If $\gcd(m,n) = 1$ then $(2^m-1)(2^n-1)$ divides $2^{mn} - 1$ because each of $2^m-1$, $2^n-1$ divide $2^{mn}-1$ and $\gcd(2^m-1, 2^n-1) = 2^{\gcd(m,n)}-1 = 1$. How about the converse? If $\gcd(m,n) > 1$, can one show that $(2^m-1)(2^n-1)$ does not divide $(2^{mn}-1)$?
$(2^m -1)(2^n-1)$ divides $(2^{mn} -1)$ if and only if $\gcd(m,n) = 1$.
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elementary-number-theory