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This isn't a duplicate! the "duplicate" question doesn't use my method of proof but other methods and I made it clear that I want this method in specific.

I want to prove the following claim using the following method.

$(ax,bx)=x(a,b)$

What I have done so far:

let's call $(a,b)$ by $d$, then:

$d|a, d|b$ so $xd|ax, xd|bx$ then $xd|(ax,bx)$

How can I continue from here? I didn't prove equality...

zoro
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    You used the common divisor part of $(a,b)$ to get the common divisor part of $(ax,bx)$. Now use the greatest part of $(a,b)$ to figure the greatest part of $(ax,bx)$. – Aaron Goldsmith Nov 14 '22 at 21:22
  • @AaronGoldsmith sorry I didn't understand anything from what you wrote... I need to prove equality not GCD – zoro Nov 14 '22 at 21:31
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    The equality you are trying to prove is about GCD, correct? There are two things needed for an integer to be the GCD: it must be a common divisor and it must be the greatest. You have proved that $x(a,b)$ is indeed a common divisor of $ax$ and $bx$. You will have equality once you show it is the greatest. – Aaron Goldsmith Nov 14 '22 at 21:45
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    @AaronGoldsmith thanks it's clear now and that's what I tried to do but failed :( – zoro Nov 14 '22 at 22:25
  • @zoro Welcome to MSE! What do you mean by "failed" with Aaron's method? That is to say, where did you get stuck? – Eric Snyder Nov 14 '22 at 22:27
  • @EricSnyder dead end. – zoro Nov 15 '22 at 16:31
  • The claim in your edit is false - the first "Theorem" in this answer in the dupe gives a divisibility based proof continuing that in your question. Most of the common proofs are there. – Bill Dubuque Nov 15 '22 at 16:44
  • @BillDubuque it's not even clear what he's trying to prove. Sorry I'm not professor in mathematics. I'm looking for easy to understand method just like what I started... – zoro Nov 17 '22 at 20:51

1 Answers1

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Continue on by writing out $xd\mid (xa,xb)$ as an equation and see what this says about the integers $\frac{(xa,xb)}{x}$, $a$, and $b$.

Chris JS
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