If $\gcd(b,c) = 1$, $b$ and $c$ divides $a$, then $bc$ divides $a$ [duplicate]

Question

this lemma is a corollary of the following lemma: if $\gcd(b,c)=1$ $gcd(a,bc)=gcd(a,c)gcd(a,b)$

Answer 1

$$b|a\implies (\exists k\in \Bbb N)\;:\; a=kb$$

$$c|a\implies (\exists k'\in\Bbb N)\;:\;a=k'c$$

So $$kb=k'c$$

which means that $ \;b|k'c \;$ but, by Gauss's Theorem, as $ gcd(b,c)=1 $, we conclude that $\; b|k' \;$ or

$$(\exists k''\in \Bbb N)\;:\; k'=k''b$$

thus $$a=k''bc$$ which says that $\; bc|a$. Done.