Proof: Clearly, we have $\left(s_{1}, n\right)=1$, hence by Lemma 4, there must exist $u_{0}, v_{0}$ satisfying
$$0<\left|u_{0}\right| \leqslant \sqrt{n}, \quad 0<\left|v_{0}\right|<\sqrt{n}$$

and
$$s_{1} u_{0} \equiv v_{0}(\bmod n)$$

From equation (37), we get
$$2 \leqslant u_{0}^{2}+v_{0}^{2}<2 n$$

Since $s_{1}$ satisfies the congruence equation (35), from equation (38) we can deduce
$$u_{0}^{2}+v_{0}^{2} \equiv 0(\bmod n)$$

From the above two equations, we obtain
$$u_{0}^{2}+v_{0}^{2}=n$$

Thus, $u_{0}, v_{0}$ is a solution to (20). Next, we prove that it is primitive, i.e., $d=\left(u_{0}, v_{0}\right)=1$. From equation (39), we have $d^{2} \mid n$, and from equation (38), we get
$$s_{1}\left(u_{0} / d\right) \equiv v_{0} / d(\bmod n / d)$$

Therefore,
$$\frac{n}{d^{2}}=\left(\frac{u_{0}}{d}\right)^{2}+\left(\frac{v_{0}}{d}\right)^{2} \equiv\left(\frac{u_{0}}{d}\right)^{2}+s_{1}^{2}\left(\frac{u_{0}}{d}\right)^{2} \equiv 0\left(\bmod \frac{n}{d}\right)$$

Here, we used $s_{1}^{2} \equiv-1(\bmod n / d)$. The above equation holds only when $d=1$. Therefore, $u_{0}, v_{0}$ is a primitive solution to (20).

Finally, when $u_{0}, v_{0}$ have the same sign, take $y_{1}=\left|u_{0}\right|, x_{1}=\left|v_{0}\right|$; when $u_{0}, v_{0}$ have opposite signs, take $x_{1}=\left|u_{0}\right|, y_{1}=\left|v_{0}\right|$. We can easily verify that $x_{1}, y_{1}$ is a non-negative (actually positive) primitive solution to (20) and satisfies equation (36) (left to the reader). The lemma is proved.