Find all rational triples (a, b, c) for which a, b, c are the roots of x3 + ax2 + bx + c = 0.
Solution
Straightforward.
Answer: (0, 0, 0); (1, -1, -1), (1, -2, 0).
We require (1) a + b + c = -a, (2) ab + bc + ca = b, and (3) abc = -c.
From (3), either c = 0, or ab = -1. If c = 0, then (1) becomes b = -2a, and (2) becomes b(a - 1) = 0. Hence either a = b = 0, or a = 1, b = -2.
So assume c ≠ 0, and ab = -1. (1) becomes c = - b - 2a. Substituting in (2), we get: -1 - (2a + b)(a + b) = b, so -a2 -2a4 + 3a2 - 1 = -a, or 2a4 - 2a2 - a + 1 = 0. So a = 1, or 2a3 + 2a2 - 1 = 0 (*). The first possibility gives a = 1, b = -1, c = -1. Suppose a = m/n is a root of (*) with m, n relatively prime integers. Then 2m3 + 2m2n - n3 = 0. So any prime factor of n must divide 2 and any prime factor of m must divide 1. Hence the only possibilities are a = 1, -1, 1/2, -1/2, and we easily check that these are not solutions. So (*) has no rational roots.
© John Scholes
jscholes@kalva.demon.co.uk
15 Sep 1999