Number of positive integer solutions of $\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}$ for distinct primes $p$ and $q$ 
Let $p$ and $q$ be distinct primes. Then find the number of positive integer solutions of the equation $$\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}$$

We get $pq=\frac{xy}{x+y}$
Now $x+y$ must divide $xy$ as L.H.S. is a positive integer with two prime factors but how do we make sure the same on R.H.S. ?
Given options are $3$ or $4$ or $8$ or $9$.
 A: Since $x$ and $y$ are positive, we have $\frac1x<\frac1{pq}$ and $\frac1y<\frac1{pq}$, which implies $x>pq$ and $y>pq$. This suggests the substitution $x:=pq+a$ and $y:=pq+b$ with positive integers $a$ and $b$.
Under this substitution, the given equation can be rewritten into the equivalent equation $ab=p^2q^2$. As $p^2q^2$ has nine positive divisors ($1$, $p$, $p^2$, $q$, $pq$, $p^2q$, $q^2$, $pq^2$, and $p^2q^2$), there are exactly nine positive integer solutions.
A: Starting hint: rewrite as 
$$
pq (x+y) = xy
$$
Can you say anything about the prime factors of $x$ and $y$? 
A: Since
$$
\begin{align}
(x-pq)(y-pq)
&=xy-pq(x+y)+p^2q^2\\
&=p^2q^2
\end{align}
$$
By breaking up the $9$ factors of $p^2q^2$, and solving for $x$ and $y$, we get
$$
\begin{align}
\frac1{pq}
&=\frac1{pq+p^2q^2}+\frac1{pq+1}\\
&=\frac1{pq+pq^2}+\frac1{pq+p}\\
&=\frac1{pq+q^2}+\frac1{pq+p^2}\\
&=\frac1{pq+q}+\frac1{pq+p^2q}\\
&=\frac1{pq+pq}+\frac1{pq+pq}
\end{align}
$$
So there are $9$ solutions because the last is symmetric.
