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corollary of B\'ezout's lemma

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\;If \,$\gcd(a,\,c) = 1$\, and \,$c|ab$, \,then \,$c|b$.
{\em Proof.} \,\PMlinkname{B\'ezout's lemma}{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 $c$ .

\textbf{Note.} \,A similar theorem holds in all \PMlinkname{B\'ezout domains}{BezoutDomain}, also in {\em B\'ezout rings}.