# 1986 AHSME Problems/Problem 30

## Problem

The number of real solutions $(x,y,z,w)$ of the simultaneous equations $2y = x + \frac{17}{x}, 2z = y + \frac{17}{y}, 2w = z + \frac{17}{z}, 2x = w + \frac{17}{w}$ is $\textbf{(A)}\ 1\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$

## Solution

Consider the cases $x>0$ and $x<0$, and also note that by AM-GM, for any positive number $a$, we have $a+\frac{17}{a} \geq 2\sqrt{17}$, with equality only if $a = \sqrt{17}$. Thus, if $x>0$, considering each equation in turn, we get that $y \geq \sqrt{17}, z \geq \sqrt{17}, w \geq \sqrt{17}$, and finally $x \geq \sqrt{17}$.

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 $x \geq \sqrt{17}$, 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 $\boxed{B}$.

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 