nt.number theory - Polynomial with prime powers values

Actually you can prove a lot more.

Theorem

For any non-constant polynomial $p(x)inmathbb{Z}[x]$ and any positive integer $k$ there is an integer $n$ such that $p(n)$ is divisible by at least $k$ distinct primes.

Proof

If we prove that there exist integers $n_1,ldots,n_k$ and distinct primes $p_1,ldots,p_k$ such that $p(n_i)equiv 0 bmod{p_i}$ then we are done, because there exists an $n$ such that $nequiv n_ibmod{p_i}$ by the Chinese Remainder Theorem, and any such $n$ satisfies $f(n)equiv f(n_i)equiv 0 bmod{p_i}$, as desired.

Now by contradiction suppose that $p$ is only divisible by the primes $p_1,ldots,p_l$, $lle k-1$. Since we have that $p(0)neq 0$, let $p(0)=pm p_1^{alpha_1}ldots p_l^{alpha_l}$, $xequiv 0bmod{p_1^{alpha_1+1}ldots p_l^{alpha_l+1}}$.

Then $p(x)equiv p(0)bmod{p_i^{alpha_i+1}}$, $1leq lleq k-1$, so that the greatest power of $p_i$ that divides $p(x)$ is $p_i^{alpha_i}$.

But by hypothesis $p(x)$ is only divisible by the $p_i$, so we conclude that $p(x)=pm p(0)$. Using the pigeonhole principle and the fact that a non-constant polynomial can only assume a value a finite amount of times this is a contradiction, as desired.