If $m\mid n$ then $p^m-1\mid p^n-1$

Question

I know $m$ ,$n$ are two positive integer numbers such that $m\mid n$. If $p$ is a prime number, I want to show $p^m-1\mid p^n-1$.

Which lets you reduce to the question of showing $q-1\mid q^d-1$. — Thomas Andrews, Apr 09 '15 at 14:20

score 2 · Answer 1 · edited Apr 09 '15 at 14:28

2

Rewrite $n=d\cdot m$ and observe that we are looking to show

$p^m-1|(p^m)^d-1$.

This boils down to showing that $x-1|x^d-1$, which should not be a problem.

$\textbf{Note}$: We don't need $p$ to be prime.

edited Apr 09 '15 at 14:28

robjohn

answered Apr 09 '15 at 14:26

W2701

Sorry about the edit. I misread and have now rolled back – robjohn Apr 09 '15 at 14:29
@robjohn No problem, it happens. – W2701 Apr 09 '15 at 14:29

1 Answers1