Local Mathematical Competitions

Clearly a constant non-null polynomial $P$ fulfills the requirements, so we shall assume in the sequel $\deg P \ge 1$. Before further proceeding with the solution, we state the following "folklore" preliminaries.
(1) Dirichlet's Theorem. The arithmetic progression $(an + b)_{n \in \mathbb{N}}$ with $a \in \mathbb{N}^*$, $b \in \mathbb{Z}^*$, $(a, b) = 1$, contains infinitely many primes.
(2) "Fundamental" property of integer polynomials. For any $f \in \mathbb{Z}[X]$ and any $a, b \in \mathbb{Z}$ we have $a - b \mid f(a) - f(b)$.
Claim. $P(0) = 0$. We proceed by contradiction, thus assume $P(0) \ne 0$.
We will prove by induction on $k$ the following statement
*For any $k \in \mathbb{N}^*$ there exists a prime $p > |P(0)|$ such that $P(p)$ has at least $k$ distinct prime factors.*
For $k=1$, there exists a prime $p_1 > |P(0)|$ such that $|P(p_1)| \ne 1$, otherwise for all such primes $p$ we would have $P(p)^2 = 1$ and thus the polynomial $P(X)^2 - 1$ has infinitely many roots, thus it is the null polynomial, hence $P(X) \equiv 1 \pmod{p}$ or $P(X) \equiv -1 \pmod{p}$, contradicting $\deg P \ge 1$. Therefore $P(p_1)$ has at least one prime factor, so the statement is proven for $k=1$.
Let $p_k$ be a prime with the property $p_k > |P(0)|$ and $P(p_k)$ has at least $k$ distinct prime factors. Let us notice that $(p_k, P(p_k)) = 1$, since from the "fundamental" property we have $P(p_k) \equiv P(0) \pmod{p_k}$ and since $p_k > |P(0)|$ and it is a prime, the conclusion follows. The arithmetic progression $(nP(p_k)^2 + p_k)_{n \in \mathbb{N}}$ (according to Dirichlet's Theorem) contains infinitely many primes. Let $p_{k+1} = sP(p_k)^2 + p_k$ be such a prime. Then $P(p_{k+1}) \equiv P(p_k) \pmod{P(p_k)^2}$, thus there exists $t$ such that $P(p_{k+1}) = P(p_k)(1 + tP(p_k))$. From this it is obvious that $P(p_{k+1})$ has at least one more prime factor than $P(p_k)$. Since $P(p_k)$ had at least $k$ distinct prime factors, it follows that $P(p_{k+1})$ has at least $k+1$ prime factors.
Thus the statement is proven and we get that the assumption $P(0) \ne 0$ is false. From $P(0) = 0$ we deduce there exists $m$ so that $P(X) = X^m Q(X)$ with $Q(0) \ne 0$. If $Q$ would be nonconstant, arguing the same as in the above, we again obtain a contradiction.
We conclude the only solutions are $P(X) = cX^m$, for some $m \in \mathbb{N}$ and $c \in \mathbb{Z}^*$.