# Graphing Equations Is Useful, III

Here is one problem suggested by Bulgaria for the 1967 International Mathematical Olympiad [Compendium, p.42]:

Solve the system:

$\begin{array}{3,2} x^{2}+x-1=y \\ y^{2}+y-1=z \\ z^{2}+z-1=x. \end{array}$

Solution

### References

1. D. Djukic et al, The IMO Compendium, Springer, 2011 (Second edition)

Solve the system:

$\begin{array}{3,2} x^{2}+x-1=y \\ y^{2}+y-1=z \\ z^{2}+z-1=x. \end{array}$

The system is naturally related to the question of existence of either fixed points or cycles for the iterations (discrete dynamic system) defined by function $f(x)=x^{2}+x-1$. Elsewhere we considered the system defined by the function $\displaystyle f(x)=\frac{2x^{2}}{1+x^{2}}$ and by the function $f(x)=1-x^{2}$. There is also a more general problem that was offered at the 1968 IMO. In all cases it is useful to visualize the graphs of $y=f(x)$ and $y=x$, for this is a convenient way to see a solution at a glance

It is immediate from the diagram (and is easily verified algebraically) that the equation $f(x)=x$ has two solutions: $x=\pm 1$. Accordingly, the given system has two solutions $(1,1,1)$ and $(-1,-1,-1)$. The task is to establish that there are no other solutions. (Note in passing that the very same consideration will apply to a similar system with more than three equations.)

It helps to observe that $x=1$ is a repelling, while $x=-1$ is an attractive, fixed points. In part, the iterations that start with $x\gt 1$ produce an increasing sequence, for then $x^{2}+x-1\gt x$. As an implication for the given system,

$x \lt f(x) = y \lt f(y) = z \lt f(z) = x$

shows that the system may not have a solution in which one of the variables exceeds $1$.

It is also true that no solution may have one of the variables below $-2$, because $x\lt -2$ implies $f(x)\gt 1$.

On the other hand, $t\gt f(t)$, for $t\in(-1,1)$, such that $-1\lt x, y, z\lt 1$ leads to a contradiction:

$x \gt f(x) = y \gt f(y) = z \gt f(z) = x$

So, for the last step, assume that one of them, say $x$, is less than $-1$: $x\in(-2,-1)$. What then?

First, if $x$ is part of a solution, then $x=f(z)\ge -\frac{5}{4}$, which is the minimum of $f$. We thus can restrict $x$ further and assume that $x\in[-\frac{5}{4},-1)$. But this requires $z\in (-1,0)$ and also implies $y\in (-1,0)$. But this is a contradiction, because, for $y\in (-1,0)$, $z=f(y)\in [-\frac{5}{4},-1)$ and $[-\frac{5}{4},-1)\cap (-1,0)=\emptyset$.

Thus the only remaining possibilities are $x=y=z=1$ and $x=y=z=-1$.