How do I prove that $\gcd(a,b)=\gcd(-a,b)$

Question

How do I prove that $\gcd(a,b)=\gcd(-a,b)$?

Here is what I have tried: Let $d=\gcd(a,b)$ and $d'=\gcd(-a,b)$

Then $d|a$, $d|b$, $d'|(-a)$, $d'|b$ $\implies d'|(b-a)$ and $d|(a+b)$. Then there is some $k,t\in\mathbb{Z}$ such that $b-a=kd'$ and $a+b=td$, but am now unsure what to do, and am not sure whether or not I am overcomplicating potentially trivial proof.

Please search for answers before posting questions. – Bill Dubuque Aug 20 '23 at 06:55 — Bill Dubuque, Aug 20 '23 at 06:55

score 1 · Accepted Answer · answered Aug 20 '23 at 06:09

1

$d' | -a \implies d'| a\;$ and $\;d'|b$ . Therefore, $d'|d$.

similarly, $d|a \implies d|-a\;$ and $\; d|b\;$ therefore, $d|d'$

$\implies d=d'$

answered Aug 20 '23 at 06:09

Arfin

1,460

1

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Aug 20 '23 at 06:47

How do I prove that $\gcd(a,b)=\gcd(-a,b)$

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