If there is one perfect square in an arithmetic progression, then there are infinitely many Consider the following positive integers:
$$a,a+d,a+2d,\dots$$
Suppose there is a perfect square in the above list of numbers. Then prove that there are infinitely many perfect square in the above list. How can I do this?
At first I started in this way:
Let the $n$th term is perfect square. Therefore,
$$t_n=a+(n-1)d=m^2.$$
Then I think that I will put values at the position of $n$. But I failed to find anything from this level. Can somebody help me?
 A: Let $n^2$ be the known square.
Then $n^2+kd=m^2$ is equivalent to $kd=(m-n)(m+n)$. You can take $m-n$ to be any multiple of $d$, and $k$ follows.
A: Note that $$(m+d)^2=m^2+(2m+d)\cdot d$$
A: Starting from where you left, suppose the $k$th term is a perfect square such that $$a_k = a_1 + (k-1)d = p^2$$
Now add $2pmd + m^2d^2 $ to both sides where $m$ is a natural number giving us, $$\Rightarrow a_1 + (k-1)d + 2pmd + m^2d^2 = p^2 + 2pmd + m^2d^2 $$ $$\Rightarrow a_1 + [(k-1)+2pm+m^2]d = (p+md)^2$$
The RHS is a perfect square, and the left side
is the $(k-1+2pm+m^2)$th term for infinitely many values of $m $. Hope it helps. 
A: Suppose that given $a_n = a+n*d = x_n^2$ there is theoretically another value $a_n' = a+n'*d = x_n'^2$ where all numbers are whole.
Taking the difference between these number yields:
$$
a_n' - a_n = a_n'- a_n\\
(a+n'*d) - (a+n*d) = x_n'^2 - x_n^2\\
(n' - n)d = (x_n' - x_n)(x_n' + x_n)
$$
Setting the outer expressions equal to each other, and the inner expressions equal to each other yields:
$$x_n' - x_n = d
\\x_n' = d + x_n$$
and
$$n' - n = x_n'+x_n\\n' = n + x_n'+x_n\\n' = n + d + 2 x_n$$

Therefore: If $a_n = a + n*d = x_n^2$ is a square, you can use theses values to generate a new larger square within the sequence where: $a_n' = a + n'd = a + (n+d+2*x_n)*d$.  This will be: $x_n'^2 = (d+x_n)^2$.
