Infiniteness of the set of primes such $f$ have root $\bmod p$

Question

Let $f \in \mathbb{Z}[x]$ be non constant. How to prove that there exist infinitely many primes $p$ such $f$ have root in $\mathbb{Z}/p\mathbb Z$?

Answer 1

By contradiction using the same argument as Euclid' proof of infinitude of primes.

First you can assume that $f$ has no root in $\Bbb Z$ otherwise it's obvious that $f$ has a root in every $\Bbb Z_p$.

Take $g(x)=\dfrac{f(xf(0))}{f(0)}$ (this is a well defined polynomial over $\Bbb Z$), and we have also $g(0)=1$.

Now assume that $g$ has a root modulo only finitely many primes $p_1,\dots,p_k$. This means that for every integer $n$ the only primes which are allowed to divide $g(n)$ are only $p_1,\dots,p_k$.

Consider $m=\prod_{i=1}^kp_i$. Let $t\neq 0 $ be an integer. We have $g(tm)\equiv g(0)\equiv 1\bmod m$, so no prime from $p_1,\dots,p_k$ divides $g(tm)$, hence $g(tm)=1$ for all values of $t\neq 0$ and finally $g$ is a constant because it's a polynomial and here we have a contradiction as the polynomial $g$ is assumed to be non-constant.