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

Let $a$, $b$ an element of $\mathbb{Z}$ with $a$ and $b$ not both zero and let $c$ be a nonzero integer. Prove that $$(ca,cb) = |c|(a,b)$$

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Please don't post in the imperative mode, giving orders. If you have a question, then please ask. – Arturo Magidin Mar 6 '11 at 4:31

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.

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

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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.
Moreover, you have not offered your motivation which absolutely will make the post better.

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