smc

Consider the cases $x>0$ and $x<0$, and also note that by AM-GM, for any positive number $a$, we have $a+\frac{17}{a}\ge 2\sqrt{17}$, with equality only if $a=\sqrt{17}$. Thus, if $x>0$, considering each equation in turn, we get that $y\ge \sqrt{17},z\ge \sqrt{17},w\ge \sqrt{17}$, and finally $x\ge \sqrt{17}$. Now suppose $x>\sqrt{17}$. Then $y-\sqrt{17}=\frac{x^2+17}{2x}-\sqrt{17}=(\frac{x-\sqrt{17}}{2x})(x-\sqrt{17})<\frac{1}{2}(x-\sqrt{17})$, so that $x>y$. Similarly, we can get $y>z$, $z>w$, and $w>x$, and combining these gives $x>x$, an obvious contradiction. Thus we must have $x\ge \sqrt{17}$, but $x\ngtr \sqrt{17}$, so if $x>0$, the only possibility is $x=\sqrt{17}$, and analogously from the other equations we get $x=y=z=w=\sqrt{17}$; indeed, by substituting, we verify that this works. As for the other case, $x<0$, notice that $(x,y,z,w)$ is a solution if and only if $(-x,-y,-z,-w)$ is a solution, since this just negates both sides of each equation and so they are equivalent. Thus the only other solution is $x=y=z=w=-\sqrt{17}$, so that we have $2$ solutions in total, and therefore the answer is $\boxed{2}$.