Show that $M_a\neq M_b$. Let $f\in \mathbb Z[x]$ be a non-constant polynomial .Show that as $a$ varies  over $\mathbb Z$,the set of divisors of $f(a)$ includes infinitely many different primes.
My try:
Let $f(x)=a_0+a_1x+a_2x^2+\cdot+\cdot +a_nx^n$, 
$f(a)=a_0+a_1a+a_2a^2+\cdot\cdot\cdot\cdot +a_na^n$.Suppose that $f(a)$  has the form $f(a)=(a-p_1)^{k_1}(a-p_2)^{k_2}\cdot\cdot\cdot(a-p_m)^{k_m}$ for each $a\in \mathbb Z$.
Let $M_a$ be the set of different divisors of $f(a)$. If we can show that $M_a\neq M_b$ for each $a,b\in \mathbb Z$ then as $\mathbb Z$ is countable then $M=\{M_a:a\in \mathbb Z\}$ is infinite and hence the proof.
I am stuck to show that $M_a\neq M_b$.Please help.

Hint: Suppose that the set of prime divisors of the $f(a), a\in \mathbb{Z}$ is finite. Then there exists $a\in \mathbb{Z}$ such that $f(a)\not =0$ has the maximum of prime divisors possible. Fix such an $a$, and put $f(a+x)=\sum_{j=0}^n b_jx^j$, with the $b_j\in \mathbb{Z}$ ($b_n\not = 0$, and $b_0=f(a))$. Now put $x=mf(a)^2$, for $m\in \mathbb{Z}$. Show that $f(a+mf(a)^2)$ has as prime divisors all the prime divisors of $f(a)$, with in addition, if $m$ is large, another prime divisor, a contradiction.   
