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, $1\le n\le 4$. $$\text{If }[x]=n=1,\text{ then }{x}^2-4[x]+1=x^2-4+1=x^2-3=0,\text{ so }x=\sqrt{3}.$$ $$\text{If }[x]=n=2,\text{ then }{x}^2-4[x]+1=x^2-8+1=x^2-7=0,\text{ so }x=\sqrt{7}.$$ $$\text{If }[x]=n=3,\text{ then }{x}^2-4[x]+1=x^2-12+1=x^2-11=0,\text{ so }x=\sqrt{11}.$$ $$\text{If }[x]=n=4,\text{ then }{x}^2-4[x]+1=x^2-16+1=x^2-15=0,\text{ so }x=\sqrt{15}<4,\text{ contrary to }[x]=4.$$