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

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

• Where is the question? Commented Oct 21, 2020 at 21:02

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