Exercise 5.3.6 (Consequences of $6 uv = w^2$ (where $u \wedge v = 1$))

Proof. Let $u,v$ be positive integers. We assume that $6\mathit{uv}=w^2(w>0)$ is a perfect square, where $u\wedge v=1$. Then $3\mid w^2$. Since $3$ is prime, $3\mid w$. Similarly $2\mid w$, thus $6\mid w$, so $w=6w^{\prime }$ for some positive integer $w^{\prime }$. Therefore $\mathit{uv}=6{w{}^{\prime }}^2$.

Since $6\mid \mathit{uv}$, then $3\mid u$ or $3\mid v$, and $2\mid u$ or $2\mid v$. This gives fours cases:

• if $2\mid u,3\mid u$, then $u=6u^{\prime },v=v^{\prime }$;
• if $2\mid u,3\mid v$, then $u=2u^{\prime },v=3v^{\prime }$;
• if $3\mid u,2\mid v$, then $u=3u^{\prime },v=2v^{\prime }$;
• if $2\mid v,3\mid v$, then $u=u^{\prime },v=6v^{\prime }$

for some positive integers $u^{\prime },v^{\prime }$.

In each of these four cases, there are positive integers $d,e$ such that

Then $\mathit{de}{u}^{\prime }{v}^{\prime }=6{w{}^{\prime }}^2$, so $u^{\prime }{v}^{\prime }=w^2$. Moreover, $u^{\prime }\mid u,v^{\prime }\mid v$, and $u\wedge v=1$, therefore $u^{\prime }\wedge v^{\prime }=1$. Then Lemma 5.4 shows that $u^{\prime }=r^2,v^{\prime }=s^2$ for some positive integers $r,s$.

In conclusion, if $6\mathit{uv}$ is a perfect square, then there are positive integers $r,s,d,e$ such that

□