If $\{a_n\}$ is an arithmetic progression such that $a_1^2,a_2^2,a_3^2$ belong to $\{a_n\}$, every $a_n$ is an integer Let $\{a_n\}$ be an arithmetic progression such that $a_1^2,a_2^2,a_3^2$ belong to  $\{a_n\}$. Prove that every $a_n$ is integer.

I try to write with common difference $d$:
$a_2=a_1+d$
$a_3=a_1+2d$
and square
$a_2^2=a_1^2+2a_1d+d^2$
$a_3^2=a_1^2+4a_1d+4d^2$
and take difference:
$a_2^2-a_1^2=d(2a_1+d)$
and here I have been stuck. What to do?
 A: Let $d$ be the common difference. 
If $d=0$, then we have either $a_n=0$ or $a_n=1$.
In the following, $d\not=0$.
Since $a_1^2,a_2^2,a_3^2$ belong to $\{a_n\}$, there exist integers $s,t,u$ such that
$$a_1^2=a_1+sd\tag1$$
$$(a_1+d)^2=a_1+td\tag2$$
$$(a_1+2d)^2=a_1+ud\tag3$$
From $(2)-(1)$, we have
$$2a_1d+d^2=td-sd\implies 2a_1+d=t-s\tag4$$
From $(3)-(2)$, we have
$$2a_1d+3d^2=ud-td\implies 2a_1+3d=u-t\tag5$$
From $(3)-(1)$, we have
$$4a_1d+4d^2=ud-sd\implies 4a_1+4d=u-s\tag6$$
From $(5)-(4)$, we have
$$2d=u-2t+s\in\mathbb Z\tag7$$
From $(6)(7)$, we have
$$4a_1=u-s-2(u-2t+s)\in\mathbb Z$$
So, there exist integers $b,c$ such that
$$a_1=\frac b4,\qquad d=\frac c2$$
Then, $(1)$ is equivalent to
$$b^2=2(2b+4sc)$$
It follows from this that $b$ is even. 
So, there is an integer $f$ such that $a_1=\frac f2$.
Then, $(3)$ is equivalent to
$$f^2=2(f-2fc-2c^2+uc)$$
It follows from this that $f$ is even.
Now, $(2)$ is equivalent to
$$c^2=-f^2+2f-2fc+2tc$$
It follows from this that $c$ is even.
Since both $a_1$ and $d$ are integers, the claim follows.
