Find all ordered triples of non-negative integers $(a, b, c)$ such that $a^{2}+2 b+c, b^{2}+2 c+a$, and $c^{2}+2 a+b$ are all perfect squares.
We have the trivial solutions $(a, b, c)=(0,0,0)$ and $(a, b, c)=(1,1,1)$, as well as the solution $(a, b, c)=(127,106,43)$ and its cyclic permutations.

The case $a=b=c=0$ works. Without loss of generality, $a=\max \{a, b, c\}$. If $b$ and $c$ are both zero, it's obvious that we have no solution. So, via the inequality

$$
a^{2}<a^{2}+2 b+c<(a+2)^{2}
$$

we find that $a^{2}+2 b+c=(a+1)^{2} \Longrightarrow 2 a+1=2 b+c$. So,

$$
a=b+\frac{c-1}{2} .
$$

Let $c=2 k+1$ with $k \geq 0$; plugging into the given, we find that

$$
b^{2}+b+2+5 k \quad \text { and } \quad 4 k^{2}+6 k+3 b+1
$$

are both perfect squares. Multiplying both these quantities by 4 , and setting $x=$ $2 b+1$ and $y=4 k+3$, we find that

$$
x^{2}+5 y-8 \text { and } y^{2}+6 x-11
$$

are both even squares.

We may assume $x, y \geq 3$. We now have two cases, both of which aren't too bad:

- If $x \geq y$, then $x^{2}<x^{2}+5 y-8<(x+3)^{2}$. Since the square is even, $x^{2}+5 y-8=$ $(x+1)^{2}$. Then, $x=\frac{5 y-9}{2}$ and we find that $y^{2}+15 y-38$ is an even square. Since $y^{2}<y^{2}+15 y-38<(y+8)^{2}$, there are finitely many cases to check. The solutions are $(x, y)=(3,3)$ and $(x, y)=(213,87)$.
- Similarly, if $x \leq y$, then $y^{2}<y^{2}+6 x-11<(y+3)^{2}$, so $y^{2}+6 x-11=(y+1)^{2}$. Then, $y=3 x-6$ and we find that $x^{2}+15 x-38(!)$ is a perfect square. Amusingly, this is the exact same thing (whether this is just a coincidence due to me selecting the equality case to be $x=y$, I'm not sure). Here, the solutions are $(x, y)=(3,3)$ and $(x, y)=(87,255)$.

Converting back, we see the solutions are $(0,0,0),(1,1,1)$ and $(127,106,43)$ and its cyclic permutations.