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\begin{center}
\vskip 1cm{\LARGE\bf A Note on Arithmetic Progressions on Quartic Elliptic Curves}

\vskip 1cm
\large
Maciej Ulas \\
Jagiellonian University \\
Institute of Mathematics\\
Reymonta 4 \\
30-059 Krak\'{o}w \\
Poland \\
\href{mailto:Maciej.Ulas@im.uj.edu.pl}{Maciej.Ulas@im.uj.edu.pl}
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\vskip .2 in
\begin{abstract}
G. Campbell 
described a technique for
producing infinite families  of quartic elliptic curves containing
a length-9 arithmetic progression. He also gave an example of a
quartic elliptic curve containing a length-12 arithmetic
progression. In this note we give a construction of an infinite
family of quartics on which there is an arithmetic progression  of
length 10. Then we show that there exists an infinite family of
quartics containing a sequence of length 12.
\end{abstract}


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% A Note on Arithmetic Progressions on Elliptic Curves
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%    General info
% \subjclass{11G05, 11B25}

% \date{May 19, 2005}

% \keywords{Elliptic Curves, Arithmetic Progression}




% ------------ SECTION 1:  INTRODUCTION -----------------------------

\section{Introduction}

Let us consider a curve $E:\; y^2=f(x)$, where $f \in \Q[x]$ and
$f$ is not a square of a polynomial. We say that points
$P_{i}=(x_{i},\;y_{i}),\;i=1,\ldots,\;k$ on the curve $E$ form an
{\it arithmetic progression of length $k$} if the sequence
$x_{1},\;x_{2},\;\ldots,\;x_{k}$ form an arithmetic progression.

G. Campbell \cite{GC} proved the following theorems:

\begin{theorem}
There are infinitely many elliptic curves of the form $y^2=w(x)$,
with $w(x)$ a quartic, containing 9 points in arithmetic
progression.
\end{theorem}

\begin{theorem}
There exists an elliptic curve in the form $y^2=w(x)$, with $w(x)$
a quartic, containing 12 points in arithmetic progression.
\end{theorem}

In this paper we propose two different ways to obtain an infinite
family of quartic elliptic curves with a sequence of length 10.
Then we present a family of quartics (parametrized by the rational
points on an elliptic curve with a non zero rank) with an
arithmetic progression of length 12.

% ------------ SECTION 2: ARITHMETIC PROGRESSIONS OF LENGTH 10 -----------------------------

\section{Arithmetic Progressions of Length 10}
The proof of the first theorem is similar to the one given by
G. Campbell \cite{GC}.

\begin{theorem}\label{T.1}
There are infinitely many quartic elliptic curves $y^2=f(x)$
containing 10 points in arithmetic progression.
\end{theorem}
\begin{proof}
Let us consider the following polynomial
\begin{equation}\label{R.1}
P_{t}(x)=(x^2-9x-4t)\prod_{i=0}^{9}(x-i)\in \Q(t)[x].
\end{equation}
Then, we have
\begin{align*}
P_{t}(x)=Q_{t}(x)^2-F_{t}(x),
\end{align*}
where $Q_{t}$ is the unique monic polynomial defined over $\Q(t)$
such that $F_{t}$ has degree 4.

The discriminant of the polynomial $F_{t}(x)$ is non zero for
$t\in\Q \setminus S$, where
\begin{center}
$S=\{\pm1, \pm2, \pm4, -5,-6,-8,-11\}$.
\end{center}
 Hence, for such
parameters $t$, the quartic elliptic curve
\begin{equation}\label{R.2}
E_{t}:\;y^2=F_{t}(x)
\end{equation}
 contain the points $P_{i}=(i,\;Q_{t}(i))$ for $i=0,\ldots,\;9$ which
 form an arithmetic progression of length 10 on the curve $E_{t}$.
 \end{proof}

