A question regarding the proof of $\gcd(a^m-1, a^n-1) = a^{\gcd(m,n)}-1$

Question

I have a problem trying to understand the proof:

Theorem $\boldsymbol{1.1.5.}$ For natural numbers $a,m,n$, $\gcd\left(a^m-1,a^n-1\right)=a^{\gcd(m,n)}-1$

Outline. Note that by the Euclidean Algorithm, we have $$ \begin{align} \gcd\left(a^m-1,a^n-1\right) &=\gcd\left(a^m-1-a^{m-n}\left(a^n-1\right),a^n-1\right)\\ &=\gcd\left(a^{m-n}-1,a^n-1\right) \end{align} $$ original image

We can continue to reduce the exponents using the Euclidean Algorithm, until we ultimately have $\gcd\left(a^m-1,a^n-1\right)=a^{\gcd(m,n)}-1$. $\square$

I see that the next iteration (if possible) is $\gcd(a^{m-2n}, a^n-1)$. However, why is the conclusion is true by euclidean algorithm?

@Saad I am a new user so I currently don't have the privilege to. — , Jul 07 '20 at 09:32
As a new user, you are allowed to write out your question. — Angina Seng, Jul 07 '20 at 09:36
@AnginaSeng I personally think it would be contextually better if I used an image instead. — , Jul 07 '20 at 09:41
I think what they mean is that you should be using MathJax to format your math equations, rather than images. — robjohn, Jul 07 '20 at 09:42
Does this answer your question? Proving that $\gcd(2^m - 1, 2^n - 1) = 2^{\gcd(m,n )} - 1$ — Angina Seng, Jul 07 '20 at 09:50
This proof is not wrong, but the idea is not well described. What is meant that we subtract repeatedly $n$ until we have equal exponents or a smaller still positive exponent. Then, we can change the roles and continue. Anyway, I think, there are better proofs. — Peter, Jul 07 '20 at 09:50

score 1 · Answer 1 · answered Jul 07 '20 at 10:18

Let $a,m,n$ be positive integers, $a>1$. Then

$$ \gcd(a^m-1,a^n-1) = a^{\gcd(m,n)}-1. $$

Write $g=\gcd(a^m-1,a^n-1)$. Suppose $m=qn+r$, $0 \le r <n$. From $g \mid (a^n-1)$ we have $g \mid (a^{qn}-1)$. Therefore,

$$ g \mid \big((a^m-1)-a^r(a^{qn}-1)\big) = (a^r-1). $$

Therefore, $g \mid \gcd(a^n-1,a^r-1)$. We have deduced this analogous to deducing $\gcd(m,n) \mid \gcd(n,r)$ in Euclid's algorithm to find $\gcd(m,n)$.

Suppose the steps in Euclid's algorithm are as follows: \begin{eqnarray*} m & = & qn+r, \: 0<r<n, \\ n & = & q_1r + r_1, \: 0 < r_1 < r, \\ r & = & q_2r_1 + r_2, \: 0 < r_2 < r_1, \\ \vdots & = & \vdots \\ r_{k-2} & = & q_kr_{k-1} + r_k, \: 0 < r_k < r_{k-1}, \\ r_{k-1} & = & q_{k+1}r_k. \end{eqnarray*}

Using the same argument repeatedly leads to $g \mid \gcd(a^{r_{k-1}}-1,a^{r_k}-1)$, and since $r_k \mid r_{k-1}$, to $g \mid (a^{r_k}-1) = (a^{\gcd(m,n)}-1)$.

Conversely, $\gcd(m,n)$ divides both $m$ and $n$, and so $a^{\gcd(m,n)}-1$ divides both $a^m-1$ and $a^n-1$. But then $a^{\gcd(m,n)}-1$ divides $g$.

This completes the proof. $\blacksquare$

score 0 · Answer 2 · answered Jul 08 '20 at 12:34

Suppose :

$gcm(m, n)= k$

$m=m_1 k$

$n=n_1 k$

Then we have:

$a^m-1=(a^K)^{m_1}-1=(a^k-1)[(a^k)^{m_1-1}+(a^k)^{m_1-2}+ . . . $

$a^n-1=(a^K)^{n_1}-1=(a^k-1)[(a^k)^{n_1-1}+(a^k)^{n_1-2}+ . . . $

That means:

$gcd(a^m-1, a^n-1)=a^k-1=a^{gcm(m, n)}-1$

Notice that the factor $(a^k-1)$ exists for both cases where m and n are both odd or even.

A question regarding the proof of $\gcd(a^m-1, a^n-1) = a^{\gcd(m,n)}-1$

2 Answers2