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Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$
Let $p$ and $q$ be distinct primes.
I wonder is the following statement always true? $$\gcd(x^p-1, x^q-1) \stackrel{?}{=} x-1$$
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user706071
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Related : http://math.stackexchange.com/questions/7473/prove-that-gcdan-1-am-1-a-gcdn-m-1 – lab bhattacharjee Jan 02 '13 at 16:44
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More generally, we have $\gcd(x^n-1,x^m-1)=x^{\gcd(n,m)}-1$.
Hagen von Eitzen
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I have just found http://www2.hawaii.edu/~robertop/Courses/Math_454/Handouts/HW_Nov_30_sols.pdf Do you know a proof? – user706071 Jan 02 '13 at 15:50
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@user706071 Strong induction works, since for $n\geq m$, $\gcd(x^n-1, x^m-1) = \gcd(x^n-x^m, x^m-1) = \gcd(x^{n-m} -1, x^m-1)$ – Calvin Lin Jan 02 '13 at 16:24