Prove that $\phi (2^n-1)$ is a multiple of $n$ for any $n>1$

Asked Mar 29 '15 at 22:19

Active Jan 05 '16 at 13:50

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Prove that $\phi (2^n-1)$ is a multiple of $n$ for any $n>1$.

I'm not really sure how to start this proof.

edited Jan 05 '16 at 13:50

Martin Sleziak

asked Mar 29 '15 at 22:19

atl

But basically, $2^n\equiv 1\pmod{2^n-1}$. Can $2^m\equiv 1\pmod{2^n-1}$ for any $m<n$? – Thomas Andrews Mar 29 '15 at 22:34

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