If x² + y² + z² = 3, where x, y and z are integers, then how many values can (xy + yz + zx) have?

Integer solutions of x² + y² + z² = 3 require each of x², y², z² = 1, so each of x, y, z is ±1. Cases: (i) all three same sign (1,1,1) or (-1,-1,-1): xy+yz+zx = 3. (ii) two same, one opposite (e.g., 1,1,-1): xy+yz+zx = 1 - 1 - 1 = -1. So xy + yz + zx can take 2 distinct values: 3 and -1.