* Let $n$ be a given positive integer, find the number of positive integer solutions $(x, y)$ for $\frac{1}{n}=\frac{1}{x}+\frac{1}{y}, x \neq y$.
From the original equation, it can be seen that $x>n, y>n$. Therefore, let
$$
x=n+r, y=n+s, r \neq s .
$$

Substituting them into the original equation yields $n^{2}=r \cdot s$. The number of different divisors $r$ of $n^{2}$ is $d\left(n^{2}\right)$, and since $r \neq s$, we know $r \neq$ n. Therefore, the number of positive integer solutions $(x, y)$ of the original equation is $d\left(n^{2}\right)-1$.