Find all $(x, y, z) \in \mathbb{R^3}$ satisfying: $$x^2 + 4y^2 = 4xz \tag1$$ $$y^2 + 4z^2 = 4xy \tag2$$ $$z^2 + 4x^2 = 4yz \tag3$$
This is a very difficult problem. I added $-4(1) + (3)$ to get:
$$z^2 - 16y^2 = 4yz - 16xz \tag4 $$
But I get stuck after.
$(1)+(2)+(3)\\\implies\left(x^2-4xy+4y^2\right)+\left(y^2-4yz+4z^2\right)+\left(z^2-4zx+4x^2\right)=0\\\implies(x-2y)^2+(y-2z)^2+(z-2x)^2=0\\\implies (x=2y)\land(y=2z)\land(z=2x)\\\implies x=y=z=0$