So I'm working on a proof that for positive integers $a$, $b$ and $c$, $c>1$, that $GCD(c^a-1,c^b-1)=c^{GCD(a,b)}-1$. Not sure what I should think about doing first. My guess is I should start by showing the right hand side divides the numbers of the left. Thoughts?
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Yes, that would show that the LHS $\gcd$ is at least that big. – Joffan Feb 08 '15 at 21:16
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$\!\begin{eqnarray}{\bf Hint}\quad \rm mod\,\ d\!:\ \ c^A,\:c^B\equiv 1&\iff&\rm ord(c)\ |\ A,B\color{}\iff ord(c)\ |\ (A,B)\iff c^{\,(A,B)}\equiv 1\\ \rm i.e.\ \ \ d\ |\ c^A\!-\!1,\:c^B\!-\!1\ &\iff&\rm d\ |\ c^{\,(A,B)}\!-\!1,\qquad\ \ where \rm\quad\! (A,B)\, :=\, gcd(A,B) \end{eqnarray}$
Bill Dubuque
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How do you show the first step? The divisibility of A by ord(c)? – timelessGiant92 Feb 09 '15 at 04:15
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@time It's a basic result in group/number theory that $\ c^n=1\iff {\rm ord}(c)\mid n,,$ e.g. see here. $\ \ $ – Bill Dubuque Feb 09 '15 at 04:37