Prove: Using the notation in Lemma 2, take $d=2$. It is easy to see that,
$$\begin{array}{c} 1 \leqslant t_{j}=2 j<p / 2, \quad 1 \leqslant j<p / 4 \\ p / 2<t_{j}=2 j<p, \quad p / 4<j<p / 2 \end{array}$$

From the second equation, we know
$$n=\frac{p-1}{2}-\left[\frac{p}{4}\right]$$

Therefore, we have
$$n=\left\{\begin{array}{ll} l, & p=4 l+1 \\ l+1, & p=4 l+3 \end{array}\right.$$

From this and Lemma 2, we get
$$\left(\frac{2}{p}\right)=(-1)^{n}=\left\{\begin{array}{ll} 1, & p \equiv \pm 1(\bmod 8) \\ -1, & p \equiv \pm 3(\bmod 8) \end{array}\right.$$

This is the conclusion we want to prove. Equation (3) shows that 2 is a quadratic residue modulo $p$ if and only if the prime $p \equiv \pm 1(\bmod 8)$.

The significance of Lemma 2 lies in its linking whether $d$ is a quadratic residue modulo $p$ to the parity of the number $n$, where $n$ is the number of $d_{j}$ $(1 \leqslant j \leqslant(p-1) / 2)$, the smallest positive residues modulo $p$ of the numbers $1,2, \cdots,(p-1) / 2$ multiplied by $d$, that do not lie in $1,2, \cdots,(p-1) / 2$. However, the properties of $n$ itself are not well understood, nor is there an explicit formula for it. For this reason, further analysis of Lemma 2 and its proof is required. Using the notation of the integer part $[x]$, equation (1) can be expressed as
$$j d=p\left[\frac{j d}{p}\right]+t_{j}, \quad 1 \leqslant j<p / 2$$

Summing both sides over $j$ gives
$$d \sum_{j=1}^{(p-1) / 2} j=p \sum_{j=1}^{(p-1) / 2}\left[\frac{j d}{p}\right]+\sum_{j=1}^{(p-1) / 2} t_{j}=p T+\sum_{j=1}^{(p-1) / 2} t_{j}$$

Here, $T=\sum_{j=1}^{(p-1) / 2}\left[\frac{j d}{p}\right]$. From the proof of Lemma 2, we know
$$\begin{aligned} \sum_{j=1}^{(p-1) / 2} t_{j}= & s_{1}+\cdots+s_{k}+r_{1}+\cdots+r_{n} \\ = & s_{1}+\cdots+s_{k}+\left(p-r_{1}\right)+\cdots+\left(p-r_{n}\right)-n p \\ & +2\left(r_{1}+\cdots+r_{n}\right) \\ = & \sum_{j=1}^{(p-1) / 2} j-n p+2\left(r_{1}+\cdots+r_{n}\right) \end{aligned}$$

From the above two equations, we get
- $\quad \frac{p^{2}-1}{8}(d-1)=p(T-n)+2\left(r_{1}+\cdots+r_{n}\right)$.

When $d=2$, it is clear that $T=0$, and $n \equiv\left(p^{2}-1\right) / 8(\bmod 2)$, from which and Lemma 2, we again derive Theorem 3. When $(d, 2 p)=1$, we have
$$T \equiv n(\bmod 2)$$

Thus, we obtain an explicit expression for $n$. From this and Lemma 2, the proof is completed.