51st Ukrainian National Mathematical Olympiad, 3rd Round

a) Take $x=-y$, we get $$3x^2+z^2=z^3.$$ Take $x=tz$, we get: $3z^2{t}^2+z^2=z^3$, or $3t^2+1=z$, therefore $$z=3t^2+1,x=t(3t^2+1),y=-t(3t^2+1)$$

b) Note that $t^3\equiv t(\mathrm{mod}3)$. Suppose that there exists solution and denote and consider equation modulo $3$. We have $$\begin{array}{rl}a+1& \equiv x+y+z+1\equiv x^3+y^3+z^3+1\equiv x^2+y^2+z^2-xy-yz-zx\equiv \\ & \equiv x^2+y^2+z^2+2(xy+yz+zx)\equiv (x+y+z{)}^2\equiv a^2,\end{array}$$ so $a^2-a-1=3$ which is impossible.