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\begin{center}
\vskip 1cm{\LARGE\bf Arithmetic Progressions on $y^2 = x^3 + k$
}
\vskip 1cm
\large
Pallab Kanti Dey and Bibekananda Maji  \\
Harish-Chandra Research Institute\\
Allahabad - 211019\\
India\\
\href{mailto:pallabkantidey@gmail.com}{\tt{pallabkantidey@gmail.com}} \\
\href{mailto:bibekmaji@hri.res.in}{\tt bibekmaji@hri.res.in}\\
\end{center}

\vskip .2 in

\begin{abstract}
Many authors have studied the problem of finding sequences  of
rational points on elliptic curves such that  either the abscissae
or  the ordinates of these points are in arithmetic progression. In
this paper we obtain upper bounds for the lengths of sequences of
rational points  on curves  of the type $y^2=x^3+k$,  $k \in
\mathbb{Q} \setminus \{0\}$,  such that   the ordinates  of the points
are in arithmetic progression,
and also when both  the abscissae and the ordinates of the points are
separately the terms of two arithmetic progressions.
\end{abstract}

\section{Introduction} 
  
 Let there exist $n$ rational points $P_i=(x_i,y_i),~i=1,2,\ldots,n$ on an elliptic  curve $E$ over $\mathbb{Q}$ such that 
either the $n$ numbers $x_i,\;i=1,2,\ldots,n$, or the $n$ numbers $y_i,\;i=1,2,\ldots,n$, are the terms of an arithmetic progression. 
We say that there exists an $x$-arithmetic progression, or a $y$-arithmetic progression, of length $n$ on the elliptic curve $E$ depending on whether 
the abscissae or the ordinates of the points $P_i$  are in arithmetic progression. Further, we say there  exists a simultaneous 
 arithmetic progression of length $n$ on the  curve $E$ if the $n$ numbers $x_i,\;i=1,2,\ldots,n$ are in arithmetic progression and simultaneously, 
the $n$ numbers $y_i,\;i=1,2,\ldots,n$ are also, in some order, the terms of an arithmetic progression. 

The problem of finding $x$-arithmetic progressions on various models of elliptic curves has been studied by several mathematicians 
\cite{AB1,AB2,GC,AC,AM,DM1,DM1,MU1}. 
 The existence of simultaneous arithmetic progressions on the cubic model of elliptic curves has also been studied \cite{GT}. 

This paper is concerned with the existence of  $y$-arithmetic progressions as well as 
simultaneous arithmetic progressions on Mordell curves which  are   elliptic curves defined by equations of the type,
\begin{equation}
 y^2=x^3+k,\label{Mordell}
\end{equation}
where $k$ is a rational number. Mohanty \cite{SPM} found an $y$-arithmetic progression of four consecutive integers on the curve \eqref{Mordell}
when $k= 1025$ and conjectured  \cite{SPM1} that the length of an $x$-arithmetic progression on a Mordell curve cannot exceed 4.
Later, Lee and V\'{e}lez \cite{LV} found infinitely many  families of Mordell  curves with $x$-arithmetic progressions
of length 4 and $y$-arithmetic progressions of length 6. 

In this paper we show that the maximum length of a $y$-arithmetic progression on a Mordell curve is  6, and we explicitly find  infinitely many Mordell 
curves on which there  exist $y$-arithmetic progressions of length  6. Further, we show that 
the maximum length of a  simultaneous arithmetic progression on a Mordell curve is  3, and we explicitly find  infinitely many Mordell 
curves on which there  exist simultaneous arithmetic progressions of length  3.




\section{Notation and preliminaries}

Now we will introduce some basic notation.
We denote by $S_x(E)$ (respectively, $S_y(E)$) as the maximal number length of $x$-arithmetic progression (respectively, $y$-arithmetic progression) on the elliptic curve $E$. Also, let $S_{x,y}(E)$ denote the maximal length of simultaneous arithmetic progression on the elliptic curve $E$


To prove one of our results we need the following theorem of Bremner \cite{AB}.

