Prove that the number of distinct integer roots of $P^2(x)-1$ is at most $d+2$. 
Let $P(x)$ be a polynomial with integer coefficients of degree $d>0$. Prove that the number of distinct integer roots of $P^2(x)-1$ is at most $d+2$.

My approach: Note that $P^2(x)-1=(P(x)-1)(P(x)+1), \forall x.$ Now since $\deg P(x)=d\implies \deg(P(x)-1)=\deg(P(x)+1)=d.$ Also note that the leading coefficient of $P(x)-1$ and $P(x)+1$ are the same; let it be $a\in\mathbb{Z}$. So, let that $\alpha_1,\alpha_2,\dots,\alpha_d$ be the roots of $P(x)-1$ and $\beta_1,\beta_2,\dots,\beta_d$ be the roots of $P(x)+1$. This implies that $$P^2(x)-1=a^2(x-\alpha_1)\cdots(x-\alpha_d)(x-\beta_1)\cdots(x-\beta_d),$$ that is $\deg(P^2(x)-1)=2d$ and $\alpha_1,\dots,\alpha_d,\beta_1,\dots,\beta_d$ are its roots and $a^2$ is it's leading coefficient. Also, since $P(x)\in\mathbb{Z}[X]$, implies that, $P^2(x)-1\in\mathbb{Z}[X]$. Next note that if $\gamma$ is a root of $P(x)-1$, then $\gamma$ is not a root of $P(x)+1$. This implies that $\alpha_i\neq \beta_j, \forall 1\le i,j\le d.$
Now note that if $\alpha_i,\beta_j\in\mathbb{Z}$ for some $1\le i,j\le d,$ then $2$ is divisible by $|\alpha_i-\beta_j|$, that is $\alpha_i-\beta_j=-2,-1,1,2.$
So, for the sake of contradiction let us assume that $P^2(x)-1$ has at least $d+3$ distinct integer roots.
I was not able to make any significant progress after this. So, can someone help me proceed?
 A: You found all the necessary properties for the proof. For the sake of contradiction
$$P^2(x)-1=(P(x)-1)(P(x)+1)$$
has at least $d+3$ distinct integral roots, and each of $P(x)\pm 1$ has at most $d$ distinct integral roots, hence each of the two factors must have at least $3$ disctinct integral roots (otherwise one of the $P(x)\pm 1$ would have at least $d+1$ roots and so it would have to be a zero polynomial, which means $P(x)=\pm 1$ is a constant polynomial, ruled out by assumptions). This also immediately rules out polynomials with degree $d \leq 2$, so continue with $d \geq 3$.
So we can take $3$ distinct integral roots $\alpha_1 < \alpha_2 < \alpha_3$ of $P(x)-1$, and similarly $3$ distinct integral roots $\beta_1 < \beta_2 < \beta_3$ of $P(x)+1$. Then clearly
$$
\alpha_1-\beta_3<\alpha_2 - \beta_3< \alpha_3 - \beta_3 < \alpha_3 - \beta_2  < \alpha_3 - \beta_1
$$
are $5$ distinct values, but that is impossible as you have found $\alpha_i-\beta_j$ can only have one of $4$ distinct values $-2,-1,1$ and $2$.
