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 \markboth{\qquad A Diophantine system \hfill}{\hfill Andrew Bremner
\qquad}

 \title{A Diophantine system}
 \author{Andrew Bremner}
 \date{\em Dedicated to Peter Swinnerton-Dyer on \\
 the occasion of his seventy-fifth birthday}

\begin{document}
\maketitle
\addcontentsline{toc}{chapter}{Andrew Bremner, A Diophantine system}
% \tableofcontents

This note concerns the Diophantine system
\begin{eqnarray}
\label{surface}
x_1^2+x_2^2+x_3^2 & = & y_1^2+y_2^2+y_3^2 \nonumber \\
x_1^3+x_2^3+x_3^3 & = & y_1^3+y_2^3+y_3^3 \\
x_1^4+x_2^4+x_3^4 & = & y_1^4+y_2^4+y_3^4 \nonumber
\end{eqnarray}
which represents a surface of degree 24. 
The system is of interest in that Palam\`a~\cite{Pa}
in 1951 showed that the only {\it real} points on (\ref{surface}) in the
positive quadrant ($x_i>0$, $y_i>0$ for $i=1,2,3$) are trivial points,
that is, points where $(x_1,x_2,x_3)$ is a permutation of $(y_1,y_2,y_3)$.
Geometrically, the only real points on (\ref{surface}) in the positive 
quadrant lie upon a finite number of planes.
Choudhry~\cite{Ch} discovers the nontrivial rational point (which we also
refer to as a nontrivial {\it solution})
 \[
 (x_1,x_2,x_3;y_1,y_2,y_3) = (358, -815, 1224; -776, 1233, -410),
 \]
and we observe that this point also satisfies $x_1+x_2+y_1+y_2 = 0$.
In this note, we investigate the section of the surface (\ref{surface})
cut by the plane
 \begin{equation}
\label{plane}
 x_1+x_2 = t(y_1+y_2) 
 \end{equation}
for $t=\pm1$. There are only finitely many nontrivial points on 
the section with $t=1$, but infinitely many nontrivial points on the 
section with $t=-1$.

First, make the substitution
\begin{eqnarray}
\label{param}
x_1 = a_1 m+b_1 n, & & y_1=a_1 m-b_1 n \nonumber \\
x_2 = a_2 m+b_2 n, & & y_2=a_2 m-b_2 n\\
x_3 = a_3 m+b_3 n, & & y_3=a_3 m-b_3 n \nonumber
\end{eqnarray}
where $m n \neq 0$, so that (\ref{plane}) above becomes
 \[
 m(t-1)(a_1+a_2) = n(t+1)(b_1+b_2).
 \]
\subsection*{Case I: $t=1$}
Then $b_2=-b_1$, and substituting (\ref{param}) into (\ref{surface})
gives:
\begin{eqnarray}
\label{mn}
a_1 b_1-a_2 b_1+a_3 b_3 & = & 0 \nonumber \\
3(a_1^2 b_1-a_2^2 b_1+a_3^2 b_3)m^2+b_3^3 n^2 & = & 0\\
(a_1^3 b_1-a_2^3 b_1+a_3^3 b_3)m^2+(a_1 b_1^3-a_2 b_1^3+a_3 b_3^3)n^2 & = & 0. \nonumber
\end{eqnarray}
Nontrivial solutions demand $a_1 \neq a_2$, $a_3 \neq 0$. 
Eliminating $m$, $n$ in (\ref{mn}) and using $b_3=(-a_1+a_2) b_1/a_3$
gives
\begin{multline*}
a_1^4-a_1^3 a_2-a_1 a_2^3+a_2^4-3 a_1^3 a_3+3 a_1^2 a_2 a_3+3 a_1 a_2^2
a_3 \\ -3a_2^3 a_3+2a_1^2 a_3^2-4a_1 a_2 a_3^2+2a_2^2 a_3^2+3a_1
a_3^3+3a_2 a_3^3-3a_3^4 = 0.
\end{multline*}
This quartic curve is singular at the point $(a_1,a_2,a_3)=(1,1,0)$, and has genus 2.
Put
 \[
 a_1=u, \quad a_2=u-v, \quad a_3=w
 \]
to give
\begin{equation}
\label{uvw}
3 u^2 v^2 -3(v^3+2 v^2 w-2 w^3) u+(v^2-w^2)(v^2+3 v w+3 w^2) = 0.
\end{equation}
The discriminant (as function of $u$) being square implies
\begin{equation} 
\label{genus2}
-3(v^6-4 v^4 w^2+12 v^2 w^4-12 w^6) = \mbox{square}.
\end{equation}
This latter curve of genus $2$ has of course only finitely many
rational points. Its Jacobian is isogenous to the product of
the two elliptic curves
 \[
 \begin{matrix}
 E_1: & -3(V^3-4V^2+12V-12) &=&S_1^2, \\[4pt]
 E_2: & -3(1-4W+12W^2-12W^3)&=&S_2^2,
 \end{matrix}
 \]
both of which are of rank 1 (with generators $(0,6)$ and $(1,3)$ 
respectively).  Accordingly, Chabauty's method (see for example 
Coleman~\cite{Col}) for determining the rational points
on (\ref{genus2}) does not apply. It is possible that methods of
Flynn and Wetherell~\cite{Fl} may be effective, but we have not
pursued the calculation; see also Bruin and Elkies~\cite{BE}.
In any event, there are only finitely many solutions of the
system (\ref{surface}) satisfying $x_1+x_2 = y_1+y_2$, and their 
determination is afforded by finding all rational points on the 
curve (\ref{genus2}).  A modest computer search finds only the 
points $(u,v,w)=(1,0,0)$, $(1,0,2)$, $(0,1,1)$, $(1,1,1)$
on (\ref{uvw}), corresponding to trivial solutions of (\ref{surface}).

