Suppose $f(a)=p$, a prime, then $f$ has a t most three distinct zeros.

Question

Suppose $f(x)$ is a polynomial with integer coefficients. Show that if $f(a)=p$ for some integer $a$ and prime number $p$, then $f(x)$ has at most three integer roots; that is, there are at most three distinct integers $\alpha,\beta,\text{ and }\gamma$ such that $f(\alpha)=f(\beta)=f(\gamma) = 0$.

Answer 1

David points out the answer in a comment. Let me adapt that answer here:

Suppose $f(x)$ has four roots $\alpha,\beta,\gamma$ and $\delta$. Then $f(x)=g(x)(x-\alpha)(x-\beta)(x-\gamma)(x-\delta)$. Applying to $a$, we find $$ g(a)(a-\alpha)(a-\beta)(a-\gamma)(a-\delta)=p. $$ But $(a-\alpha),(a-\beta),(a-\gamma),(a-\delta)$ are four distinct integers, and this is impossible. This is because if $p$ is prime, the maximal amount of distinct integers we can factor it into is $3$, given by multiplying its divisors: $$ p=(-p)\cdot(-1)\cdot 1$$

Therefore $f$ has at most $3$ roots.