IMO 1968 SL 4

Origin: BUL

Problem

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

$$\begin{cases} ax_1^2 + bx_1 + c = x_2, \ ax_2^2 + bx_2 + c = x_3, \ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \ ax_{n-1}^2 + bx_{n-1} + c = x_n, \ ax_n^2 + bx_n + c = x_1, \end{cases}$$

has a unique real solution if and only if $(b - 1)^2 - 4ac = 0$.

Remark. It is assumed that $a \neq 0$.

Solution

We will prove the equivalence in the two directions separately:

$(\Rightarrow)$ Suppose ${x_1, \ldots, x_n}$ is the unique solution of the equation. Since ${x_n, x_1, x_2 \ldots, x_{n-1}}$ is also a solution, it follows that $x_1 = x_2 = \cdots = x_n = x$ and the system of equations reduces to a single equation $ax^2 + (b - 1)x + c = 0$. For the solution for $x$ to be unique the discriminant $(b - 1)^2 - 4ac$ of this quadratic equation must be $0$.

$(\Leftarrow)$ Assume $(b - 1)^2 - 4ac = 0$. Adding up the equations, we get

$$\sum_{i=1}^{n} f(x_i) = 0,$$

where

$$f(x) = ax^2 + (b - 1)x + c.$$

But by the assumed condition,

$$f(x) = a\left(x + \frac{b - 1}{2a}\right)^2.$$

Hence we must have $f(x_i) = 0$ for all $i$, and

$$x_i = -\frac{b - 1}{2a},$$

which is indeed a solution.