jmc

By inspection, $(\sqrt{17},\sqrt{17},\sqrt{17},\sqrt{17})$ and $(-\sqrt{17},-\sqrt{17},-\sqrt{17},-\sqrt{17})$ are solutions. We claim that these are the only solutions.

Let $$f(x)=\frac{1}{2}\left(x+\frac{17}{x}\right)=\frac{x^2+17}{2x}.$$Then the given equations become $f(x)=y,$ $f(y)=z,$ $f(z)=w,$ and $f(w)=x.$ Note that none of these variables can be 0.

Suppose $t>0.$ Then $$f(t)-\sqrt{17}=\frac{t^2+17}{2t}-\sqrt{17}=\frac{t^2-2t\sqrt{17}+17}{2t}=\frac{(t-\sqrt{17}{)}^2}{2t}\ge 0,$$so $f(t)\ge \sqrt{17}.$ Hence, if any of $x,$ $y,$ $z,$ $w$ are positive, then they are all positive, and greater than or equal to $\sqrt{17}.$

Furthermore, if $t>\sqrt{17},$ then $$f(t)-\sqrt{17}=\frac{(t-\sqrt{17}{)}^2}{2t}=\frac{1}{2}\cdot \frac{t-\sqrt{17}}{t}(t-\sqrt{17})<\frac{1}{2}(t-\sqrt{17}).$$Hence, if $x>\sqrt{17},$ then $$\begin{array}{rl}y-\sqrt{17}& <\frac{1}{2}(x-\sqrt{17}),\\ z-\sqrt{17}& <\frac{1}{2}(y-\sqrt{17}),\\ w-\sqrt{17}& <\frac{1}{2}(z-\sqrt{17}),\\ x-\sqrt{17}& <\frac{1}{2}(w-\sqrt{17}).\end{array}$$This means $$x-\sqrt{17}<\frac{1}{2}(w-\sqrt{17})<\frac{1}{4}(z-\sqrt{17})<\frac{1}{8}(y-\sqrt{17})<\frac{1}{16}(x-\sqrt{17}),$$contradiction.

Therefore, $(\sqrt{17},\sqrt{17},\sqrt{17},\sqrt{17})$ is the only solution where any of the variables are positive.

If any of the variables are negative, then they are all negative. Let $x^{\prime }=-x,$ $y^{\prime }=-y,$ $z^{\prime }=-z,$ and $w^{\prime }=-w.$ Then $$\begin{array}{rl}2y^{\prime }& =x^{\prime }+\frac{17}{x^{\prime }},\\ 2z^{\prime }& =y^{\prime }+\frac{17}{y^{\prime }},\\ 2w^{\prime }& =z^{\prime }+\frac{17}{z^{\prime }},\\ 2x^{\prime }& =w^{\prime }+\frac{17}{w^{\prime }},\end{array}$$and $x^{\prime },$ $y^{\prime },$ $z^{\prime },$ $w^{\prime }$ are all positive, which means $(x^{\prime },y^{\prime },z^{\prime },w^{\prime })=(\sqrt{17},\sqrt{17},\sqrt{17},\sqrt{17}),$ so $(x,y,z,w)=(-\sqrt{17},-\sqrt{17},-\sqrt{17},-\sqrt{17}).$

Thus, there are $\boxed{2}$ solutions.