14-17 Let $n$ be a natural number. How many solutions does the equation $x^{2}-\left[x^{2}\right]=(x-[x])^{2}$ have in the interval $1 \leqslant x \leqslant n$?
(Swedish Mathematical Competition, 1982)
[Solution] Clearly, $x=n$ is a solution to the equation.
Let $1 \leqslant x<n$.
Denote
$$
\begin{array}{l}
m=[x],\{x\}=\alpha, \text { then } \\
x=m+\alpha, \\
x^{2}=m^{2}+2 m \alpha+\alpha^{2}, \quad 0 \leqslant \alpha<1 .
\end{array}
$$

Substituting into the given equation, we get
$$
\begin{array}{c}
m^{2}+2 m \alpha+\alpha^{2}-\left[m^{2}+2 m \alpha+\alpha^{2}\right]=\alpha^{2}, \\
m^{2}+2 m \alpha-m^{2}-\left[2 m \alpha+\alpha^{2}\right]=0, \\
2 m \alpha-\left[2 m \alpha+\alpha^{2}\right]=0, \\
2 m \alpha=\left[2 m \alpha+\alpha^{2}\right] .
\end{array}
$$

Thus, $2 m \alpha$ is an integer.
To make $2 m \alpha$ an integer, and since $0 \leqslant \alpha<1$, then
$$
\alpha=0, \frac{1}{2 m}, \frac{2}{2 m}, \cdots, \frac{2 m-1}{2 m} .
$$

Therefore, for each fixed $m$, $\alpha$ can take $2 m$ values, i.e., there are $2 m$ solutions.
Since $m$ can take $1,2, \cdots, n-1$, then there are
$$
2[1+2+\cdots+(n-1)]=n(n-1)
$$

solutions, plus the solution $x=n$, then there are
$$
n(n-1)+1=n^{2}-n+1 \text { solutions. }
$$