Ytukyg

I have to solve this system of equations with $(x,y,z) ∈ ℝ$

$x^2 + y + z = q$

$x+ y^2 + z = q$

$x + y + z^2 = q$

for $q = -1$

So we have:

$x^2 + y + z = -1$ (1)

$x+ y^2 + z = -1$ (2)

$x + y + z^2 = -1$ (3)

I do not have an idea for an approach.
Should I subtract some equations?

Like (1) - (3):

$ x^2-x + z - z^2 = 0 $

$(x+z-1)(x-z)=0$

If one of the factors equals 0, the whole equations will be 0.

Thus:

I $x+z-1=0 => z=x-1$

II $x=z$

Can this be done without breaking the laws of mathematics? In the comments it says

We would then get

$x=y=z=-1$

Thanks for the help.

In this kind of "symetric" systems, introducing the symetric functions of $x,y,z$ is often a good idea. Let $sigma=x+y+z$. Then you can rewrite your system :
$$left{begin{matrix}x^2-x+s+1=0 \ y^2-y+s+1=0 \ z^2-z+s+1=0 \ s=x+y+z end{matrix}right.$$
Each of the first three equations is of the form $X^2-X+a=0$, so $X=frac12pmdelta$, where $delta$ is a square root of $-s-frac34$ (we'll discuss the "reality" of those solutions later). Each of $x$, $y$ and $z$ belongs to the set ${frac12-delta,frac12+delta}$.

Adding those three solutions, you find $s=x+y+z=frac32+k$, where $kin{-3delta,-delta,delta,3delta}$, so $(s-frac32)^2=-s-frac34$ or $(s-frac32)^2=-9s-frac{27}{4}$.

First equation has no real solution, second has unique solution $s=-3$. Now $x$, $y$ and $z$ are solutions of $X^2-X-2=0$, so they are either $-1$ or $2$, but as $s=-1$, only remains the solution $x=y=z=-1$.

Certainly not the easiest way to find the solutions (remains to study the complex solutions), but funny, no ?