System of polynomial equations induced by $f(x,y,z) = xyz - x^2 - y^2 - z^2$ where $f_x = f_y $ $= f_z = 0$

Question

For the function

$$f(x,y,z) := xyz - x^2 - y^2 - z^2$$

I was looking for the points where $f_x = f_y = f_z = 0$, where these denote the partial derivatives with respect to the subscripted variable. This leads to the system of 3 equations:

$$ \begin{aligned} xy-2z &= 0 \\ xz-2y &= 0 \\ yz-2x &= 0 \end{aligned} $$

It is clear that $x=0$, $y=0$, $z=0$ is one solution to this system. However, I was wondering how we find the full set of solutions for a polynomial system like this.

Using an online calculator I have found that $(2,2,2)$, $(-2,-2,2)$, $(2,-2,-2)$, $(-2,2,-2)$ are all solutions, although I wanted to see if I could see how to find these myself.

Answer 1

As you said, $(0,0,0)$ is the trivial solution. It's easy to prove than any other possible solution must have $x,y,z\ne0$. Therefore, you can isolate $z$ in the first equation, por instance, and replace it in the other two: $$xy-2z=0 \implies z=\frac{xy}{2} \implies \begin{cases} x\dfrac{xy}{2}-2y=0,\\y\dfrac{xy}{2} - 2x=0. \end{cases}$$ Finally, use those to conclude that there are only four solutions more: precisely, those that you found using the calculator.

Answer 2

Using SymPy:

>>> from sympy import *
>>> x, y, z  = symbols('x y z', real=True)
>>> f = x*y*z - (x**2 + y**2 + z**2)
>>> solve_poly_system([diff(f,x), diff(f,y), diff(f,z)], x, y, z)
[(-2, -2, 2), (-2, 2, -2), (0, 0, 0), (2, -2, -2), (2, 2, 2)]

These are the solutions you found.

Answer 3

$xy = 2z\\yz = 2x \\ xz = 2y$

$xyz = 2z^2\\xyz = 2y^2\\ xyz = 2x^2$

$x^2 = y^2 = z^2$

$|x| = |y| = |z|$

Plugging this into any of the first set of equations.

$2|z| = z^2\\ |z| = 2$

and $|xyz| > 0$

And that is enough to get the other 4 solutions.