TITLE: $(a,b)\!=\!1\!=\!(a,c)\Rightarrow (a,bc)\!=\!1$ [coprimes to $\,a\,$ are closed under products]
QUESTION [21 upvotes]: How do I go about proving this?
If $\gcd(a,b)=1$ and $\gcd(a,c)=1$, then $\gcd(a,bc)=1$. I'm very confused with gcd proofs.

REPLY [5 votes]: We can take as the definition of "$a,b$ are relatively prime" that there are $s,t$ with $1=as+bt$. Then the question becomes: given that$$1=as+bt$$and$$1=as^\prime+ct^\prime$$find $S,T$ with$$1=aS+(bc)T.$$Multiplying by $1$ is a useful trick/technique for questions of this sort.$$1=as+bt=as+\mathbf{1}bt=as+\mathbf{(as^\prime+ct^\prime)}bt$$so$$\begin{align}1&=a(s+cst^\prime)+bc(tt^\prime)\\&=a(s^\prime+bs^\prime t)+bc(tt^\prime).\end{align}$$
The second formula arises from the first by applying $(bc)(ss^\prime)(tt^\prime)$ or shifting the derivation. A single formula invariant under that permutation is
$$1=(as+bt)(as^\prime+ct^\prime)=a(ass^\prime+bts^\prime+cst^\prime)+bc(tt^\prime).$$
The symmetry is nice but the first formulas are smaller.
What I like is that uses nothing more than basic properties of rings so the proof is applicable in many rings.