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I would really appreciate help with this system of equations: $$ \left\{ \begin{array}{c} x^2 +3y=-2 \\ y^2 +3z=-2 \\ z^2+3x=-2 \end{array} \right. $$

It seems quite obvious that $x, y$ and $z$ are all equal and then the equation can be solved easily, but I don't know how to show that they are equal mathematically.

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hint, there are two solutions, and for each solution, you are correct that all three variables are equal. In each solution, x, y, and z are (negative) integers. – amWhy Nov 4 '12 at 16:19
Over the complex numbers there are solutions where x,y,z are different. – Gottfried Helms Nov 4 '12 at 22:48
up vote 8 down vote accepted

It's clear that we must have $x,y,z<0$, and that by the symmetry of the equations we can take either $x \le y \le z$ or $x \le z \le y$.

Suppose $x \le y \le z$ and $x<z$. Then $$x^2+3y=z^2+3x\ \Rightarrow\ x^2-z^2=3(x-y)$$ But $x^2-z^2 > 0$ and $x-y \le 0$, so the equation is both $>0$ and $\le 0$... clearly this can't happen, so we must have $x=z$ and hence $x=y=z$.

A similar argument works for $x \le z \le y$ by considering $x^2+3y=y^2+3z$.

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What if $x<0$, then $x<z$ can be consistent with $x^2 > z^2$? – user17762 Nov 4 '12 at 16:20
@Marvis: We know that $x$ and $z$ are negative. That's why $x < z$ implies $x^2-z^2 > 0$, not $<$. – Clive Newstead Nov 4 '12 at 16:20
Ok. Mistake on my part for glossing over the answer without reading it completely. +1. – user17762 Nov 4 '12 at 16:22
I found complex solutions (with rotations and with its complex conjugate), where the values for x,y,z are different. See my answer. – Gottfried Helms Nov 4 '12 at 21:50

The roots of the polynomial $$ f(x)= x^8 + 8 x^6 + 60 x^4 + 176 x^2 + 2187 x + 1942 $$ allow to satisfy the conditions for $x \ne y \ne z$.
We rediscover the 2 solutions $x=y=z=-1$ and $x=y=z=-2$ but also one complex solution
$(x,y,z)=(2.44944 -1.66645i,-1.74091+2.72124i, 0.791465+3.15828i) $
where the rotations $(y,z,x), (z,x,y)$ and the complex conjugates $(\overline x,\overline y, \overline z)$ are also solutions.

If we write $$g(x) = -{x^2+2\over 3}$$ then your set of equations is also $$ \begin{eqnarray} y=g(x)&;& z=g(y)=g(g(x))&;& x=g(z)=g(g(g(x))) \end{eqnarray}$$ and the equation $x = g(g(g(x))))$ leads to the polynomial $f(x)$ above.

I've computed the roots using Pari/GP and get them to the first few digits: $$ \small \begin{array} {rr} -2.00000000000 \\ -1.00000000000 \\ 2.44944037937&-1.66644590342*I \\ 2.44944037937&+1.66644590342*I \\ -1.74090540769&-2.72123992391*I \\ -1.74090540769&+2.72123992391*I \\ 0.791465028319&-3.15828086610*I \\ 0.791465028319&+3.15828086610*I \end{array}$$ We rewrite the set of equations $$ \begin{eqnarray} y = -{x^2+2\over3} & ;&z = -{y^2+2\over3}&;&x = -{z^2+2\over3} \end{eqnarray}$$ If I insert the roots I get for the successive evalutations the following ( and the last entry must equal the first for each row): $$ \small \begin{array} {rr|rr|rr|rr} x=root_r&& y=g(x) && z=g(y) && x=g(z) \\ \hline -2.00000 && -2.00000 && -2.00000 && -2.00000 \\ -1.00000 && -1.00000 && -1.00000 && -1.00000 \\ 2.44944 &-1.66645i & -1.74091&+2.72124i & 0.791465&+3.15828i & 2.44944&-1.66645i \\ 2.44944 &+1.66645i & -1.74091&-2.72124i & 0.791465&-3.15828i & 2.44944&+1.66645i \\ -1.74091&-2.72124i & 0.791465&-3.15828i & 2.44944&+1.66645i & -1.74091&-2.72124i \\ -1.74091&+2.72124i & 0.791465&+3.15828i & 2.44944&-1.66645i & -1.74091&+2.72124i \\ 0.791465&-3.15828i & 2.44944&+1.66645i & -1.74091&-2.72124i & 0.791465&-3.15828i \\ 0.791465&+3.15828i & 2.44944&-1.66645i & -1.74091&+2.72124i & 0.791465&+3.15828i \end{array} $$

We can express the 8 possible solutions of sets $(x,y,z)$ by the following indexes into the polroots, where we begin counting at 1:


The first two solutions have $x=y=z$
while the following 6 solutions have $x \ne y \ne z$ (being rotations and complex conjugates of each other)

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