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Let $m,n$ be positive integers. Prove that $(2^m-1)^2|(2^n-1)$ if and only if $m(2^m-1)|n$.

Source:Russia 1997

Hint:

$2^m-1|2^n-1 \longleftrightarrow m|n$.Let $n=km$. As $2^m-1|\frac{2^n-1}{2^m-1}$
and $2^{m(k-1)}+2^{m(k-2)}+.............+1 \equiv k \equiv 0 (mod 2^m-1)$
\[\Longleftrightarrow 2^m-1|k\]
\[\Longleftrightarrow m(2^m-1)|n\]

https://www.awesomemath.org/assets/PDFs ... _Lemma.pdf
A general version, and applicable in many criteria. Particularly, this problem becomes one liner

Masum wrote:https://www.awesomemath.org/assets/PDFs ... _Lemma.pdf
A general version, and applicable in many criteria. Particularly, this problem becomes one liner

Ithink problem 3 has a crack in the solution. How can we assume a and d relatively prime?