There is a positive integer $y$ such that for a polynomial with integer coefficients we have $f(y)$ as composite Show that if $f(x)=a_nx^n+\cdots a_1x+a_0$ with $a_i \in \mathbb{Z}$, then there is a positive integer $y$ such that $f(y)$ is composite.
To prove this, we suppose that $f(x)=p$.  Then for $f(x+kp)$ we have $f(x)+Kp$ where $K$ is a constant.  I'm stuck on a specific part of understanding this proof, why is it that since $p \mid f(x) \Rightarrow p \mid (f(x)+kP)$.  I'm also stuck understanding why $f(x+kp)$ is necessarily also prime.

Look at $(x+kp)^n$
$$
(x+kp)^n = x^n + {n \choose 1} x^{n-1} k p + {n \choose 2} x^{n-2} k^2 p^2 + \cdots \\ =
x^n + \alpha_n p \text{ where $\alpha_n$ is an integer} $$
Similarly for other powers. This gives
$$ f(x+kp) = a_n (x^n + \alpha_n p) + a_{n-1} (x^{n-1} + \alpha_{n-1} p) \cdot \\=
 a_n x^n + a_{n-1} x^{n-1} + \cdots + (a_n \alpha_n + a_{n-1} \alpha_{n-1}+\cdots) p\\
=f(x) + K p
$$
