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For a > 1 show that the gcd$(a^n - 1, a^m - 1) = a^{(m,n)} - 1$

What are some useful equalities that might help in proving this relationship? I believe the constrains for $m,n$ are all positive integers.

Arthur Collé
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Hint $\ $ The simple equivalences demonstrated below prove that both sides of your equation have the same common divisors $\rm\,d,\, $ therefore they have the same greatest common divisor. $\ \ $ QED

$$\begin{eqnarray}\rm\ \ mod\,\ d\!:\ \ a^M,\:a^N\equiv 1&\iff&\rm ord(a)\ |\ M,N\iff ord(a)\ |\ (M,N)\iff a^{\,(M,N)}\equiv 1\\ \rm i.e.\ \ \ d\ |\ a^M\!-\!1,\:a^N\!-\!1\ &\iff&\rm\ d\ |\ a^{\,(M,N)}\!-\!1,\qquad\ \ \, where \rm\quad (M,N)\, :=\, gcd(M,N) \end{eqnarray}$$

Note $ $ The conceptual structure at the heart of this simple proof is the ubiquitous order ideal. $\ $ See my post here for more on this and the more familiar additive form of a denominator ideal.

Generally $\rm\ gcd(f(m), f(n))\ =\ f(gcd(m,n))\ \ \ if\ \ \ f(n)\ \equiv\ f(n\!-\!m)\ \ (mod\ f(m)),\: $ and $\rm\: f(0)\ =\ 0.\ $ See my post here for a simple inductive proof.

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Bill Dubuque
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