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That is, given a positive integer m, is the set

$\{n\mid \gcd(m,2^n-1)=1\}$

where n is a positive integer infinite?

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1 Answer 1

We have $n>0$ and $\gcd(2^n-1,m)>1$ only if $2^n\equiv 1\pmod p$ for some $p|m$ (and only if $p$ is odd). If $k_p>0$ is minimal with $2^{k_p}\equiv 1\pmod p$, then $k_p\ge2$. At least the infinitely many $n$ with $n\equiv 1\pmod {k_p}$ for all odd $p|m$ thus have $\gcd(m,2^n-1)=1$.

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