Does there exist a nonconstant polynomial $f \in \mathbf{Z}[t]$ such that $|f(n)|$ is a power of some prime number (depending on $n$), for every $n \in \mathbf{Z}$?
My Attempt: We know that there is no polynomial with integer coefficient which is generate only prime number but What about in this case. Give some hints.
First Remark:
If this holds, then it holds for a single prime $p$ for all $n$. To see this:
Suppose $p,q$ are distinct primes such that $p\,|\,f(a)$ and $q$ divides $f(b)$ for some $a,b\in \mathbb N$.
Then $$a'\equiv a \pmod p\implies f(a')\equiv 0\pmod p$$ with a similar claim for $q,b$.
However, since $p\neq q$ we can solve $$qk\equiv (a-b)\pmod p$$ for $k$ and, with that $k$ we see that both $p,q$ divide $f(b+qk)$.
Example: taking $f(n)=n^2+n+1$ we see that $3\,|\,f(1)$ and $7\,|\,f(2)$. We solve $7k\equiv -1 \pmod 3$ to get $k=2$ and remark that $21$ divides $f(2+7\times 2)=f(16)$
Hence you are down to the case where there is only a single prime $p$.
To rule out the case of a single prime $p$, note that the constant term can not be $0$ (if it were, then we'd get $n\,|\,f(n)$ which is clearly not possible). Let $f(0)=p^k$. Then $f(p^j)$ can not be divisible by $p^{k+1}$ if $j>k$. But then there are only finitely many values $f(p^j)$ might have, and that is not possible, since a polynomial of degree $d$ can take each given value no more than $d$ times.
And we are done.
