Sum implies null product. True?

Question

Let $x, y, z \in \mathbb{C}$. Assuming that $x+y+z=x^2+y^2+z^2=0$, is it true that $x^2y^2+y^2z^2+z^2x^2=0$?

Answer 1

We have $$\left(x^2+y^2+z^2 \right)^2 = x^4 + y^4 + z^4 + 2 \left(x^2y^2+y^2z^2 + z^2x^2 \right)$$ This implies we have $$x^4 + y^4 + z^4 = -2 \left(x^2y^2+y^2z^2 + z^2x^2 \right) \,\,\,\, (\spadesuit)$$ Similarly, we have $$\left(x+y+z\right)^4 = x^4 + y^4 + z^4 + 4(x^3(y+z)+y^3(x+z)+z^3(x+y)) + 6(x^2y^2+y^2z^2+z^2x^2) + 12xyz(x+y+z)$$ This implies $$0 = x^4 + y^4 + z^4 + 4(x^3(-x)+y^3(-y)+z^3(-z)) + 6(x^2y^2+y^2z^2+z^2x^2)$$ Hence, $$x^4+y^4+z^4 = 2(x^2y^2+y^2z^2+z^2x^2) \,\,\,\, (\clubsuit)$$ Comparing $(\spadesuit)$ and $(\clubsuit)$, we obtain $$\boxed{x^4+y^4+z^4 = 0 = x^2y^2+y^2z^2+z^2x^2}$$