$3 \cdot 14$ Solve the system of equations
$$
2 y=x+\frac{17}{x}, \quad 2 z=y+\frac{17}{y}, \quad 2 w=z+\frac{17}{z}, \quad 2 x=w+\frac{17}{w}
$$

for the number of real solutions $(x, y, z, w)$.
(A) 1 .
(B) 2 .
(C) 4 .
(D) 8 .
(E) 16 .
(37th American High School Mathematics Examination, 1986)
［Solution 1］From the problem, we have
$$
\left\{\begin{array}{l}
x^{2}-2 x y+17=0, \\
y^{2}-2 y z+17=0, \\
z^{2}-2 z w+17=0, \\
w^{2}-2 w x+17=0 .
\end{array}\right.
$$

By symmetry, it is easy to see that $x=y=z=w$. Substituting this into the above equations, we get
$$
x=y=z=w= \pm \sqrt{17} \text {. }
$$

Therefore, the answer is $(B)$.
[Solution 2] From the problem, we know that $x, y, z, w$ have the same sign, and if the system of equations has a solution $\left(x_{0}, y_{0}, z_{0}\right.$, $\left.w_{0}\right)$, and $x_{0}, y_{0}, z_{0}, w_{0} \in R^{+}$, then there must be another solution $\left(-x_{0},-y_{0},-z_{0}\right.$, $-w_{0}$ ).
Without loss of generality, assume $x, y, z, w \in R^{+}$, and $\quad x \leqslant y \leqslant z \leqslant w$.
Adding the four equations from the problem, we get
$$
4 x \leqslant x+y+z+w=17\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{w}\right) \leqslant \frac{4 \cdot 17}{x},
$$

Thus, we have $x^{2} \leqslant 17$, which means $0<x \leqslant \sqrt{17}$.
On the other hand, from $2 x=w+\frac{17}{w} \geqslant 2 \sqrt{17}$, we know $x \geqslant \sqrt{17}$, hence
$$
x=\sqrt{17} \text {. }
$$

Similarly, we have
$$
y=z=w=\sqrt{17} .
$$

And the system of equations has only one set of positive real solutions $(\sqrt{17}, \sqrt{17}, \sqrt{17}, \sqrt{17})$, and one set of negative real solutions $(-\sqrt{17},-\sqrt{17},-\sqrt{17},-\sqrt{17})$.
Therefore, the answer is $(B)$.