58. Find natural numbers $a$ and $b$, such that the number, which is the reciprocal of their difference, is three times greater than the number, which is the reciprocal of their product.
58. Let $a>b$. By the condition $\frac{1}{a-b}=3 \cdot \frac{1}{a b}$, hence $a b=3 a-3 b$ and $a=\frac{3 b}{3-b}$. The number $a$ will be a natural number if $b=2$. In this case, $a=6$. Therefore, $a=6, b=2$.