Sunday, July 10, 2016

elementary number theory - Prove that $gcd(a^n - 1, a^m - 1) = a^{gcd(n, m)} - 1$

For all $a, m, n \in \mathbb{Z}^+$,



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

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