Prove that gcd$(k^a-1,k^b-1)=k^{gcd(a,b)}-1$ where $k>1$; $k$, $a$, $b$ $\in$ $\mathbb N$.
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2Unfortunately, your post doesn’t fully meet this site’s policy (see how to ask a good question). I recommend that you edit your post. (1) You should provide some context to your question. Where is this problem from? Why is it interesting and important? (2) Please tell in the post, did you try to solve it? What progress did you achieve? Where were you “stuck”? (3) This (or similar) question was probably answered already. Use search. – Aig Mar 24 '24 at 17:42