# Solving this system equations?

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.

• (1)-(3) is $x^2-z^2+z-x=(x+z-1)(x-z)=0$. Can you go from here? Nov 17, 2018 at 12:31
• Do I have to write x in function of z? Like $x = -z+1$ Nov 17, 2018 at 12:37
• Either $z=x$, or $z=1-x$. Rewrite your system in both cases. Remember $x=y=z=-1$ is a solution of your system. Nov 17, 2018 at 12:56

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 $$x=y=z=-1$$.
• I didn't think any more math was needed than to see that if all $x=y=z=-1$, then any one of them squared would yield $1$ and the other two would sum to $-2$. You did have an interesting way of showing it to be the case though. Nov 17, 2018 at 13:19