## Problem 1

For which positive integers $\mathrm{a}, \mathrm{b}$ does $\left(\mathrm{x}^{\mathrm{a}}+\ldots+\mathrm{x}+1\right)$ divide $\left(\mathrm{x}^{\mathrm{ab}}+\mathrm{x}^{\mathrm{ab}-\mathrm{b}}+\ldots+\mathrm{x}^{2 b}+\mathrm{x}^{\mathrm{b}}+1\right)$ ?


## Solution

Answer: $\mathrm{a}+1$ and $\mathrm{b}$ relatively prime.

The question is when $\left(\mathrm{x}^{\mathrm{a}+1}-1\right) /(\mathrm{x}-1)$ divides $\left(\mathrm{x}^{\mathrm{b}(a+1)}-1\right) /\left(\mathrm{x}^{\mathrm{b}}-1\right)$ or when $\left(\mathrm{x}^{\mathrm{a}+1}-1\right)\left(\mathrm{x}^{\mathrm{b}}-1\right)$ divides $\left(\mathrm{x}^{\mathrm{b}(\mathrm{a}+1)}-\right.$ 1)( $x-1)$. Now both $\left(x^{a+1}-1\right)$ and $\left(x^{b}-1\right)$ divide ( $\left.x^{b(a+1)}-1\right)$. They both have a factor $(x-1)$, so if that is their only common factor, then their product divides $\left(x^{b(a+1)}-1\right)(x-1)$. That is true if $a+1$ and $b$ are relatively prime, for the roots of $x^{k}-1$ are the kth roots of 1 . Thus if $a+1$ and $b$ are relatively prime, then the only (complex) number which is an (a+1)th root of 1 and a bth root of 1 is 1 .

But suppose $d$ is a common factor of $a+1$ and $b$, then $\exp (2 \pi i / d)$ is a root of both $x^{a+1}-1$ and $x^{b}-1$. It is a root of $\mathrm{x}^{\mathrm{b}(a+1)}-1$, but only with multiplicity 1 , so $\left(\mathrm{x}^{\mathrm{a+1}}-1\right)\left(\mathrm{x}^{\mathrm{b}}-1\right)$ does not divide $\left(\mathrm{x}^{\mathrm{b}(a+1)}-1\right)(\mathrm{x}-1)$.

