Problem 4. Determine all natural numbers $x$ and $y$ for which $\sqrt{\frac{x^{2}+20}{x^{2}-20}}$ and $\sqrt{\frac{y^{3}+\frac{24}{5}}{y^{3}-\frac{24}{5}}}$ are natural numbers, and the difference $\sqrt{\frac{x^{2}+20}{x^{2}-20}}-\sqrt{\frac{y^{3}+\frac{24}{5}}{y^{3}-\frac{24}{5}}}$ is a perfect square of a natural number.
Solution. According to the condition of the problem, $\sqrt{\frac{x^{2}+20}{x^{2}-20}}$ and $\sqrt{\frac{y^{3}+\frac{24}{5}}{y^{3}-\frac{24}{5}}}$ are natural numbers, so $\frac{x^{2}+20}{x^{2}-20}$ and $\frac{y^{3}+\frac{24}{5}}{y^{3}-\frac{24}{5}}=\frac{5 y^{3}+24}{5 y^{3}-24}$ must be perfect squares of natural numbers.

We have, $\frac{x^{2}+20}{x^{2}-20}=\frac{x^{2}-20+40}{x^{2}-20}=1+\frac{40}{x^{2}-20}$, from which, since $x^{2}-20>0$, it follows that $x^{2}=20+k$, where $k \mid 40$. Clearly, $k \in\{1,2,4,5,8,10,20,40\}$ and by direct verification, we find that only for $k=5$ the right side of the last equality is a perfect square, and in this case $x=5$ and $\sqrt{\frac{x^{2}+20}{x^{2}-20}}=3$.

On the other hand, $\frac{5 y^{3}+24}{5 y^{3}-24}=1+\frac{48}{5 y^{3}-24}$, from which, since $5 y^{3}-24>0$, it follows that $y^{3}=\frac{24+m}{5}$, where $m \mid 48$. Clearly $m \in\{1,2,3,4,6,8,12,16,24,48\}$ and by direct verification, we find that only for $m=16$ the right side of the last equality is a perfect cube, and in this case $y=2$ and $\sqrt{\frac{y^{3}+\frac{24}{5}}{y^{3}-\frac{24}{5}}}=\sqrt{\frac{5 y^{3}+24}{5 y^{3}-24}}=2$.

Finally, since

$$
\sqrt{\frac{x^{2}+20}{x^{2}-20}}-\sqrt{\frac{y^{3}+\frac{24}{5}}{y^{3}-\frac{24}{5}}}=3-2=1^{2}
$$

we find that the only natural numbers that satisfy the conditions of the problem are $x=5$ and $y=2$.

## IX Department