Find the number of positive integer solutions

There's also the standard way of converting equalities like these into diophantine equations. Multiplying each side by $xypq$ , we get,

$ypq + xpq = xy$

$xy - xpq - ypq + (pq)^2 = p^2q^2$

$(x-pq)(y-pq) = p^2q^2$

We see that $x-pq$ and $y-pq$ can be whatever positive integer factor we want, so we just want to find the number of ways to factor $p^2q^2$ into a product of two integers. We get $(2+1)(2+1) = 9$ .

The above equation can be rewritten as:

$\frac{1}{x} + \frac{1}{y} = \frac{1}{pq} \Rightarrow \frac{x+y}{xy} = \frac{1}{pq} \Rightarrow y = pq + \frac{p^{2}q^{2}}{x - pq}.$

If $p,q$ are distinct primes, then $p^{2}q^{2}$ has nine positive integer divisors. Hence there are nine ordered-pairs $(x,y)$ with $x,y \in \mathbb{N}.$