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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 $c_0 \equiv 0$ modulo $p$ 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. (See the re-reformulation below.)

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 $f(p_1,\ldots,p_n) \equiv 0$ modulo $p_i$. Is it true that $x_1 \cdots x_n$ divides $f(x_1,\ldots,x_n)$ in $\mathbb{Z}[x_1,\ldots,x_n]$?

Edit: It was pointed out to me that $f(x_1,\ldots,x_n) = x_1$ answers the question above in the negative. I wish to add that $f \in \mathbb{Z}[x_1,\ldots,x_n]$ should be strictly contained in $\mathbb{Z}[x_1,\ldots,x_n]$ in the sense that all of the variables are present. Note that $f(x_1,\ldots,x_n) = x_1$ lies in $\mathbb{Z}[x_1]$ for which the answer is proved affirmative above.

Edit$^2$: Apparently $f(x_1,\ldots,x_n) = x_1 (x_1 + \cdots + x_n)$ answers the reformulated question in the negative. For my purposes, the following would suffice.

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 $f(p_1,\ldots,p_n) \equiv 0$ modulo $p_i$. Is it true that there exists an $i \in \{1, \ldots, n\}$ such that $x_i \mid f(x_1,\ldots,x_n)$ in $\mathbb{Z}[x_1,\ldots,x_n]$?

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Is this really what you mean? It seems that the polynomial $f(x_1,\dots,x_n) = x_1$ has this property. – Greg Martin Nov 18 '11 at 2:49
@GregMartin This is some sort of degenerate case I missed but that I don't wish to consider. I'll edit the post. Perhaps there are easy examples I've missed in the new formulation, too. Thanks! – Dan Glasscock Nov 18 '11 at 3:04
Don't you still have the example $x_1(x_1+\cdots+x_n)$? – Gerry Myerson Nov 18 '11 at 3:28
@GerryMyerson I clearly need to rethink and reformulate this... Thanks for the comments! – Dan Glasscock Nov 18 '11 at 3:49
Hopefully the problem is non-trivial now. – Dan Glasscock Nov 18 '11 at 4:51

I think it may be true, and indeed follow from an observation on the one-variable case. If $p$ divides $f(p)$ for infinitely many $p$, then infinitely many $p$ divide $c_0$, so $c_0=0$, so $x$ is a factor of $f(x)$.

Now in the multi-variable case, it folows from the hypothesis that there is some $i$ such that $p_i$ divides $f(p_1,\dots,p_n)$ for infinitely many sets of distinct primes. Write $f$ as a polynomial $g(x_i)$, with coefficients polynomials in the other indeterminates. Then $p$ divides $g(p)$ for infinitely many primes $p$, so $x_i$ is a factor of $g$, hence of $f$.

Incidentally, I don't think there has been any actual use of the primeness of these primes; it looks like it may be good enough to have $k$ dividing $f(k)$ for infinitely many distinct integers $k$.

I hope someone will take a good look at this argument and say whether it holds or whether it has a hole big enough to drive a truck through it.

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Gerry, I agree that there is some $i$ (WLOG $i=1$) such that for infinitely many prime tuples $(p_1,\ldots,p_n)$, $p_i$ divides $f(p_1,\ldots,p_n)$. If we write $f$ as a polynomial in $x_i$ as you suggest, however, this is not the condition that $p$ divides $g(p)$ for infinitely many primes $p$. In what you wrote, $g(p) \in \mathbb{Z}[x_1,\ldots,x_{i-1},x_{i+1},\ldots,x_n]$, and so $p$ divides $g(p)$ would mean division in this polynomial ring. Perhaps you can refine what you said; otherwise I don't think it works. – Dan Glasscock Nov 18 '11 at 23:42

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