China Mathematical Competition

If $z=0$, then $$\{\begin{array}{l}x+y=0,\\ xy+y=0\end{array}.$$ It follows that $$\{\begin{array}{l}x=0,\\ y=0\end{array}\text{ or }\{\begin{array}{l}x=-1,\\ y=1.\end{array}$$ If $z\ne 0$, from $xyz+z=0$ we get $$xy=-1.\stackrel{◯}{1}$$ From $x+y+z=0$ we have $$z=-x-y.\stackrel{◯}{2}$$ Substituting $\stackrel{◯}{2}$ into $xy+yz+xz+y=0$, we obtain $$x^2+y^2+xy-y=0.\stackrel{◯}{3}$$ From $\stackrel{◯}{1}$ we have $x=-\frac{1}{y}$. We substitute it into $\stackrel{◯}{3}$, and then make a simplification: $$(y-1)(y^3-y-1)=0.$$ It is easy to see that $y^3-y-1$ has no rational solution, and so $y=1$. Then from $\stackrel{◯}{1}$ and $\stackrel{◯}{2}$ we get $x=-1$ and $z=0$, contradicting $z\ne 0$.

In summary, the system has exactly two solutions: $$\{\begin{array}{l}x=0,\\ y=0,\\ z=0\end{array},\{\begin{array}{l}x=-1,\\ y=1,\\ z=0\end{array}$$