Proving that $\gcd(ac,bc)=|c|\gcd(a,b)$

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Below is a proof of the distributive law for GCDs that works in every domain.

THEOREM $\rm\quad (a,b)\ =\ (ac,bc)/c\quad$ if $\rm\ (ac,bc)\ $ exists

Proof $\rm\quad d\ |\ a,b\ \iff\ dc\ |\ ac,bc\ \iff\ dc\ |\ (ac,bc)\ \iff\ d|(ac,bc)/c$

See my post here for further discussion of this property and its relationship with Euclid’s Lemma.

Let $d = (ca,cb)$ and $d' = |c|(a,b)$. Show that $d|d'$ and $d'|d$.

If $(a,b)=d$, then the equation $ax+by=dz$ has a solution for all $z \in \mathbb{N}$, and this implies that $acx+bcy=(dc)z$ admits a solution for all $z \in \mathbb{N}$. And hence we can deduce the result which must appear in every elementary number theory book.
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