Proving gcd(rs , r + s)= 1: I'm in Over My Head!

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Discussion Overview

The discussion revolves around proving that if r and s are positive integers with r > s and gcd(r, s) = 1, then gcd(rs, r + s) = 1. Participants explore various methods of proof, including the Euclidean algorithm and linear combinations, while also addressing the implications of r and s being prime numbers.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant suggests using the Euclidean algorithm or linear combinations to prove gcd(rs, r + s) = 1 but expresses difficulty in doing so.
  • Another participant proposes dropping the condition that r and s are positive integers, noting that they still need to be non-zero.
  • A participant questions how showing that d divides certain expressions leads to the conclusion that gcd(rs, r + s) = 1, suggesting that more needs to be shown about divisors.
  • One participant introduces the idea of primes, stating that if r and s are different primes, then gcd(rs, r + s) = 1 needs to be shown.
  • Another participant argues that if a prime divides both rs and r + s, it must divide either r or s, leading to a discussion about whether r or s can divide r + s.
  • A later reply suggests that gcd(r^s, s^r) = 1 can be reasoned similarly, noting that r divides r^s but not s, and vice versa.
  • Another participant asserts that for distinct primes r and s, gcd(r^m, s^n) = 1 is clear for any non-negative integers m and n due to prime factorization.

Areas of Agreement / Disagreement

Participants express varying levels of understanding and approaches to the problem, with some agreeing on the implications of r and s being primes while others remain uncertain about the proof methods. No consensus is reached on a definitive proof.

Contextual Notes

Some participants note the importance of the conditions under which the gcd is being evaluated, particularly regarding the nature of r and s as primes and the implications for divisibility.

Who May Find This Useful

This discussion may be useful for individuals interested in number theory, particularly those exploring properties of gcd and prime numbers.

buddyholly9999
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I saw this on a website. Prove that if r and s are positive integers with
r > s and gcd(r,s)=1, then gcd(rs , r + s)= 1. I can think of two ways to show this. Using the Euclidean algorithm or by showing that 1 can be a linear combination of rs and r + s. Funny thing...I couldn't do either them. I'm ashamed of myself and now I'm looking for some answers...
 
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You can drop the conditions that r, s are positive (but I suppose you still need r, s != 0) and r > s.

Let d = gcd(rs, r + s).

d divides both rs - r(r + s) and rs - s(r + s).
 
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Ok..I'm afraid I'm not following...d = 1 in this case. 1 divides anything. So how is that 1 dividing rs - r(r + s) and rs - s(r + s) show that the
gcd(rs, r + s) = 1. Wouldn't we have to show that 2 isn't a divisor, then 3 isn't a divisor?
 
Suppose p is a prime and p divides both rs and r+s. WLOG p divides r... the rest is left to you.
 
Wait wait...I think someone got something wrong...probably me...

Ok, r and s are primes...oops..I might have forgotten to mention that..

that integer in my first post is typo...so...

does that change everything?

Ok OK..i'll restate it...

let r and s be two different primes...show that gcd(rs, r + s) = 1.

Again...I can't prove this..

...I type this as tears hit the keyboard... :(
 
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buddyholly9999 said:
let r and s be two different primes...show that gcd(rs, r + s) = 1.

Again...I can't prove this..

...I type this as tears hit the keyboard... :(

Suppose you have a prime that divides both rs and r+s. It divides rs, so it must be either r or s. Can either r or s divide r+s?
 
Oh yeah...makes sense now...

you da man shmoe...now how about...

gcd(r^s, s^r) = 1 same scenario...

Could I say something like, r divides r^s, but s does not...and s dvides s^r but r does not..thus the gcd is 1?
 
with r and s still distinct primes gcd(r^m,s^n)=1 should be pretty clear for any non-negative integers m and n. You have the prime factorization of two numbers sitting in front of you, finding the gcd is trivial.
 
Thanks for the help...ever need a back rub, foot massage...someone dealt with...you know where i'll be..
 

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