Proof for the arithmetic progression So I was going through a few olympiad questions, and here is a question I found

Now, I found the three terms of the progression in terms of a and b, and arrived at $a^2$+ 2 b + 1 = 0.  
However, I'm stuck at this point. What do I do now?
 A: As you wrote, we have
$$P(0)^2-P(-1)^2=P(1)^2-P(0)^2\implies 2b=-a^2-1\tag1$$
Since $P(-1)^2,P(0)^2=b^2$ and $P(1)^2$ are integers, we have that
$$4b^2-1=(2b)^2-1=(-a^2-1)^2-1=a^2(a^2+2)\in\mathbb Z\tag2$$
and that
$$P(-1)^2-P(1)^2=2a(a^2-1)\in\mathbb Z\tag3$$
Now let $b^2=m$ where $m$ is a positive integer($m=0$ leads that $a^2=-1$). Then,
$$b^2=\left(\frac{-a^2-1}{2}\right)^2=m\implies 4m=(a^2+1)^2\implies 2\sqrt m=a^2+1$$
Also, from $(3)$, there exists an integer $n$ such that
$$2a(a^2-1)=n\implies 2a(2\sqrt m-2)=n\implies a=\frac{n}{4\sqrt m-4}$$
where we may assume that $\sqrt m\not=1$ ($\sqrt m=1$ leads that $a^2=1,b=-1$).
Therefore,
$$(a^2=)\ 2\sqrt m-1=\left(\frac{n}{4\sqrt m-4}\right)^2\implies (2\sqrt m-1)(\sqrt m-1)^2=\frac{n^2}{16}\in\mathbb R$$
$$(2\sqrt m-1)(m-2\sqrt m+1)\in\mathbb R\implies 2m\sqrt m+2\sqrt m+2\sqrt m\in\mathbb R\implies \sqrt m\in\mathbb R$$
It follows from this that $a,b$ are rational numbers.
So, we can set $b=-\frac pq$ where $p,q$ are coprime positive integers, and we have that $b^2=\frac{p^2}{q^2}\in\mathbb Z$ from which $q=1$ follows. Therefore, $b$ is an integer.
It follows from this that $a^2$ is an integer. Similarly, we get that $a$ is an integer.
