Lemma 1   If $a\mid bc$ and $\gcd(a,b)=1$ then $a\mid c$ .

Proof. By assumption $\gcd(a,b)=1$ , thus we can use Bezout's lemma to find integers $x,y$ such that $ax+by=1$ . Hence $c\cdot(ax+by)=c$ and $acx+bcy=c$ . Since $a\mid a$ and $a \mid bc $ (by hypothesis), we conclude that $a \mid acx + bcy =c $ as claimed.