Let $ m$ a positive integer and $ p$ a prime number, both fixed. Define $ S$ the set of all $ m$-uple of positive integers $ \vec{v} \equal{} (v_1,v_2,\ldots,v_m)$ such that $ 1 \le v_i \le p$ for all $ 1 \le i \le m$. Define also the function $ f(\cdot): \mathbb{N}^m \to \mathbb{N}$, that associates every $ m$-upla of non negative integers $ (a_1,a_2,\ldots,a_m)$ to the integer $ \displaystyle f(a_1,a_2,\ldots,a_m) \equal{} \sum_{\vec{v} \in S} \left(\prod_{1 \le i \le m}{v_i^{a_i}} \right)$. 
Find all $ m$-uple of non negative integers $ (a_1,a_2,\ldots,a_m)$ such that $ p \mid f(a_1,a_2,\ldots,a_m)$.
[i](Pierfrancesco Carlucci)[/i]
1. Define \( S \) as the set of all \( m \)-tuples of positive integers \(\vec{v} = (v_1, v_2, \ldots, v_m)\) such that \(1 \le v_i \le p\) for all \(1 \le i \le m\).
2. Define the function \( f(\cdot): \mathbb{N}^m \to \mathbb{N} \) by:
   \[
   f(a_1, a_2, \ldots, a_m) = \sum_{\vec{v} \in S} \left(\prod_{1 \le i \le m} v_i^{a_i} \right)
   \]
3. Let \( S_k = 1^{a_k} + 2^{a_k} + \ldots + p^{a_k} \). Observe that \( f(a_1, a_2, \ldots, a_m) \) factors as \( S_1 S_2 \ldots S_m \).

4. Therefore, \( p \mid f(a_1, \ldots, a_m) \) if and only if \( p \) divides one of the terms in the product \( S_1 S_2 \ldots S_m \).

5. Clearly, if some \( a_k \) is 0, then \( p \mid S_k \). Thus, we may suppose each of \( a_1, a_2, \ldots, a_m \) are nonzero.

6. **Lemma**: If \( a_k > 0 \), and \( p - 1 \mid a_k \), then \( p \mid S_k + 1 \). Otherwise, if \( a_k > 0 \) then \( p \mid S_k \).

   **Proof**:
   - If \( p - 1 \mid a_k \), the result follows easily from Fermat's Little Theorem, which states that for any integer \( a \) not divisible by \( p \), \( a^{p-1} \equiv 1 \pmod{p} \).
   - Otherwise, let \( g \) be a generator modulo \( p \). Observe that:
     \[
     S_k \equiv \sum_{i=1}^{p-1} i^{a_k} \equiv \sum_{i=0}^{p-2} (g^i)^{a_k} \equiv \frac{(g^{a_k})^{p-1} - 1}{g^{a_k} - 1} \pmod{p}
     \]
     Since \( g^{a_k} \not\equiv 1 \pmod{p} \) (because \( p - 1 \nmid a_k \)), and by Fermat's Little Theorem \( (g^{a_k})^{p-1} - 1 \) is a multiple of \( p \), \( S_k \) is a multiple of \( p \), as desired.

7. With the lemma, it follows that \( p \mid f(a_1, \ldots, a_m) \) if and only if some one of \( a_1, a_2, \ldots, a_m \) is 0 or not a multiple of \( p - 1 \).

\(\blacksquare\)

The final answer is \( \boxed{ (a_1, a_2, \ldots, a_m) } \) such that at least one \( a_i = 0 \) or \( a_i \) is not a multiple of \( p - 1 \).