Solve $\frac1x+\frac1y=\frac1{pq}$

For $x,y\in\mathbb{N}$ how many ordered pairs $\left(x,y\right)$ satisfy $\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{pq}$ where $p,q$ are distinct primes?

The "easy way" to deal with $\frac 1x+\frac 1y=\frac 1N$ is to set $x=N+a, y=N+b$ and clear fractions to obtain $ab=N^2=p^2q^2$

Now $p^2q^2$ has $(2+1)(2+1)$ factors so there should be nine solutions for $a$ - from which $b$ is determined.

Let's start by writing $x=au$, $y=av$ where $a=\gcd(x,y)$. Then, $$ \frac{1}{pq}=\frac{1}{au}+\frac{1}{av}=\frac{u+v}{auv} \iff pq=uv\cdot\frac{a}{u+v} $$ where we must have $u+v \mid a$ since $u+v$ is coprime with both $u$ and $v$. Options for $(u,v)$ are then all combinations of $u,v\in\{1,p,q,pq\}$ so that $uv|pq$, with $a=(u+v)\cdot pq/uv$. So given distinct primes $p$ and $q$, there are 9 options for $(x,y)$.

I my previous solution, I had managed to overlook that alternative that $(u,v)$ be either $(pq,1)$ or $(1,pq)$, which add another two solutions.