Exercise 5.3.4 (Consequences of $uv = w^2$)

Proof. We suppose that $\mathit{uv}=w^2>0$ is a perfect square, and let $g=u\wedge v>0$. Then $u=u^{\prime }g,v=v^{\prime }g$, where $u^{\prime }\wedge v^{\prime }=1$. Then $g^2\mid w^2$, and the unique factorization theorem shows that $g\mid w$, so there is some integer $w^{\prime }$ such that $w=g{w}^{\prime }$. Then ${u{}^{\prime }}^2{v{}^{\prime }}^2={w{}^{\prime }}^2$, where $u^{\prime }\wedge v^{\prime }=1$. By Lemma 5.4, $u^{\prime },v^{\prime }$ are perfect squares, so $u^{\prime }=r^2,v^{\prime }=s^2$ for some positive integers $r,s$.

There exist positive integers $r,s$ such that $u=g{r}^2$ and $v=g{s}^2$. □