Which are the pairs of natural numbers, the product of which is equal to five times their difference?
Solution. Let the two sought numbers be denoted by $x$ and $y$; then $x y=5(x-y)$. The product on the right side of the equation is non-negative (since $x$ and $y$ are natural numbers), so $x-y \geq 0$. Rearrange the equation:

$$
\begin{gathered}
x y=5 x-5 y \\
x(5-y)=5 y
\end{gathered}
$$

$y=5$ does not provide a solution, so $x=\frac{5 y}{5-y}$.

Since $x \geq 0$, it follows that $5-y>0$, or $0 \leq y<5$, thus $y=0,1,2,3,4$ can be possible. Calculating the corresponding value of $x$ for each case:

| $y$ | 0 | 1 | 2 | 3 | 4 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| $x$ | 0 | 1.25 | $\frac{10}{3}$ | 7.5 | 20 |

Based on the above, two pairs of natural numbers satisfy the requirements of the problem: $(0 ; 0)$ and $(20 ; 4)$.