What is the proof for GCD(n.m)=1 if 2^m-1 and 2^n-1 have a common divisor of 1?

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Discussion Overview

The discussion revolves around proving the equivalence of the greatest common divisor of two integers \( n \) and \( m \) being 1, and the greatest common divisor of \( 2^n - 1 \) and \( 2^m - 1 \) also being 1. The focus is on theoretical aspects of number theory.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant seeks to prove that \( \text{GCD}(n, m) = 1 \) if and only if \( \text{GCD}(2^n - 1, 2^m - 1) = 1 \).
  • Another participant notes that the polynomial \( x^k - 1 \) is divisible by \( x - 1 \) and suggests that this property can be applied to \( 2^n - 1 \) to show divisibility relationships.
  • A different participant argues that if an integer \( r \) divides both \( 2^m - 1 \) and \( 2^n - 1 \), then it follows that \( 2^m \equiv 2^n \equiv 1 \mod r \), leading to a conclusion about the relationship between \( m \) and \( n \) under the Euclidean algorithm.

Areas of Agreement / Disagreement

Participants express varying degrees of certainty about the proof, with some providing reasoning and others indicating they are not fully convinced. No consensus is reached on the validity of the claims or the proof itself.

Contextual Notes

Some assumptions regarding the integers \( n \) and \( m \) are not explicitly stated, and the discussion does not resolve the mathematical steps involved in the proof.

HussamLawen
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Hi

i need help to prove this theory:
GCD(n.m)=1 <=> GCD((2^n)-1,(2^m)-1)=1

n,m real numbers
 
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sorry n,m Integers****
 
I don't have yet a proof (thus I'm not completely convinced of the fact yet), but I believe this is a start:

The polynomial x^k-1 is divisible by x-1; actually, x^k-1 factors as (x-1)(x^{k-1}+x^{k-2}+ ... +x^3+x^2+x+1). Applying this with x = 2^n you deduce that 2^n-1 divides 2^{k n}-1.
 
If an integer r divides 2^m-1 and 2^n-1, then 2^m\equiv2^n\equiv1\bmod r

By the Euculidean algorithm, since m and n are relatively prime, there exists a and b such that am+bn=1.

Thus 2^{am+bn}\equiv2\equiv(2^{am})(2^{bn})\equiv(1^a)(1^b)\equiv1\bmod r Thus r equals 1.
 
Last edited:

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