Saudi Arabia Mathematical Competitions

The system is equivalent to $$\{\begin{array}{l}x+y+z=-(w+1)\\ xy+yz+zx=-\left(w^2-w+1\right)\\ xyz=w^3-1\end{array}$$ Multiplying the first two equations of the system we get $$(x+y+z)(xy+yz+zx)=w^3+1.$$ Eliminating $w$ between this relation and the last equation, it follows $$(x+y+z)(xy+yz+zx)-xyz=2.$$ The equation (1) is equivalent to $$(x+y)(y+z)(z+x)=2.$$ Taking into account the symmetry of equation (3) we have to consider only the following situations: $$\{\begin{array}{l}x+y=2\\ y+z=-1\\ z+x=-1\end{array}\{\begin{array}{l}x+y=2\\ y+z=1\\ z+x=1\end{array}\{\begin{array}{l}x+y=-2\\ y+z=1\\ z+x=-1\end{array}$$ Because $w^2-w+1>0$, from the second equation of (1) it follows that the solutions to the above system must satisfy $$xy+yz+zx<0.$$ The first system has solution $(x,y,z)=(1,1,-2)$. Replacing in the first equation of (1) we obtain $w=-1$. In this case we get solutions $(1,1,-2,-1)$, $(1,-2,1,-1)$, $(-2,1,1,-1)$. The solutions to the second and third system do not satisfy condition $xy+yz+zx<0$, hence they will not give solutions to our problem.

Solution 2: $x,y,z$ are the roots of the polynomial $$P(t)=t^3+(w+1)t^2-\left(w^2-w+1\right)t-w^3+1$$ We have $$P(t)=(t+w+1)\left(t^2-w^2+w-1\right)+2.$$ Since $P(x)=0$, it follows $$(x+w+1)\left(x^2-w^2+w-1\right)=-2$$ There are four possibilities \left\{ \begin{array}{l} x + w + 1 = 1 \\ x^{2} - w^{2} + w - 1 = -2 \end{array} \quad \left\{ \begin{array}{l} x + w + 1 = -1 \\ x^{2} - w^{2} + w - 1 = 2 \end{array} \right. \left\{ \begin{array}{l} x + w + 1 = 2 \\ x^{2} - w^{2} + w - 1 = -1 \end{array} \quad \left\{ \begin{array}{l} x + w + 1 = -2 \\ x^{2} - w^{2} + w - 1 = 1 \end{array} \right. The second and the fourth system have no solutions in integers. From the first and from the second system we get $w=-1$ and $w=1$. If $w=-1$, the polynomial $P$ is $P(t)=t^3-3t+2$ with roots $x=y=1,z=-2$. We get solutions $(1,1,-2,-1)$, $(1,-2,1,-1)$, $(-2,1,1,-1)$. If $w=1$, the polynomial $P$ is $P(t)=t^3+2t^2-t$ and not all the roots are integers.