One, find all real numbers $x$ that satisfy $\left[x^{2}-2 x\right]=[x]^{2}-2[x]$, where [ $x$ ] denotes the greatest integer not exceeding $x$.


Let $[x]=n, x-[x]=y$. Then $n \in Z$, $y \in[0,1)$. The original equation can be transformed into $\left[y^{2}+2(n-1) y\right]$ $=0$. Therefore,
$$
0 \leqslant y^{2}+2(n-1) y<1 \text {. }
$$

When $n \leqslant 0$, i.e., $n$ is a non-positive integer, it is evident that $y=0$ satisfies (1).
When $n \geqslant 1$, solve the inequality (1) for $y$.
In summary, all real numbers that satisfy the equation are: non-negative integers $\left[n, \sqrt{(n-1)^{2}+1}+1\right](n=1,2$, $3, \cdots)$.