IMO 1968 · Problem 3

Consider the system of equations \[ \begin{aligned} ax_1^2+bx_1+c&=x_2\\ ax_2^2+bx_2+c&=x_3\\ &\cdots\\ ax_{n-1}^2+bx_{n-1}+c&=x_n\\ ax_n^2+bx_n+c&=x_1, \end{aligned} \] with unknowns \(x_1,x_2,\ldots,x_n\), where \(a,b,c\) are real and \(a\ne0\). 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.

Official text of the problem. Official problem paper (PDF)

Solution

Set \(x_{n+1}=x_1\). For any real solution, the equations give \[ x_{i+1}-x_i=ax_i^2+(b-1)x_i+c \qquad (1\le i\le n). \] Completing the square yields \[ 4a(x_{i+1}-x_i)=(2ax_i+b-1)^2-\Delta. \] Summing over all \(i\), the left-hand side telescopes to zero. Hence every solution satisfies \[ \boxed{\sum_{i=1}^{n}(2ax_i+b-1)^2=n\Delta.} \]

(a) If \(\Delta<0\), the right-hand side is negative, whereas the left-hand side is a sum of squares of real numbers. Thus no real solution exists.

(b) If \(\Delta=0\), every square in the sum must vanish. Since \(a\ne0\), this forces \[ x_1=x_2=\cdots=x_n=r, \qquad r=\frac{1-b}{2a}. \] This tuple is indeed a solution, because \[ ar^2+(b-1)r+c =c-\frac{(b-1)^2}{4a} =-\frac{\Delta}{4a}=0, \] so \(ar^2+br+c=r\). Therefore there is exactly one solution.

(c) If \(\Delta>0\), the quadratic equation \[ at^2+(b-1)t+c=0 \] has two distinct real roots \[ r_-=\frac{1-b-\sqrt{\Delta}}{2a}, \qquad r_+=\frac{1-b+\sqrt{\Delta}}{2a}. \] Each root satisfies \(ar_\pm^2+br_\pm+c=r_\pm\). Consequently, both \[ (x_1,\ldots,x_n)=(r_-,\ldots,r_-) \quad\text{and}\quad (x_1,\ldots,x_n)=(r_+,\ldots,r_+) \] solve the system. These tuples are distinct, so there is more than one solution.

The solution was checked against John Scholes' IMO solution archive (Kalva).