Let $c_n$ be sequence defined for $n\ge 1$ as follows: $$ c_n=a^n-1.$$
Prove that this sequence satifies strong divisibility condition, that is: $$ \gcd(c_n, c_m)=c_{\gcd(n,m)}.$$
where $\gcd$ denotes greatest common divisor. Any hints?
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Can you show that $a^{(m,n)}-1$ divides the two at least? – Sean Nemetz May 13 '17 at 16:21
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Hints? Well $a^k - 1 = (a -1)(a^{k-1} + ..... + 1$ so $a^{dm} - 1 = (a^d - 1)(a^{(m-1)d} + .... + 1$. So $a^d - 1$ will divide all $a^{dk} - 1$ – fleablood May 13 '17 at 16:27