a, b, c are real with a non-zero. x1, x2, ... , xn satisfy the n equations:
axi2 + bxi + c = xi+1, for 1 ≤ i < n
axn2 + bxn + c = x1
Prove that the system has zero, 1 or >1 real solutions according as (b - 1)2 - 4ac is <0, =0 or >0.
Solution
Let f(x) = ax2 + bx + c - x. Then f(x)/a = (x + (b-1)/2a)2 + (4ac - (b-1)2)/4a2. Hence if 4ac - (b-1)2 > 0, then f(x) has the same sign for all x. But f(x) > 0 means ax2 + bx + c > x, so if {xi} is a solution, then either x1 < x2 < ... < xn < x1, or x1 > x2 > ... > xn > x1. Either way we have a contradiction. So if 4ac - (b-1)2 > 0 there cannot be any solutions.
If 4ac - (b-1)2 = 0, then we can argue in the same way that either x1 ≤ x2 ≤ ... ≤ xn ≤ x1, or x1 ≥ x2 ≥ ... ≥ xn ≥ x1. So we must have all xi = the single root of f(x) = 0 (which clearly is a solution).
If 4ac - (b-1)2 < 0, then f(x) = 0 has two distinct real roots y and z and so we have at least two solutions to the equations: all xi =y, and all xi = z. We may, however, have additional solutions. For example, if a = 1, b = 0, c = -1 and n is even, then we have the additional solution x1 = x3 = x5 = ... = 0, x2 = x4 = ... = -1.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
4 Oct 1998
Last corrected/updated 4 Oct 1998