China Mathematical Competition (Extra Test)

Let $p=x+z$, $q=xz$. The second to fourth equations of the system become $$\begin{array}{rl}{p}^2& =x^2+z^2+2q,\\ p^3& =x^3+z^3+3pq,\\ p^4& =x^4+z^4+4p^{2}q-2q^2.\end{array}$$ Similarly, let $s=y+w$, $t=yw$. The second to fourth equations of the system become $$s^2=y^2+w^2+2t,$$ $$s^3=y^3+w^3+3st,$$ $$s^4=y^4+w^4+4s^{2}t-2t^2.$$ Also, the first equation in the system can now be expressed as $$p=s+2.\stackrel{◯}{1}$$ Therefore $$p^2=s^2+4s+4,$$ $$p^3=s^3+6s^2+12s+8,$$ $$p^4=s^4+8s^3+24s+32s+16.$$ Substituting the expressions of $p^2$, $p^3$, $p^4$ and $s^2$, $s^3$, $s^4$ obtained previously into the original system, we get $$x^2+z^2+2q=y^2+w^2+2t+4s+4,$$ $$x^3+z^3+3pq=y^3+w^3+3st+6s^2+12s+8,$$ $$x^4+z^4+4p^{2}q-2q^2=y^4+w^4+4s^{2}t-2t^2+8s^3+24s+32s+16.$$ Using the second to the fourth equations in the system to simplify the above, we get $$q=t+2s-1,\stackrel{◯}{2}$$ $$pq=st+2s^2+4s-4,\stackrel{◯}{3}$$ $$2p^{2}q-q^2=2s^{2}t-t^2+4s^3+12s^2+16s-25.\stackrel{◯}{4}$$ Substituting ① and ② into ③, we get $$t=\frac{s}{2}-1.\stackrel{◯}{5}$$ Substituting ⑤ into ②, $$q=\frac{5}{2}s-2.\stackrel{◯}{6}$$ Substituting ①, ⑤, ⑥ into ④, we get $s=2$. Therefore $t=0$, $p=4$, $q=3$. Consequently, $x,z$ and $y,w$ are the roots of equations $X^2-4X+3=0$ and $Y^2-2Y=0$ respectively. That means $$\{\begin{array}{l}x=3,\\ z=1\end{array}\text{ or }\{\begin{array}{l}x=1,\\ z=3\end{array}$$ and $$\{\begin{array}{l}y=2,\\ w=0\end{array}\text{ or }\{\begin{array}{l}y=0,\\ w=2.\end{array}$$ Specifically, the system of equations has 4 solutions: $$x=3,y=2,z=1,w=0;$$ $$x=3,y=0,z=1,w=2;$$ $$x=1,y=2,z=3,w=0;$$ $$x=1,y=0,z=3,w=2.$$