 \begin{theorem}\label{Bremner}
 The only general rational solution of the surface
 \begin{equation}
 x^3 + y^3 + c z^3 = c \,\, (c \neq 1)
 \end{equation}
are given, up to symmetry, by
$$
 (i)~  ( \lambda , - \lambda, 1)~\mbox{and}~~
(ii) \left( \frac{9}{c} {\lambda}^4 - 3\lambda, -\frac{9}{c} {\lambda}^4, \frac{9}{c}{\lambda}^3 -1 \right)
$$
and the additional case when $c=2$
\begin{equation}
\begin{split}
(iii)~ (x,y,z)=&amp; (-4 {\lambda}^2 + 6\lambda - 1, - 4{\lambda}^2 + 2\lambda + 1,  4{\lambda}^2 - 4 \lambda + 1) \\ 
(iv)~ (x,y,z)= &amp;(\frac{2}{27}(4{\lambda}^4 - 4{\lambda}^3 -6{\lambda}^2 +17\lambda-2), \\
&amp; \frac{4}{27} (2{\lambda}^4 - 8{\lambda}^3 +6{\lambda}^2 + 4\lambda - 13), \\
&amp; \frac{-1}{27} (8{\lambda}^4 - 20{\lambda}^3 +24{\lambda}^2 + 16\lambda - 37))
\end{split}
\end{equation}
with $\lambda \in \mathbb{Q}$.

 
 \end{theorem}


\section{Main results} 
 We obtained an upper bound for $ S_y(E) $ and $S_{x,y} (E)$ where $E : y^2 = x^3 + k$ with $k\in \mathbb{Q^{*}}$. Indeed, we prove the following results.

\begin{theorem}\label{thm2}
 Let $E: y^2 = x^3+k$ be an elliptic curve for some $k\in\mathbb{Q}$. Then $S_{y}(E) \leq 6.$ Moreover, there exist infinitely many such elliptic curves $E : y^2 = x^3 + k$ 
 with $S_{y}(E) = 6$.
\end{theorem}

\begin{proof}


 At first we will construct an infinite
family of elliptic curves of the form $E: y^2 = x^3 + k$ with six rational points such that the $y$-coordinates of these six rational points are in arithmetic
progression. After that,  we will show that there exist at most six rational points whose $y$-coordinates are in arithmetic progression i.e., $S_{y}(E) \leq 6$.

 Let $r$ and $d$ be two arbitrary non-zero rational numbers  and $k = r^2 - d^3$. Consequently $(d, \pm r)$ 
 lie on $E$. Now we want $(x_1 , \pm 3r)$ and $ (x_2, \pm 5r)$ (with $x_1, x_2 \in \mathbb {Q}$) to be on $E$ and that would give,
\begin{equation*}\label{equn42}
 8 r^2 + d^3 = x_1^3,~~24 r^2 + d^3 = x_2^3.
 \end{equation*}
Hence,
\begin{equation}\label{equn44}
16 r^2 =  x_2^3 - x_1^3
\end{equation}
and 
\begin{equation}\label{equn45}
 2 d^3 = 3 x_1^3 - x_2^3. 
\end{equation}
Considering $X_1 = \frac{x_1}{d}$ and $X_2 = \frac{x_2}{d}$ in $\eqref{equn45}$, we get
\begin{equation}\label{equn46}
 3 X_1^3 - X_2^3 = 2.
\end{equation}


Now  $(X_1,X_2) = (\frac{1}{4}, \frac{-5}{4})$ is a point on \eqref{equn46} 
and correspondingly one gets $x_1 = \frac{d}{4} $ and $x_2 =  \frac{-5d}{4}$. 
Now putting the values of $x_1$ and $x_2$ in $\eqref{equn44}$ we get
\begin{equation}\label{equn48}
 2^{10} r^2 = - 126 d^3.
\end{equation}
At this point if we choose $d = - 14 q^2$ for any $q \in {\mathbb{Q}}^*$, then $r = \pm \frac{147}{8} q^3$. Hence
\begin{equation}\label{equn49}
 k = r^2 - d^3 = \frac{197225}{64} q^6.
\end{equation}
Therefore corresponding to every rational number $q$ we can find $d, r $ and $k$; i.e., we can find infinitely 
many elliptic curves of the form $ y^2 = x^3 + k$  with at least six rational points whose $y$-coordinates are
in arithmetic progression.


Next we will show that there do not exist more than six rational
points whose $y$-coordinates are in arithmetic progression.

