IMO 1967 LL ROM43

Origin: ROM

Problem

The equation x5 + 5\lambdax4 −x3 + (\lambda\alpha −4)x2 −(8\lambda + 3)x + \lambda\alpha −2 = 0 is given. (a) Determine \alpha such that the given equation has exactly one root inde- pendent of \lambda. (b) Determine \alpha such that the given equation has exactly two roots inde- pendent of \lambda.

Solution

We can write the given equation in the form x5 −x3 −4x2 −3x −2 + \lambda(5x4 + \alphax2 −8x + \alpha) = 0. A root of this equation is independent of \lambda if and only if it is a common root of the equations x5 −x3 −4x2 −3x −2 = 0 and 5x4 + \alphax2 −8x + \alpha = 0. The first of these two equations is equivalent to (x −2)(x2 + x + 1)2 = 0 and has three different roots: x1 = 2, x2,3 = (−1 \pm i \sqrt 3)/2. (a) For \alpha = −64/5, x1 = 2 is the unique root independent of \lambda. (b) For \alpha = −3 there are two roots independent of \lambda: x1 = \omega and x2 = \omega2.