Is $\gcd(a^n-1,a^nm) = \gcd(a^n-1,m)?$

Asked Jul 08 '23 at 14:51

Active Jul 08 '23 at 19:11

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I came across this procedure while trying to solve $gcd((2^{100})-1,(2^{120})-1))$, the procedure yielded correct result. E.g $gcd((2^{120})-1,2^{100}-1))$ =$gcd((2^{100})-1,2^{120}-2^{100}))$ = $gcd((2^{100})-1,2^{20}-1))$ And I kept solving like this using repeated subtraction and the property mentioned in the question.

The answer is $2^{20}-1$ So, by using Euclidean algorithm and other properties of gcd can we prove that $gcd((a^n)-1,a^n(m)) = gcd((a^n)-1,m))$

edited Jul 08 '23 at 19:11

Bill Dubuque

272,048

asked Jul 08 '23 at 14:51

Optimus

1

Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Jul 08 '23 at 14:58
$\gcd(u,v)=1\implies\gcd(u,vm)=\gcd(u,m).$ – Anne Bauval Jul 08 '23 at 15:15
Thanks,this works for the given scenario. Edited the question,hope it's understandable now. I believe the general equality given in the question holds because a^n and (a^n)-1 are coprimes. – Optimus Jul 08 '23 at 15:19
You believe right, this was precisely the meaning of my comment and link to the duplicate. – Anne Bauval Jul 08 '23 at 15:36
Your edit should really be improved: format your formulas (look at the way I edited your title) – Anne Bauval Jul 08 '23 at 15:37

Is $\gcd(a^n-1,a^nm) = \gcd(a^n-1,m)?$

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