2. Let $m$ be a positive integer, the equation $x^{2}-\left[x^{2}\right]=\{x\}^{2}$ has $(\quad)$ solutions on $[1, m]$.
A. 0
B. $m(m-1)+1$
C. $m^{2}$
D. infinitely many
2. B Let $[x]=a,\{x\}=b$, then from $x^{2}-\left[x^{2}\right]=\{x\}^{2}$ we get $(a+b)^{2}-\left[(a+b)^{2}\right]=b^{2}$, which means $(a+b)^{2}-a^{2}-\left[b^{2}+2 a b\right]=b^{2}$. Therefore, $2 a b=\left[b^{2}+2 a b\right]$.
From this, we know $2 a b \in \mathbf{Z}$, and when $2 a b \in \mathbf{Z}$, by $0 \leqslant b<1$ we know $0 \leqslant b^{2}<1$, so at this time $2 a b=\left[b^{2}+2 a b\right]$, which means $x$ is a solution of the equation $\Leftrightarrow 2 a b \in \mathbf{Z}$. For any $a \in\{1,2, \cdots, m-1\}$, there are $a$ values of $b$ that satisfy the requirement, plus $x=m$, there are $m(m-1)+1$ solutions.