Number Theory: Prove that $m | \phi(a^m - 1)$

Question

$a$ and $m$ are natural numbers and $a>1$. Prove that $m | \phi(a^m - 1)$.

Any hints how can I prove this statement?

Answer 1

From $\gcd(a,a^m-1)=1$ we have $$a^{\varphi(a^m-1)}\equiv 1 \pmod{a^m-1}$$ also $$a^{m}\equiv 1 \pmod{a^m-1}$$ If we assume $\varphi(a^m-1)=mq+r, 0<r<m$ then $$a^{\varphi(a^m-1)}\equiv a^{mq+r} \equiv (a^m)^qa^r \equiv (1)^qa^r \equiv a^r\equiv 1 \pmod{a^m-1}$$ which is a contradiction (because of $0<r<m$), so $r=0$.