Symmetric System of Equations

We understand that $x+y+z=0$ by adding all the three equations.

Adding the two equations $x^2 - y^2 = z$ and $y^2 - z^2 = x$ , we get $x^2 - z^2 = x+z$ .

So, $(x+z)(x-z) = x+z$ which implies that either $x+z = 0$ or $x-z = 1$ .

If $x+z=0$ , then y=0. So, x and z can be (-1,1) or (0,0).

If $x-z=1$ , then the values of x and z can be (1,0) or (0,-1).

So, the possible solutions will be $(0,0,0) (-1,0,1) (1,-1,0) (0,1,-1)$ .

Add x, y, and z. x^2 - y^2 + y^2 - z^2 + z^2 - x^2 = 0 x + y + z = 0 This gives one solution of (0, 0, 0). If there is one 0, the other two must be equal and opposite. If you add 2 of the 3 equations, you get:

A. (x+z)(x-z) = x+z

If x = 0, z = -1

If z = 0, x = 1

B. (y+x)(y-x) = y+x

If x = 0, y = 1

If y = 0, x = -1

C. (z+y)(z-y) = z+y

If y = 0, z = 1

If z = 0, y = -1

which give solutions of: (0, 1, -1), (1, -1, 0), and (-1, 0, 1) respectively.

And thus there are 4 possible solutions of (0, 0, 0), (0, 1, -1), (1, -1, 0), and (-1, 0, 1).

By adding all three equations and cancelling out on the left, we get $x+y+z=0.$ Then the first equation becomes $(x-y)(x+y)=z,$ which can be rewritten as $(y-x)z=z$ . Similarly, $(z-y)x=x,$ $(x-z)y=y$ . If all the numbers are nonzero, this implies that $y-x=1,\ z-y=1,\ x-z=1.$ Adding these up, we get $0=3$ , a contradiction. So at least one of the numbers is zero. Since their sum is zero, either all three are zero (which is one solution) or exactly one of them is zero. Suppose $x=0,\ y\neq 0,\ z\neq 0.$ Plugging in and dividing by the non-zero numbers, we get $y=1$ , $z=-1$ . Similarly, we get two more solutions: $(-1,0,1)$ and $(1,-1,0).$ Hence there are a total of 4 solutions.

Using Mathematica from my Calculus days I was able to graph the 3 functions using the plot function. From the graph I was unable to clearly see what points clearly intersected and at which points. Using some helpful hints from forums I found a way using a linear solve to find the 4 points.

This should be longer, but as my first worked solution I hope this suffices.