real analysis - There is no polynomial $f(x)$ with integer coefficients such that $f(n)$ is prime for all non-negative integers $n$.

I am currently working on a homework problem from my Real Analysis class:

So far, I was able to prove part(a), that is, $f(kp) \mid p$. I'm quite confused as to how I can use this result to prove part (b). Also, I don't see the intuition in how $f(x)-p$ would be of any help to reach a contradiction. Any suggestions would be helpful.

Answer

Like paw88789, I'm assuming that the problem asks you to prove that there is no nonconstant $f$ that satisfies the described property. So suppose otherwise that there is a nonconstant $f(x)=\sum_{i=0}^r a_ix^i$ such that $f(k)$ is prime for all nonnegative integers $k$.

From (a), $f(kp)$ is divisible by $p$. But for $k\geq 0$, $f(kp)$ is prime so that $f(kp)=p$ for all $k\geq 0$. But this means the polynomial $g(x)=f(xp)-p$ has infinitely many roots $x=0,1,2,\ldots$ By the theorem that you are allowed to assume, we infer that $g(x)$ is identically $0$. But
$$g(x)=\sum_{i=1}^r(a_ip^i)x^i$$
so we have $a_i=0$ for all $i\geq 1$. This means $f(x)$ is the constant polynomial $f(x)=p$. Contradiction.