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.
Starting hint: rewrite as $$ pq (x+y) = xy $$ Can you say anything about the prime factors of $x$ and $y$?
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.