corollary of Bézout’s lemma

Theorem.

If  $\gcd (a,c)=1$  and  $c|ab$,  then  $c|b$.

Proof.  Bézout’s lemma (http://planetmath.org/BezoutsLemma) gives the integers $x$ and $y$ such that $xa+yc=1$.  This implies that  $xab+ybc=b$,  and because here the both summands are divisible by $c$, so also the sum, i.e. $b$, is divisible by $c$ .

Note.  A similar theorem holds in all Bézout domains (http://planetmath.org/BezoutDomain), also in Bézout rings.