51st Ukrainian National Mathematical Olympiad, 3rd Round

We have $x=\frac{[x^2]+1}{2}$. Hence, $x=t$, or $x=t+\frac{1}{2}$, where $t$ is an integer.

If $x=t$, we have $$[x^2]-2x+1=[t^2]-2t+1=t^2-2t+1=(t-1{)}^2=0\iff t=1\iff x=1.$$

If $x=t+\frac{1}{2}$, then $$[x^2]-2x+1=[(t+\frac{1}{2}{)}^2]-2(t+\frac{1}{2})+1=[t^2+t+\frac{1}{4}]-2t-1+1.$$ Since $t^2+t+\frac{1}{4}$ is not an integer, $[t^2+t+\frac{1}{4}]=t^2+t$. So, $$[x^2]-2x+1=t^2+t-2t-1+1=t^2-t.$$ Thus, $$t^2-t=0\iff t=0\text{ or }t=1\iff x=\frac{1}{2}\text{ or }x=\frac{3}{2}.$$

Therefore, the solutions are $x=1$, $x=\frac{1}{2}$, and $x=\frac{3}{2}$.