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If $f(x)$ and $g(x)$ are polynomials over $\mathbb{Z}$ and $f(n)|g(n)$ for all $n\in \mathbb{Z}$ then does it follow that $f$ divides $g$ in $\mathbb{Z}[x]?$

I'm pretty sure the answer is yes and that the proof should be easy (in fact I think it should be true with far weaker assumptions), but can't seem to get too far. If it helps, note in particular that $f$ has no integral roots.

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up vote 4 down vote accepted

Nope, but for silly reasons. Take $f(x) = 2, g(x) = x^2 + x$.

If $f$ is monic, the proof is straightforward: using the division algorithm we may assume WLOG that $g$ has degree strictly less than $f$ and then $\frac{g}{f}$ eventually tends to zero, so must itself be zero.

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Thanks! Do you know of necessary/sufficient conditions that would make it true? –  Jon May 27 '12 at 3:53
    
@Jon: not beyond what I've already said. –  Qiaochu Yuan May 27 '12 at 4:00

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