## Task B-3.2.

Determine two natural numbers that are divisible by four and for which the difference of their cubes is a four-digit number divisible by 91.
## Solution.

Let $x=4m$ and $y=4n, m, n \in \mathbb{N}$. It holds that $x^{3}-y^{3}=91k, k \in \mathbb{N}$.

Then $64m^{3}-64n^{3}=91k$, or $64(m^{3}-n^{3})=91k, k \in \mathbb{N}$. This equality is possible only if the number $k$ is a multiple of 64. Since the number $91k$ must be a four-digit number, the only multiple for which this holds is $k=64, (91 \cdot 64=5824)$. From this, it follows that

$$
m^{3}-n^{3}=91 \Leftrightarrow (m-n)(m^{2}+mn+n^{2})=13 \cdot 7
$$

We have the following possibilities, or systems of equations:

1. $m-n=13, m^{2}+mn+n^{2}=7$ or
2. $m-n=7, m^{2}+mn+n^{2}=13$ or
3. $m-n=1, m^{2}+mn+n^{2}=91$ or
4. $m-n=91, m^{2}+mn+n^{2}=1$.

Since $m$ and $n$ are natural numbers, it is necessary that

$$
m-n < m^{2}+mn+n^{2}
$$

Therefore, it is sufficient to consider the second and third cases. (In the first and fourth cases, the system reduces to a quadratic equation that has no real solutions.)

System 2 reduces to the quadratic equation $n^{2}+7n+12=0$, whose solutions are negative numbers -3 and -4.

System 3 gives an equation that has one positive solution $n=5$. Then $m=n+1=6$.

We conclude that the sought numbers $x=4 \cdot 6=24$ and $y=4 \cdot 5=20$ are the only solution to the problem.