Show that if $a$ is a divisor of $bc$ and $(a, b) = 1$, then $a$ is a divisor of $c$. Suppose I set $ax+by=1$ such that $x, y$ are integers and $a$ and $b$ are elements. 
The greatest common divisor of $a, b$ is $1$. 
If $a=bc$, then $bcx+by=1$. 
But then what do I do?

We have $$ax+by=1\Rightarrow acx+bcy=c$$
But $a|bc\Rightarrow bc=ak$, $k\in\Bbb{Z}$
Thus $$acx+aky=c\\a(cx+ky)=c\\am=c\\a|c$$
