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\vskip 1cm{\LARGE\bf An Arithmetic Progression on Quintic Curves}

\vskip 1cm \large
Alejandra Alvarado \\
Department of Mathematics\\
University of Arizona\\
617 N. Santa Rita Ave.\\
Tucson, AZ 85721\\
USA \\
\href{mailto:alvarado@math.arizona.edu}{alvarado@math.arizona.edu}
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\vskip .2 in
\begin{abstract}
Consider a degree five curve of the form $y^2=f(x)$ where $f(x)\in
\mathbb{Q} [x]$.  Ulas previously showed the existence of an
infinite family of curves $C$ which contain an arithmetic
progression (AP) of length 11.  The author also found an example of
said curve which contains 12 points in AP.  In this paper, we
construct an infinite family of curves with an AP of length 12.
\end{abstract}

\newtheorem{theorem}{Theorem}[section]
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% An Arithmetic Progression on Quintic Curves
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%    General info
% \subjclass{11G05, 11B25, 14H45}

% \date{June 1, 2009}

% \keywords{quintic curves, arithmetic progression, elliptic curves}




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

\section{Introduction}

An \emph{arithmetic progression} (AP) is a sequence of numbers such
that the difference between any two consecutive numbers is constant.
When we talk about an AP on a curve $y^2=f(x)$, we mean an AP in the
$x$-coordinates.

The study of solutions to diophantine equations has been around
throughout history.  But it was not until the twentieth century that
remarkable theories and results flourished.  The problem of finding
consecutive integer solutions to a diophantine equation can be
traced back to Mohanty~\cite{MO}.  On the curve $y^2=x^3+k$, Mohanty
investigated integral AP's in the $x$ and $y$-coordinates.

Let $f$ be an irreducible degree five polynomial over the rationals.
Consider the hyperelliptic curve $y^2=f(x)$.  Previously,
Ulas~\cite{UL} had found an infinite family of curves with length 11
AP.  By computer search, he found one example with length 12.  We
will show the existence of an infinite family of curves which
contain an AP of length 12.  In order to construct an infinite
family of curves, we make use of Mestre's Theorem \cite{ME}:

\begin{theorem}\label{Mestre}
Let $P(x)$ be a monic polynomial of degree $2n$ defined over a field
$K$.  Then there exist unique polynomials $Q(x)$ and $R(x)$ defined
over $K$ such that
\begin{enumerate}
\item
$P(x)=Q(x)^2-R(x)$
\item
the degree of $R(x)$ is strictly less than $n$.
\end{enumerate}
\end{theorem}

In this paper, we use the above theorem to construct an infinite
family of curves.

% ------------ SECTION 2: ARITHMETIC PROGRESSIONS OF LENGTH 12 -----------------------------

\section{Arithmetic Progressions of Length 12}

In this section, we will prove the following theorem, whose proof
uses similar techniques as Campbell~\cite{GC}.

\begin{theorem}\label{Kai}
There exists an infinite family of curves of the form $y^2=f(x)$
which contain an arithmetic progression of length 12, where $f$ is a
degree five polynomial.  The curves are defined over the rationals.
\end{theorem}
\begin{proof}
Consider the following polynomial in $\mathbb{Q}(u)[x]$.

