Find all pairs of prime numbers $p\,,q$ for which:
\[p^2 \mid q^3 + 1 \,\,\, \text{and} \,\,\, q^2 \mid p^6-1\]

[i]Proposed by P. Boyvalenkov[/i]
1. **Initial Assumptions and Setup:**
   We are given two conditions involving prime numbers \( p \) and \( q \):
   \[
   p^2 \mid q^3 + 1 \quad \text{and} \quad q^2 \mid p^6 - 1
   \]
   We need to find all pairs of prime numbers \( (p, q) \) that satisfy these conditions.

2. **Using Lifting the Exponent Lemma (LTE):**
   We will use the Lifting the Exponent Lemma (LTE) to analyze the given conditions. The LTE lemma is useful for finding the highest power of a prime \( p \) dividing expressions of the form \( x^n - y^n \).

3. **Case 1: \( \gcd(p, 3) = 1 \) and \( \gcd(q, 6) = 1 \):**
   - Assume \( p \neq 3 \) and \( q \neq 2, 3 \).
   - By LTE, for \( p^2 \mid q^3 + 1 \), we have:
     \[
     v_p(q^3 + 1) = v_p(q + 1) + v_p(3) \geq 2
     \]
     Since \( p \neq 3 \), \( v_p(3) = 0 \), so:
     \[
     v_p(q + 1) \geq 2 \implies p^2 \mid q + 1
     \]
   - For \( q^2 \mid p^6 - 1 \), we have:
     \[
     v_q(p^6 - 1) = v_q(p - 1) + v_q(6) \geq 2
     \]
     Since \( q \neq 2, 3 \), \( v_q(6) = 0 \), so:
     \[
     v_q(p - 1) \geq 2 \implies q^2 \mid p - 1
     \]
   - This implies:
     \[
     p^2 \mid q + 1 \quad \text{and} \quad q^2 \mid p - 1
     \]
     However, this leads to a contradiction because \( p \) and \( q \) are primes and the conditions \( p^2 \mid q + 1 \) and \( q^2 \mid p - 1 \) cannot hold simultaneously for distinct primes \( p \) and \( q \).

4. **Case 2: \( \gcd(p, 3) \neq 1 \) (i.e., \( p = 3 \)):**
   - If \( p = 3 \), then:
     \[
     9 \mid q^3 + 1
     \]
     This implies:
     \[
     q^3 \equiv -1 \pmod{9}
     \]
     Testing small primes, we find \( q = 2 \) works because:
     \[
     2^3 + 1 = 9 \implies 9 \mid 9
     \]
   - For \( q = 2 \):
     \[
     2^2 \mid 3^6 - 1
     \]
     This is true because:
     \[
     3^6 - 1 = 728 = 2^3 \cdot 7 \cdot 13 \implies 4 \mid 728
     \]
   - Therefore, \( (p, q) = (3, 2) \) is a valid solution.

5. **Case 3: \( \gcd(q, 6) \neq 1 \) (i.e., \( q = 2 \) or \( q = 3 \)):**
   - If \( q = 2 \), we already considered this in Case 2.
   - If \( q = 3 \), then:
     \[
     9 \mid p^6 - 1
     \]
     This implies:
     \[
     p^6 \equiv 1 \pmod{9}
     \]
     Testing small primes, we find \( p = 2 \) works because:
     \[
     2^6 - 1 = 63 \implies 9 \mid 63
     \]
   - For \( p = 2 \):
     \[
     2^2 \mid 3^3 + 1
     \]
     This is true because:
     \[
     3^3 + 1 = 28 = 2^2 \cdot 7 \implies 4 \mid 28
     \]
   - Therefore, \( (p, q) = (2, 3) \) is a valid solution.

Conclusion:
The only pairs of prime numbers \( (p, q) \) that satisfy the given conditions are \( (3, 2) \) and \( (2, 3) \).

The final answer is \( \boxed{ \{(3, 2), (2, 3)\} } \)