The 58th question: Given that $p$ is an odd prime, if a natural number $x$ satisfies that for any natural number $t, p \nmid\left(x-t^{2}\right)$, then $x$ is called a quadratic non-residue modulo $p$. Let the set $A=\{a \mid a \in Z, 1 \leq a<p, a$ and $4-$ $\mathrm{a}$ are both quadratic non-residues modulo $\mathrm{p}\}$. Find the remainder of the product of all elements in $\mathrm{A}$ divided by $\mathrm{p}$.
The 58th question,
Solution: The elements below are all elements modulo $\mathrm{p}$ (i.e., elements in $\mathrm{Z}_{\mathrm{p}}$).
Let $U$ be the set of all quadratic non-residues modulo $p$, and $V$ be the set of all quadratic residues modulo $p$ (excluding 0). Let
$$
A=\left\{a \in Z_{p} \mid a \in U, 4-a \in U\right\}, B=\left\{b \in Z_{p} \mid b \in V, 4-b \in V\right\}
$$

Clearly, $0, 4 \notin \mathrm{B}$.
According to the above definitions, $\mathrm{a} \in \mathrm{A} \Leftrightarrow 4-\mathrm{a} \in \mathrm{A}, \mathrm{b} \in \mathrm{B} \Leftrightarrow 4-\mathrm{b} \in \mathrm{B}$. Therefore,
$$
\begin{array}{l}
4-y^{2} \in V \quad x(4-x) \in V \quad \begin{array}{c}
x \neq 2 \\
x(4-x) \in V
\end{array} \\
\end{array}
$$

Notice that, $x(4-x) \in V$ if and only if $x \in A \cup B$, and $2 \in A \cup B$. Hence, according to $(*)$ we know
$$
\begin{array}{l}
2 \prod_{\mathrm{b} \in \mathrm{B}} \mathrm{b} \equiv \underset{\substack{1 \leq x \leq p-1 \\
x(4-x) \in V}}{ } x \equiv \prod_{\mathrm{x} \in \mathrm{A}} x \cdot \prod_{\mathrm{x} \in \mathrm{B}} x(\bmod p) \\
\Rightarrow \prod_{\mathrm{x} \in \mathrm{A}} \mathrm{x} \equiv 2(\bmod p)
\end{array}
$$

In summary, the product of all elements in $\mathrm{A}$ modulo $\mathrm{p}$ is 2.