Belarusian Mathematical Olympiad

Answer: $\sqrt{3}$, $\sqrt{7}$, $\sqrt{11}$.
Since $8[x]$ is an integer number, we can rewrite the initial equation as
$[x^2 - 4[x] + 1] - [-x^2 - 4[x] + 1] = 0$.
Set $y = x^2 - 4[x] + 1$, then $[y] = [-y]$.
It is easy to see that for $y > 0$ we have $0 \le [y] = [-y] \le -1$, which is impossible.
Similarly, for $y < 0$ we have $-1 \ge [y] = [-y] \ge 0$.
So $y = 0$, i.e. $x^2 - 4[x] + 1 = 0$.
Note that for $x < 1$ we have $[x] \le 0$, therefore $x^2 - 4[x] + 1 \ge 1 > 0$.
Thus $x \ge 1$.
Let $x = n + a$, where $n \in \mathbb{N}$, $0 \le a < 1$.
Then $0 = x^2 - 4[x] + 1 = n^2 + 2na + a^2 - 4n + 1 = n(n - 4) + 2na + a^2 \ge n(n - 4)$, so $n \le 4$.
Therefore, $1 \le n \le 4$.
$$\text{If } [x] = n = 1, \text{ then } x^2 - 4[x] + 1 = x^2 - 4 + 1 = x^2 - 3 = 0, \text{ so } x = \sqrt{3}.$$
$$\text{If } [x] = n = 2, \text{ then } x^2 - 4[x] + 1 = x^2 - 8 + 1 = x^2 - 7 = 0, \text{ so } x = \sqrt{7}.$$
$$\text{If } [x] = n = 3, \text{ then } x^2 - 4[x] + 1 = x^2 - 12 + 1 = x^2 - 11 = 0, \text{ so } x = \sqrt{11}.$$
$$\text{If } [x] = n = 4, \text{ then } x^2 - 4[x] + 1 = x^2 - 16 + 1 = x^2 - 15 = 0, \text{ so } x = \sqrt{15} < 4, \text{ contrary to } [x] = 4.$$