1)\alpha}{t^2+1}.
\end{cases}
\end{equation}
Let $v=xy/\alpha$. Then we obtain
\begin{equation}\label{E4}
v^2=(t^2-2t+3)(3t^2-2t+1).
\end{equation}
Let $\mathcal{C}$ be the quartic curve defined by~\eqref{E4}. Since $\mathcal{C}$ has a rational point $(1,2)$, it is an elliptic curve. MAGMA~\cite{MA} shows that a Weierstrass model of $\mathcal{C}$ is
\begin{equation}\label{E9}
y^2 = x^3 - x^2 - 2x,
\end{equation}
which has the Mordell--Weil group isomorphic to $(\Z/2\Z)^2$. Note that $\mathcal{C}$ with model~\eqref{E9} is labeled as 96A in Table~I in the classical book of Birch and Kuyk~\cite{Birch}. The table also shows that $\mathcal{C}$ has rank~0 with the Mordell--Weil group isomorphic to $(\Z/2\Z)^2$.

Since equation~\eqref{E4} has four rational solutions $(t,v)=(\pm 1,\pm 2)$, they are all rational solutions of~\eqref{E4}. It follows from~\eqref{E3} that if neither $\alpha$ nor $3\alpha$ is a rational square, then~\eqref{E1} admits no rational solutions, completing the proof of Theorem~\eqref{T1}.
\end{proof}

\section*{Acknowledgments}

The author is indebted to the anonymous referee for pointing out the error in the previous version of the paper and sincerely appreciates the suggestion, which helped correct the error and improve the presentation of the paper.

\section*{Declaration of interests}

The author does not work for, advise, own shares in, or receive funds from any organization that could benefit from this article, and has declared no affiliations other than their research organizations.

\printbibliography

\end{document}