Numerical polynomials by means of prime powers

Question

Let $\mathbb{E}$ be the set of prime powers (except $1$). Let $f \in \mathbb{Q}[x]$ be a rational polynomial with $f(\mathbb{E}) \subseteq \mathbb{Z}$. Does it follow that $f$ is numerical, i.e. satisfies $f(\mathbb{Z}) \subseteq \mathbb{Z}$?

Notice that the set $\mathbb{P}$ of prime numbers is not enough, as the example $(x^3+x+2)/4$ shows.

Example. If $q$ is a prime power, then $(q^{10}-q^{5}-q^2+q)/10$ is the number of monic irreducible polynomials over $\mathbb{F}_q$ of degree $10$. Can we conclude from this directly that $(q^{10}-q^{5}-q^2+q)/10$ is an integer, for every integer $q$ (which I also know to be true for other reasons)?

Answer 1

No.

The polynomial $p(x)=\frac{x(x-1)(x-2)(x-3)(x-4)(x-5)(x-7)}{4}$ takes integer values with $p(x) \equiv 0 \bmod 8$ for all $x \not\equiv 6 \bmod 8$, but $p(6) \equiv 4 \bmod 8$. If $x$ is a prime power, then $x \not\equiv 6 \bmod 8$. It follows that $\frac{p(x)}{8}$ takes integer values at all prime powers, but not at $6$.

Answer 2

Following Jyrki Lahtonen's suggestion, let's write $f(x)=xg(x)+t$ and try to prove that $f(x)=0$ (mod m) for all $x \in \{0,1,\ldots,m-1\}$. If $gcd(x,m)=1$, then we can find a prime $p=g_q$ (mod q) for all prime powers $q|m$ and $g_q$ being a primitive root mod q. Hence we can write $x=p^n$ (mod m) for some $n$ and we are done. If $gcd(x,m)>1$, we first observe that $t=0$(mod m) (by setting $x=q$ for $q$ a prime power dividing $m$) and repeat the argument for $m'=\frac{m}{gcd(m,x)}$.