
Problem 8.1. Find all natural numbers $x$ and $y$ such that:
a) $\frac{1}{x}-\frac{1}{y}=\frac{1}{3}$
b) $\frac{1}{x}+\frac{1}{y}=\frac{1}{3}+\frac{1}{x y}$.


Solution: a) The equation is equivalent to $3 y-3 x=x y$, so $x=\frac{3 y}{y+3}=\frac{3 y+9-9}{y+3}=3-\frac{9}{y+3}$. Therefore $y+3=9$ and thus $y=6$. Hence there is a unique solution $x=2, y=6$.

b) Let $x \geq y$ be a solution of the problem. Now $\frac{1}{3}=\frac{1}{x}+\frac{1}{y}-$ $\frac{1}{x y} \leq \frac{2}{y}-\frac{1}{x y}=\frac{2 y-1}{x y}<\frac{2 y}{x y}=\frac{2}{x}$, giving $x<6$. When $x=1, x=2$ or $x=3$, no solution exists. When $x=4$, it follows that $y=9$, and $x=5$ implies $y=6$. If $y \geq x$, we apply the same reasoning. The
problem has four solutions:

$$
x=4, y=9 ; x=5, y=6 ; x=6, y=5 ; x=9, y=4
$$
