## Problem II - 2

Determine all integers $n$ such that

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

is an integer.
Solution.

Let

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

(the last equality is verified by squaring the expression given and called $p$). Then

$$
n=\left(\frac{p^{2}-25}{2}\right)^{2}
$$

and $p$ is an odd number greater than or equal to 5.

If $p \geq 9$, then $n>\frac{625}{4}$ and the expression for $p$ would not be real.

The possible values of $p$ are 5 and 7, which give the solutions

$$
n=0 \text { and } 144
$$