4.41 Given that $n$ is a positive integer, determine the number of solutions in ordered pairs of positive integers $(x, y)$ for the equation $\frac{x y}{x+y}=n$.
(21st Putnam Mathematical Competition, 1960)
［Solution］ Given the equation can be transformed into
$$
(x-n)(y-n)=n^{2} \text {. }
$$

Obviously, if $0<n, \quad y>n .
$$

Let the standard prime factorization of $n$ be

then
$$
\begin{aligned}
n & =\prod_{i=1}^{k} p_{i}^{\alpha_{i}}, \\
n^{2} & =\prod_{i=1}^{k} p_{i}^{2 \alpha_{i}} .
\end{aligned}
$$
The number of integer solutions to (1) is equal to the number of different ways to decompose $n^{2}$ into the product of two ordered factors, which is
$$
\left(2 \alpha_{1}+1\right)\left(2 \alpha_{2}+1\right) \cdots\left(2 \alpha_{k}+1\right) .
$$