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I was working on: https://math.stackexchange.com/questions/161578/kind-of-functional-eq-in-integers

I found a sort of way...

but I need to show that:

$p(x) \mid q(x)$ for infinite values of $x$ (integer) implies $p(n) \mid q(n) \quad \forall n \in \mathbb{Z}$

I was told that it is always true, but I can show it just for polynomials. Is it a well known theorem, or is there an easy proof?

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    Without any hypothesis on $p$ and $q$, it is trivially false.2012-06-22
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    In fact I had some doubts. Does it become true with functions like http://math.stackexchange.com/questions/161578/kind-of-functional-eq-in-integers ?2012-06-22
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    What are you interested in exactly ? And it is false even for polynomials ($p(x) = 2$, $q(x) = x$).2012-06-22
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    I'm trying to solve that problem, and I was reasoning with some primes, when I got that proving that if for infitely many primes $p^d-1 \mid p^n-1$ iff $d \mid n$ would solve the problem2012-06-22
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    This is easy since $p^n-1 = (p^d - 1)\left(1+p^d+p^{2d}+\dotsb+p^{d\left(\frac nd-1\right)}\right)$.2012-06-22
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    See http://math.stackexchange.com/questions/156646/division-of-qn-1-by-qm-1-in-wedderburns-theorem2012-06-22

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