Let $d$ be any non-zero rational number. 
Let us assume that there exist seven rational points on $E$, whose $y$-coordinates are in arithmetic progression. Without loss of generality, we may take the points to be
$$
 P_1 = (x_1, y_1), P_2 = (x_2, y_1 + d), P_3 = (x_3 , y_1 - d), P_4 = (x_4, y_1 + 2d),
 $$
$$ 
P_5 = (x_5, y_1 - 2d), P_6 = (x_6, y_1 + 3d), P_7 = (x_7, y_1 - 3d). 
$$  
One has
\begin{equation}\label{equn50}
 y_1^2 = x_1^3 + k,
\end{equation}
\begin{equation}\label{equn51}
 (y_1 + d)^2 = x_2^3 + k,
\end{equation}
\begin{equation}\label{equn52}
 (y_1 - d)^2 = x_3^3 + k,
\end{equation}
\begin{equation}\label{equn53}
 (y_1 + 2d)^2 = x_4^3 + k,
\end{equation}
\begin{equation}\label{equn54}
 (y_1 -2 d)^2 = x_5^3 + k,
\end{equation}
\begin{equation}\label{equn55}
 (y_1 + 3d)^2 = x_6^3 + k,
\end{equation}
\begin{equation}\label{equn56}
 (y_1 -3 d)^2 = x_7^3 + k.
\end{equation}
Now calculating $\left[\eqref{equn51} - \eqref{equn52}\right], \left[\eqref{equn53} - \eqref{equn54}\right],\,\, \textrm{and}\,\,\left[ \eqref{equn55} -\eqref{equn56}\right]$ one has 
\begin{equation}\label{equn57}
 4 y_1 d = x_2^3 - x_3^3,
\end{equation}
\begin{equation}\label{equn58}
 8 y_1 d = x_4^3 - x_5^3,
\end{equation}
\begin{equation}\label{equn59}
 12 y_1 d = x_6^3 - x_7^3.
\end{equation}


\noindent {\bf Case 1:}~~ In this case we are considering $y_1 \neq 0$. From
\eqref{equn58} it is clear that one of $x_4$ and 
$x_5$ has to be non-zero as $d\neq 0$ and $y_1\neq 0$. So,
without loss of generality  we can assume that  $x_5$ is non-zero.

Now, eliminating $y_1$ from the equations \eqref{equn57} and \eqref{equn58}, we have
\begin{equation}\label{equn60}
\begin{split}
&amp; 2(x_2^3 - x_3^3) = x_4^3 - x_5^3\\
\Rightarrow &amp; 8(x_2^3 - x_3^3) = 4(x_4^3 - x_5^3)\\
\Rightarrow &amp; -8 x_2^3 + 8 x_3^3 + 4 x_4^3  =  4 x_5^3\\
\Rightarrow &amp; \left(\frac{-2 x_2}{x_5}\right)^3 + \left(\frac{2 x_3}{x_5}\right)^3 + 4 \left(\frac{x_4}{x_5}\right)^3 = 4\\
\Rightarrow &amp; X^3 + Y^3 + 4 Z^3 = 4
\end{split}
\end{equation}
where $X = \frac{-2 x_2}{x_5}, Y = \frac{2 x_3}{x_5}, $ and $ Z = \frac{x_4}{x_5}$. 

Again, eliminating $y_1$ from the equations \eqref{equn59} and \eqref{equn58}, we get
\begin{equation}\label{equn61}
 \begin{split}
&amp; 2(x_7^3 - x_6^3) = 3(x_5^3 - x_4^3)\\ 
 \Rightarrow &amp;  8 x_7^3 - 8 x_6^3 + 12 x_4^3 = 12 x_5^3\\
 \Rightarrow &amp; \left( \frac{2 x_7 }{x_5}\right)^3 + \left( \frac{-2 x_6}{x_5} \right)^3 + 12 \left( \frac{x_4}{x_5} \right)^3 = 12\\
 \Rightarrow &amp; U^3  + V^3 + 12 W^3 = 12 \\
\end{split} 
\end{equation}
where $ U = \frac{2 x_7 }{x_5}, V = \frac{-2 x_6}{x_5}, $ and $ W = \frac{x_4}{x_5}$.



To find rational solution of \eqref{equn60} and \eqref{equn61} we use Bremner's result.
Using Theorem (\ref{Bremner}) we get all possible rational solutions of \eqref{equn60} and \eqref{equn61}.
The rational solutions of \eqref{equn60} are
$$
 (i)~  ( \lambda , - \lambda, 1)~\mbox{and}~~
(ii) \left( \frac{9}{4} {\lambda}^4 - 3\lambda, -\frac{9}{4} {\lambda}^4, \frac{9}{4}{\lambda}^3 -1 \right)
$$
with $\lambda \in \mathbb{Q}$.


