Find all functions $f:\mathbb{N}_0 \to \mathbb{Z}$ such that $$f(k)-f(l) \mid k^2-l^2$$ for all integers $k, l \geq 0$.
1. **Initial Assumptions and Simplifications:**
   - Given the function \( f: \mathbb{N}_0 \to \mathbb{Z} \) such that \( f(k) - f(l) \mid k^2 - l^2 \) for all \( k, l \geq 0 \).
   - We can assume without loss of generality (wlog) that \( f(0) = 0 \). This is because if \( f \) works, then \( f + c \) for any constant \( c \in \mathbb{Z} \) also works.
   - We also assume \( f \) is injective, otherwise the left-hand side (LHS) of the divisibility condition would not make sense.

2. **Assertion and Initial Values:**
   - Let \( P(k, l) \) be the assertion of the given functional equation.
   - Consider \( P(p, 0) \) where \( p \) is a prime number:
     \[
     f(p) \mid p^2 \implies f(p) \in \{-p^2, -p, -1, 1, p, p^2\}
     \]
   - Since \( f \) works, so does \( -f \). Therefore, we can assume \( f(1) \geq 0 \).
   - Using \( P(1, 0) \):
     \[
     f(1) \mid 1^2 \implies f(1) = 1
     \]

3. **Considering Prime Values:**
   - Using \( P(p, 1) \):
     \[
     f(p) - 1 \mid p^2 - 1 \implies f(p) - 1 \mid (p-1)(p+1)
     \]
   - This implies \( f(p) \in \{-p, -1, p, p^2\} \).

4. **Case Analysis:**
   - **Case 1:** There exist infinitely many primes \( p \) such that \( f(p) = p^2 \).
     - Using \( P(k, p) \) with an arbitrarily large prime \( p \):
       \[
       f(k) - p^2 \mid k^2 - p^2 \implies f(k) - p^2 \mid k^2 - f(k)
       \]
       - For \( f(k) = k^2 \):
         \[
         k^2 - p^2 \mid k^2 - k^2 \implies 0 \mid 0
         \]
       - Thus, \( f(x) = x^2 \) for all \( x \in \mathbb{N}_0 \).

   - **Case 2:** There exist infinitely many primes \( p \) such that \( f(p) = p \).
     - Using \( P(k, p) \) with an arbitrarily large prime \( p \):
       \[
       f(k) - p \mid k^2 - p^2 \implies f(k) - p \mid k^2 - f(k)^2
       \]
       - For \( f(x) = (-1)^{g(x)} x \) where \( g(x) \in \{0, 1\} \):
         \[
         (-1)^{g(k)} k - p \mid k^2 - ((-1)^{g(k)} k)^2 \implies (-1)^{g(k)} k - p \mid 0
         \]
       - Thus, \( f(x) = (-1)^{g(x)} x \) for all \( x \in \mathbb{N}_0 \).

   - **Case 3:** There exist infinitely many primes \( p \) such that \( f(p) = -p \).
     - Using \( P(p, k) \) with an arbitrarily large prime \( p \):
       \[
       p + f(k) \mid p^2 - k^2 \implies p + f(k) \mid f(k)^2 - k^2
       \]
       - For \( f(x) = (-1)^{g(x)} x \) where \( g(x) \in \{0, 1\} \):
         \[
         p + (-1)^{g(k)} k \mid 0
         \]
       - Thus, \( f(x) = (-1)^{g(x)} x \) for all \( x \in \mathbb{N}_0 \).

   - **Case 4:** There are only finitely many primes \( p \) with \( f(p) = p \), \( f(p) = p^2 \), or \( f(p) = -p \).
     - Let \( \mathcal{P} \) be the set of such finite primes \( p \), \( \mathcal{Q} \) the set of such finite primes \( q \), and \( \mathcal{S} \) the set of such finite primes \( s \).
     - Let \( \mathbb{P} \) be the set of all prime numbers. For any prime \( r \) in \( \mathbb{P} \) but not in \( \mathcal{P} \), \( \mathcal{Q} \), or \( \mathcal{S} \), we have \( f(r) = -1 \).
     - However, there are infinitely many such \( r \), while \( f(r) = -1 \) can be held by at most one prime, leading to a contradiction.

5. **Conclusion:**
   - The function \( f \) must be of the form \( f(x) = (-1)^{g(x)} x + c \), \( f(x) = x^2 + c \), or \( f(x) = -x^2 + c \) for all \( x \in \mathbb{N}_0 \) with \( g(x) \in \{0, 1\} \) and a constant \( c \in \mathbb{Z} \).

The final answer is \( \boxed{ f(x) = (-1)^{g(x)} x + c } \), \( f(x) = x^2 + c \), or \( f(x) = -x^2 + c \) for all \( x \in \mathbb{N}_0 \) with \( g(x) \in \{0, 1\} \) and a constant \( c \in \mathbb{Z} \).