$4 \cdot 31$ Solve the equation $\left[x^{2}-2 x\right]=[x]^{2}-2[x]$ for real numbers.

Find the real solutions to the equation $\left[x^{2}-2 x\right]=[x]^{2}-2[x]$.
[Solution] Let $y=x-1$, then $\left[y^{2}\right]=[y]^{2}$.
If $y \geqslant 0$, then $y^{2} \geqslant[y]^{2}>y^{2}-1$.
So $[y] \leqslant y<\sqrt{1+[y]^{2}}$.
That is, $x \in\left[n+1,1+\sqrt{1+n^{2}}\right)$, for any $n \in N \cup\{0\}$.
If $y<0$, then $[y]^{2} \geqslant y^{2} \geqslant\left[y^{2}\right]$,

So $y$ must be an integer to have $\left[y^{2}\right]=[y]^{2}$, that is, $x$ is a negative integer or 0.