Prove $\gcd(a^m - 1, a^n - 1)=a^{\gcd(m,n)} - 1$ for any natural number $a>1$.

Asked Dec 26 '19 at 17:58

Active Dec 26 '19 at 18:05

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Let $n$, $m$ - natural numbers, at least one $>0$. Let $a$ - natural number $>1$. Prove $\gcd(a^m - 1, a^n - 1)=a^{\gcd(m,n)} - 1$.

It is probably easy, but I cant catch it. I thought abut treating these guys as polynomials of variable "$a$". Looking that way, I know that $a^d-1$ divides $a^n - 1$ (both treated as polynomials) if and only if $d$ divides $n$. Therefore, I see that guy on the right divides guys on the left. But how I can it is the biggest common divisor? I don't see it. Looking from the ideal point of view is also pointless. That exercise does not have hint at the end of book I'm working with. Thanks in advance for help and Merry Christmas.

edited Dec 26 '19 at 18:05

Angina Seng

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asked Dec 26 '19 at 17:58

robin3210

Can you show that if $m>n$ then $\gcd(a^m-1,a^n-1)=\gcd(a^{m-n}-1,a^n-1)$? – Angina Seng Dec 26 '19 at 18:05
iterate the process $a(a^n-1)+(a-1)$ – Dec 26 '19 at 18:09
1

@LordSharktheUnknown At that moment I can't :( But I understand it is enough - by induction on $m$ with $n$ constant; here You reduce $m$ to $m-n$ which is smaller. – robin3210 Dec 26 '19 at 18:14
Correct (+1). Following that line of thinking you can basically switch the running of Euclid's algorithm to the exponents :-) – Jyrki Lahtonen Dec 26 '19 at 18:16

Prove $\gcd(a^m - 1, a^n - 1)=a^{\gcd(m,n)} - 1$ for any natural number $a>1$.

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