## Task B-2.3.

Determine two positive numbers, divisible by four, such that the difference of their cubes is a four-digit number divisible by 91.
## Solution.

Let the required numbers be $x$ and $y$, with $x > y$. Since they are divisible by 4, we have $x = 4a$, $y = 4b$, for some natural numbers $a$ and $b$, with $a > b$.

According to the condition of the problem, we have

$$
(4a)^3 - (4b)^3 = 64(a^3 - b^3) = 64 \cdot 91k, \quad k \in \mathbb{N}
$$

Since the number must be a four-digit number, and $64 \cdot 91 = 5824$, it follows that $k = 1$.

Then, $a^3 - b^3 = 91$, or $(a - b)(a^2 + ab + b^2) = 7 \cdot 13$.

We have $a^2 + ab + b^2 = (a - b)^2 + 3ab > (a - b)^2 > a - b$ for $a - b > 1$. Therefore, the factor $a - b$ must be smaller than the factor $a^2 + ab + b^2$, so it is sufficient to solve these systems of equations:

$$
\begin{array}{lr}
a - b = 1, & a^2 + ab + b^2 = 91 \\
a - b = 7, & a^2 + ab + b^2 = 13
\end{array}
$$

Solving system (1) gives the equation $b^2 + b - 30 = 0$, which has the positive solution $b = 5$. Then, $a = 6$.

Solving system (2) gives the equation $b^2 + 7b + 12 = 0$, which has no positive solutions. Therefore, the required numbers are $x = 4 \cdot 6 = 24$ and $y = 4 \cdot 5 = 20$.