# Graphing Equations Is Useful, IV

Here is Problem 3 from the 1968 International Mathematical Olympiad [Compendium, p. 49]:

Let $a$, $b$, $c$ be real numbers, $a\ne 0$. Prove that the system of equations

$\left\{ \begin{array}{5,3} ax^{2}_{1}+bx_{1}+c &=x_{2} \\ ax^{2}_{2}+bx_{2}+c &=x_{3} \\ \cdots\cdots\cdots \\ ax^{2}_{n-1}+bx_{n-1}+c &=x_{n} \\ ax^{2}_{n}+bx_{n}+c &=x_{1} \end{array} \right.$

1. has no real solutions if $(b-1)^{2}-4ac\lt 0$;
2. has a unique real solution if $(b-1)^{2}-4ac=0$;
3. has more than one real solution if $(b-1)^{2}-4ac\gt 0$.

Solution

### References

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

Let $a$, $b$, $c$ be real numbers, $a\ne 0$. Prove that the system of equations

$\left\{ \begin{array}{5,3} ax^{2}_{1}+bx_{1}+c &=x_{2} \\ ax^{2}_{2}+bx_{2}+c &=x_{3} \\ \cdots\cdots\cdots \\ ax^{2}_{n-1}+bx_{n-1}+c &=x_{n} \\ ax^{2}_{n}+bx_{n}+c &=x_{1} \end{array} \right.$

1. has no real solutions if $(b-1)^{2}-4ac\lt 0$;
2. has a unique real solution if $(b-1)^{2}-4ac=0$;
3. has more than one real solution if $(b-1)^{2}-4ac\gt 0$;

Interestingly, the problem has been offered by Bulgaria, and originally included only part (b). The other two parts have been added later on.

The inclusion of the discriminant of the quadratic equation $ax^{2}+(b-1)x+c=0$ gives the solution away; for it suggests the equation $ax^{2}+bx+c=x$ is relevant to the problem. Indeed it is as we've seen in other pages. The main observation here is that solving the system is equivalent to finding fixed points or cycles for the iterations $t_{k+1}=at_{k}^{2}+bt_{k}+c$.

Any intersections of the graph of $y=ax^{2}+bx+c$ with the diagonal $y=x$ gives a solution to the system. For a parabola, there may be two, one, or no intersections. The later would mean that there are no fixed points and no cycles as the iterations diverge for any starting point. Having two solutions means "more than one", although there may be more - actually any number - but this in part depends on the unspecified parameter $n$, the number of equations.

Assume the parabola is tangent to the diagonal (i.e., $(b-1)^{2}-4ac=0$) at point $(\alpha ,\alpha)$. Then, obviously, there is one solution $(\alpha ,\alpha ,\ldots ,\alpha)$. And there is only one solution of this sort; for, there is only one point of tangency for a line and a parabola. Point $(\alpha ,\alpha)$ is a repelling fixed point so that the iterations diverge (actually grow or decrease - depending on the sign of $a$ - without bound). In a more formal way, $f(t)=at^{2}+(b-1)t+c = a(t + \frac{b-1}{2a})^2$, due to the condition $(b-1)^{2}-4ac=0$. Summing up all the equations gives $\displaystyle\sum_{k=1}^{n}f(x_{k})=0$, and, since $f(t)\ge 0$, $f(x_{k})=0$, for all $k=1,2,\ldots ,n$, implying that all of them are equal to $-\frac{b-1}{2a}$.

On the other hand, if $(x_{1}, x_{2}, \ldots , x_{n-1},x_{n})$ is a solution, so is $(x_{2}, x_{3}, \ldots , x_{n},x_{1})$ such that, for the solution to be unique, all the variables need to be equal.