Find all triples $(p,q,n)$ that satisfy
\[q^{n+2} \equiv 3^{n+2} (\mod p^n) ,\quad p^{n+2} \equiv 3^{n+2} (\mod q^n)\]
where $p,q$ are odd primes and $n$ is an positive integer.
To find all triples \((p, q, n)\) that satisfy the given congruences:
\[ q^{n+2} \equiv 3^{n+2} \pmod{p^n} \]
\[ p^{n+2} \equiv 3^{n+2} \pmod{q^n} \]
where \(p\) and \(q\) are odd primes and \(n\) is a positive integer, we proceed as follows:

1. **Case when \(p = q = 3\):**
   - If \(p = q = 3\), then the congruences become:
     \[ 3^{n+2} \equiv 3^{n+2} \pmod{3^n} \]
     \[ 3^{n+2} \equiv 3^{n+2} \pmod{3^n} \]
   - These are trivially true for any positive integer \(n\).
   - Therefore, \((3, 3, n)\) is a solution for any positive integer \(n\).

2. **Case when \(p, q, 3\) are pairwise distinct:**
   - Assume \(p > q\) without loss of generality.
   - From the given congruences, we have:
     \[ p^n \mid q^{n+2} - 3^{n+2} \]
     \[ q^n \mid p^{n+2} - 3^{n+2} \]
   - This implies:
     \[ p^n q^n \mid p^{n+2} + q^{n+2} - 3^{n+2} \]
   - Since \(p\) and \(q\) are primes, we have:
     \[ p^n q^n \leq p^{n+2} + q^{n+2} - 3^{n+2} < 2p^{n+2} \]
   - Therefore:
     \[ q^n < 2p^2 \]
     \[ p^2 > q^{n-1} \]
   - Additionally, from \(q^{n+2} - 3^{n+2} \geq p^n\), we get:
     \[ q^{n+2} > p^n > q^{n(n-1)/2} \]
   - Thus:
     \[ n+2 > \frac{n(n-1)}{2} \]
   - Solving this inequality, we find:
     \[ n \leq 3 \]

3. **Subcase \(n = 2\):**
   - For \(n = 2\), the congruences become:
     \[ p^2 \mid q^4 - 81 = (q^2 + 9)(q + 3)(q - 3) \]
   - Since \(p > q\), \(p \nmid q - 3\). If \(p \mid q + 3\), then \(q + 3 < 2p\), so \(p = q + 3\), which is impossible.
   - Therefore, \(p^2 \mid q^2 + 9\). Since \(q^2 + 9 < 2p^2\), we have:
     \[ q^2 + 9 = p^2 \]
     \[ (p + q)(p - q) = 9 \]
   - Solving this, we get \(p + q = 9\) and \(p - q = 1\), leading to \(p = 5\) and \(q = 4\), which is a contradiction since 4 is not a prime.

4. **Subcase \(n = 3\):**
   - For \(n = 3\), the congruences become:
     \[ p^3 \mid q^5 - 243 \]
     \[ q^3 \mid p^5 - 243 \]
     \[ p^3 q^3 \mid p^5 + q^5 - 243 \]
   - If \(p = 5\), then \(q^3 \mid 5^5 - 3^5 = 2 \cdot 11 \cdot 131\), which is impossible.
   - Therefore, \(p \neq 5\) and similarly \(q \neq 5\).

5. **Further analysis:**
   - Assume \(q \mid p - 3\). Note that \(p^3 \leq q^5 - 243 < q^6\), so \(p < q^2\). Thus, \(p - 3\) is not divisible by \(q^2\).
   - We have \(q^3 \mid p^5 - 243\), so:
     \[ q \mid p^4 + 3p^3 + 9p^2 + 27p + 81 \]
     \[ 0 \equiv p^4 + 3p^3 + 9p^2 + 27p + 81 \equiv 5 \cdot 3^4 \pmod{q} \]
   - Thus, \(q = 3\) or \(5\), which is impossible.

6. **Conclusion:**
   - Therefore, the only solution is when \(p = q = 3\).

The final answer is \( \boxed{ (3, 3, n) } \) for any positive integer \(n\).