If $x+1/x = y$, $y+1/y=z$, $z+1/z=x$, then find $x+y+z$. Is there any way to do so without taking out the values of $x$, $y$, $z$?

Please help,


Let us first write down the equations: $$x+\frac{1}{x}=y, \; y + \frac{1}{y}=z,\; z+\frac{1}{z}=x$$

We now add these equations together: $$x+\frac{1}{x}+ y + \frac{1}{y}+z+\frac{1}{z}=x +y+z$$

So we can see cancelling out $x+y+z$ gives us something, which was our motivation in adding them up:

$$\frac{1}{x} + \frac{1}{y}+\frac{1}{z}=0$$

Expanding, we get $xy+yz+xz=0$.

Here, we can note the identity $(x+y+z)^2 = x^2 +y^2+z^2 + 2(xy+yz+zx)$. Hence, if we can find $x^2+y^2+z^2$, we are done.

This can be done by realizing that by multiplying $x$ on both sides of $x+\frac{1}{x} = y$, gives us $x^2+1=xy$. Doing this with other equations and adding them all, we have:


Substituing $xy+yz+xz=0$, we have $x^2 +y^2+z^2 = -3$. So,

$$x+y+z=\pm\sqrt{x^2 +y^2+z^2 + 2(xy+yz+zx)} = \pm\sqrt{-3+2(0)} = \pm\sqrt{-3} = \pm\sqrt{3}i$$

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    $\begingroup$ Nice. If complex solutions allowed, you have both $\pm \sqrt3 i$. $\endgroup$ – Macavity Nov 5 '13 at 14:29
  • $\begingroup$ Oh, yes. Forgot that. $\endgroup$ – Sawarnik Nov 5 '13 at 14:36
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    $\begingroup$ Nice solution Sawarnik $\endgroup$ – juantheron Nov 9 '13 at 14:50

Adding the three equations, you get $\frac1x + \frac1y + \frac1z=0$ or $xy + yz + zx = 0$.

The equations can be written also as $x^2+1 = xy, y^2+1 = yz, z^2+1 = zx$, so $x^2+y^2+z^2 = -3$ which implies there are no solutions possible among real numbers.

Not sure if that qualifies as your answer as you didn't want to solve for $x, y, z$ first...

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    $\begingroup$ Nice solution Macavity..... $\endgroup$ – juantheron Nov 9 '13 at 14:50

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