# 1968 IMO Problems/Problem 3

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## Problem

Consider the system of equations $$ax_1^2 + bx_1 + c = x_2$$ $$ax_2^2 + bx_2 + c = x_3$$ $$\cdots$$ $$ax_{n-1}^2 + bx_{n-1} + c = x_n$$ $$ax_n^2 + bx_n + c = x_1$$ with unknowns $x_1, x_2, \cdots, x_n$ where $a, b, c$ are real and $a \neq 0$. Let $\Delta = (b - 1)^2 - 4ac$. Prove that for this system

(a) if $\Delta < 0$, there is no solution,

(b) if $\Delta = 0$, there is exactly one solution,

(c) if $\Delta > 0$, there is more than one solution.

## Solution

Adding the $n$ equations together yields $$\sum_{i=1}^{n} (ax_i^2+(b-1)x_i+c)=0$$

Let $s_i=ax_i^2+(b-1)x_i+c$.

(a) If $\Delta<0$, then there is no solution to the quadratic equation $ax^2+(b-1)x+c$, as the determinant is negative. This implies that either $s_i>0$ for all $i$, or $s_i<0$ for all $i$. In either case the above summation cannot be 0, which implies that there are no solutions to the given system of equations. $\blacksquare$

(b) If $\Delta=0$, then there is exactly one solution to the quadratic equation $ax^2+(b-1)x+c$ (let it be $r$), and either $s_i\geq 0$ for all $i$, or $s_i\leq 0$ for all $i$. The only way that the above summation is 0 is if $s_i=0$ for all $i$. As there is exactly one $x_i$ that makes $s_i=0$ (namely $x_i=r$), then the only possible solution to the system of equations is $(x_1,x_2,\dots , x_n)=(r, r, \dots , r)$. It's not hard to show that this works, so when $\Delta=0$ the system of equations has exactly one solution. $\blacksquare$

(c) If $\Delta>1$, then there are exactly two solutions to the quadratic equation $ax^2+(b-1)x+c$. Let the roots be $r_1$ and $r_2$. If $x_i=r_1$ for all $i$, then $ax_i^2+bx_i+c=ar_1^2+br_1+c=r_1=x_{i+1}$. This shows that $(x_1,x_2,\dots ,x_n)=(r_1,r_1,\dots, r_1)$ is a solution. We can show that $(x_1,x_2,\dots ,x_n)=(r_2,r_2,\dots, r_2)$ is another solution using the same reasoning, which shows that the equation has more than one solution. $\blacksquare$