Find the number of distinct integer roots of $P^2 (x)-1$ 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 $d+2$.
 A: If $d=1$ (or $d=2$) the statement is true, since the degree of $p^2-1$ is $4$ (2) and $d+2=4$ ($d+2=3$).
I will not post a complete answer, but the end is probably not too hard for you:
We assume that $d\geq 3$. Also, assume (to get a contradiction) that you have at least $d+3$ zeros. Then, since $p^2-1=(p-1)(p+1)$ both $p-1$ and $p+1$ must have at least three distinct zeros (since each factor can have at most $d$ distinct integer zeros).
Now, let $\alpha_1$, $\alpha_2$ and $\alpha_3$ be distinct integer zeros of $p-1$, i.e. $p(\alpha_1)=p(\alpha_2)=p(\alpha_3)=1$. Also, denote by $\beta_1$, $\beta_2$ an $\beta_3$ three distinct integer zeros of $p+1$ (also, distinct from the alphas). From the factor theorem,
$$
p(x)+1=(x-\beta_1)(x-\beta_2)(x-\beta_3)q(x),
$$
where $q$ is a polynomial of degree $d-3$ with integer coefficients. Inserting $x=\alpha_1$, $x=\alpha_2$ and $x=\alpha_3$, respectively, gives
$$
\begin{aligned}
2=p(\alpha_1)+1&=(\alpha_1-\beta_1)(\alpha_1-\beta_2)(\alpha_1-\beta_3)q(\alpha_1)\\
2=p(\alpha_2)+1&=(\alpha_2-\beta_1)(\alpha_2-\beta_2)(\alpha_2-\beta_3)q(\alpha_2)\\
2=p(\alpha_3)+1&=(\alpha_3-\beta_1)(\alpha_3-\beta_2)(\alpha_3-\beta_3)q(\alpha_3)\\
\end{aligned}
$$
Since $2$ is a prime, all factors in the right-hand side has to be $\pm 1$ or $\pm2$. Now, I leave it to you to argue that you cannot have distinct integers $\alpha_j$ and $\beta_k$ with this property.
A: My approach was the following:
$P^2(x)-1=0$, implies that$\{P(x)+1\}\{P(x)-1\}=0$ i.e. 
either $P(x)+1=0$ or $P(x)-1=0$ .
From this we can find two roots $\alpha$ and $\beta$ such that $P(\alpha)=1$ and $P(\beta)=-1$.
So we can deduce easily that $|\beta -\alpha|$divides $P(\beta)-P(\alpha)=1-(-1)=2$ . 
So $|\beta -\alpha|$ must be $1$ or $2$ .
Now let the degree is $2$ . Now, assume that one of the integer roots is $n$ .
So other integer roots can be $(n-1)$ or $(n+1)$.
One can say that $(n\pm 2)$ can also be other integer roots of this polynomial, but then difference between $(n-1)$ and $(n+2)$ will be $3$ that does not divide $2$. 
So the roots are $n$, $(n-1)$ and $(n+1)$ . 
So the number roots are $3$ which is definitely less than $2+2=4$ $($recall that the number of distinct integer roots will be less than $d+2$$)$. 
Thus, we can do it for $d=3$. Then, analogously, the integer roots will be $n$, $(n\pm 1)$ . 
So we can't find more than three distinct integer roots.
Now the lowest degree can be $d=1$. 
So, here the claim follows, as the number of integer roots can not exceed $3$ i.e. $1+2$ i.e. $min\{d\} +2$ .
