Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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.

share|cite|improve this answer

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

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.