% ------------ SECTION  3:  THE SEQUENCE OF LENGTH 12 -----------------------------

\section{The sequence of length 12}
There doesn't seem to be a way to construct an infinite family
 of quartics with an arithmetic progression of length 12 from
 family of curves (\ref{R.2}).
 In this section we shall construct another
family which will be better for our purposes. First let us
consider the polynomial $f\in\mathbb{Q}[p,\;q,\;r,\;s,\;t][x]$
\begin{equation*}
f(x)=\;\sum_{i=0}^{4}a_{i}x^{i},
\end{equation*}
where
\begin{align*}
a_{0}=\;&5p^2-10q^2+10r^2-5s^2+t^2,\\
a_{1}=\;&\frac{1}{12}(-77p^2+214q^2-234r^2+122s^2-25t^2),\\
a_{2}=\;&\frac{1}{24}(71p^2-236q^2+294r^2-164s^2+35t^2),\\
a_{3}=\;&\frac{1}{12}(-7p^2+26q^2-36r^2+22s^2-5t^2),\\
a_{4}=\;&\frac{1}{24}(p^2-4q^2+6r^2-4s^2+t^2).
\end{align*}
We have
\begin{center}
$f(1)=p^2,\quad f(2)=q^2,\quad f(3)=r^2,\quad f(4)=s^2,\quad
f(5)=t^2$,
\end{center}
so we see that
\begin{center}
$E:\;\;y^2=f(x)$
\end{center}
 is a five parameter family of quartics containing
an arithmetic progression of length 5. In order to obtain a family
with a sequence of length 10 we have to consider the following
system of equations
\begin{equation}\label{R.3}
\begin{cases}
f(6)=p^2-5q^2+10r^2-10s^2+5t^2=P^2, \\
f(7)=5p^2-24q^2+45r^2-40s^2+15t^2=Q^2, \\
f(8)=15p^2-70q^2+126r^2-105s^2+35t^2=R^2,\\
f(9)=35p^2-160q^2+280r^2-224s^2+70t^2=S^2,\\
f(10)=70p^2-315q^2+540r^2-420s^2+126t^2=T^2,
\end{cases}
\end{equation}
in integers $p,\;q,\;r,\;s,\;t,\;P,\;Q,\;R,\;S,\;T$. Since the
general solution is hard to obtain we will look for particular
solutions with $P=t,\;Q=s,\;R=r,\;S=q,\;T=p$. Then, in this case
it is easy to realize that (\ref{R.3}) is equivalent to
\begin{equation}\label{R.4}
\begin{cases}
p^2=15r^2-35s^2+21t^2, \\
q^2=5r^2-9s^2+5t^2.
\end{cases}
\end{equation}
Making a substitution $(p,\;r,\;s,\;t)=(p,\;a+p,\;b+p,\;c+p)$ we
get a parametrized solution of the first equation in (\ref{R.4})
\begin{equation}\label{R.5}
\begin{cases}
p=15a^2-35b^2+21c^2, \\
r=-15a^2+70ab-35b^2-42ac+21c^2,\\
s=15a^2-30ab+35b^2-42bc+21c^2,\\
t=15a^2-35b^2-30ac+70bc-21c^2.
\end{cases}
\end{equation}
Now inserting $r,\;s,\;t$ from (\ref{R.5}) to the second equation
in (\ref{R.4}), we obtain
\begin{align}\label{R.6}
q^2=\;&441c^4-168(15a-7b)c^3+2(675a^2+2520ab-2303b^2)c^2\nonumber\\
&+40(45a^3-189a^2b+63ab^2+49b^3)c\\
&+25(9a^4-96a^3b+278a^2b^2-224ab^3+49b^4)\nonumber.
\end{align}
Moreover, if we take $b=3(a+1),\;c=2(a+1)$, we get that
$q=3(56a^2+142a+91)$. Finally we have
\begin{center}
\begin{equation}\label{R.7}
\begin{cases}
p=T= -3(72a^2+154a+77),\\
q=S=  3(56a^2+142a+91),\\
r=R= -3(40a^2+112a+77),\\
s=Q=  3(24a^2+68a+49),\\
t=P= -3(8a^2+6a-7),
\end{cases}
\end{equation}
\end{center}
which is a solution of (\ref{R.3}). Specializing the polynomial
$f$ as given by (\ref{R.7}) we obtain
\begin{equation*}
f_{a}(x)=\;\sum_{i=0}^{4}a_{i}x^{i},
\end{equation*}
where
\begin{align*}
a_{0}=&\;9(7744a^4+25216a^3+22544a^2-784a-5831),\\
a_{1}=&-66(a+1)(16a+21)(24a^2-52a-119),\\
a_{2}=&\;3(a+1)(16a+21)(48a^2-709a-1085),\\
a_{3}=&\;66(a+1)(5a+7)(16a+21),\\
a_{4}=&-3(a+1)(5a+7)(16a+21),\\
\end{align*}
and the discriminant $R_{a}$ of the polynomial $f_{a}$ is
\begin{align*}
R_{a}=&-419904(a+1)^4(2a+3)^2(4a+5)^2(2a+7)^2(4a+7)^2(5a+7)^3\\
          &\times(6a+7)^2(12a+17)^2(16a+21)^4(1008a^2+1831a+623).
\end{align*}
 Then for
$a\in\mathbb{Q}\setminus W$, where
\begin{center}
$W:=\{-1,-3/2,-5/4,-7/2,-7/4,-7/5,-7/6,-17/12,-21/16\}$,
\end{center} we get a nontrivial quartic elliptic curve
\begin{equation*}
C_{a}:\;y^2=f_{a}(x)
\end{equation*}
containing an arithmetic progression of length 10.

