$\textbf{N5.}$ Find all $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for all $a,b,c \in \mathbb{N}$
$f(a)+f(b)+f(c)-ab-bc-ca \mid af(a)+bf(b)+cf(c)-3abc$
1. **Initial Casework**: We start by determining the values of \( f(1) \), \( f(2) \), and \( f(3) \).

   - For \( a = b = c = 1 \):
     \[
     f(1) + f(1) + f(1) - 1 \cdot 1 - 1 \cdot 1 - 1 \cdot 1 \mid 1 \cdot f(1) + 1 \cdot f(1) + 1 \cdot f(1) - 3 \cdot 1 \cdot 1 \cdot 1
     \]
     \[
     3f(1) - 3 \mid 3f(1) - 3
     \]
     This is always true, so no new information is gained.

   - For \( a = 2, b = c = 1 \):
     \[
     f(2) + f(1) + f(1) - 2 \cdot 1 - 1 \cdot 1 - 1 \cdot 2 \mid 2f(2) + f(1) + f(1) - 3 \cdot 2 \cdot 1 \cdot 1
     \]
     \[
     f(2) + 2f(1) - 4 \mid 2f(2) + 2f(1) - 6
     \]
     Let \( f(1) = 1 \):
     \[
     f(2) + 2 - 4 \mid 2f(2) + 2 - 6
     \]
     \[
     f(2) - 2 \mid 2f(2) - 4
     \]
     This implies \( f(2) = 4 \).

2. **Using \( P(a,1,1) \) and \( P(a,2,2) \)**:
   - For \( P(a,1,1) \):
     \[
     f(a) + f(1) + f(1) - a \cdot 1 - 1 \cdot 1 - 1 \cdot a \mid af(a) + f(1) + f(1) - 3a
     \]
     \[
     f(a) + 2 - 2a \mid af(a) + 2 - 3a
     \]
     \[
     f(a) - 2a + 2 \mid af(a) - 3a + 2
     \]
     Simplifying, we get:
     \[
     f(a) - 2a + 2 \mid (a-1)f(a) - 3a + 2
     \]

   - For \( P(a,2,2) \):
     \[
     f(a) + f(2) + f(2) - a \cdot 2 - 2 \cdot 2 - 2 \cdot a \mid af(a) + 2f(2) - 6a
     \]
     \[
     f(a) + 8 - 4a \mid af(a) + 8 - 6a
     \]
     \[
     f(a) - 4a + 8 \mid af(a) - 6a + 8
     \]

3. **Analyzing the Divisibility Conditions**:
   - From \( f(a) - 2a + 2 \mid 2(a-1)^2 \)
   - From \( f(a) - 4a + 8 \mid 4(a-2)^2 \)

4. **Considering \( a = p+1 \) for some prime \( p \)**:
   - From the first condition, if \( a = p+1 \):
     \[
     f(a) \in \{a^2, 4a-3, 2a+1, 2a, 3a-2, a, 2a-2, 1, 2a^2-2a+1\}
     \]

5. **Using Euclidean Algorithm**:
   - From the second condition, \( f(a) \in \{a^2, 4a-3\} \) if \( p > 61 \).

6. **Using \( P(a,a,1) \)**:
   - We use the Euclidean algorithm to eliminate the \( f(a) = 4a-3 \) case.
   - Conclusion: For \( p > 61 \) a prime, we have \( f(p+1) = (p+1)^2 \).

7. **Using \( P(a,p+1,p+1) \) for large primes \( p \)**:
   - We get:
     \[
     (f(a) - a^2) + (p+1-a)^2 \mid 2(p+1)(p+1-a)^2
     \]
   - Thus for all large \( p \):
     \[
     (f(a) - a^2) + (p+1-a)^2 \mid 2(p+1)(a^2 - f(a))
     \]
   - If \( f(a) \neq a^2 \), the last part causes a contradiction when \( p \) becomes sufficiently large in comparison to \( a \).

Thus, \( f(a) = a^2 \) which works clearly. Proved.

The final answer is \( \boxed{ f(a) = a^2 } \)