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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.

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  • $\begingroup$ (1)-(3) is $x^2-z^2+z-x=(x+z-1)(x-z)=0$. Can you go from here? $\endgroup$
    – player100
    Nov 17, 2018 at 12:31
  • $\begingroup$ Do I have to write x in function of z? Like $x = -z+1$ $\endgroup$ Nov 17, 2018 at 12:37
  • $\begingroup$ Either $z=x$, or $z=1-x$. Rewrite your system in both cases. Remember $x=y=z=-1$ is a solution of your system. $\endgroup$ Nov 17, 2018 at 12:56

1 Answer 1

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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 ?

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  • $\begingroup$ 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. $\endgroup$
    – poetasis
    Nov 17, 2018 at 13:19

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