TITLE: Why if $n \mid m$, then $(a^n-1) \mid (a^m-1)$?
QUESTION [6 upvotes]: My Number Theory book says that for $n, m$ be positive integers and $a>1$, then

$(a^n -1)\mid(a^m -1)$ if and only if $n\mid m$.

I understand the proof for only if part, but in if part the autor says "it is clear". However a tried to prove that but a get stuck. Can you give a hint?

REPLY [2 votes]: If $n|m$, say $m = dn$, then
$$
a^m-1 = {a^n}^d - 1 = (a^n - 1)((a^n)^{d-1} + (a^n)^{d-2} + \ldots + a^n + 1)
$$
can be checked directly or follows from setting $\alpha = a^n$ and $N = d$ in
$$
\alpha^N-1 = (\alpha-1)(\alpha^{N-1} + \alpha^{N-2} + \ldots + \alpha + 1).
$$
Alternatively, you could also just note that all the complex roots $\zeta$ of $x^n - 1$ must be roots of $x^m - 1$ if $n\mid m$, since $\zeta^n = 1 \implies \zeta^m = (\zeta^n)^d = 1$.