Given $f(x) \in \mathbb{Z} [x] $ a polynomial, that evaluated in any $a \in \mathbb{N} $, results always in a multiple of 101 or a multiple of 107 (both prime numbers). Prove then, that $f(x)$ is always divisible by 101 for all the values of $a$, or $f(x)$ is always divisible by 107 for all $a$.
Any suggestions on how should I start?
You have a very valuable hypothesis that $f\in\Bbb Z[x]$: the coefficients are integers (if this were not given you could only prove them to be rational). This allows reducing the polynomial itself modulo any number$~n$, and conclude that the evaluated polynomial $f(k)$ modulo$~n$ depends only on the congruence class of the value$~k$ (at which it was evaluated) modulo$~n$. Now for a proof by contradiction assume $f(k)$ is not divisible by$~107$ and $f(l)$ is not divisible by$~101$. Using that $107$ and $101$ are relatively prime (which is all that matters here), can you find a value to evaluate at the will give a contradiction?