$$g(u,x)=(x-u)^2\prod_{1\leq i\leq 5}(x^2-i^2)$$
We choose this form of polynomial to maximize the use of symmetry.
Note that $g$ vanishes identically at $x=\pm 1,\ldots, \pm 5.$  By
Mestre's theorem, there exist unique polynomials $h, f
\in\mathbb{Q}(u)[x]$ of degree 6 and 5, respectively such that
$$f(u,x)=h(u,x)^2-g(u,x).$$ We will write
$$h(u,x)=x^6+h_5x^5+h_4x^4+h_3x^3+h_2x^2+h_1x+h_0.$$  Since $f$
is of degree 5, we easily find the coefficients $h_i$.
\begin{equation*}
\begin{split}
f(u,x)&amp; =h(u,x)^2-g(u,x)\\
&amp; = 2(h_5+u)x^{11} + (h_5^2+2h_4-u^2+55)x^{10} + 2(h_5h_4+h_3
-55u)x^9 \\
&amp; \quad + (2h_3h_5 +h_4^2+2h_2 +55u^2- 1023)x^8 + 2(h_2h_5 +
h_3h_4 + h_1+1023u)x^7 \\
&amp; \quad + (h_3^2 +  + 2h_1h_5 + 2h_2h_4 + 2h_0-1023u^2 + 7645)x^6 +
2(h_0h_5 + h_1h_4\\
&amp; \quad + h_2h_3-7645u)x^5 + (2h_0h_4
+ 2h_1h_3 + h_2^2 + 7645u^2 - 21076)x^4 \\
&amp; \quad + (h_0h_3 + h_1h_2+21076u)x^3 + (2h_0h_2+ h_1^2-
21076u^2 + 14400)x^2 \\
&amp; \quad + 2(h_0h_1-14400u)x + (h_0^2+14400u^2)
\end{split}\end{equation*}\newline where
\begin{align*}
h_5&amp; = -u &amp; h_4&amp; = \frac{-55}{2}\\
h_3&amp; = \frac{55}{2}u &amp; h_2&amp; = \frac{1067}{8}\\
h_1&amp; = \frac{-1067}{8}u &amp; h_0&amp; = \frac{-2475}{16}
\end{align*}
We can then write $f$ as
\begin{equation*}\begin{split}
f(u,x)&amp; = \frac{-2475}{8}ux^5 + \left(\frac{2475}{8}u^2 +
\frac{334125}{64}\right) x^4 - \frac{61875}{32}ux^3 +
\left(\frac{-210375}{64}u^2 - \frac{1719225}{64}\right)x^2 \\
&amp; \quad + \frac{797625}{64}ux + \left( 14400u^2 +
\frac{6125625}{256}\right)
\end{split}\end{equation*}
By construction, $f(u,x)=h(u,x)^2$ for $x=\pm 1,\ldots, \pm 5.$  We now
add the constraint that $f$ also be square at $x=0$, so then we have
an arithmetic progression of length at least 11 on the curve $C:
p^2=f(u,x)$. We want $f(u,0)=p^2$. In other words,

$$(120u)^2+(\frac{2475}{16})^2=p^2.$$
Without loss of generality, multiply through by
$(\frac{16}{2475})^2$, then replace $\frac{16}{2475}p$ with $p_1$.
Rewrite as $$(\frac{128}{165}u)^2+1=p_1^2.$$ We have a
parametrization
\begin{align*}
\frac{128}{165}u &amp; =\frac{2t}{t^2-1} &amp; p_1 &amp; =\frac{t^2+1}{t^2-1}\\
\end{align*}
Thus, $u=u(t)=\frac{165t}{64(t^2-1)}$.  Substituting back for $u$ in
$f(u,x)$, and removing any squares, we obtain the curve
\begin{equation*}
\begin{split} P^2 &amp; = (-929280t^3 + 929280t)x^5 \\
&amp; \quad + (6082560t^4 - 9769320t^2 + 6082560)x^4 \\
&amp; \quad + (-5808000t^3 + 5808000t)x^3 \\
&amp; \quad + (-31297536t^4 + 37139697t^2 - 31297536)x^2 \\
&amp; \quad + (37435200t^3 - 37435200t)x \\
&amp; \quad + (27878400t^4 + 55756800t^2 + 27878400)
\end{split}
\end{equation*}
where $P=\frac{2^{18}(t-1)^2(t+1)^2}{15^2}p$.  The above curve has
an arithmetic progression of length at least 11, for
$t\in\mathbb{Q}$ except $t\neq\pm1$.  The arithmetic progression of
the $x$ coordinates is $\{-5,-4,\ldots,4,5\}$. To find an arithmetic
progression of length at least 12 on $C$, we need to determine
whether $x=6$ or $x=-6$ gives us a point on the curve. At $x=-6$ we
have
\begin{equation}\label{quar}
\left(\frac{P}{6}\right)^2 =188449024t^4 + 229333280t^3 -
313007023t^2 - 229333280t + 188449024\end{equation} We find that
$t=\frac{5}{6}$ gives $p=\frac{34001}{6}$.  Note that because of
symmetry, $t=-\frac{6}{5}$ also gives a point on the curve. Now we
have at least one curve with an arithmetic progression of length at
least 12.  In particular, at $t=\frac{5}{6}$, the curve
\begin{equation*}
y^2 = 70400x^5+663960x^4+440000x^3-6128751x^2-2836000x+23814400
\end{equation*}
contains the $x$ arithmetic progression $\{-6,-5,-4,\ldots,4,5\}$

Returning to the quartic curve (\ref{quar}), since a rational point
exists, the curve is birationally equivalent to an elliptic curve.
The quartic curve is commonly called a \emph{quartic elliptic
curve}~\cite{UL}.  With the aid of MAGMA~\cite{MG}, we found the
minimal model of the quartic elliptic curve.  The output is the
elliptic curve, $E$,
\begin{equation*}
\begin{split}Y^2 + XY + Y &amp; =
X^3
+ X^2 \\
&amp; \quad - 14206480669846430X \\
&amp; \quad + 651651670263534709965275.
\end{split}
\end{equation*}

