Solve the following equation:

$$
\sqrt{[\sqrt{[x]]}}=[\sqrt{[\sqrt{x}]}] .
$$
If $x$ is a non-negative solution to the equation, let $h$ and $k$ be the non-negative integers for which

$$
k^{2} \leqq h \leqq x < h+1 \leqq (k+1)^{2}
$$

that is, $h = [x]$, and $k^{2}$ is the largest square number not greater than $h$. Thus, the equation is equivalent to

$$
\sqrt{k} = [\sqrt{k}]
$$

which holds if and only if $k$ is a square number, i.e., $k = n^{2}$; hence, the solutions to our equation are those $x$ for which there exists a non-negative integer $n$ such that

$$
n^{4} \leqq x < \left(n^{2}+1\right)^{2} = n^{4} + 2n^{2} + 1
$$

Réti Zoltán (Budapest, Berzsenyi D. Gym., III. o. t.)

Remark. The right-open intervals $[0,1)$ and $[1; 4)$ derived for $n=0$ and $n=1$ can be combined into the interval $[0; 4)$. The following are: $[16; 25)$, $[81; 100)$, $[256; 289)$, etc.