Let $p$ and $q$ be distinct primes. $\frac{1}{x}+\frac{1}{y}=\frac{1}{pq}$

Question

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

I have no idea how to start. Any hints would be helpful.

Answer 1

Multiply with $xypq$: $$\tag1 ypq+xpq=xy$$ It follows that $ypq$ is a multiple of $x$ and $xpq$ is a multiple of $y$. Say, $ypq=ax$, $xpq=by$. Then $yp^2q^2=axpq=aby$, i.e., there are only few possibilities for $a,b$, namely $a\in\{1,p,p^2,q,pq,p^2q,q^2,pq^2,p^2q^2\}$ and $b=p^2q^2/a$. In particular, dividing $(1)$ by $x$ or $y$ gives $a+pq=y$, $pq+b=x$. All in all, we have found nine solutions: $$\begin{align}\frac1{pq}&=\frac1{pq+1}+\frac1{pq+p^2q^2}\\ &=\frac1{pq+p}+\frac1{pq+pq^2}\\ &=\frac1{pq+p^2}+\frac1{pq+q^2}\\ &=\frac1{pq+q}+\frac1{pq+p^2q}\\ &=\frac1{pq+pq}+\frac1{pq+pq}\\ &=\frac1{pq+p^2q}+\frac1{pq+q}\\ &=\frac1{pq+q^2}+\frac1{pq+p^2}\\ &=\frac1{pq+pq^2}+\frac1{pq+p}\\ &=\frac1{pq+p^2q^2}+\frac1{pq+1}\\ \end{align}$$

Answer 2

Observe that we have $x | pqy$ and $y | pqx$. So write $pqy=tx$ and $pqx=sy$. From this find all possibilities of $s,t$. Then along with these equations and $pq(x+y)=xy$. Show that there are only 4 distinct solutions $x=p(1+q), y=qx$ etc. (2 if you don't distinguish between $x,y$).