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\vskip 1cm{\LARGE\bf 14-term Arithmetic Progressions \\
\vskip .1in
on Quartic Elliptic Curves}

\vskip 1cm
\large
Allan J. MacLeod\\
Department of Mathematics and Statistics\\
University of Paisley\\
High St.\\
Paisley, Scotland PA1 2BE\\
UK\\
\href{mailto:allan.macleod@paisley.ac.uk}{allan.macleod@paisley.ac.uk}\\
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\bigskip

\begin{abstract}
Let $P_4(x)$ be a rational quartic polynomial which is not the
square of a quadratic. Both Campbell and Ulas considered the
problem of finding an rational arithmetic progression
$x_1,x_2,\ldots,x_n$, with $P_4(x_i)$ a rational square for $1
\le i \le n$. They found examples with $n=10$ and $n=12$. By
simplifying Ulas' approach, we can derive more general parametric
solutions for $n=10$, which give a large number of examples with
$n=12$ and a few with $n=14$.
\end{abstract}

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\newtheorem{theorem}{Theorem}[section]
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\section{Introduction}
Let $P_4(x)=a x^4+b x^3+c x^2+d x+e$ be a rational polynomial
which is not the square of a quadratic. If $P_4(s)=t^2$ for
rational $(s,t)$ then $P_4$ is birationally equivalent to an
elliptic curve. Campbell \cite{camp} investigated the possibility
of such curves having a rational arithmetic progression
$x_1,x_2,\ldots,x_n$ of arguments such that $P_4(x_i)=y_i^2,
i=1,\ldots,n$, and provided an example with $n=12$.

Ulas \cite{ulas} approached the problem in a different manner and
succeeded in deriving a parametric solution for $n=10$, and used a
specific elliptic curve to derive a family of solutions for
$n=12$. He also introduced the name {\it quartic elliptic curve}
for such curves.

In this short note, we show that Ulas' approach can be
considerably simplified. This allows us to find a more general
parametric solution for $n=10$, which generates a large number of
solutions for $n=12$ and a few for $n=14$, thus answering the open
question at the end of Ulas' paper.

\section{Algebraic Formulation}
Ulas considers the arithmetic progression ( AP for short )
$\{1,2,3,\ldots,9,10\}$, and fits $P_4$ at $\{1,2,3,4,5\}$ to the
set of values $\{p^2,q^2,r^2,s^2,t^2\}$. He then forces $P_4$ at
$\{6,7,8,9,10\}$ to fit the set $\{t^2,s^2,r^2,q^2,p^2\}$.

It is a simple observation that this is equivalent to enforcing
$P_4$ to be symmetric about $x=5.5$. But we can easily translate
the x-arguments so that the quartic is symmetric about $x=0$,
which makes $P_4$ an even function. Thus, the new approach is to
assume $P_4(x)=a x^4+bx^2+c$.

Set $P_4(1)=P_4(-1)=p^2$, $P_4(3)=P_4(-3)=q^2$, and
$P_4(5)=P_4(-5)=r^2$. It is standard linear algebra to find
\begin{displaymath}
a=\frac{2p^2-3q^2+r^2}{384}
\end{displaymath}

\begin{displaymath}
b=-\frac{34p^2-39q^2+5r^2}{192}
\end{displaymath}

\begin{displaymath}
c=\frac{150p^2-25q^2+3r^2}{128}
\end{displaymath}

Enforcing $P_4(7)=P_4(-7)=s^2$ implies that we must have
\begin{displaymath}
s^2=5p^2-9q^2+5r^2
\end{displaymath}

Since $(1,1,1,1)$ is an obvious solution, we can parameterize as
follows
\begin{displaymath}
p=-5u^2-9v^2+18uv+5w^2-10uw
\end{displaymath}

\begin{displaymath}
q=5u^2-10uv+9v^2-10vw+5w^2
\end{displaymath}

\begin{displaymath}
r=5u^2-10uw-9v^2+18vw-5w^2
\end{displaymath}

Now, setting $P_4(9)=P_4(-9)=t^2$, gives
\begin{equation}
t^2=25u^4+Eu^3+Fu^2+Gu+H
\end{equation}
with
\begin{itemize}
\item[(i)]$E=40(15w-7v)$,
\item[(ii)]$F=2(347v^2-680vw+25w^2)$,
\item[(iii)]$G=-8(63v^3+65v^2w-235vw^2+75w^3)$,
\item[(iv)]$H=81v^4+1440v^3w-2330v^2w^2+800vw^3+25w^4$
\end{itemize}

We consider equation $(1)$ as a quartic in $u$, with $v,w$ as free
parameters. Investigations show simple rational values of $u$
which give rational $t$. One such is $u=9v/5-w$ which gives
$t=2(3v-5w)(6v-5w)/5$. The existence of this point shows that the
quartic is birationally equivalent to an elliptic curve.

