China Mathematical Competition (Extra Test)

Let $p = x + z$, $q = xz$. The second to fourth equations of the system become
$$\begin{aligned} p^2 &= x^2 + z^2 + 2q, \\ p^3 &= x^3 + z^3 + 3pq, \\ p^4 &= x^4 + z^4 + 4p^2q - 2q^2.\end{aligned}$$
Similarly, let $s = y + w$, $t = yw$. The second to fourth equations of the system become
$$s^2 = y^2 + w^2 + 2t,$$
$$s^3 = y^3 + w^3 + 3st,$$
$$s^4 = y^4 + w^4 + 4s^2t - 2t^2.$$
Also, the first equation in the system can now be expressed as
$$p = s + 2. \qquad \textcircled{1}$$
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
$$x^2 + z^2 + 2q = y^2 + w^2 + 2t + 4s + 4,$$
$$x^3 + z^3 + 3pq = y^3 + w^3 + 3st + 6s^2 + 12s + 8,$$
$$x^4 + z^4 + 4p^2q - 2q^2 = y^4 + w^4 + 4s^2t - 2t^2 + 8s^3 + 24s + 32s + 16.$$
Using the second to the fourth equations in the system to simplify the above, we get
$$q = t + 2s - 1, \qquad \textcircled{2}$$
$$pq = st + 2s^2 + 4s - 4, \qquad \textcircled{3}$$
$$2p^2q - q^2 = 2s^2t - t^2 + 4s^3 + 12s^2 + 16s - 25. \qquad \textcircled{4}$$
Substituting ① and ② into ③, we get
$$t = \frac{s}{2} - 1. \qquad \textcircled{5}$$
Substituting ⑤ into ②,
$$q = \frac{5}{2}s - 2. \qquad \textcircled{6}$$
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
$$\begin{cases} x = 3, \\ z = 1 \end{cases} \text{ or } \begin{cases} x = 1, \\ z = 3 \end{cases}$$
and
$$\begin{cases} y = 2, \\ w = 0 \end{cases} \text{ or } \begin{cases} y = 0, \\ w = 2. \end{cases}$$
Specifically, the system of equations has 4 solutions:
$$x = 3, y = 2, z = 1, w = 0;$$
$$x = 3, y = 0, z = 1, w = 2;$$
$$x = 1, y = 2, z = 3, w = 0;$$
$$x = 1, y = 0, z = 3, w = 2.$$