The first equation rewrites as $x=\frac{w+z}{1-w z}$, which is a fairly strong reason to consider trigonometric substitution. Let $x=\tan (a), y=\tan (b), z=\tan (c)$, and $w=\tan (d)$, where $-90^{\circ}<a, b, c, d<90^{\circ}$. Under modulo $180^{\circ}$, we find $a \equiv c+d ; b \equiv$ $d+a ; c \equiv a+b ; d \equiv b+c$. Adding all of these together yields $a+b+c+d \equiv 0$. Then $a \equiv c+d \equiv-a-b$ so $b \equiv-2 a$. Similarly, $c \equiv-2 b ; d \equiv-2 c ; d \equiv-2 a$. Hence, $c \equiv-2 b \equiv 4 a, d \equiv-2 c \equiv-8 a$, and $a \equiv-2 d \equiv 16 a$, so the only possible solutions are $(a, b, c, d) \equiv(t,-2 t, 4 t,-8 t)$ where $15 t \equiv 0$. Checking, these, we see that actually $5 t \equiv 0$, which yields 5 solutions. Our division by $1-y z$ is valid since $1-y z=0$ iff $y z=1$, but $x=y+z+x y z$ so $y=-z$, which implies that $y z \leq 0<1$, which is impossible. (The solutions we have computed are in fact $(0,0,0,0)$ and the cyclic permutations of $\left.\left(\tan \left(36^{\circ}\right), \tan \left(-72^{\circ}\right), \tan \left(-36^{\circ}\right), \tan \left(72^{\circ}\right)\right).\right)$
Compute the number of real solutions $(x, y, z, w)$ to the system of equations: $$\begin{array}{rlrl} x & =z+w+z w x & z & =x+y+x y z \\ y & =w+x+w x y & w & =y+z+y z w \end{array}$$