Now we are ready to prove the following:

\begin{theorem}\label{T.2}
There are infinitely many quartic elliptic curves $y^2=f(x)$
containing 12 points in arithmetic progression.
\end{theorem}
\begin{proof} The curve $C_{a}$ defined above contains the 10 points with $x=1,\;\ldots,\;10$. Observe that
\begin{equation*}
f_{a}(0)=f_{a}(11)=9(7744a^4+25216a^3+22544a^2-784a-5831).
\end{equation*}

Now, consider the curve:
\begin{equation*}
E:\;Y^2=9(7744a^4+25216a^3+22544a^2-784a-5831).
\end{equation*}
 A short computer
search reveals that $P=(-1,\;15)$ is a rational point on the
quartic $E$. Using the program APECS \cite{IC} we found that $E$
is birationally equivalent to the elliptic curve
\begin{equation*}
E':\;y^2=x^3-x^2-33433x+2213737.
\end{equation*}
  For the curve $E'$ we have
\begin{equation*}
\operatorname{Tors}E(\mathbb{Q})=\{\mathcal{O},\;(127,\;0)\},
\end{equation*}
and with the help of {\tt mwrank} \cite{JC} we see that the free
part of $E'$ is generated by
\begin{center}
$G_{1}=(77,\;-300),\;G_{2}=(-193,\;-1200),\;G_{3}=(-48,\;-1925)$.
\end{center}
Hence the curve $E$ has an infinite number of rational points and
it is clear that all but finitely many of them leads to the
quartic $C_{a}$ containing arithmetic progression of length 12.
\end{proof}

It is natural to state the following question:

\begin{question}
Is there an quartic elliptic curve $E$ containing a length 13
arithmetic progression?
\end{question}
% ------------ SECTION  4:  ACKNOWLEDGMENT-----------------------------

\section{Acknowledgment}

I would like to thank anonymous referee for his/her valuable
comments.

\vskip 1cm
 \nocite{*}
\begin{thebibliography}{10}


\bibitem{BR}
A. Bremner,
\newblock On arithmetic progressions on elliptic curves.
\newblock {\em Experiment. Math.} {\bf 8} (1999), 409--413.

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

\bibitem{IC}
I. Connel, {\sc APECS}, available from
\href{ftp.math.mcgill.ca/pub/apecs/}{\tt ftp.math.mcgill.ca/pub/apecs/}.

\bibitem{JC}
J. Cremona, {\tt mwrank} program, available from \hfil\break
\href{http://www.maths.nottingham.ac.uk/personal/jec/ftp/progs/}{\tt http://www.maths.nottingham.ac.uk/personal/jec/ftp/progs/}.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: 11G05,
11B25.

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


\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received November 11 2004;
revised version received May 21 2005. 
Published in {\it Journal of Integer Sequences}, May 24 2005.

\bigskip
\hrule
\bigskip



\noindent Return to \htmladdnormallink{Journal of Integer
Sequences home page}{http://www.math.uwaterloo.ca/JIS/}.


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