TITLE: Infiniteness of set of primes such $f$ have root $\mod p$
QUESTION [5 upvotes]: Let $f \in \mathbb{Z}[x]$ be non constant. How to prove that exists infinitely many primes such $f$ have root in $\mathbb{Z/_{(p)}}$. I spent much time, but with no benefits.

REPLY [6 votes]: 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 $g(x)=\frac{f(xf(0))}{f(0)}$ (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 primes : $p_1,\cdots,p_k$, this means in other words that the only primes which are allowed to divide $g(n)$ for any integer $n$ are only $p_1,\cdots,p_k$
Consider $m=\prod_{i=1}^kp_i$, Let $t\neq 0 $ be an integer we have $g(tm)\equiv g(0)\equiv 1\mod m$ so no prime from $p_1,\cdots,p_k$ divides $g(tm)$ but the only primes which can divide $g(x)$ are $p_1,\cdots,p_k$ 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.