How can I find $\gcd(n^a-1,m^a-1)$?

Asked Oct 23 '18 at 18:37

Active Oct 24 '18 at 14:02

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From Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ , we have $$\gcd(a^n-1,a^m-1)=a^{\gcd(n,m)}-1$$ for every positive integers $a,n,m$.

I reversed $a$ with $n,m$, and I had this question:

Find $\gcd(n^a-1,m^a-1)$ for every positive integers $a,n,m$

My attempt is to find the greatest common divisor of $n^p-1$ and $m^p-1$ for every prime $p$ such that $p|a$, but I couldn't get any further.

How can I find the general formula of $\gcd(n^a-1,m^a-1)$ ?

(sorry for my grammar mistake, English is my second language)

edited Oct 23 '18 at 20:34

asked Oct 23 '18 at 18:37

2

I don't think there is any reason to expect a nice answer for $\gcd(n^a-1, m^a-1)$ in general. Of course if $a+1$ is prime and $n$ and $m$ are coprime to $a+1$, $a+1$ divides that gcd. – Robert Israel Oct 23 '18 at 19:41

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