2A. Solve the equation $\frac{1}{x}+\frac{1}{y}=\frac{1}{p}$, where $p>1$ is a given prime number, in the set of natural numbers.
Solution. The equation can be sequentially written in the form

$$
\begin{aligned}
& p x+p y=x y \\
& x y-p x-p y+p^{2}=p^{2} \\
& x(y-p)-p(y-p)=p^{2} \\
& (x-p)(y-p)=p^{2}
\end{aligned}
$$

Since $p$ is a prime number, the following cases are possible from the last equation:

1) $x-p=1, y-p=p^{2}$,
2) $x-p=p, y-p=p$ and
3) $x-p=p^{2}, y-p=1$.

From 1) $x=p+1, y=p^{2}+p$, from 2) $x=2 p, y=2 p$, and from 3) $x=p^{2}+p, y=p+1$. It is not difficult to verify that the obtained solutions are solutions to the original equation.