Let $f,g \in \mathbb{Z}[x]$ be monic polynomials such that $f(n)$ divides $g(n)$ for infinitely many values of positive integers $n.$ Prove that $f$ divides $g$ in $\mathbb{Z}[x].$
I first wrote out $f(x) = a_0 + a_1x + \cdots + a_{n+1} x^n$ and $g(x) = b_0 + b_1x + \cdots + b_{m+1} x^m.$ However, I am unsure where to go from here.
Write $g(x) = f(x)q(x) + r(x)$ for $\deg r < \deg f$ and $q, r \in \mathbb{Z}[x].$ Rewrite this as $$\frac{g(n)}{f(n)} = q(n) + \frac{r(n)}{f(n)},$$ which is true whenever $f(n) \neq 0,$ which will happen only finitely many times.
Since $g(n)/f(n)$ is an integer infinitely often, and since $q(n)$ is always an integer, we find that $r(n)/f(n)$ is an integer infinitely often. But $\deg r < \deg f,$ so for sufficiently large $n$ we must have $|r(n)/f(n)| < 1.$ Thus $r(n)/f(n)$ is zero infinitely many times, which means the polynomial $r$ must be the zero polynomial.
