The 36th question, given that $\mathrm{p}$ is a prime number, the set $\mathrm{F}=\{0,1,2, \ldots, \mathrm{p}-1\}$, if a subset $\mathrm{A}$ of $\mathrm{F}$ satisfies: for any $a$ and $b \in A$, it holds that $a b+1(\bmod p) \in A$, then $A$ is called a good set. Find the possible number of elements in a good set.
The 36th question,
Answer: When $\mathrm{p}=2$, it is obvious that the set $\mathrm{A}=\{0,1\}$ satisfies the condition. Now, assume $\mathrm{p}$ is an odd prime.
(i) If $0 \in S$, then $0 \times 0+1=1 \in \mathrm{A}, 1 \times 1+1=2 \in \mathrm{A}, 1 \times 2+1=3 \in \mathrm{A}, 1 \times 3+1=4 \in A, \cdots$, and so on, $A=\{0,1,2, \ldots, p-1\}$. Clearly, $A=\{0,1,2, \ldots, p-1\}$ satisfies the condition.
(ii) If $0 \notin A$. Let $A=\left\{r_{1}, r_{2}, \ldots, r_{k}\right\}$.
(1) If $k=1$, then $r_{1}^{2}+1 \equiv r_{1} \Rightarrow (2 r_{1})^{2}+3 \equiv 0(\bmod p)$, hence $p=3$ or $p \equiv 1(\bmod 3)$. Conversely, if $p=3$ or $p \equiv 1(\bmod 3)$, there exists $r$ such that $r^{2}+1 \equiv r(\bmod p)$, so taking $A=\{r\}$ satisfies the condition. This indicates that when $p=3$ or $p \equiv 1(\bmod 3)$, $A$ can be a singleton set.
(2) If $k \geq 2$. According to the condition, we have:
$$
\begin{array}{l}
\left\{r_{1}, r_{2}, \ldots, r_{k}\right\}(\bmod p) \\
=\left\{r_{1} \cdot r_{1}+1, r_{1} \cdot r_{2}+1, \ldots, r_{1} \cdot r_{k}+1\right\}(\bmod p) \\
=\left\{r_{2} \cdot r_{1}+1, r_{2} \cdot r_{2}+1, \ldots, r_{2} \cdot r_{k}+1\right\}(\bmod p) \\
\end{array}
$$

Thus, we know:
$$
\begin{array}{l}
\sum_{i=1}^{k} r_{i} \equiv \sum_{i=1}^{k}\left(r_{1} \cdot r_{i}+1\right) \equiv \sum_{i=1}^{k}\left(r_{2} \cdot r_{i}+1\right)(\bmod p) \\
\Rightarrow\left\{\begin{array}{c}
\left(r_{1}-r_{2}\right) \cdot\left(\sum_{i=1}^{k} r_{i}\right) \equiv 0(\bmod p) \\
\left(1-r_{1}\right) \cdot\left(\sum_{i=1}^{k} r_{i}\right) \equiv\left(1-r_{2}\right) \cdot\left(\sum_{i=1}^{k} r_{i}\right) \equiv k(\bmod p)
\end{array}\right. \\
\Rightarrow \sum_{i=1}^{k} r_{i} \equiv k \equiv 0(\bmod p) \\
\Rightarrow k=p \\
\end{array}
$$

Therefore, $|\mathrm{A}|=k=p$, which indicates $\mathrm{A}=\{0,1,2, \ldots, \mathrm{p}-1\}$. This contradicts $0 \notin \mathrm{A}$.
In summary, the possible values of $|A|$ are $\left\{\begin{array}{c}1 \text { or } p, \text { when } p=3 \text { or } p \equiv 1(\bmod 3) \text { } \\ p, \text { when } p \equiv 2(\bmod 3) \text { }\end{array}\right.$.