## Task A-2.2.

Determine all ordered pairs $(a, b)$ of natural numbers such that $V(a, b)-D(a, b)=\frac{a b}{5}$.
## First solution.

Let $V(a, b)=m$ and $D(a, b)=d$. We know that $a b=m d$.

From the equality $m-d=\frac{m d}{5}$, we get $5 m-5 d=m d$.

By rearranging, we have

$$
\begin{aligned}
25+5 m-5 d-m d & =25 \\
(5+m)(5-d) & =25
\end{aligned}
$$

The divisors of 25 are $\pm 1, \pm 5$, and $\pm 25$. Since $m>0$, it must be that $5+m>5$, so the only possibility is $5+m=25, 5-d=1$.

Therefore, $m=20$ and $d=4$.

Now we need to determine $a$ and $b$ such that $D(a, b)=4$ and $V(a, b)=20$. Since both $a$ and $b$ must be divisible by 4, and their product is $a b=m d=80$, the only possible solutions are $a=4, b=20$ and $a=20, b=4$.

Note: From $5 m-5 d=m d$, we can write $m=\frac{5 d}{5-d}=\frac{25}{5-d}-5$, so we conclude that $5-d$ must be a positive divisor of 25. This means that $5-d=1$, and we complete the solution in the same way.