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

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.

Answer 1

Let $$ 3T:=a^2+b^2+c^2-18 = 2\sqrt{2a+6} + 2\sqrt{2b+6} + 2\sqrt{2c+6} $$

If $$ a= {\rm max}\ \{ a,b,c\},\ b = {\rm min}\ \{ a,b,c\},\ a>b$$ then $$T>b^2-6=2\sqrt{2a+6} > T$$

Contradiction.

Answer 2

$$(b^2-6)^2-(c^2-6)^2=8(a-b)\Leftrightarrow (b^2-c^2)(b^2+c^2-12)=8(a-b)$$ $$(b^2-6)^2-(a^2-6)^2=8(a-c)\Leftrightarrow (b^2-a^2)(b^2+a^2-12)=8(a-c)$$ $$(c^2-6)^2-(a^2-6)^2=8(b-c)\Leftrightarrow (c^2-a^2)(c^2+a^2-12)=8(b-c)$$ 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.

Answer 3

Surprisingly, this system does have solutions in $\mathbb{R}$ where $a\neq b\neq c$. Unfortunately, it involves a deg-$54$ equation. Given,

$$ (b^2-6)^2=4(2a+6)\tag1$$ $$(c^2-6)^2=4(2b+6)\tag2$$ $$(a^2-6)^2=4(2c+6)\tag3$$

Do the substitutions:

$$a =\tfrac{1}{2}(-6+x^2)\tag4$$

$$b =\tfrac{1}{8}(12-12c^2+c^4)\tag5$$

$$c =\tfrac{1}{128}(-240 - 288x^2 + 168 x^4 - 24 x^6 + x^8)\tag6$$

This satisfies $(2),(3)$. However, $(1)$ is satisfied if a factorable $2+2+6+54=64$-deg eqn in $x$ is solved. The quadratic roots are $-1\pm\sqrt{3},\,1\pm\sqrt{7}$ which yield $a=b=c$, the sextic has all complex roots, but the $54$-deg has 12 real roots. (You can see that monster equation in all its glory here: WolframAlpha resultant.)

From the definitions of $(4),(5),(6)$, if $x$ is real, then $a,b,c$ are also real and $a\neq b\neq c$. We give one root to 50 decimal places,

$$x =-2.9849962763167580809615620439502727527653816890554$$

so,

$$\begin{align} a &= 1.4551013848124557801133370412617382434937978383740\dots\\ b &= -0.1732265781180354496841227929738397618222596966\dots\\ c &= -1.1155995411074882152529545257291565862139304073\dots\\ \end{align}$$