Let $P$ be a polynomial of degree $n \geq 5$ with $n$ distinct integer roots such that $P(0)=0$. Determine the number of integer roots of $P(P(X))$.

## 2 Solution
We will show that there are $n$ of them: the roots of $P$ are roots of $P(P(X))$ since $P(0)=0$, so it suffices to show that these are the only ones. Let $r_{1}=0, r_{2}, \ldots, r_{n}$ be the roots of $P(P(X))$ and suppose that $r$ is a root of $P(P(X))$ different from the $r_{i}$. Then $P(r)$ is a root of $P$ and is not 0. We denote it by $r_{k}$. Without loss of generality (by replacing $P$ with $-P(-X)$ if necessary), we can assume $r_{k} \geq 0$. Therefore,

$$
r\left(r-r_{2}\right) \ldots\left(r-r_{n}\right)=r_{k}
$$

and in particular $r\left(r-r_{k}\right)$ divides $r_{k}$. In particular, $r \leq r_{k}$ and $r_{k}-r \leq r_{k}$, so $00$ so $r \mid s$ and $s \mid r$ thus $r=s=1$ or 2. In the first case, we have $r_{k}=2$ so

$$
-\prod_{i \neq 1, k} r-r_{i}=2
$$

but this product contains at least three factors different from $r$ and $r-r_{k}$, hence different from 1 and -1, so the absolute value of the left-hand side is at least 8, which is absurd.

In the second case, we have

$$
2 \times(-2) \prod_{i \neq 1, k} r-r_{i}=4
$$

but the product on the left contains at least 3 distinct factors, so at least one is not 1 or -1, and we obtain a new contradiction.

## 4 Group D

## 1 Statement