TITLE: Prove that $n$ divides $\phi(a^n -1)$ where $a, n$ are positive integer without using concepts of abstract algebra
QUESTION [5 upvotes]: I need to show that $n$ divides $\phi(a^n -1)$ where $a, n$ are positive integer without using concepts of abstract algebra
I know that $$a^n\equiv 1\pmod {a^n-1}$$
How do I proceed from there?

REPLY [5 votes]: In general (for $\gcd (a,m)=1)$, if $a_k≡1 \mod m $, then $ord_m(a) ~∣~k.$ In particular, since $a^{ϕ(m)}≡1(\mod m)$, we have $ord_m(a)∣ϕ(m)$. We have $ord_{(a^n−1)}a=n.$
Proof: Let there exist $d< n$ such that $ a^d ≡ 1 \mod(a^n-1) $ .
=> $ (a^n-1)~ |~ (a^d -1)$ which is not possible since $d < n $.
=> $ord _{a^n−1} =n.$
=> $n~|~ \phi(a^n-1)$