$ k$ is a given natural number. Find all functions $ f: \mathbb{N}\rightarrow\mathbb{N}$ such that for each $ m,n\in\mathbb{N}$ the following holds:	\[ f(m)\plus{}f(n)\mid (m\plus{}n)^k\]
Given the function \( f: \mathbb{N} \rightarrow \mathbb{N} \) such that for each \( m, n \in \mathbb{N} \), the following holds:
\[ f(m) + f(n) \mid (m + n)^k \]

We need to find all such functions \( f \).

1. **Step 1: Analyze the function for small values of \( m \) and \( n \)**

   Let's start by considering \( P(1, 1) \):
   \[ f(1) + f(1) \mid 2^k \]
   This implies:
   \[ 2f(1) \mid 2^k \]
   Since \( f(1) \) is a natural number, \( f(1) \) must be a power of 2. Let \( f(1) = 2^a \) for some non-negative integer \( a \).

2. **Step 2: Consider \( P(2, 2) \)**

   \[ f(2) + f(2) \mid 4^k \]
   This implies:
   \[ 2f(2) \mid 4^k \]
   Since \( 4^k = (2^2)^k = 2^{2k} \), \( f(2) \) must be a power of 2. Let \( f(2) = 2^b \) for some non-negative integer \( b \).

3. **Step 3: Consider \( P(1, 2) \)**

   \[ f(1) + f(2) \mid 3^k \]
   Substituting \( f(1) = 2^a \) and \( f(2) = 2^b \):
   \[ 2^a + 2^b \mid 3^k \]
   Since \( 3^k \) is not a power of 2, the only way this can hold is if \( 2^a + 2^b = 1 \). However, this is impossible for non-negative integers \( a \) and \( b \) unless \( a = b = 0 \). But \( f(1) \) and \( f(2) \) cannot be zero since they are natural numbers.

   Therefore, we must have \( a = b = 1 \), so:
   \[ f(1) = 2 \quad \text{and} \quad f(2) = 2 \]

4. **Step 4: Generalize for all \( m \)**

   Consider \( P(m, 1) \):
   \[ f(m) + f(1) \mid (m + 1)^k \]
   Substituting \( f(1) = 2 \):
   \[ f(m) + 2 \mid (m + 1)^k \]

   Similarly, consider \( P(m, 2) \):
   \[ f(m) + f(2) \mid (m + 2)^k \]
   Substituting \( f(2) = 2 \):
   \[ f(m) + 2 \mid (m + 2)^k \]

   Since \( f(m) + 2 \) must divide both \( (m + 1)^k \) and \( (m + 2)^k \), and since \( f(m) \) must be a natural number, the simplest solution that satisfies these conditions is \( f(m) = m \).

5. **Step 5: Verify the solution**

   We need to check if \( f(m) = m \) satisfies the original condition:
   \[ f(m) + f(n) = m + n \]
   \[ (m + n)^k \]
   Clearly, \( m + n \mid (m + n)^k \) for any natural numbers \( m \) and \( n \).

Therefore, the only function that satisfies the given condition is:
\[ \boxed{f(m) = m \text{ for all } m \in \mathbb{N}} \]