For prime $p$, define the sets
$$
\begin{array}{l} 
S_{1}(p)=\left\{(a, b, c) \in \mathbf{Z}^{3} \mid a^{2} b^{2}+b^{2} c^{2}+\right. \\
\left.c^{2} a^{2}+1 \equiv 0(\bmod p)\right\}
\end{array}
$$

and $S_{2}(p)=\left\{(a, b, c) \in \mathbf{Z}^{3} \mid a^{2} b^{2} c^{2}\left(a^{2} b^{2} c^{2}+\right.\right.$
$$
\left.\left.a^{2}+b^{2}+c^{2}\right) \equiv 0(\bmod p)\right\} .
$$

Find all primes $p$ such that
$$
S_{1}(p) \subseteq S_{2}(p) .
$$
All prime numbers $p$ that satisfy the conditions are $2, 3, 5, 13, 17$.
(1) First, verify that when $p=2, 3, 5, 13, 17$, the conditions are met.
(i) When $p=2$, for any $(a, b, c) \in S_{1}(p)$, $a, b, c$ are all odd or two are odd and one is even. In this case,
$(a, b, c) \in S_{2}(p)$.
Thus, $S_{1}(p) \subseteq S_{2}(p)$.
(ii) When $p=3$, by the property that squares modulo 3 are 0 or 1, we have $a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+1 \equiv 1$ or $2 \pmod{3}$.
Therefore, $S_{1}(p)=\varnothing \subseteq S_{2}(p)$.
(iii) When $p=5$, if
$(a, b, c) \in S_{1}(p)$
$\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+1 \equiv 0 \pmod{5}$.
By the property that $a^{2}, b^{2}, c^{2}$ modulo 5 are 0 or $\pm 1$, we get that $a^{2}, b^{2}, c^{2}$

cannot all be 1 or -1 modulo 5. In this case, there must be
$a^{2} b^{2} c^{2}\left(a^{2} b^{2} c^{2}+a^{2}+b^{2}+c^{2}\right) \equiv 0 \pmod{5}$
$\Rightarrow (a, b, c) \in S_{2}(p)$.
Thus, $S_{1}(p) \subseteq S_{2}(p)$.
(iv) When $p=13$, if
$(a, b, c) \in S_{1}(p)$
$\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+1 \equiv 0 \pmod{13}$.
By the property that $a^{2}, b^{2}, c^{2}$ modulo 13 are 0 or $\pm 1$ or $\pm 3$ or $\pm 4$, we find that one of $a^{2}, b^{2}, c^{2}$ is 0 or -1 modulo 13. In this case, there must be
$a^{2} b^{2} c^{2}\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \equiv 0 \pmod{13}$
$\Rightarrow a^{2} b^{2} c^{2} \equiv 0 \pmod{13}$ or $\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \equiv 0 \pmod{13}$
$\Rightarrow a^{2} b^{2} c^{2} \equiv 0 \pmod{13}$ or $a^{2} b^{2} c^{2}+a^{2}+b^{2}+c^{2} \equiv 0 \pmod{13}$
$\Rightarrow (a, b, c) \in S_{2}(p)$.
Thus, $S_{1}(p) \subseteq S_{2}(p)$.
(v) When $p=17$, if
$(a, b, c) \in S_{1}(p)$
$\Rightarrow a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+1 \equiv 0 \pmod{17}$.
By the property that $a^{2}, b^{2}, c^{2}$ modulo 17 are 0 or $\pm 1$ or $\pm 2$ or $\pm 4$ or $\pm 8$, we find that one of $a^{2}, b^{2}, c^{2}$ is 0 or -1 modulo 17. In this case, there must be
$a^{2} b^{2} c^{2}\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \equiv 0 \pmod{17}$
$\Rightarrow a^{2} b^{2} c^{2} \equiv 0 \pmod{17}$ or $\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \equiv 0 \pmod{17}$
$\Rightarrow a^{2} b^{2} c^{2} \equiv 0 \pmod{17}$ or $a^{2} b^{2} c^{2}+a^{2}+b^{2}+c^{2} \equiv 0 \pmod{17}$
$\Rightarrow (a, b, c) \in S_{2}(p)$.
Thus, $S_{1}(p) \subseteq S_{2}(p)$.
(2) Proof: When $p \equiv 3 \pmod{4}$ and $p>3$, the conditions are not met.
It suffices to show that there exists $(a, b, c) \in S_{1}(p)$, but $(a, b, c) \notin S_{2}(p)$.
In fact, since $p>3$, there exists an integer $c$ such that $c^{2} \equiv 0, \pm 1 \pmod{p}$.
By $p \equiv 3 \pmod{4} \Rightarrow x^{2}+c^{2} \equiv 0 \pmod{p}$ has no solution.
Define a function $f:\left\{0,1, \cdots, \frac{p-1}{2}\right\} \rightarrow \mathbf{Z}$,
$f(a)=-\frac{a^{2} c^{2}+1}{a^{2}+c^{2}}$.
If $f(a)=f\left(a^{\prime}\right)$, then
$\left(a^{2}-a^{\prime 2}\right)\left(c^{4}-1\right)=0 \Rightarrow a=a^{\prime}$.
Thus, $f$ is injective (modulo $p$).
Therefore, the range of $f$ contains $\frac{p+1}{2}$ values.
By the pigeonhole principle, there exists an integer $b$ such that
$f(a)=b^{2} \Leftrightarrow (a, b, c) \in S_{1}(p)$.
Note that $b \neq 0$ (otherwise, $a^{2} c^{2}+1=0$ contradicts $p \equiv 3 \pmod{4}$), and $a \neq 0$ (otherwise, $b^{2}=f(0)=-\left(c^{-1}\right)^{2}$ contradicts $p \equiv 3 \pmod{4}$).
If $(a, b, c) \in S_{2}(p)$, then by
$a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+1 \equiv 0 \pmod{p}$,
$\left\{a^{2} b^{2} c^{2}+a^{2}+b^{2}+c^{2} \equiv 0 \pmod{p}\right.$
$\Rightarrow \left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \equiv 0 \pmod{p}$
and $a^{2} \neq -1, b^{2} \neq -1, c^{2} \neq -1$, thus
$(a, b, c) \notin S_{2}(p) \Rightarrow S_{1}(p) \not \subset S_{2}(p)$.
Therefore, when $p \equiv 3 \pmod{4}$ and $p>3$, the conditions are not met.
(3) Proof: When $p \equiv 1 \pmod{4}$ and $p>17$,
Let $c=a+b$, where $(a, b) \in S$, and $S$ is the set of solutions to $a^{2}+a b+b^{2}=x$. Then
$a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}+1$
$=(a b+b c+c a)^{2}+1$
$=\left(a^{2}+a b+b^{2}\right)^{2}+1=x^{2}+1=0$.
Thus, $(a, b, c) \in S_{1}(p)$.
If $(a, b, c) \in S_{2}(p)$, then there are the following four cases:
(i) $a=0 \Rightarrow b^{2}=x$ has at most two values $(a, b)$.
(ii) $b=0 \Rightarrow a^{2}=x$ has at most two values $(a, b)$.
(iii) $c=0 \Rightarrow a+b=0$ and $b^{2}=x$ has at most two values $(a, b)$.
(iv) $a^{2} b^{2} c^{2}+a^{2}+b^{2}+c^{2} \equiv 0$, in this case,
$a^{2}+b^{2}+c^{2}=2\left(a^{2}+a b+b^{2}\right)=2 x$.