31st Hellenic Mathematical Olympiad

For $x,y,z\in \mathbb{R}$, such that $xyz\ne 0$, the system is written: $$x^{3}yz=z^2-2y^2(1),y^{3}zx=x^2-2z^2(2),z^{3}xy=y^2-2x^2(3)$$ Using summation by parts we find: $$xyz(x^2+y^2+z^2)=-(x^2+y^2+z^2)\iff (x^2+y^2+z^2)(xyz+1)=0.$$ Since $xyz\ne 0$ we have $x^2+y^2+z^2>0$, and so: $xyz=-1$ (4) Using equation (4) in the system of (1)-(3) we get: $$x^2=-z^2+2y^2(5),y^2=-x^2+2z^2(6),z^2=-y^2+2x^2(7)$$ From (5) and (6) we get $y^2=z^2$, while from (6) and (7) we get $x^2=z^2$, and so: $$x^2=y^2=z^2\iff x=y=\pm z\text{or}x=-y=\pm z.(8)$$ Finally from equations (8) and (4) we have the solutions: $$(x,y,z)=(-1,-1,-1),(x,y,z)=(1,1,-1),(x,y,z)=(1,-1,1),(x,y,z)=(-1,1,1).$$