Find all non-negative integers $n$ such that there are $a,b \in \mathbb Z$ satisfying $n^2=a+b$ and $n^3=a^2+b^2$.

[i]Lucian Dragomir[/i]
1. We start with the given equations:
   \[
   n^2 = a + b
   \]
   \[
   n^3 = a^2 + b^2
   \]

2. We express \( n \) in terms of \( a \) and \( b \) using the given equations. First, we solve for \( n \) from the second equation:
   \[
   n = \frac{(n^2)^2}{n^3} = \frac{(a + b)^2}{a^2 + b^2}
   \]

3. Simplify the expression:
   \[
   n = \frac{a^2 + 2ab + b^2}{a^2 + b^2} = 1 + \frac{2ab}{a^2 + b^2}
   \]

4. Let \( k = \frac{2ab}{a^2 + b^2} \). Then:
   \[
   n = 1 + k
   \]

5. Since \( n \) is a non-negative integer, \( k \) must be an integer. We need to determine the possible values of \( k \). Note that:
   \[
   -1 \leq \frac{2ab}{a^2 + b^2} \leq 1
   \]
   This is because \( \frac{2ab}{a^2 + b^2} \) is bounded by -1 and 1 by the trivial inequality.

6. Therefore, \( k \) can only be -1, 0, or 1. This gives us the possible values for \( n \):
   \[
   n - 1 = k \implies n - 1 \in \{-1, 0, 1\}
   \]
   \[
   n \in \{0, 1, 2\}
   \]

7. We verify each value of \( n \):
   - For \( n = 0 \):
     \[
     n^2 = 0 \implies a + b = 0
     \]
     \[
     n^3 = 0 \implies a^2 + b^2 = 0
     \]
     The only solution is \( a = 0 \) and \( b = 0 \).

   - For \( n = 1 \):
     \[
     n^2 = 1 \implies a + b = 1
     \]
     \[
     n^3 = 1 \implies a^2 + b^2 = 1
     \]
     Possible solutions are \( (a, b) = (1, 0) \) or \( (0, 1) \).

   - For \( n = 2 \):
     \[
     n^2 = 4 \implies a + b = 4
     \]
     \[
     n^3 = 8 \implies a^2 + b^2 = 8
     \]
     Possible solutions are \( (a, b) = (2, 2) \).

Thus, the non-negative integers \( n \) that satisfy the given conditions are \( \{0, 1, 2\} \).

The final answer is \(\boxed{n \in \{0, 1, 2\}}\).