$a =\gcd(2^m - 1; 2^n - 1)$. Why $2^n ≡ 1$ (mod $a$) and $2^m ≡ 1$ (mod $a$)?
closed as off-topic by user21820, Arnaud D., kimchi lover, 1 0, postmortes Nov 6 at 6:13
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If $a=\gcd(2^m-1;2^n-1)$, then in particular $a$ divides $2^m-1$ and $2^n-1$. That is, there exists integers $i,j$ such that $2^m-1=ai$ and $2^n-1=aj$. Thus,
$$2^m-1\equiv ai=0\pmod a$$
$$2^m\equiv 1\pmod a,$$
etc. Recall that, by definition, $c\equiv d\pmod p$ if and only if $p$ divides $c-d$.