Infinitely many prime divisors of $f(a)$ Let $f(x)\in \mathbb{Z}[x]$ be a non constant polynomial with integer coefficients. Show that as $a$ varies over the integers, the set of divisors of $f(a)$ includes infinitely many primes...
To be frank, I have no idea where to start... 
Trivial case is when constant term of $f(x)$ is zero.
In case of $f(x)=x(a_nx^n+\cdots+a_1)$  we have $p$ divides $f(p)$ for all primes $p$...
Other than this i have no idea...
Please give only hints..
 A: Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+……+a_1x+a_0\in \mathbb Z[x]$.
If $a_0=0$ it is evident that $p$ divides $f(p)$ for arbitrary primes so we make $a_0\ne 0$. Assume for the absurd, that there is only a finite number of prime divisors $p_1,p_2,p_3,……,p_N$ for $ f (k);\space k\in \mathbb Z $, and make the product $P=p_1p_2p_3\cdot\cdot\cdot p_N$.
We have
$$f(a_0P)=a_0\left(a_na_0^{n-1}P^n+a_{n-1}a_0^{n-2}P^{n-1}+….+ \space a_2a_0P^2+a_1P+1\right)$$
It is clear that non prime divisor of the factor $$ a_na_0^{n-1}P^n+a_{n-1}a_0^{n-2}P^{n-1}+….+ \space a_2a_0P^2+a_1P+1$$ can be one of the $p_1,p_2, p_3,……,p_N$. This is a contradiction.
A: We will show that for any $H$, however huge, there is a prime $p\gt H$ that divides some $f(a)$.
As you observed, we can assume that $f(x)$ has non-zero constant term $a_0$. Consider $f(a_0x)=a_0(1+xg(x))$. 
The equations $1+xg(x)=1$ and $1+xg(x)=-1$ have only finitely many solutions. Let $N\ge H$ be chosen so that $N!$ is greater than any of these solutions. Then $1+N!g(N!)$ cannot be $1$ or $-1$, so is divisible by some prime $p$.
If $p\le N$, then $p$ divides $N!$ so $p$ cannot divide $1+N!g(N!)$. Thus $p\gt N\ge H$. Since $p$ divides $1+N!g(N!)$ it follows that $f(a_0N!)\equiv 0\pmod{p}$. 
