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 $\mathbb{Z}/p\mathbb{\mathbb Z}$.
I got a hint that Taylor's theorem is to be used. But I could not understand it properly. Any help is appreciated.
On
I don't see how to use Taylor's theorem here, but here is an alternative solution: $f$ has a zero in $\mathbb{Z}/p\mathbb{Z}$ 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)
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 $q\mathbb{Q}[x] = dq\mathbb{Q}[x]$, 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 $\mathcal{D} := \{ p \in \mathbb{Z} \mid p \in \mathbb{P} \land \exists a \in \mathbb{Z}: q(a) \in p\mathbb{Z} \}$ 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 $1 + y^kb \in p\mathbb{Z}$ and hence $q(a_0y^k) \in p\mathbb{Z}$, 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.