If $f \in \mathbb{Z}[x]$ is such that $p \mid f(p)$ for all primes $p$, then $x \mid f(x)$ in $\mathbb{Z}[x]$. This follows by writing $f(x) = \sum \limits_{i=0}^d c_i x^i$ and noting that $p$ divides $c_0$ for every prime $p$ implies that $c_0 = 0$.

There are perhaps several ways to generalize this to several variables; I am particularly interested in the following.

Let $f \in \mathbb{Z}[x_1,\ldots,x_n]$ be such that for every set of distinct primes $\{p_1,\ldots,p_n\}$, there exists an $i \in \{1,\ldots,n\}$ such that $p_i$ divides $f(p_1,\ldots,p_n)$. Is it true that there exists an $i \in \{1, \ldots, n\}$ such that $x_i$ divides $f(x_1,\ldots,x_n)$ in $\mathbb{Z}[x_1,\ldots,x_n]$?

I strongly believe this to be true and think there must be a nice, elementary way to show it.

a proofis not a problem, but to have it "nice and elementary" is a bit harder. Is using the fact that you have about $x/\log x$ primes up to $x$ OK? (I do not need the PNT, just Chebyshev). If so, I'll post an answer. If not, I'll have to think more. $\endgroup$ – fedja Nov 19 '11 at 3:48