5th Irish 1992 Problem A2 How many (x, y, z) satisfy x 2 + y 2 + z 2 = 9, x 4 + y 4 + z 4 = 33, xyz = -4?
12 solutions (x,y,z) = (2,2,-1), (2,-2,1), (-2,2,1), (-2,-2,-1) (2,1,-2), (2,-1,2), (-2,1,2), (-2,-1,-2) (1,2,-2), (1,-2,2), (-1,2,2), (-1,-2,-2) Solution Put X = x 2 , Y = y 2 , Z = z 2 . Then X+Y+Z = 9, X 2 +Y 2 +Z 2 = 33, XYZ = 16. Hence XY+YZ+ZX = ((X+Y+Z) 2 - (X 2 +Y 2 +Z 2 ))/2 = 24. So X, Y, Z are roots of the polynomial k 3 - 9k 2 + 24k - 16 = 0. The polynomial factorises as (k-4) 2 (k-1), so the roots are 1, 1, 4. Thus x, y, z ∈ {1, -1, 2, -2}. To get xyz = -4, we must have one of |x|, |y|, |z| = 1 and the others = 2. Also an odd number must be negative. It is easy to check that we then satisfy the other equations. 5th Irish 1992 © John Scholes jscholes@kalva.demon.co.uk 4 February 2004 Last updated/corrected 4 Feb 04