Three, (25 points) Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let $\{x\}=x-[x]$. Find all positive numbers $x$ such that $[x]$ and $\{x\}$ satisfy the relation $[x]^{2}=x\{x\}$.

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Three, from the problem, we have $x=[x]+\{x\}$. Therefore, the original equation can be transformed into
$$
[x]^{2}=([x]+\{x\})\{x\} \text {. }
$$
(1) Suppose $\{x\}=0$, then $[x]=0$. Thus, $x=0$, which contradicts the given that $x$ is a positive number.
(2) Suppose $\{x\} \neq 0$, then $0<\{x\}<1$. From $[x]^{2}=([x]+\{x\})\{x\}$, we have
$[x]^{2}=x\{x\}$.
If $[x]=1$, then $1=x\{x\}$. Since $0<\{x\}<1$, we have $1<x<2$.
If $[x] \geqslant 2$, the equation (1) can be transformed into
$[x]([x]-\{x\})=\{x\}^{2}$.
Since $0<\{x\}<1$, we have $[x]-\{x\}>1$. Therefore,
$[x]([x]-\{x\})>2 \times 1=2$.
Thus, $[x]([x]-\{x\}) \neq\{x\}^{2}$.
Therefore, equation (2) does not hold.
In summary, we have $[x]=1$.
At this point, equation (1) can be transformed into $\{x\}^{2}+\{x\}-1=0$.
Solving this, we get $\{x\}=\frac{\sqrt{5}-1}{2}$.
Thus, $x=[x]+\{x\}=1+\frac{\sqrt{5}-1}{2}=\frac{\sqrt{5}+1}{2}$.
(Li Yaowen, Dongzhuang No. 40 Middle School, Shandong Province, 277200)