Prove that GCD$(n^a - 1,n^b -1)= n^{GCD(a,b)} -1$

Question

I used Euclid's theorem $a=bq +r$.
$n^a -1 = ((n^b)^q )n^r -1$ I don't know how to move forward.

Recommended intermediate step: show that $n^i-1\mid n^j-1\iff i\mid j$. — Arthur, Oct 02 '17 at 20:21
See the list of most voted elementary number questions. You will find this question there with any answers — Vidyanshu Mishra, Oct 02 '17 at 20:36

score 0 · Answer 1 · answered Oct 02 '17 at 20:34

By symmetry, we can assume $b\geq a$. Now observe that

$n^b-1=n^{b-a}(n^a-1)+n^{b-a}-1$. So

$gcd(n^b-1,n^a-1)=gcd(n^{b-a}(n^a-1)+n^{b-a}-1,n^a-1)=gcd(n^{b-a}-1,n^a-1)$

Go on this way, we find that the exponential ends up with $gcd(a,b)$ by Euclidean algorithm.

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