Prove that when $n=1$, all solutions of the indeterminate equation (20) are
$$\{ \pm 1,0\}, \quad\{0, \pm 1\} ;$$

When $n=2$, all solutions are
$$\{ \pm 1, \pm 1\} .$$

This leads to equation (23). The primitive solutions $x, y$ of the indeterminate equation (20) must satisfy
$$(x, y)=(n, x y)=1$$

Therefore, when $n>1$, it must be that
$$|x| \geqslant 1, \quad|y| \geqslant 1$$

and $|x|,|y|$ are non-negative primitive solutions of (20). Thus, when $n>1$, the non-negative primitive solutions $x_{1}, y_{1}$ of the indeterminate equation (20) must be positive, and $\pm x_{1}, \pm y_{1}$ give four different primitive solutions of (20). This proves equation (24). Finally, to prove equation (25). Suppose $x, y$ are solutions of (20), $(x, y)=d$. Then, it must be that $d^{2} \mid n$, and $u=x / d, v=y / d$ are
$$u^{2}+v^{2}=n / d^{2}$$

primitive solutions, and different solutions $\{x, y\}$ correspond to different solutions $\{u, v\}$. Conversely, let $d \geqslant 1$, $d^{2} \mid n$. If $u, v$ are a set of primitive solutions of (27), then $x=d u, y=d v$ are solutions of (20), and different solutions $\{u, v\}$ correspond to different solutions $\{x, y\}$. This proves equation (25). Q.E.D.