IMO 1973 Problem A3 a and b are real numbers for which the equation x 4 + ax 3 + bx 2 + ax + 1 = 0 has at least one real solution. Find the least possible value of a 2 + b 2 .
Put y = x + 1/x and the equation becomes y 2 + ay + b - 2 = 0, which has solutions y = -a/2 ±√(a 2 + 8 - 2b)/2. We require |y| ≥ 2 for the original equation to have a real root and hence we need |a| + √(a 2 + 8 - 4b) ≥ 4. Squaring gives 2|a| - b ≥ 2. Hence a 2 + b 2 ≥ a 2 + (2 - 2|a|) 2 = 5a 2 - 8|a| + 4 = 5(|a| - 4/5) 2 + 4/5. So the least possible value of a 2 + b 2 is 4/5, achieved when a = 4/5, b = -2/5. In this case, the original equation is x 4 + 4/5 x 3 - 2/5 x 2 + 4/5 x + 1 = (x + 1) 2 (x 2 - 6/5 x + 1). s 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. 15th IMO 1973 © John Scholes jscholes@kalva.demon.co.uk 10 Oct 1998