Last edited by milfner; May 5th 2008 at 08:42 PM.
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$\displaystyle \gcd(a^n-1,a^m -1)\text{lcm}(a^n-1,a^m-1) = (a^n-1)(a^m-1)$. Thus, $\displaystyle \text{lcm}(a^n-1,a^m-1)|(a^n-1)$. But, $\displaystyle (a^n-1)|( a^{\text{lcm}(n,m)}-1)$. Thus, the proof follows.
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