Polynomial with prime powers values It's well known that there are no non-constant polynomials with integer coefficients whose values at integer points are primes. Could this result be generalized to the case of prime powers?
The question is whether there exists a polynomial $p(x) \in \mathbb{Z}[x]$ with degree at least one such that for all $x \in \mathbb{Z}$ $|p(x)|$ is prime power.
 A: With integer coefficients the answer is surely no: let $f(x)\in\mathbb{Z}[x]$ a degree $d$ polynomial, and let $a\in \mathbb{Z}$ be such that $f(a)\neq0,1,-1$. Then there exist a prime $p$ such that $f(a)=0 \mod p$. Now consider the sequence $x_n=f(a+np)$. Since $x_n=0\mod p$, by the assumption on the values of $f$, we must have $x_n=p^{\alpha_n}$ for some positive integer $\alpha_n$. Moreover, since $x_n$ is definitely strictly increasing (it is no loss of generality assuming the leading coefficient of $f$ is positive), then so is $\alpha_n$. Therefore $p^{\alpha_n}$ has at least an exponential behaviour. But $x_n=O(n^d)$.
For rational coefficients I don't know, but I expect a negative answer in that case, too.
A: Actually you can prove a lot more.
Theorem
For any non-constant polynomial $p(x)\in\mathbb{Z}[x]$ 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 $n\equiv n_i\bmod{p_i}$ 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 $p_1,\ldots,p_l$, $l\le k-1$. Since we have that $p(0)\neq 0$, let $p(0)=\pm p_1^{\alpha_1}\ldots p_l^{\alpha_l}$, $x\equiv 0\bmod{p_1^{\alpha_1+1}\ldots p_l^{\alpha_l+1}}$.
Then $p(x)\equiv p(0)\bmod{p_i^{\alpha_i+1}}$, $1\leq l\leq k-1$, 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.
