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 $X=\frac12\pm\delta$, 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 $k\in\{-3\delta,-\delta,\delta,3\delta\}$, 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 ?