If $P(x)$ is a non-zero constant polynomial, the result is obvious. So, it suffices to prove that if $\mathbb S := \{p\in\mathbb P : \exists\text{ }n \text{ s.t. }p\mid P(n)\}$ is finite, then $P(x)$ is a non-zero constant polynomial. If $P(x)$ has no constant term, $x|P(x)$ and we are led to a contradiction. Therefore, $P(x)$ has a constant term. Set $n=\prod_{p\in{\mathbb S}}p^N$ for $N$ large enough. Then $$P(n)=\sum_{i=0}^{deg{P}} {n^i}*a_i=\prod_{p_i\in{\mathbb S}}{p_i}^{b_i}$$If it is not constant, $P(n) \longrightarrow \infty$ (or $-\infty$, the argument remains the same) and the only prime divisors of $P(n)$ are those belonging to $\mathbb S$, which are finite. Note that the prime divisors of $P(n)$ must all divide $a_0$, but their exponents cannot exceed the ones of the respective prime divisors of $a_0$. So, the $LHS$ can grow infinitely large, while the $RHS$ cannot, which is absurd. So, $P(x)$ is constant and its constant term is not zero. Note: The problem is actually asking to prove Schur's Theorem which states that there are infinitely many primes dividing at least one non-zero term of the sequence $P(1),P(2),P(3), ... ,$ where $P(x)$ is a non-constant polynomial with integer coefficients.

bsf714 wrote: Let $P(x)$ be a polynomial with integer coefficients. We will denote the set of all prime numbers by $\mathbb P$. Show that the set $\mathbb S := \{p\in\mathbb P : \exists\text{ }n \text{ s.t. }p\mid P(n)\}$ is finite if and only if $P(x)$ is a non-zero constant polynomial. How can they ask such an obvious Schur Theorem at the test??

MatBoy-123 wrote: How can they ask such an obvious Schur Theorem at the test?? I don't think "obvious" is correct here. Maybe you mean "well-known". Certainly, if you haven't seen this before, it is a not-so-easy and nice problem, actually. So if they are sure that none of the students in the competition knew the result (which I suppose they were), then it actually makes sense to propose this (although of course it is always nice to have original problems).