Let a,b and n be positive integers with n ≥ 2. Prove that: gcd((n^a) - 1, (n^b) - 1)= n^(gcd(a,b)) - 1. Help please!! I have no idea how to do this!
Follow Math Help Forum on Facebook and Google+
A starting point might be similar to: $\displaystyle x^6-1 = (x^2-1)(x^4+x^2+1) $ so if n = dk, with d = gcd(n,m) $\displaystyle x^n-1 = (x^d-1)(x^{d*(k-1)}+...) $ and the same for m = dq $\displaystyle x^m-1 = (x^d-1)(x^{d*(q-1)}+...) $
View Tag Cloud