Olympiad problem :Integer roots of $P(P(x))$ in function of the roots of $P(x)$ Here is the original problem : 

A polynomial $P(x)$ of degree $n \geq 5$ with integer coefficients and
  $n$ distinct integer roots is given. Find all integer roots of $P(P(x))$
  given that $0$ is a root of $P(x)$.

Here is my solution, however i'm not sure of it , could you guys please check it ?
Solution :
It's easy to see that $P(P(x))=0$ for all $x=x_1,..,x_n$. The other roots that  $P(P(x))$ may have are the eventual values $X_j$ for which  $P(X_j)=x_j, j\neq 1$ (and those $X_j$ must be integers.) We have $P(0)=0$ so obviously $P(x)=a_nx^n+..+a_1x$.
For all integers $a,b$ :
 $a-b\mid P(a)-P(b)$, since $P\in\mathbb Z[x]$.
Taking $a=X_j$ and $b=x_j$ (for $i=2,..,n$, this is just :
$X_j-x_j\mid x_j$.
This means there exists some integer $k>1$ such that: $X_j=kx_j$, for all $2\le i\le n$.
We have : $P(x_j)=a_nx_j^n+..+a_1x_j$.
Thus : $P(X_j)=P(kx_j)=a_nk^nx_j^n+..+a_1kx_j=x_j$$\Longleftrightarrow$
$a_nk^nx_j^{n-1}+..+a_1k=1$.
Thus :
$a_nk^{n-1}x_j^{n-1}+..+a_1=\frac{1}{k}$. But $P$ has integer coefficients and $k,x_j$ are integers, so $\frac{1}{k}$ must be an integer, so $k=1\Longrightarrow X_j=x_j$, so the integer roots of $P(P(x))$ are the same as those of $P(x)$.
Thanks a lot !
 A: So $P(x)=x\prod_{i=2}^n (x-x_i)$ where $x_2,\ldots, x_n$ are the nonzero integral roots of $P$. Now suppose there is a a nonzero root $y$ of $P(P(x))$ that is distinct from $0,x_2,\ldots, x_n$. Then $y\prod_{i=2}^n(y-x_i)$ must be in $\{x_2,x_3,\ldots, x_n\}$. We show that this is impossible for integral $y \not = 0,x_2,\ldots, x_n$ via Claim 1 below:

Claim 1: Let us use the notation as above, and let us write $x_n$ be the root of $P$ with the largest modulus i.e., $|x_n| \ge |x_i|$ for each $i=2,3,\ldots, n$. Let $y$ be a nonzero integer. Then $|P(y)| > |x_n|$.

Case 1: $0< |y| \le \frac{|x_n|}{2}$. Then $|y\prod_{i=2}^n (y-x_i)| > |x_n|$. Indeed, that $y$ is distinct from $0,x_2,\ldots, x_n$ implies $|y|$ and $|y-x_i|$ for each $i=2, \ldots, n-2$ is at least 1, and that at least 2 of $\{|y|, |y-x_i|; i=2,\ldots, n-1\}$ is at least 2, since the degree $n$ of $P$ is at least 5. So $|y\prod_{i=2}^{n-1}(y-x_i)|$ is at least 4. But then indeed as $|y| \le \frac{|x_n|}{2}$ it follows that $|y-x_n|$ is at least $\frac{|x_n|}{2}$, this implies that 
$$|y\prod_{i=2}^{n} (y-x_i)| \ge \frac{|x_n|}{2} \times |y\prod_{i=2}^{n-1}(y-x_i)|$$ 
$$\ge \frac{|x_n|}{2} \times 4 > |x_n|.$$
So Claim 1 follws for Case 1.
Case 2: $|y| \ge \frac{|x_n|}{2}$. Then $|y\prod_{i=2}^n (y-x_i)| > |x_n|$. Indeed, that $y$ is distinct from $0,x_2,\ldots, x_n$ implies $|y-x_i|$ for each $i=2, \ldots, n$ is at least 1, and that $|y-x_i| \ge 2$ for at least 2 of the other $i$s in $2,\ldots, n$, since the degree $n$ of $P$ is at least 5. So $|\prod_{i=2}^{n}(y-x_i)|$ is at least 4. But then indeed as $|y|$ is at least $|x_n/2|$, this implies that 
$$|y\prod_{i=2}^{n} (y-x_i)| \ge \frac{|x_n|}{2} \prod_{i=2}^{n}|(y-x_i)|$$
$$\ge \frac{|x_n|}{2} \times 4  > |x_n|.$$ 
So Claim 1 follows for the remaining Case 2 as well, thus Claim 1 follows.