The rational solutions of \eqref{equn61} are
$$
 (i)~ ( \mu , - \mu, 1)~\mbox{and}~~
(ii) \left( \frac{3}{4} {\mu}^4 - 3\mu, -\frac{3}{4} {\mu}^4, \frac{3}{4}{\mu}^3 -1 \right)
$$
with $ \mu \in {\mathbb Q}$. 

One can see
from \eqref{equn60} and \eqref{equn61} that,
\begin{equation}\nonumber
 Z = W = \frac{x_4}{x_5}.
\end{equation}
Clearly, 
$$
\frac{9}{4}{\lambda}^3 -1  =
1 \quad \textrm{and} \quad \frac{3}{4}{\mu}^3 -1 =1 
$$
cannot  hold for any $\lambda, \mu  \in {\mathbb Q}$. 

Now if possible let us assume  that for some $\lambda$ and $\mu$,
 $$
 \frac{9}{4}{\lambda}^3 -1 = \frac{3}{4}{\mu}^3 -1. 
 $$
That gives $ 3\lambda^3 = \mu^3$, which is again not possible as $\lambda, \mu \in \mathbb{Q}$.

Thus we get
$
Z = W  = 1
$
and that implies $x_4 = x_5$.
Now from \eqref{equn53} and \eqref{equn54} we can write, 
$$( y_1 + 2d) = \pm (y_1 - 2d ).$$ 
This forces $y_1 = 0$ as $ d \neq 0$, which is a contradiction.

\bigskip

\noindent {\bf Case 2:}~~ In this case we consider $y_1=0$.
Thus our points are
$$ 
P_1 = (x_1, 0), P_2 =(x_2, d), P_3 = (x_3 , -d ), P_4 = (x_4, 2d),
$$
$$
P_5= (x_5, -2d), P_6= (x_6,3d), P_7= (x_7, -3d).
$$
One can easily check that, $ k = (-x_1)^3, x_2 = x_3,$ and $ x_6 = x_7$. Thus,
\begin{equation}\label{equn62}
 d^2 = x_2^3 - x_1^3.
\end{equation}
\begin{equation}\label{equn63}
 4 d^2 = x_4^3 - x_1^3.
\end{equation}
\begin{equation}\label{equn64}
 9 d^2 = x_6^3 - x_1^3. 
\end{equation}
From \eqref{equn62} it is clear that one of $x_1$ and 
$x_2$ has to be non-zero as $d\neq 0$. So,
without loss of generality  we can assume that  $x_1$ is non-zero.

Eliminating $d$ from the equations \eqref{equn62}, \eqref{equn63} and \eqref{equn64}, we have
\begin{equation}\label{equn65}
\begin{split}
 &amp; x_6^3 - x_4^3 = 5(x_2^3 - x_1^3)\\
 \Rightarrow &amp; \left( \frac{x_4}{x_1} \right)^3 + \left( \frac{-x_6}{x_1} \right)^3 + 5 \left( \frac{x_2}{x_1} \right)^3 = 5\\ 
 \Rightarrow &amp; L^3 + M^3 + 5 N^3 = 5 
\end{split}
\end{equation}
where $L= \frac{x_4}{x_1}, M= \frac{-x_6}{x_1}, $ and $N= \frac{x_2}{x_1}$. 
Again using Bremner's result, the solution set of the equation
\eqref{equn65} is
$$
  (i) ~ (\nu, -\nu, 1)~\mbox{and}~~ (ii)  \left( \frac{9}{5} {\nu}^4 -
  3 \nu , -\frac{9}{5} {\nu}^4, \frac{9}{5} {\nu}^3 -1 \right), $$
where $\nu \in \mathbb{Q}$.

