Example 7 Let $n$ be a positive integer. How many positive integer solutions does the equation $\frac{x y}{x+y}=n$ have?
(21st Putnam Mathematical Competition)
Solve: Write the equation as $x y=n(x+y)$,
$$
x y-n x-n y=0 \text {. }
$$

Transform it into
$$
(x-n)(y-n)=n^{2} \text {. }
$$

Since we want to obtain positive integer solutions, it must be that $x>n, y>n$. Otherwise, if $0<x<n$ and $0<y<n$, it would imply $(x-n)(y-n)<n^{2}$.
$$
n^{2}=p_{1}^{2 a_{1}} p_{2}^{2 a_{2}} \cdots p_{k}^{2 a_{k}} \text {. }
$$

Since each divisor of $n^{2}$ determines a solution, the number of solutions to the above equation is
$$
\left(2 \alpha_{1}+1\right)\left(2 \alpha_{2}+1\right) \cdots\left(2 \alpha_{k}+1\right) .
$$