Problem - 4183
Show that for any positive integer $n$, $\varphi(2^n-1)$ is a multiple of $n$ where $\varphi(n)$ is Euler's totient function.
By # 4182, we have $n$ is the multiplicative order of $2$ modulo $(2^n-1)$. Meanwhile, by Euler's theorem, we have $2^{\varphi(2^n-1)}\equiv 1\pmod{2^n -1}$. Therefore, we have $n\mid \varphi(2^n-1)$.