Note that $N \neq 1$, as $N = 1$ implies $ d = 0$, which is not possible.
Let $x_1$ be any non-zero rational number, say $q$. Then we have
$$
x_2  = \left( \frac{9}{5} {\nu}^3 -1 \right) q ,~
x_4  = \left( \frac{9}{5} {\nu}^4 - 3 \nu \right) q, ~
x_6  = \left( \frac{9}{5} {\nu}^4 \right) q .
$$
Substituting in the values of $x_2$ and $x_6$  into \eqref{equn62} and \eqref{equn64} respectively, we have
$$
 d^2  = \left[ \left( \frac{9}{5} {\nu}^3 - 1 \right)^3 -1 \right]q^3, $$
$$ 
 9 d^2 = \left[ \left( \frac{9}{5} {\nu}^4 \right)^3 - 1 \right]q^3 . 
 $$

Therefore we get
$$
9  \left[ \left( \frac{9}{5} {\nu}^3 - 1 \right)^3 -1 \right] = \left[ \left( \frac{9}{5} {\nu}^4 \right)^3 - 1 \right],
$$
which leads us to
\begin{equation*}
 9 ( 9 t -5 )^3 = 9^3 t^4 + 1000,
 \end{equation*}
 where ${\nu}^3 =t $.

We will prove that this equation does not have any rational solution.
Putting $s = 3t$ we have the following equation:
\begin{equation}\label{equn66}
 9 s^4 - 243 s^3 + 1215 s^2 - 2025 s + 2125 =0.  
\end{equation}


Using the rational root theorem, we can see that \eqref{equn66} does not have any rational solution. Therefore our system
of equations $[\eqref{equn50}\, \textrm{to}\, \eqref{equn56}]$  do not have any rational solution. Hence there do not exist seven rational points on $y^2 = x^3 + k$ with
$y$-coordinates in arithmetic progression, i.e., $S_{y}(E) \leq 6 $.

\end{proof}

Now we will state the result related to simultaneous arithmetic progression.
\begin{theorem}\label{thm1}
 Let $E: y^2 = x^3+k$ be an elliptic curve for some $k\in\mathbb{Q}$. Then $S_{x,y}(E) \leq 3.$ Moreover, there exist infinitely many such elliptic curves $E : y^2 = x^3 + k$ 
 with $S_{x,y}(E) = 3$.
\end{theorem}

\begin{proof}

Let $d$ and $d'$ be given nonzero rational numbers. Suppose  $P_1= (x_1, y_1)$ is a rational point on $E: y^2 = x^3+k$ for some $k\in \mathbb Q$. Therefore, $P_1$ satisfies 
\begin{equation}\label{equn1}
 y_1^2 = x_1^3 + k.
\end{equation}
Suppose $P_2$ and $P_3$ are other rational points on $E$ such that $P_1, P_2$ and $P_3$ form a  simultaneous arithmetic progression with differences $d$ and $d'$ of length $3$. Therefore, we let
$P_2 = (x_1 + d, y_1 + d')$ and  $P_3 = (x_1 - d, y_1 - d')$ and we get
\begin{equation}\label{equn2}
(y_1 + d')^2 = (x_1 + d)^3 + k
\end{equation}
and
\begin{equation}\label{equn3}
(y_1 - d')^2 = (x_1 - d)^3 + k.
\end{equation}
By subtracting (\ref{equn3}) from (\ref{equn2}),  we get
\begin{equation}\label{equn4}
4 y_1 d' = 6 x_1^2 d + 2d^3.
\end{equation}
From \eqref{equn1} and \eqref{equn2}, we get
\begin{equation}\label{equn5}
2 y_1 d' + {d'}^2 = 3 x_1^2 d + 3 x_1 d^2 + d^3.
\end{equation}
Therefore,  from (\ref{equn4}) and (\ref{equn5}),  we get
\begin{equation}\label{equn6}
2{d'}^2 = 6 x_1 d^2  \  \iff \  x_1 = \frac{1}{3} \frac{{d'}^2}{d^2}.
\end{equation}
Once we get $x_1$ as a function of $d$ and $d'$, we can get $y_1$ and $k$ as a function of $d$ and $d'$. Since this is true for a given non-zero rational numbers $d$ and $d'$, we conclude that there are 
infinitely many elliptic curves $E$ of the form $y^2 = x^3+k$ which admits $S_{x,y}(E)\geq 3$.  



In order to finish the proof of the theorem, now it is enough to prove that $S_{x, y}(E) \leq 3$ for all elliptic curves $E: y^2 = x^3+k$.  


Let $E$ be one such curve. If possible, let  $P_1, P_2, P_3$ and $P_4$ be the rational points on $E$ which form a simultaneous arithmetic progression of length $4$ with some differences $d$ and $d'$ for the $x$ and $y$ coordinates respectively. Now, we need to consider several cases depending on the arrangement of coordinates of $P_i$'s.

