Balti Tee 1993 · Ülesanne 4

Let

$$p=\sqrt{\frac{25}{2}+\sqrt{\frac{625}{4}-n}}+\sqrt{\frac{25}{2}-\sqrt{\frac{625}{4}-n}}=\sqrt{25+2 \sqrt{n}} .$$

Then $n=\left(\frac{p^{2}-25}{2}\right)^{2}$ and obviously $p$ is an odd number not less than 5. If $p \geq 9$ then $n>\frac{625}{4}$ and the initial expression would be undefined. The two remaining values $p=5$ and $p=7$ give $n=0$ and $n=144$ respectively.