Find all ordered triples of non-negative integers $(a,b,c)$ such that $a^2+2b+c$, $b^2+2c+a$, and $c^2+2a+b$ are all perfect squares.

[i]Proposed by Matthew Babbitt[/i]
We need to find all ordered triples of non-negative integers \((a, b, c)\) such that \(a^2 + 2b + c\), \(b^2 + 2c + a\), and \(c^2 + 2a + b\) are all perfect squares.

1. **Assume \(a \leq b \leq c\):**
   - We start by analyzing the expression \(c^2 + 2a + b\). Since \(a \leq b \leq c\), we have \(2a + b \leq 3c\). Therefore,
     \[
     c^2 \leq c^2 + 2a + b \leq c^2 + 3c.
     \]
     This implies that \(c^2 + 2a + b\) must be a perfect square between \(c^2\) and \((c+1)^2\). Hence,
     \[
     c^2 \leq c^2 + 2a + b < (c+1)^2.
     \]
     The only possibility is \(c^2 + 2a + b = c^2\), which implies \(2a + b = 0\). Since \(a\) and \(b\) are non-negative integers, the only solution is \(a = b = 0\). Thus, \(c\) can be any non-negative integer.

2. **Check the other expressions:**
   - For \(a = b = 0\), we have:
     \[
     a^2 + 2b + c = c,
     \]
     \[
     b^2 + 2c + a = 2c,
     \]
     \[
     c^2 + 2a + b = c^2.
     \]
     Since \(c\) is a non-negative integer, \(c\) must be a perfect square. Therefore, \(c = 0\) is the only solution.

3. **Consider the case where \(2a + b = 2c + 1\):**
   - If \(c^2 + 2a + b = (c+1)^2\), then:
     \[
     c^2 + 2a + b = c^2 + 2c + 1,
     \]
     \[
     2a + b = 2c + 1.
     \]
   - Similarly, for \(b^2 + 2c + a\), we have:
     \[
     b^2 \leq b^2 + 2c + a \leq (b+1)^2,
     \]
     \[
     b^2 \leq b^2 + 2c + a < (b+1)^2.
     \]
     The only possibility is \(b^2 + 2c + a = b^2\), which implies \(2c + a = 0\). Since \(a\) and \(c\) are non-negative integers, the only solution is \(a = c = 0\). Thus, \(b\) can be any non-negative integer.

4. **Combining the conditions:**
   - From the above analysis, we have:
     \[
     2a + b = 2c + 1,
     \]
     \[
     2c + a = 2b + 1.
     \]
   - Solving these equations, we get:
     \[
     b = 3a - 2,
     \]
     \[
     2c = 5a - 3.
     \]
   - Substituting these into the first expression, we get:
     \[
     a^2 + \frac{17}{2}a - \frac{11}{2} = (a+1)^2, (a+2)^2, (a+3)^2, (a+4)^2.
     \]
     Solving these, we find the non-negative integer solutions for \(a\) are \(a = 43, 1\). These give the solution sets \((43, 127, 106)\) and \((1, 1, 1)\).

The final answer is \(\boxed{(0, 0, 0), (1, 1, 1), (43, 127, 106)}\).