6 Determine all triples \((p, q, n)\) that simultaneously satisfy
$$
q^{n+2} \equiv 3^{n+2}\left(\bmod p^{n}\right), p^{n+2} \equiv 3^{n+2}\left(\bmod q^{n}\right)
$$

where \(p, q\) are odd primes, and \(n\) is an integer greater than 1.
It is evident that $(3,3, n)(n=2,3, \cdots)$ are all sets that meet the requirements.
Assume $(p, q, n)$ is another set that meets the requirements, then $p \neq q, p \neq 3$, $q \neq 3$. Without loss of generality, let $q > p \geq 5$. If $n=2$, then $q^{2} \mid\left(p^{4}-3^{4}\right)$, which means
$$
q^{2} \mid\left(p^{2}-3^{2}\right)\left(p^{2}+3^{2}\right)
$$

Since $q$ does not simultaneously divide $p^{2}-3^{2}$ and $p^{2}+3^{2}$, hence
$$
\left.q^{2} \text { | } p^{2}-3^{2}\right) \text { or } q^{2} \mid\left(p^{2}+3^{2}\right)
$$

But $0 < p < q$, so
$$
q^{2} \leqslant p^{2}+3^{2} < 34
$$

This is a contradiction. If $n \geqslant 3$, then
$$
q^{n} \leqslant p^{n+2}-3^{n+2} < p^{n+2}
$$

Thus, $q < p$. This is a contradiction. Therefore, $n=2$.
By $p^{3} \mid \left(q^{5}-3^{5}\right)$, we know $p \mid \left(q^{5}-3^{5}\right)$.
By Fermat's Little Theorem, we know $p \mid \left(q^{p-1}-3^{p-1}\right)$. Therefore,
$$
p \mid\left(q^{(5, p-1)}-3^{(5, p-1)}\right)
$$

If $(5, p-1)=1$, then $p \mid (q-3)$. By
$$
\begin{aligned}
\frac{q^{5}-3^{5}}{q-3}= & q^{4}+q^{3} \times 3+q^{2} \times 3^{2}+q \times 3^{3}+3^{4} \equiv \\
& 5 \times 3^{4}(\bmod p)
\end{aligned}
$$

and $p > 5$, we know

Contradiction.
$$
\begin{array}{c}
p \times \frac{q^{5}-3^{5}}{q-3} \\
p^{3} \mid (q-3)
\end{array}
$$

Therefore,
By $q^{3} \mid \left(p^{5}-3^{5}\right)$, we know
$$
q^{3} \leqslant p^{5}-3^{5} \quad (p \geqslant 7)
$$

and $q^{3} \mid \left(p^{5}-3^{5}\right)$, we know
$$
q^{3} \left\lvert\, \frac{p^{5}-3^{5}}{p-3}\right.
$$

Thus, $q^{3} \leqslant \frac{p^{5}-3^{5}}{p-3}=p^{4}+p^{3} \times 3+p^{2} \times 3^{2}+p \times 3^{3}+3^{4}$
By 5 | $(p-1)$ and 5 | $(q-1)$, we know $p \geqslant 11, q \geqslant 31$, then
$$
\begin{array}{c}
q^{3} \leqslant p^{4}\left(1+\frac{3}{p}+\left(\frac{3}{p}\right)^{2}+\left(\frac{3}{p}\right)^{3}+\left(\frac{3}{p}\right)^{4}\right)\left(\frac{8}{11}\right)^{\frac{1}{4}} q^{\frac{3}{4}}
\end{array}
$$

Thus, $\quad \frac{p^{5}+q^{5}-3^{5}}{p^{3} q^{3}}<\frac{p^{2}}{q^{3}}+\frac{q^{2}}{p^{3}}<\frac{1}{q}+\left(\frac{11}{8}\right)^{\frac{3}{4}} \frac{1}{31^{\frac{1}{4}}}<1$ which contradicts equation (1), i.e., $p^{3} q^{3} \mid \left(p^{5}+q^{5}-3^{5}\right)$.
In summary, $(3,3, n)(n=2,3, \cdots)$ are all the sets that meet the requirements.