\bigskip

\noindent {\bf{Case 1:}} Let $P_1 = (x_1 - d, y_1 - d'), P_2 = (x_1, y_1), P_3 = (x_1 + d, y_1 + d')$ and $P_4 = (x_1 + 2d, y_1 + 2d')$.


Since $P_1,P_2$ and $P_3$ are in simultaneous arithmetic progression, by the previous discussion, we conclude that $x_1 = \displaystyle\frac{1}{3} \frac{{d'}^2}{d^2}$. Since $P_4$ is a rational point on $E$, we get 
\begin{equation}\label{equn7}
(y_1 + 2d')^2 = (x_1 + 2d)^3 + k.
\end{equation}
From the equations  (\ref{equn7}) and (\ref{equn1}),  we get
\begin{equation}\label{equn8}
4 y_1 d' + 4{d'}^2 = 6 x_1^2 d + 12 x_1 d^2 + 8d^3.
\end{equation}
Now, by putting $x_1 = d'^2/3d^2$ in \eqref{equn8}, we arrive at  $d = 0$, which is a contradiction. Thus, we conclude, in this case, $S_{x,y}(E) \leq 3$. 

\bigskip

\noindent {\bf{Case 2:}} Let $P_1= (x_1,y_1), P_2 = (x_1 + d, y_1 + d') , P_3= (x_1 + 2d, y_1 + 3d'), P_4= (x_1 + 3d, y_1 + 2d')$. 

Since these are rational points on $E$, we have
\begin{equation}\label{equn9}
 y_1^2 = x_1^3 + k.
\end{equation}
\begin{equation}\label{equn10}
 (y_1 + d')^2 = (x_1 + d)^3 + k,
\end{equation}
\begin{equation}\label{equn11}
 (y_1 + 3d')^2 = (x_1 + 2d)^3 + k,
\end{equation}
and
\begin{equation}\label{equn12}
 (y_1 + 2d')^2 = (x_1 + 3d)^3 + k.
\end{equation}
Eliminating $y_1$ and $k$, from the equations (\ref{equn9}), (\ref{equn10}) and (\ref{equn11}),  we get
\begin{equation}\label{equn13}
 6 {d'}^2 = -3 x_1^2 d + 3 x_1 d^2 + 5 d^3.
\end{equation}
Again, eliminating $y_1$ and $k$, from the equations (\ref{equn9}), (\ref{equn11}) and \eqref{equn12}, we have
\begin{equation}\label{equn14}
 6{d'}^2 = -15 x_1^2 d - 57 x_1 d^2 -65 d^3 .
\end{equation}
Now eliminating $d'$ from the equations \eqref{equn13} and \eqref{equn14},  we 
get
\begin{equation}\label{equn15}
 6 x_1^2 + 30 x_1 d + 35 d^2 = 0 .
\end{equation}
Clearly, the quadratic equation \eqref{equn15} in $x_1$ does not have any rational solutions and hence we conclude, in this case, $S_{x, y}(E) \leq 3$. 

The remaining $22$ cases can be proved similarly and we list them in Table 1. 


From Table 1,  we can see that  the discriminant $D$ is not a perfect square for a rational number  $d$ in all the cases. Hence, we conclude that $x_1$ cannot be rational, which is a contradiction. Therefore, in all the cases, we get
$S_{x,y}(E) \leq 3$.   This completes the  proof of Theorem  \ref{thm1}.


\end{proof}







\section{Some open problems} While we obtained  upper bounds for the lengths of $y-$arithmetic progressions and  simultaneous arithmetic progressions 
on Mordell curves, we could not obtain an upper bound for the length of $x-$arithmetic progressions on Mordell curves. It would be of interest 
to determine such an upper bound. Similarly it would be of interest to determine upper bounds for the length of $x-$arithmetic progressions, 
$y-$arithmetic progressions and simultaneous arithmetic progressions on  various models of elliptic curves. 


