IMO 1982 SL 4

Problem

A4 (BUL 2) Determine all real values of the parameter a for which the equation 16x4 −ax3 + (2a + 17)x2 −ax + 16 = 0 has exactly four distinct real roots that form a geometric progression.

Solution

Suppose that a satisfies the requirements of the problem and that x, qx, q2x, q3x are the roots of the given equation. Then x ̸= 0 and we may assume that |q| > 1, so that |x| < |qx| < |q2x| < |q3x|. Since the equation is symmetric, 1/x is also a root and therefore 1/x = q3x, i.e., q = x−2/3. It follows that the roots are x, x1/3, x−1/3, x−1. Now by Vieta’s formula we have x+x1/3 +x−1/3 +x−1 = a/16 and x4/3 +x2/3 + 2 +x−2/3 +x−4/3 = (2a + 17)/16. On setting z = x1/3 + x−1/3 these equations become z3 −2z = a/16, (z2 −2)2 + z2 −2 = (2a + 17)/16. Substituting a = 16(z3 −2z) in the second equation leads to z4 −2z3 − 3z2 + 4z + 15/16 = 0. We observe that this polynomial factors as (z + 3/2)(z −5/2)(z2 −z −1/4). Since |z| = |x1/3 +x−1/3| \geq2, the only viable value is z = 5/2. Consequently a = 170 and the roots are 1/8, 1/2, 2, 8.