We thus have the maps to and from the quartic and cubic curve. If
this curve has rank at least one, and if we can find at least one
point on the curve of infinite order, then $C$ will have an
arithmetic progression of length at least 12, namely,
$$x=-6,-5,-4,\ldots,5.$$

The following commands were entered into SAGE~\cite{SG}, to
determine the rank of this curve and find its generators.

\begin{verbatim}
E=mwrank_EllipticCurve([1,1,1,
                        -14206480669846430,651651670263534709965275])
E.rank() E.gens()
\end{verbatim}
SAGE found the rank of $F$ to be three, and its (possible)
generators were found to be
\begin{center}$[-118512027,
-818950617977, 1]$\end{center}\begin{center}$[578945454,
404307680999, 8]$\end{center}\begin{center}$[68136369, -1117555865,
1]$\end{center}

So this constructs a three parameter family of degree five curves,
containing an arithmetic progression of length 12.  The equation of
the family of curves is,
\begin{equation}\label{quin}
\begin{split} y^2 &amp; = -929280(t^3 - t)x^5+ 3960(1536t^4 - 2467t^2 + 1536)x^4\\
&amp; \quad -5808000(t^3 - t)x^3 -9(3477504t^4 - 4126633t^2 + 3477504)x^2 \\
&amp; \quad + 37435200(t^3 - t)x + 27878400(t^4 + 2t^2 + 1)
\end{split}
\end{equation}
with $x$-AP,
\begin{multline*}
\{(-6,6\sqrt{188449024t^4+229333280t^3-313007023t^2
-229333280t+188449024}),\\
\quad(-5,15(3680t^2+2079t-3680)), (-4,12(2744t^2+1485t-2744)),
(-3,3(5152t^2+2915t-5152)),\\
\quad(-2,6(16t^2+1155t-16)), (-1,3(544t^2-3135t-544)), (0,5280(t^2+1)),\\
\quad(1,3(544t^2+3135t-544)),(2,6(16t^2-1155t-16)),
(3,3(5152t^2-2915t-5152)),\\
\quad(4,12(2744t^2-1485t-2744)), (5,(3680t^2-2079t-3680)) \}.
\end{multline*}
Since we have a map from $C$ to $E$, we can express $t$ in terms of
the points on $E$,
\begin{equation*}
t=\dfrac{27124113X+1405Y-1429181772291724}
{2(18185324X+843Y-1413723166396761)}.
\end{equation*}

\end{proof}

% ------------ SECTION  3:  THE SEQUENCE OF LENGTH 13 -----------------------------

\section{An Arithmetic Progression of Length 13}

It is natural to state the following question:
\begin{question}
Can we find a quintic curve containing a length 13 arithmetic
progression?
\end{question}

By computer search, we attempted to find an example of length 13 AP
on the curve (\ref{quin}).  By modifying the polynomial $g(u,x)$,
found at the beginning of section two, we attempted to construct an
example of length 13.  Thus far, by ranging the degree of $g$ from
12 to 14, we have not found an example.


% ------------ SECTION  4:  ACKNOWLEDGMENT-----------------------------

\section{Acknowledgment}

I would like to thank anonymous referee for his/her valuable
comments, and A. Bremner for guiding me towards this problem.

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

\bibitem{MG}
W. Bosma, J. Cannon, and C. Playoust, {\sc MAGMA {\tt 2.14-1}},
available from \href{http://magma.maths.usyd.edu.au/}{\tt
http://magma.maths.usyd.edu.au/}.

\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 J. Integer Sequences} {\bf 6} (2003), 
Paper 03.1.3.

\bibitem{ME}
J. Mestre,
\newblock Construction d'une courbe elliptique de rang $\geq 12$,
\newblock {\em C. R. Acad. Sci. Paris S\'er. I Math.} {\bf 295}
(1982), 643--644.

\bibitem{MO}
S. Mohanty,
\newblock On consecutive integer solutions for $y\sp{2}-k=x\sp{3}$,
\newblock {\em Proc. Amer. Math. Soc.} {\bf 48} (1975), 281--285.

\bibitem{SG}
W. Stein,
\newblock SAGE:  Software for Algebra and Geometry Experimentation.
\newblock available from \href{http://www.sagemath.org} {\tt
http://www.sagemath.org}

\bibitem{UL}
M. Ulas,
\newblock On arithmetic progressions on genus two curves,
\newblock {\em Rocky Mountain J. Math.} {\bf 39} (2009), 971--980.

\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2000 {\it Mathematics Subject Classification}: 11G05,
11B25, 14H45.

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

\bigskip
\hrule
\bigskip

\vspace*{+.1in} \noindent Received August 16 2009;
revised version
received October 19 2009. Published in {\it Journal of Integer
Sequences}, October 21 2009.

\bigskip
\hrule
\bigskip

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