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 $9$.

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.
