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If $m$ is an odd natural number, show that $$m\mid 2^{\phi(m)}-1,$$ where $\phi$ is the Euler totient function.


Can someone provide me some hints.

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The result you want to prove is a straight consequence of Euler's totient theorem. On a quick look at the theorem, you will find that:

If n and a are coprime positive integers, then

$${\displaystyle a^{\varphi (n)}\equiv 1{\pmod {n}}}$$

Since m is odd and 2 is even so m and 2 are coprime and hence you can apply Euler's totient theorem here, without any hesitation.

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