<s>[INST] Solving the system $a^2-6=2\sqrt{2c+6}, \, b^2-6=2\sqrt{2a+6}, \, c^2-6=2\sqrt{2b+6}$ Question:

Solve the following system for $a,b,c\in \mathbb{R}$:
  $$\begin{cases}
b^2-6=2\sqrt{2a+6}\\
c^2-6=2\sqrt{2b+6}\\
a^2-6=2\sqrt{2c+6}
\end{cases}$$

I found the following:$$ (b^2-6)^2=4(2a+6)$$
$$(c^2-6)^2=4(2b+6)$$
$$(a^2-6)^2=4(2c+6)$$
Then maybe $a=b=c$ is one case.
Thank you.
 [/INST] $$(b^2-6)^2-(c^2-6)^2=8(a-b)\Leftrightarrow (b^2-c^2)(b^2+c^2-12)=8(a-b)$$
$$(b^2-6)^2-(a^2-6)^2=8(a-c)\Leftrightarrow (b^2-a^2)(b^2+a^2-12)=8(a-c)$$
$$(c^2-6)^2-(a^2-6)^2=8(b-c)\Leftrightarrow (c^2-a^2)(c^2+a^2-12)=8(b-c)$$
Assume that $a\geq b\geq c\geq 0$ then 
$$b^2+c^2\geq 12$$
on the other hand
$$b^2+a^2\leq12$$
and $$c^2+a^2\leq12$$
which would yield $$b^2+a^2\leq b^2+c^2\Leftrightarrow a^2\leq c^2\Leftrightarrow a\leq c$$ and 
$$c^2+a^2\leq c^2+b^2\Leftrightarrow a^2\leq b^2\Leftrightarrow a\leq b$$
Therefore $a=b=c$.
It remains to check when not all of them are positive.
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