Using the method described in Mordell \cite{mord}, we find that
the elliptic curve can be written in the form
\begin{equation}
J^2=K^3-(137v^2-680vw+475w^2)K^2
\end{equation}
\begin{displaymath}
+84(54v^4-495v^3w+1425v^2w^2-1625vw^3+625w^4)K
\end{displaymath}
with the transformation
\begin{equation}
u=\frac{2K(7v-15w)+J-42(27v^3-195v^2w+325vw^2-125w^3)}{5K-210(3v^2-20vw+25w^2)}
\end{equation}

This elliptic curve has an obvious point of order 2, namely
$(0,0)$. Numerical investigations suggest this is the only
torsion point, but this would appear to be difficult to prove.

These numerical investigations, using small integer values of $v$
and $w$, also indicate that the curve has rank at least $2$, and
often $3$ or higher, with many integer points of small height.

Investigations found several algebraic formulae for points on the
curves. Two of these are
\begin{displaymath}
( \; 2(3v-5w)(6v-5w) \; , \; \pm 10(v-5w)(3v-5w)(6v-5w) \; )
\end{displaymath}
and
\begin{displaymath}
( \; 6(3v-w)^2 \; , \;  \pm 6(3v-w)(3v^2-106vw+91w^2) \; )
\end{displaymath}

It is possible to use these and the previous algebraic
expressions to derive parametric formulae for quartic elliptic
curves with 10-term APs, but these expressions will only cover a
small fraction of possible curves. In the next section we employ
a simple search procedure for 12-term and 14-term curves.

\section{Curve Searching}
As was stated before, the elliptic curves $(2)$ contain a large
number of integer points of small height. For a selected pair of
$(v,w)$ values, we searched for rational points with $|K| &lt; 10^7$.
From these points, we generated $u$, then $(p,q,r)$ and finally
$(a,b,c)$.

For the quartic produced, we tested if $P_4(11)$ was a square,
giving a 12-term AP. We found several hundred such curves with
$(v,w)$ in the range $|v|+|w|&lt;100$.

For the successful curves, $P_4(13)$ was tested to be square. This
discovered a total of $4$ quartics giving a 14-term AP. These are

\begin{enumerate}
\item[(a)] $-17x^4+3130x^2+8551$,
\item[(b)] $2002x^4-226820x^2+18168514$,
\item[(c)] $3026x^4-222836x^2+3709234$,
\item[(d)] $34255x^4-1436006x^2+447963175$
\end{enumerate}

Quartic (a) was derived from $3$ different $(v,w)$ pairs while
(c) came from $2$ different pairs. Given the paucity of 14-term
curves compared to the abundance of 12-term curves, we would be
very unlikely to find a 16-term curve with this method.

\section{Acknowledgement}
The author would like to express his sincere thanks to the
anonymous referee for several very useful comments on the
original submission.


\begin{thebibliography}{20}

\bibitem{camp} G. Campbell,
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL6/Campbell/campbell4.html}{A
note on arithmetic progressions on elliptic curves}, 
{\it J. Integer Sequences}, Paper 03.1.3, 2003.

\bibitem{mord} L. J. Mordell, {\it Diophantine Equations}, Academic
Press, New York, 1969.

\bibitem{ulas} M. Ulas, 
\href{http://www.cs.uwaterloo.ca/journals/JIS/VOL8/Ulas/ulas7.html}{A note on arithmetic progressions on
quartic elliptic curves}, {\it J. Integer Sequences}, Paper
05.3.1, 2005.

\end{thebibliography}

\bigskip
\hrule
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\noindent 2000 {\it Mathematics Subject Classification}:
Primary 11G05; Secondary 11B25.

\noindent \emph{Keywords: } arithmetic progression, quartic, elliptic curve.

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received October 25 2005;
revised version received  November 18 2005.
Published in {\it Journal of Integer Sequences}, November 18 2005.

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