\section{Acknowledgments}
We want to thank Ajai Choudhry for suggesting and discussing this problem.
We also want to thank 
Kalyan Chakraborty and R. Thangadurai for going through this paper carefully. Finally, we would like to thank the anonymous referee for his/her valuable comments.  

\begin{table}[H]
\label{table1}
\centerline{\bf Table 1}
\bigskip
\scalebox{0.83}{
\begin{tabular}{ |c|c|c|c| } 
 
   \hline
 Cases &amp; Points &amp; Final Equation &amp; Discriminant(D)\\
 \hline 
 Case 3 &amp; $ (x_1, y_1), (x_1 + d, y_1+ 2d')$          &amp;  $ 3 x_1^2 + 9 x_1 d + 8 d^2 = 0$  &amp;  $-15 d^2$\\
        &amp; $(x_1 + 2d, y_1 + d'),(x_1+3d, y_1 + 3d')$  &amp;  &amp;\\
\hline 
 Case 4 &amp; $ (x_1, y_1), (x_1 + d, y_1+ 2d')$          &amp;  $ 24 x_1^2 + 84 x_1 d + 86 d^2 = 0$ &amp;  $-1200 d^2$\\
        &amp; $(x_1 + 2d, y_1 + 3d'),(x_1+ 3d, y_1 + d')$ &amp;  &amp;\\
 \hline 
 Case 5 &amp; $ (x_1, y_1), (x_1 + d, y_1+ 3d')$          &amp;  $ 3 x_1^2 + 14 x_1 d + 28 d^2 = 0$ &amp;  $-140 d^2$\\
        &amp; $(x_1 + 2d, y_1 + d'),(x_1+3d, y_1 + 2d')$  &amp;  &amp;\\
 \hline
 Case 6 &amp; $(x_1, y_1), (x_1 + d, y_1 + 3d')$             &amp; $6 x_1^2 + 24 x_1 d + 29 d^2 = 0$  &amp;  $- 120 d^2$  \\
        &amp; $( x_1 + 2d, y_1 + 2d'), ( x_1 +3d, y_1 + d')$ &amp;  &amp;\\
 \hline
 Case 7 &amp; $(x_1, y_1 + d'), (x_1 + d , y_1 + 2 d'),$  &amp;  $3 x_1^2 - 3 x_1 d - 8 d^2 = 0$ &amp;  $105 d^2$\\
        &amp; $(x_1 + 2 d, y_1), (x_1 + 3 d, y_1 +3 d')$  &amp;  &amp; \\ 
 \hline
 Case 8 &amp; $(x_1, y_1 + d'), (x_1 + d, y_1 + 2d'),$    &amp;  $ 6 x_1^2 + 12 x_1 d + 11 d^2 = 0$   &amp; $- 120 d^2$\\
        &amp; $(x_1 + 2 d, y_1+ 3d'), (x_1 + 3 d , y_1)$  &amp;  &amp; \\
 \hline
 Case 9 &amp; $(x_1, y_1 + d'), (x_1 + d, y_1)   $        &amp;  $ 6 x_1^2 + 6 x_1 d - d^2 = 0$  &amp;   $60 d^2$    \\
        &amp; $(x_1 +2d, y_1 + 2d'), (x_1+3d, y_1 +3d') $ &amp;  &amp;\\
 \hline
 Case 10 &amp;  $ (x_1, y_1 + d'), (x_1 + d, y_1)   $      &amp; $ 12 x_1^2 + 36 x_1 d + 37 d^2 = 0$  &amp; $ -480 d^2$\\
        &amp;  $(x_1 +2d, y_1 + 3d'), (x_1+3d, y_1 +2d')$  &amp;  &amp;\\
 \hline
 Case 11 &amp; $ (x_1, y_1 + d'), (x_1 + d, y_1+3d') $     &amp; $ 15 x_1^2 + 45 x_1 d + 44 d^2 = 0$  &amp; $- 615 d^2$ \\
        &amp; $(x_1 +2d, y_1), (x_1+3d, y_1 +2d')$         &amp;  &amp;\\
 \hline
 Case 12 &amp; $ (x_1, y_1 + d'), (x_1 + d, y_1+3d') $     &amp; $ 12 x_1^2 + 30 x_1 d + 25 d^2 = 0$ &amp; $ -300 d^2$\\
        &amp; $(x_1 +2d, y_1+2d'), (x_1+3d, y_1)$          &amp;  &amp;\\
 \hline
 Case 13 &amp; $ (x_1, y_1 + 2d'), (x_1 + d, y_1) $       &amp; $12 x_1^2 + 30 x_1 d + 25 d^2 = 0$ &amp;  $ -300 d^2$\\
        &amp; $ (x_1 +2d, y_1 + d'), (x_1+3d,y_1 +3 d')$  &amp;  &amp;\\
 \hline
