x, y, z are complex numbers satisfying x(x - 1) + 2yz = y(y - 1) + 2zx = z(z - 1) + 2xy. Prove that there are only finitely many possible values for the triple (x - y, y - z, z - x) and enumerate them.
Solution
Answer: (0, 0, 0), (1, 0, -1), (-1, 1, 0), (0, -1, 1).
Easy.
Rearranging x(x - 1) + 2yz = y(y - 1) + 2zx we get (x2 - y2) = (x - y) + 2z(x - y), so either x = y or (y - z) = 1 + (z - x) (1). Similarly, either y = z or (z - x) = 1 + (x - y) (2).
So we must look at 4 possibilities. Obviously if x = y and y = z, then the triple is (0, 0, 0). If x = y and (z - x) = 1 + (x - y), then (x - y) = 0, (z - x) = 1, (y - z) = -1, so the triple is (0, -1, 1). If (y - z) = 1 + (z - x) and y = z, then (y - z) = 0, (z - x) = -1, (x - y) = 1, so the triple is (1, 0, -1). The final case is (y - z) = 1 + (z - x) and (z - x) = 1 + (x - y), or rearranging: 1 + 2z = x + y, 1 + 2x = y + z. Subtracting: 3(z - x) = 0, so the triple is (-1, 1, 0).
© John Scholes
jscholes@kalva.demon.co.uk
30 Sep 1999