To prove the evident, we only need to consider the absolute minimal reduced residue system modulo $p$:
$$-\frac{p-1}{2},-\frac{p-1}{2}+1, \cdots,-1,1, \cdots, \frac{p-1}{2}-1, \frac{p-1}{2}$$

$d$ is a quadratic residue modulo $p$ if and only if
$$\begin{aligned} d \equiv & \left(-\frac{p-1}{2}\right)^{2},\left(-\frac{p-1}{2}+1\right)^{2}, \cdots,(-1)^{2}, 1^{2}, \cdots, \\ & \left(\frac{p-1}{2}-1\right)^{2} \text { or }\left(\frac{p-1}{2}\right)^{2}(\bmod p) . \end{aligned}$$

Since $(-j)^{2} \equiv j^{2}(\bmod p)$, $d$ is a quadratic residue modulo $p$ if and only if
$$d \equiv 1^{2}, \cdots,\left(\frac{p-1}{2}-1\right)^{2} \text { or }\left(\frac{p-1}{2}\right)^{2}(\bmod p) .$$

When $1 \leqslant i<j \leqslant(p-1) / 2$,
$$i^{2} \not \equiv j^{2}(\bmod p),$$

Therefore, equation (7) gives all the quadratic residues modulo $p$, totaling $(p-1) / 2$. Since the reduced residue system modulo $p$ has $p-1$ numbers, the other $(p-1) / 2$ must be quadratic non-residues modulo $p$, which proves the first half of the conclusion. When $d$ is a quadratic residue modulo $p$, by equations (7) and (8), there must be a unique $i, 1 \leqslant i \leqslant(p-1) / 2$, such that $x \equiv i(\bmod p)$ is a solution to (5). Consequently, in the reduced residue system (6), there are exactly $x \equiv \pm i(\bmod p)$ as solutions to (5), meaning the number of solutions to (5) is 2. Proof completed.