Croatia_2018

From the first equation we get $x+y=z-1$. By plugging this into the second equation we get: $$\begin{array}{rl}{x}^2-y^2& =1-z^2,\\ (x+y)(x-y)& =(1-z)(1+z),\\ (z-1)(x-y)& =-(z-1)(1+z),\\ (z-1)(x-y+z+1)& =0.\end{array}$$ There are now two cases: $z=1$ or $x-y+z+1=0$.

If $z=1$, then $x+y=0$, i.e. $x=-y$. Plugging this into the third equation we get $2y^3=-2$, i.e. $y=-1$, $x=1$.

If $x-y+z+1=0$, then plugging $z=-x+y-1$ into the first equation yields $2x=-2$, i.e. $x=-1$. Now we have $z=-1+y-1=y$, which we can combine with the third equation to get $2y^3=-2$, i.e. $y=z=-1$.

Thus, the possible solutions to the given system of equations are $(1,-1,1)$ and $(-1,-1,-1)$. A direct computation confirms that both are indeed valid solutions of the system.