Euclid's Lemma

Theorem: Let $$a,b,c \in \mathbb{Z}$$. If $$a|bc$$, where $$a$$ and $$b$$ are relatively prime, then $$a|c$$.

Proof
Since $$gcd(a,b)=1$$, we may write $$ax+by=1$$ for some $$x,y \in \mathbb{Z}$$. Upon multiplication by $$c$$, we see that $$c=c(ax+by)=cax+cby$$. Since $$a|ac$$ and $$a|bc$$, it is clear that $$a|(cax+cby)$$. However, $$cax+cby=c(ax+by)=c \cdot 1=c$$. Therefore, $$a|c$$.

QED