# Polynomial divisibility over the integers

A quick check of some particular situations shows that the following makes sense. I'm not sure if it is true though. So, any insight welcomed.

Let $a_1,...,a_m$ and $b_1,...,b_n$ be positive integers such that for any integer $q$ the number of $a_{i}$'s divisible by $q$ is greater than the number of $b_{i}$'s divisible by $q$. Then, $$\left.\prod_i (f(x)^{a_{i}}-x) \mathrel{}\middle|\mathrel{} \prod_j (f(x)^{b_{j}} - x)\right..$$

• Do you mean "strictly greater" or "greater or equals"? – penartur Apr 24 '12 at 5:22
• ... because if it's "strictly greater", there would have to be infinitely many $a_i$. – Robert Israel Apr 24 '12 at 5:29
• But your conclusion does not seem to make sense, e.g. if $f$ is a polynomial of degree $\ge 2$ and $\sum_i a_i > \sum_i b_j$. – Robert Israel Apr 24 '12 at 5:32
• Perhaps you should exhibit some of the particular situations that you quickly checked. That would have been a good idea in any case, but especially so now since there are some questions about what you mean. Under any interpretation of the question that I can make sense of, the statement seems to be false -- except, that is, for the fact that it's trivially true because, as Robert pointed out, no finite sets of positive integers could fulfill the condition as written. – joriki Apr 24 '12 at 7:13
• @Anna: Please make whatever clarifications need to be made in the question itself; people shouldn't have to burrow down to the end of the comments to get a correct statement of the question. Also, if you're still interested in an answer, please give some examples as I suggested above. If you're no longer interested in an answer and/or don't believe the claim is true anymore, please indicate that clearly in the question, or consider deleting it. – joriki Apr 25 '12 at 3:39

## 1 Answer

How about $f(x) = x\,$, $a_1 = 6\;$ and $b_1 = 3\;$?

($x+1$ divides $x^3-x$, but not $x^6-x$, except in characteristic 2)