2nd USAMO 1973 Problem 4 Find all complex numbers x, y, z which satisfy x + y + z = x 2 + y 2 + z 2 = x 3 + y 3 + z 3 = 3. Solution
: 1, 1, 1. We have (x + y + z) 2 = (x 2 + y 2 + z 2 ) + 2(xy + yz + zx), so xy + yz + zx = 3. (x + y + z) 3 = (x 3 + y 3 + z 3 ) + 3(x 2 y + xy 2 + y 2 z + yz 2 + z 2 x + zx 2 ) + 6xyz, so 8 = (x 2 y + xy 2 + y 2 z + yz 2 + z 2 x + zx 2 ) + 2xyz. But (x + y + z)(xy + yz + zx) = (x 2 y + xy 2 + y 2 z + yz 2 + z 2 x + zx 2 ) + 3xyz, so xyz = -1. Hence x, y, z are the roots of the cubic w 3 - 3w 2 + 3w - 1 = (w - 1) 3 . Hence x = y = z = 1. 2nd USAMO 1973 © John Scholes jscholes@kalva.demon.co.uk 11 May 2002