For a given prime $ p > 2$ and positive integer $ k$ let \[ S_k \equal{} 1^k \plus{} 2^k \plus{} \ldots \plus{} (p \minus{} 1)^k\] Find those values of $ k$ for which $ p \, |\, S_k$.
To solve the problem, we need to determine the values of \( k \) for which \( p \mid S_k \), where \( S_k = 1^k + 2^k + \ldots + (p-1)^k \) and \( p \) is a prime number greater than 2.

1. **Fermat's Little Theorem**:
   By Fermat's Little Theorem, for any integer \( a \) such that \( 1 \leq a \leq p-1 \), we have:
   \[
   a^{p-1} \equiv 1 \pmod{p}
   \]
   This implies that \( a^k \equiv a^{k \mod (p-1)} \pmod{p} \).

2. **Reduction of \( k \) modulo \( p-1 \)**:
   Since \( a^{p-1} \equiv 1 \pmod{p} \), it follows that \( S_k \equiv S_{k \mod (p-1)} \pmod{p} \). Therefore, it suffices to consider \( k \) in the range \( 0 \leq k \leq p-2 \).

3. **Elementary Symmetric Polynomials**:
   Let \( e_n \) be the \( n \)-th elementary symmetric polynomial evaluated at the numbers \( 1, 2, \ldots, p-1 \). The polynomial \( P(x) = x^{p-1} - 1 \) has roots \( 1, 2, \ldots, p-1 \) modulo \( p \). Thus, \( P(x) \) can be factored as:
   \[
   P(x) = (x-1)(x-2) \ldots (x-(p-1)) \pmod{p}
   \]
   Expanding this product, we get:
   \[
   x^{p-1} - e_1 x^{p-2} + e_2 x^{p-3} - \ldots + (-1)^{p-1} e_{p-1} \equiv x^{p-1} - 1 \pmod{p}
   \]
   Comparing coefficients, we find that \( e_i \equiv 0 \pmod{p} \) for \( 1 \leq i \leq p-1 \).

4. **Newton's Sums**:
   By Newton's identities, we have:
   \[
   S_n + \sum_{i=1}^{n-1} (-1)^i e_i S_{n-i} + (-1)^n n e_n = 0
   \]
   for all \( n \geq 1 \). Since \( e_i \equiv 0 \pmod{p} \) for \( 1 \leq i \leq p-1 \), the equation simplifies to:
   \[
   S_n \equiv 0 \pmod{p}
   \]
   for \( 1 \leq n \leq p-2 \).

5. **Conclusion**:
   Therefore, \( S_k \equiv 0 \pmod{p} \) if and only if \( k \equiv 0 \pmod{p-1} \). This means that \( p \mid S_k \) if and only if \( p-1 \mid k \).

The final answer is \( \boxed{ k \equiv 0 \pmod{p-1} } \).