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!

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- Dec 9th 2009, 10:40 AMlawrence.nelson49Proof of integers
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! - Dec 10th 2009, 02:33 PMqmech
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)}+...)

$