IMO 1968 SL 23

Origin: CZS

Problem

Find all complex numbers m such that polynomial x3 + y3 + z3 + mxyz can be represented as the product of three linear trinomials.

Solution

We may assume w.l.o.g. that in all the factors the coefficient of x is 1. Suppose that x + ay + bz is one of the linear factors of p(x, y, z) = x3 + y3 + z3 + mxyz. Then p(x) is 0 at every point (x, y, z) with z = −ax−by. Hence x3 + y3 + (−ax −by)3 + mxy(−ax −by) = (1 −a3)x3 −(3ab + m)(ax+by)xy+(1−b3)y3 \equiv0. This is obviously equivalent to a3 = b3 = 1 and m = −3ab, from which it follows that m \in{−3, −3\omega, −3\omega2}, where \omega = −1+i \sqrt . Conversely, for each of the three possible values for m there are exactly three possibilities (a, b). Hence −3, −3\omega, −3\omega2 are the desired values.