For odd $n$, there is an $m$ such that $n \mid 2^m-1$

Question

I am really stuck with this question:

Suppose $n$ is an odd positive integer. Prove that there exists a positive integer $m$ such that (2^m − 1)\n . (Here, “divides” means that when 2^m − 1 is divided by n.)

Answer 1

$\newcommand{\Z}{\mathbb{Z}}$It's standard thing.

$\gcd(2, n) = 1$, so $2$ is invertible in $\Z/n\Z$. The group $G$ of the invertible elements of $\Z/n\Z$ is finite, of order $m = \varphi(n)$. Thus $2^{m} \equiv 1 \pmod{n}$.