 Case 14 &amp; $ (x_1, y_1 + 2d'), (x_1 + d, y_1) $        &amp;  $15 x_1^2 + 45 x_1 d + 44 d^2 = 0  $ &amp;  $- 615 d^2$\\
        &amp; $ (x_1 +2d, y_1 + 3d'), (x_1+3d,y_1 + d')$   &amp;  &amp;\\
 \hline  
 Case 15 &amp; $ (x_1, y_1 + 2d'), (x_1 + d, y_1 +d') $    &amp; $6 x_1^2 + 12 x_1 d + 11 d^2 = 0$ &amp; $- 120 d^2$ \\
        &amp; $ (x_1 +2d, y_1), (x_1+3d,y_1 + 3 d')$      &amp;  &amp;\\
 \hline
 Case 16 &amp; $ (x_1, y_1 + 2d'), (x_1 + d, y_1 +d') $   &amp;  $3 x_1^2 - 3 x_1 d - 8 d^2 = 0$ &amp;  $105 d^2$\\
        &amp; $ (x_1 +2d, y_1+ 3d'), (x_1+3d,y_1 )$      &amp;  &amp;  \\
 \hline
 Case 17  &amp; $ (x_1, y_1 + 2d'), (x_1 + d, y_1 +3d') $  &amp; $12 x_1^2 + 36 x_1 d + 37 d^2 = 0$   &amp; $-480 d^2$\\
         &amp; $ (x_1 +2d, y_1), (x_1+3d,y_1 +d') $         &amp;  &amp;\\
 \hline
 Case 18 &amp; $ (x_1, y_1 + 2d'), (x_1 + d, y_1 +3d') $  &amp; $6 x_1^2 + 6 x_1 d - d^2 = 0$  &amp;  $60 d^2$\\
        &amp; $ (x_1 +2d, y_1+d'), (x_1+3d,y_1 ) $        &amp;  &amp;\\
 \hline
 Case 19 &amp; $ (x_1, y_1 + 3d'), (x_1 + d, y_1 )$       &amp; $6 x_1^2 + 24 x_1 d + 29 d^2 = 0$  &amp;  $-120 d^2$\\
        &amp; $ (x_1 +2d, y_1+d'), (x_1+3d,y_1 + 2d')$     &amp;  &amp;\\
 \hline
 Case 20 &amp;  $ (x_1, y_1 + 3d'), (x_1 + d, y_1 )$      &amp; $3 x_1^2 + 21 x_1 d + 28 d^2 = 0$ &amp; $105 d^2$\\
        &amp; $ (x_1 +2d, y_1 + 2d'), (x_1+3d,y_1 + d')$  &amp;  &amp;\\
 \hline
 Case 21 &amp; $ (x_1, y_1 + 3d'), (x_1 + d, y_1 +d')$    &amp; $12 x_1^2 + 42 x_1 d + 43 d^2 = 0  $ &amp; $-300 d^2$ \\
        &amp; $ (x_1 +2d, y_1), (x_1+3d,y_1 + 2d')$      &amp;  &amp;\\
 \hline 
 Case 22 &amp; $ (x_1, y_1 + 3d'), (x_1 + d, y_1 +d')$    &amp; $9 x_1^2 + 27 x_1 d + 24 d^2 = 0   $  &amp; $-135 d^2$ \\
        &amp; $ (x_1 +2d, y_1 + 2d'), (x_1+3d,y_1)$       &amp;  &amp;\\
 \hline
 Case 23 &amp; $ (x_1, y_1 + 3d'), (x_1 + d, y_1 +2d')$    &amp; $6 x_1^2 +30 x_1 d + 35 d^2 = 0 $ &amp; $60 d^2 $\\
        &amp; $ (x_1 +2d, y_1 ), (x_1 + 3 d ,y_1 + d') $    &amp;  &amp;\\
 \hline
 Case 24 &amp; $ (x_1, y_1 + 3d'), (x_1 + d, y_1 +2d')$    &amp;  $d=0$ &amp; \\
        &amp; $ (x_1 +2d, y_1 + d' ), (x_1 + 3 d ,y_1) $  &amp;   &amp;  \\
 \hline
 \end{tabular}
}
\end{table}

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\end{thebibliography}

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\noindent 2010 {\it Mathematics Subject Classification}: Primary 11G05; Secondary 11B25.

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

\bigskip
\hrule
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\vspace*{+.1in}
\noindent
Received September 19 2015;
revised versions received  September 24 2015; May 8 2016; September
3 2016.
Published in {\it Journal of Integer Sequences}, September 4 2016.

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