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Find the possible values of k: x+1/y=y+1/z=z+1/x=k

lfdahl

Well-known member
Nov 26, 2013
719
$x, y$ and $z$ are real numbers (not all equal), that obey
\[x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = k\]
Find the possible values of $k$.

Source: Nordic Math. Contest
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,679
$x, y$ and $z$ are real numbers (not all equal), that obey
\[x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = k\]
Find the possible values of $k$.

Source: Nordic Math. Contest
Let $f(x) = k - \frac1x$, and define $f^{(2)}(x) = f(f(x))$ and $f^{(3)}(x) = f(f^{(2)}(x))$. We want to find $x$ such that $f^{(3)}(x) = x$ but $f(x) \ne x$.

The condition for $f(x) \ne x$ is $k - \frac1x \ne x$, or $x^2 - kx + 1 \ne 0$.

Next, \(\displaystyle f^{(2)}(x) = k - \frac1{k - \frac1x} = k - \frac x{kx-1} = \frac{k^2x - x - k}{kx-1}, \) and \(\displaystyle f^{(3)}(x) = k - \frac{kx-1}{k^2x - x - k} = \frac{k^3x - kx - k^2 - kx + 1}{k^2x - x - k}.\) The condition $f^{(3)}(x) = x$ then becomes $k^3x - kx - k^2 - kx + 1 = x(k^2x - x - k)$, or $(k^2-1)x^2 - (k^3-k)x + k^2 - 1 = 0.$ So $(k^2-1)(x^2 - kx + 1) = 0$. But we already know that $x^2 - kx + 1 \ne 0$, and so $k^2 -1 = 0$.

So the only possible values of $k$ are $\pm1$.

But if $(x,y,z) = (2,-1,\frac12)$ then $x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = 1$, and if $(x,y,z) = (-2,1,-\frac12)$ then $x + \frac{1}{y} = y + \frac{1}{z} = z + \frac{1}{x} = -1$. So the values $k = \pm1$ can both occur, and these are the only possible values of $k$.

 

lfdahl

Well-known member
Nov 26, 2013
719
Thankyou, Opalg for your participation and another exemplary solution!