9th APMC 1986 Problem 5 Find all real solutions to: x 1 2 + x 2 2 + x 3 2 + x 4 2 = 4; x 1 x 3 + x 2 x 4 + x 3 x 2 + x 4 x 1 = 0; x 1 x 2 x 3 + x 1 x 2 x 4 + x 1 x 3 x 4 + x 2 x 3 x 4 = -2; x 1 x 2 x 3 x 4 = -1.
x 1 ,x 2 ,x 3 ,x 4 = 1,1,1,-1 in any order Solution The second equation is (x 1 +x 2 )(x 3 +x 4 ) = 0. Suppose x 1 = -x 2 = a. Then the other equations are x 3 2 + x 4 2 = 4 - 2a 2 , x 3 + x 4 = 2/a 2 , x 3 x 4 = 1/a 2 . Hence 4/a 4 = (x 3 + x 4 ) 2 = 4 - 2a 2 + 2/a 2 , so (a 2 -2)(a 2 +1)(a 2 -1) = 0. If a = ±1, then x 3 + x 4 = 2, x 3 x 4 = 1, so x 3 = x 4 = 1. If a 2 = 2, then x 3 + x 4 = 1, x 3 x 4 = 1/2, and x 3 , x 4 are not real. If a 2 = -1, then x 2 , x 2 are not real. Thanks to Suat Namli 9th APMC 1986 © John Scholes jscholes@kalva.demon.co.uk 14 February 2004 Last corrected/updated 14 Feb 04