# 2007 Indonesia MO Problems/Problem 6

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## Problem

Find all triples $(x,y,z)$ of real numbers which satisfy the simultaneous equations

$$x = y^3 + y - 8$$

$$y = z^3 + z - 8$$

$$z = x^3 + x - 8.$$

## Solution

To start, since all three equations have a similar form, we can let $x = y = z$ to see if there are any solutions. Doing so results in \begin{align*} x &= x^3 + x - 8 \\ 0 &= x^3 - 8 \\ 0 &= (x-2)(x^2 + 2x + 4). \end{align*} Note that $x^2 + 2x + 4 = 0$ has complex solutions, so the solution where $x = y = z$ is $(2,2,2)$.

Additionally, note that $x^3$ and $x$ are monotonically increasing functions, so $x^3 + x - 8$ is a monotonically increasing function. Thus, we can suspect that $(2,2,2)$ is the only solution. To prove this, we can use proof by contradiction.

Assume that $x > 2$. Thus, $x^3 > 8$, so $x^3 - 8 > 0$ and $x^3 + x - 8 > x$, so $y > x > 2$. By doing the same steps, we can show that $z > y$ and $x > z$. However, that would mean that $x > x$, which does not work, so there are no solutions where $x > 2$.

Similarly, assume that $x < 2$. Thus, $x^3 < 8$, so $x^3 - 8 < 0$ and $x^3 + x - 8 < x$, so $y < x < 2$. By doing the same steps, we can show that $z < y$ and $x < z$. However, that would mean that $x < x$, which does not work, so there are no solutions where $x < 2$.

Thus, we proved that $x = 2$ is the only solution, and by substituting the value into the original equations, we get the only solution of $\boxed{(2,2,2)}$.