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 $


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


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


Thanks for the help.

  • $\begingroup$ (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


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 ?

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.