Exercise 1.2.43 ($a \mid bc$ if and only if $\frac{a}{(a,b)} \mid c$)

Proof. If $a\mid \mathit{bc}$, there is some integer $\lambda$ such that $\mathit{bc}=\mathit{\lambda a}$. Since $a\wedge b\mid a$ and $a\wedge b\mid b$, we may write this equality under the form $\frac{b}{a\wedge b}c=\lambda \frac{a}{a\wedge b}$, where $\frac{b}{a\wedge b}$ and $\frac{a}{a\wedge b}$ are integers (if $a\wedge b\ne 0$), thus

By Theorem 1.7,

Then Theorem 1.10 shows that

Conversely, if $\frac{a}{a\wedge b}\mid c$, then $a\mid (a\wedge b)c$, and $a\wedge b\mid b$, thus $a\mid \mathit{bc}$.

To conclude, if $a\wedge b\ne 0$ (that is, if $a\ne 0$ or $b\ne 0$),

□