Prove that : $n \mid \varphi (a^{n}-1)$ $a,n$ positive integers wih $a>1$
I know that $a$ has multiplicative order $n$ in the ring of integers modulo $a^{n}−1$ and the order of the group of units modulo $a^{n}−1$ is $\varphi (a^{n}-1)$.
How can I prove that $a^{\varphi (a^{n}-1)}= 1 $mod$(a^{n}-1)$?
Thaks for your help.
Let $m=a^n-1$. Then the order of $a$ in $\left(\mathbb{Z}_{/(m\mathbb{Z})}\right)^*$ is exacly $n$, since $a^n\equiv 1\pmod{m}$ but for every $1<k<n$ we have $1<a^k<m$. By the Lagrange's theorem for groups, the order of $a$ is a divisor of the order of $\left(\mathbb{Z}_{/(m\mathbb{Z})}\right)^*$, hence: $$ n\mid \varphi(a^n-1) $$ follows.
