Example 4 Find $n \in \mathbf{N}_{+}$, such that
$$
n-[n\{\sqrt{n}\}]=2 \text{. }
$$
(2002, Bulgarian Mathematical Olympiad)
【Analysis】Let
$$
\sqrt{n}=A+\alpha(A=[\sqrt{n}], \alpha=\{\sqrt{n}\}) \text {. }
$$

Then $n=(A+\alpha)^{2}$.
When $\alpha=0$, $n=A^{2}=2$. At this time, $A$ cannot be an integer.
When $\alpha \in(0,1)$,
$$
\begin{aligned}
2 & =n-[n\{\sqrt{n}\}]=n-[n \sqrt{n}-A n] \\
& =n+A n-[n \sqrt{n}],
\end{aligned}
$$

which means
$$
(A+1) n=2+[n \sqrt{n}] \text {. }
$$

Let $n=(A+1)^{2}-k$.
When $k \geqslant 2$,
$$
\begin{aligned}
& n \sqrt{n+2} \leqslant n \sqrt{n+k}=(A+1) n \\
& =[n \sqrt{n}]+2 \leqslant n \sqrt{n}+2 \\
& \Rightarrow n(\sqrt{n+2}-\sqrt{n}) \leqslant 2 \\
& \Rightarrow \frac{2 n}{\sqrt{n+2}+\sqrt{n}} \leqslant 2 \\
\Rightarrow & n \leqslant \sqrt{n+2}+\sqrt{n}<2 \sqrt{n+1} \\
\Rightarrow & n^{2}<4 n+4 \\
\Rightarrow & n<5 .
\end{aligned}
$$

Upon inspection, only when $n=2$, the original equation holds. When $k=1$,
$$
\begin{array}{l}
n \sqrt{n+1}=(A+1) n=2+[-n \sqrt{n}] \\
<2+n \sqrt{n} \\
\Rightarrow n(\sqrt{n+1}-\sqrt{n})<2 \\
\Rightarrow n<2(\sqrt{n+1}+\sqrt{n})<4 \sqrt{n+1} \\
\Rightarrow n^{2}<16 n+16 \\
\Rightarrow A^{2}+2 A=n<17 .
\end{array}
$$

When $A=1,2,3$, the corresponding $n=3,8,15$.
Upon inspection, $n=8,15$ satisfy the original equation.
In summary, $n=2,8,15$ are the solutions to the original equation.