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)

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Copyright © 1996-2017 Alexander Bogomolny

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

graph of the function y= x^2+x-1

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 \(f(x)\gt 1\).

graph of the function y= x^2+x-1 and extra info

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 \(x=y=z=-1\).

Systems of Iterated Equations

  1. Iterations on Monotone Functions
  2. Graphing Equations Is Useful
  3. Graphing Equations Is Useful, II
  4. Graphing Equations Is Useful, III
  5. Graphing Equations Is Useful, IV
  6. Graphing Equations Is Useful, V
  7. Tangential Chaos
  8. Equation in Radicals as a System of Equations
  9. Two Conditions for a Triangle to Be Equilateral

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Copyright © 1996-2017 Alexander Bogomolny

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