 \subsection*{Case II: $t=-1$}
Now $a_1+a_2=0$, and 
\begin{eqnarray}
\label{case2}
a_1 b_1-a_1 b_2+a_3 b_3 & = & 0, \nonumber \\
3 (a_1^2 b_1+a_1^2 b_2+a_3^2 b_3)m^2+(b_1^3+b_2^3+b_3^3)n^2 & = & 0,\\
(a_1^3 b_1-a_1^3 b_2+a_3^3 b_3)m^2+(a_1 b_1^3-a_1 b_2^3+a_3 b_3^3)n^2 & = & 0. \nonumber
\end{eqnarray}
Nontrivial solutions demand $b_1 \neq b_2$, $a_3 \neq 0$.
Eliminating $m$, $n$ at (\ref{case2}) and using $b_3=(-b_1+b_2)a_1/a_3$
gives
 \[
 a_1(b_1-b_2)(a_1b_1-a_3b_1-a_1b_2-a_3b_2)P(a_1,a_3,b_1,b_2) = 0,
 \]
where
 \begin{multline*}
 P(a_1,a_3,b_1,b_2) = 
a_1^4 b_1^2-2 a_1^3 a_3 b_1^2+2 a_1 a_3^3 b_1^2-a_3^4 b_1^2-2
a_1^4 b_1 b_2 \\ -8 a_1^2 a_3^2 b_1 b_2+a_3^4 b_1 b_2
+a_1^4 b_2^2+2 a_1^3 a_3 b_2^2-2 a_1 a_3^3 b_2^2-a_3^4 b_2^2.
\end{multline*}

Consequently, either $a_1=0$, or $b_1=b_2$, or $(a_1-a_3)b_1 =
(a_1+a_3)b_2$, all of which lead to trivial solutions of the original
system, or
 \[
 (a_1-a_3)^3 (a_1+a_3)b_1^2-(2 a_1^4+8 a_1^2 a_3^2-a_3^4)b_1
b_2+(a_1-a_3)(a_1+a_3)^3 b_2^2 = 0.
 \]
 The discriminant of the latter is
 \[
 3 a_3^2 (2 a_1^2+a_3^2)^2 (4 a_1^2-a_3^2)
 \]
which accordingly is square precisely when
 \[
 4 a_1^2-a_3^2 = 3 \times (\mbox{square}).
 \]
Put
 \[
 a_1 = 3 u^2+v^2 \quad\hbox{and}\quad a_3 = 6 u^2-2 v^2,
 \]
so that (without loss of generality, on changing the sign of $v$ if necessary)
 \[
 \frac{b_1}{b_2} = -\frac{(u-v)(3u+v)^3}{3(3u-v)(u+v)^3}
 \]
and
 \[
 \frac{b_3}{b_2} = \frac{2(3u^2+v^2)(3u^2+2 u v+v^2)}{3(3u-v)(u+v)^3}.
 \]
 Then from (\ref{case2}),
 \[
 -9(3u-v)^2(u+v)^6m^2+b_2^2(9u^4-24u^3 v-26u^2v^2-8u v^3+v^4)n^2 = 0,
 \]
that is,
\begin{equation}
\label{ellcurve}
U^2 = V^4-8 V^3-26 V^2-24 V+9,
\end{equation}
where 
\[ V=v/u, \qquad U=\frac{3(3u-v)(u+v)^3}{b_2 u^2} m/n. \]
A Weierstrass model for (\ref{ellcurve}) is 
\begin{equation}
\label{ellWeier}
y^2 = x^3 + x^2 -4 x + 32,
\end{equation}
and using the APECS program~\cite{Co} of Ian Connell, or the tables 
of Cremona (\cite{Cr}), where (\ref{ellWeier}) is numbered as the curve
552E1, we discover that (\ref{ellcurve}) is of rank~1, with generator 
$P(V,U)=(-9/4, -111/16)$. Accordingly, we can construct infinitely many
rational points (equivalently, integer points) on (\ref{surface}).
Indeed, from
\begin{gather*}
 a_1=3u^2+v^2, \qquad a_2=-3u^2-v^2, \qquad a_3=6u^2-2v^2, \\
 b_1=-(u-v)(3u+v)^3, \qquad b_2=3(3u-v)(u+v)^3 \\
 \hbox{and}\quad b_3 = 2(3u^2+v^2)(3u^2+2u v+v^2),
\end{gather*}
with
\[ U^2 = V^4-8 V^3-26 V^2-24 V +9, \qquad V=\frac{v}{u}, 
\qquad U=\frac{1}{u^2} \; m/n, \]
we obtain
\begin{eqnarray}
x_1 & = & (3+V^2)U-27+18V^2+8V^3+V^4, \nonumber\\
x_2 & = & -(3+V^2)U+9+24V+18V^2-3V^4, \nonumber\\
x_3 & = & 2(3-V^2)U+18+12V+12V^2+4V^3+2V^4, \nonumber\\
y_1 & = & (3+V^2)U+27-18V^2-8V^3-V^4,\\
y_2 & = & -(3+V^2)U-9-24V-18V^2+3V^4, \nonumber\\
y_3 & = & 2(3-V^2)U-18-12V-12V^2-4V^3-2V^4. \nonumber
%x_1 & = & -27+(3+V^2)U+18V^2+8V^3+V^4, \nonumber\\
%x_2 & = & 9-(3+V^2)U+24V+18V^2-3V^4, \nonumber\\
%x_3 & = & -2(-9-(3-V^2)U-6V-6V^2-2V^3-V^4), \nonumber\\
%y_1 & = & 27+(3+V^2)U-18V^2-8V^3-V^4,\\
%y_2 & = & -9-(3+V^2)U-24V-18V^2+3V^4, \nonumber\\
%y_3 & = & -2(9-(3-V^2)U+6V+6V^2+2V^3+V^4). \nonumber
\end{eqnarray}

