By Lemma 2, we know that there must be integers $x, y, 1 \leqslant x<\sqrt{p}$, $1 \leqslant|y|<\sqrt{p}$, such that for a given $m, p \nmid m$, we have
$$m x \equiv y(\bmod p)$$

In particular, taking $m=a$, where $a$ is defined as in Lemma 1, we have
$$p \mid(a x-y)$$

Thus, we also have
$$p \mid(a x-y)(a x+y)$$
which is $\square$
$$a^{2} x^{2}-y^{2} \equiv 0(\bmod p)$$

Define $x_{0}=x, y_{0}=|y|$, the above equation shows
$$a^{2} x_{0}^{2}-y_{0}^{2} \equiv 0(\bmod p)$$

and $\square$
$$1 \leqslant x_{0}<\sqrt{p}, 1 \leqslant y_{0}<\sqrt{p}$$

Substituting Lemma 1 into equation (7), we get
$$x_{0}^{2}+y_{0}^{2} \equiv 0(\bmod p)$$

which means there is a $k$ such that
$$x_{0}^{2}+y_{0}^{2}=p k,$$

But by equation (8), we know $2 \leqslant x_{0}^{2}+y_{0}^{2}<2 p$, so it must be that $k=1$, which is exactly what we need to prove.