Suppose $a$, $b$, $c$, and $d$ are non-negative integers such that
    \[(a+b+c+d)(a^2+b^2+c^2+d^2)^2=2023.\]
    Find $a^3+b^3+c^3+d^3$.

[i]Proposed by Connor Gordon[/i]
1. Given the equation:
   \[
   (a+b+c+d)(a^2+b^2+c^2+d^2)^2 = 2023
   \]
   we start by noting the prime factorization of \(2023\):
   \[
   2023 = 7 \cdot 17^2
   \]
   This implies that the only perfect square factors of \(2023\) are \(1\) and \(17^2\).

2. We consider the possible values for \((a^2 + b^2 + c^2 + d^2)^2\):
   - If \((a^2 + b^2 + c^2 + d^2)^2 = 1\), then \(a^2 + b^2 + c^2 + d^2 = 1\). This would imply:
     \[
     (a+b+c+d) \cdot 1 = 2023 \implies a+b+c+d = 2023
     \]
     However, this is not possible since \(a, b, c, d\) are non-negative integers and their squares sum to \(1\).

   - If \((a^2 + b^2 + c^2 + d^2)^2 = 17^2\), then \(a^2 + b^2 + c^2 + d^2 = 17\). This implies:
     \[
     (a+b+c+d) \cdot 17^2 = 2023 \implies a+b+c+d = 7
     \]

3. Now, we need to find non-negative integers \(a, b, c, d\) such that:
   \[
   a+b+c+d = 7 \quad \text{and} \quad a^2 + b^2 + c^2 + d^2 = 17
   \]

4. We use trial and error to find suitable values:
   - Suppose \(a = 0\), then \(b + c + d = 7\) and \(b^2 + c^2 + d^2 = 17\).
   - Trying \(b = 2\), \(c = 2\), and \(d = 3\):
     \[
     0 + 2 + 2 + 3 = 7 \quad \text{and} \quad 0^2 + 2^2 + 2^2 + 3^2 = 0 + 4 + 4 + 9 = 17
     \]
     This satisfies both conditions.

5. Therefore, one possible solution is \((a, b, c, d) = (0, 2, 2, 3)\) or any permutation of these values.

6. We now calculate \(a^3 + b^3 + c^3 + d^3\):
   \[
   0^3 + 2^3 + 2^3 + 3^3 = 0 + 8 + 8 + 27 = 43
   \]

The final answer is \(\boxed{43}\).