8. Let $[x]$ denote the greatest integer not exceeding the real number $x$, and let $\{x\}=x-[x]$. Then the sum of the squares of all natural numbers $m$ that satisfy
$$
[(2 m+1)\{\sqrt{2 m+1}\}]=m
$$

is $\qquad$
8. 0 .

Let $2 m+1=n$. Then $2[n\{\sqrt{n}\}]=n-1$.
Notice that,
$$
\begin{array}{l}
2 n\{\sqrt{n}\}-21$ when, by $n=2[n\{\sqrt{n}\}]+1$ being odd, we know $n \geqslant 3$.
Thus, $\left|4 n-(2[\sqrt{n}]+1)^{2}\right| \leqslant \frac{4}{\sqrt{n}}+\frac{1}{n}<3$.
Therefore, $4 n-(2[\sqrt{n}]+1)^{2}= \pm 1$.
This leads to
$$
2 n-1=2[\sqrt{n}]([\sqrt{n}]+1)
$$

or $n=[\sqrt{n}]([\sqrt{n}]+1)$,
which is a contradiction.
In summary, the only positive integer $n$ that satisfies the condition is 1.
Thus, $m=0$. Hence, $m^{2}=0$.