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My attempt is as follows though I'm not very confident in it:

Let $c=\gcd(|a|,|b|)$. Then $c\mid |a|$ and $c\mid |b|$, then it must also be true that $c\mid (x|a|) \space \forall x\in\mathbb{Z}.$ In particular $x=-1$. So then $c\mid |a|$ and $c\mid (-|a|)$ so it must be that $c\mid a$ so $c$ is a divisor for both $a$ and $b$.

Let us now show it is the greatest such divisor:

For sake of contradiction, assume $c \neq \gcd(|a|, |b|)$. So $\exists \space c_1$ such that $c_1 > c$ and $c_1\mid a,\space c_1\mid b,\space c_1\nmid|a|$ and $c_1\nmid|b|$. But we've already seen that $c\mid a \iff c\mid |a|$ so this $c_1$ is contradictory and therefore $\gcd(a, b) = \gcd(|a|, |b|)$.

KriSD
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  • @GregNisbet I understand that property of the gcd but I'm not sure if this proof is sufficient or correct. I am not well-practiced in validating my own proofs haha – KriSD Nov 25 '23 at 17:58
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    Same as here in the linked dupe, i.e. since $,d\mid |x|\iff d\mid x,$ we infer that $|a|,|b|$ and $,a,b,$ have the same set of common divisors so they have the same greatest common divisor. – Bill Dubuque Nov 25 '23 at 18:11
  • @BillDubuque thank you! – KriSD Nov 26 '23 at 14:13

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