And also that: $\gcd(p^m-1,p^n-1)=p^{\gcd(m,n)}-1$.
For the love of a God I don't see how to show these two identities.
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5Prove that $\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$ – Sarvesh Ravichandran Iyer May 22 '21 at 06:28
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Your last questions were duplicates. Do you search before posting? This site has a great collection of excellent answers for many questions. It is worth to look for them. – Dietrich Burde May 22 '21 at 08:13
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Yes, I should use google more, I wasn't sure it appears here. – MathematicalPhysicist May 22 '21 at 08:25
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For the first part, you can prove by induction. Assume that $p^n-1$ divides $p^{kn}-1$. Now, you can easily show that $p^n$ divides $p^{(k+1)n}-1$. Note that $p^{(k+1)n}-1 = p^{(k+1)n} -p^{kn} + p^{kn}-1$. The induction is on $k$, of course. I'm sure you can prove the base case of $k=1$, by yourself. :)
Vikrant
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3Or just use the identity $a^k - 1=(a-1)(a^{k-1}+a^{k-2}+\dots+1)$ with $a=p^n$. – dxiv May 22 '21 at 06:47
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@Vikrant How do I show that $p^n$ divides $p^{kn+n}-1$? thanks! – MathematicalPhysicist May 22 '21 at 07:18
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1So, you have to show that $p^n-1$ divides $p^{(k+1)n}-1$. Add and subtract $p^{kn}$ to and from $p^{(k+1)n}-1$. You get $p^{(k+1)n} -p^{kn} + p^{kn}-1$. We know by inductive hypothesis that $p^n-1$ divides $p^{kn}-1$. Now, we factor out $p^{kn}$ from $p^{(k+1)n}-p^{kn}$ to get $p^{kn}(p^n-1)$. – Vikrant May 22 '21 at 07:26