Austria 2010

We first note that $$\begin{array}{rl}& 4x^4-x^2(4y^4+4z^4-1)-2xyz+y^8+2y^4{z}^4+y^2{z}^2+z^8\\ & =(4x^4+y^8+z^8-4x^2{y}^4-4x^2{z}^4+2y^4{z}^4)+(x^2-2xyz+y^2{z}^2)\\ & =(2x^2-y^4-z^4{)}^2+(x-yz{)}^2.\end{array}$$ The given equation is therefore equivalent to $$(2x^2-y^4-z^4{)}^2+(x-yz{)}^2=0.$$ It therefore follows that both $x=yz$ and $2x^2-y^4-z^4=0$ must hold. Substituting $x=yz$ in the second of these equations, we obtain $2y^2{z}^2-y^4-z^4=0$, which is equivalent to $-(y^2-z^2{)}^2=0$. We see that $z=\pm y$ must hold. For $z=y=t$, we obtain $x=t^2$, and for $-z=y=t$, we obtain $x=-t^2$. It therefore follows that the set of all solutions is $$\{(t^2,t,t)\mid t\in \mathbb{R}\}\cup \{(-t^2,t,-t)\mid t\in \mathbb{R}\}.$$