\noindent
The point $P(V,U)=(-9/4, -111/16)$ pulls back to the solution
\[  (-815, 358, 1224; -776, 1233, -410), \]
and the point $2P(V,U)=(-148/33, 29219/1089)$ to the solution
 \[
 (378382959, -931219912, -156845590; 357088490, 195748463, -932263416).
 \] 

\begin{thebibliography}{1}
 \bibitem{BE} Nils Bruin and Noam D. Elkies, {\em Trinomials $a x^7+b
x+c$ and $a x^8+b x+c$ with Galois groups of order $168$ and $8 \cdot
168$}, Proceedings 5th International Symposium, ANTS-V, Sydney, 2002,
Springer-Verlag, 172--188
 \bibitem{Ch} Ajai Choudhry, {\em Equal sums of like powers}, Rocky
Mountain Journal of Mathematics {\bf31} (2001) 115--129
 \bibitem{Col} Robert F. Coleman, Effective Chabauty, Duke Math. J.
{\bf52} (1985) 765--770
 \bibitem{Co} Ian Connell, APECS, www.math.mcgill.ca/connell/public/apecs
 \bibitem{Cr} John E. Cremona, Algorithms for modular elliptic curves,
Cambridge University Press, 1997
 \bibitem{Fl} Victor Flynn and J.L. Wetherell, {\em Finding rational
points  on bielliptic genus 2 curves}, Manuscripta Math. {\bf100} (1999)
519--533
 \bibitem{Pa} Giuseppe Palam\`a, {\em Sistemi indeterminati impossibili}, 
Boll. Un. Mat. Ital. (3) {\bf6} (1951) 113--117
\end{thebibliography}

\bigskip
\noindent
Andrew Bremner, \\
Department of Mathematics and Statistics, \\
Arizona State University, \\
Tempe, AZ 85287-1804, USA \\
e-mail: bremner@asu.edu

\end{document}


editing scrap

%(a_1^4 b_1^2\!-\!2 a_1^3 a_3 b_1^2\!+\!2 a_1 a_3^3 b_1^2\!-\!a_3^4 b_1^2\!-\!2 a_1^4 b_1 b_2\!-\!8 a_1^2 a_3^2 b_1 b_2\!+\!a_3^4 b_1 b_2\\
%+\!a_1^4 b_2^2\!+\!2 a_1^3 a_3 b_2^2\!-\!2 a_1 a_3^3 b_2^2\!-\!a_3^4 b_2^2)\\


 \begin{multline*}
a_1 (b_1 - b_2) (a_1 b_1 - a_3 b_1 - a_1 b_2 - a_3 b_2)\\
(a_1^4 b_1^2-2 a_1^3 a_3 b_1^2+2 a_1 a_3^3 b_1^2-a_3^4 b_1^2-2
a_1^4 b_1 b_2-8 a_1^2 a_3^2 b_1 b_2+a_3^4 b_1 b_2\\
+a_1^4 b_2^2+2 a_1^3 a_3 b_2^2-2 a_1 a_3^3 b_2^2-a_3^4 b_2^2)
= 0.
\end{multline*}

