Jtdylktuy

Let $f(x)$ be a polynomial in $mathbb Z[X]$ such that degree of $f(x)$ is positive. I want to prove that for infinitely many $p$, $f(x)$ has a zero in $mathbb{Z}/pmathbb{mathbb Z}$.

I got a hint that Taylor's theorem is to be used. But I could not understand it properly. Any help is appreciated.

Your question can be answered with Proposition V.2.8. in the book Integer-valued polynomials by P.-J. Cahen and J.-L. Chabert. The authors state it in a different way to characterize the non-maximal prime ideals of the ring $text{Int}(mathbb{Z}) = {f in mathbb{Q}[x] mid f(mathbb{Z}) subseteq mathbb{Z} }$. But we can use the same proof to formulate it like this:

Let $q in mathbb{Q}[x]$ be irreducible. Then for infinitely many prime numbers $p in mathbb{Z}$ it holds that $q$ has a root modulo $p$.

Proof:

Let $q in mathbb{Q}[x]$ be irreducible and $d in mathbb{Z} setminus (0)$ such that $dq in mathbb{Z}[x]$. Since $qmathbb{Q}[x] = dqmathbb{Q}[x]$, we may assume that $q in mathbb{Z}[x]$. We show that for infinitely many primes $p in mathbb{Z}$ there is an $a in mathbb{Z}$ such that $p |_mathbb{Z} q(a)$.

For $q = x$ this is clear, so let $q = a_0 + a_1x + ... + a_dx^d$ with $a_0 neq 0$. Assume to the contrary that the set $mathcal{D} := { p in mathbb{Z} mid p in mathbb{P} land exists a in mathbb{Z}: q(a) in pmathbb{Z} }$ is finite, and let $y := prod_{p in mathcal{D}} p$.

$mathcal{D}$ is non-empty, since $q$ is non-constant and therefore unbounded, hence $y notin {1,-1}$. Therefore we have for non-negative integers $m neq n$ that $y^m neq y^n$. Hence there exists an integer $k > 0$ such that $q(a_0y^k) neq pm a_0$. (Otherwise $q$ would take one of the values $pm a_0$ infinitely many times and therefore would be constant.) For such $k$, it holds that $q(a_0y^k) = a_0(1 + y^kb)$ for some $b in mathbb{Z}$ and $1 + y^kb$ is no unit, since $q(a_0y^k) = a_0(1 + y^kb) neq pm a_0$. Therefore there exists a prime $p in mathbb{Z}$ such that $1 + y^kb in pmathbb{Z}$ and hence $q(a_0y^k) in pmathbb{Z}$, which implies that $p in mathcal{D}$.

But for all $p' in mathcal{D}$ it holds that $y in p'mathbb{Z}$ and hence $y^kb in p'mathbb{Z}$. Therefore $1 + y^kb notin p'mathcal{P}$ which implies $p' neq p$. So $p notin mathcal{D}$, which is a contradiction.

I don't see how to use Taylor's theorem here, but here is an alternative solution: $f$ has a zero in $mathbb{Z}/pmathbb{Z}$ if and only of $p$ is a factor of $f(n)$ for some integer $n$. So it remains to show that there are infinitely many prime factors of outputs of $f$.

There are several ways to do this; Qiaochu Yuan discusses 2 here: https://qchu.wordpress.com/2009/09/02/some-remarks-on-the-infinitude-of-primes/

(I'm making this community wiki since Qiaochu Yuan did most of the work here)