parmenides51 wrote: Determine all possible values of the expression $xy+yz+zx$ with real numbers $x, y, z$ satisfying the conditions $x^2-yz = y^2-zx = z^2-xy = 2$. $x^2-yz=y^2-zx$ $\implies$ $(x-y)(x+y+z)=0$ $x^2-yz=z^2-xy$ $\implies$ $(x-z)(x+y+z)=0$ $z^2-xy=y^2-zx$ $\implies$ $(z-y)(x+y+z)=0$ And so either $x+y+z=0$, either $x=y=z$ (which is impossible) Adding $x^2-yz=2$, $y^2-zx=2$ and $z^2-xy=2$, we get $(x+y+z)^2-3(xy+yz+zx)=6$ And so $\boxed{xy+yz+zx=-2}$ which can indeed be reached, for example with $(x,y,z)=(\sqrt 2,0,-\sqrt 2)$

Other interesting facts about the system are also deducible about the system from the above result. The solution to this system is the circle formed by intersecting the plane $x+y+z=0$ and sphere $x^2+y^2+z^2=4$. $x(t)=\frac{2\sqrt{6}}{3}\cos t$, $y(t)=\frac{2\sqrt{6}}{3}\cos\left(t+\frac{2\pi}{3}\right)$, $z(t)=\frac{2\sqrt{6}}{3}\cos\left(t+\frac{4\pi}{3}\right)$.