Hot Linked Questions

Show that $\gcd(2^m-1, 2^n-1) = 2^ {\gcd(m,n)} -1$ [duplicate]

Possible Duplicate: Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ $\gcd(b^x - 1, b^y - 1, b^ z- 1,…) = b^{\gcd(x, y, z,…)} -1$ I'm trying to figure this out: ...

asked Oct 31 '11 at 9:21

Dave

5133 bronze badges