In this post, I asked how to prove $n\mid \phi(2^n-1),(n\in \mathbb N)$. @Amr and @Abhra Abir Kundu proved more: they proved that $n\mid \phi(a^n-1),(a,n\in \mathbb N).$ The method is very nice. I quote Amr's answer in the following:
Consider $U(2^n-1)$. Clearly $2\in U(2^n-1)$. It can also be shown easily that the order of $2$ in the group $U(2^n-1)$ is $n$. By Lagrange's theorem $|2|=n$ divides $|U(2^n-1)|=\phi(2^n-1)$.
Now I want to prove a similar problem:
If $a,b,n\in \mathbb N,a>b,$ then $n\mid \phi(a^n-b^n).$
Inspired by their answers, I try to do something:
First, we assume that $GCD(a,b)=1.$ Let $ak\equiv b\pmod {a^n-b^n},k\in\mathbb N.$ Then $a^nk^n\equiv b^n\equiv a^n\pmod {a^n-b^n},a^n(k^n-1)\equiv 0\pmod {a^n-b^n},k^n\equiv 1 \pmod {a^n-b^n}.$
Consider $U(a^n-b^n)$. Clearly $k\in U(a^n-b^n)$. If we can show that the order of $k$ in the group $U(a^n-b^n)$ is $n$. By Lagrange's theorem $|k|=n$ divides $|U(a^n-b^n|=\phi(a^n-b^n)$.
Howerver, I can only prove that $k^n\equiv 1 \pmod {a^n-b^n},$ it doesn't $\implies|k|=n.$
Any help would be grateful!