382 Answers to Exercises and Solutions to Problems
- The system
p+q+r=o
up+vq+wr =0
xp+yq+%r=o
(*I
in the variables p, q, r has a nontrivial solution
Since z3 + u3 = y3 + v3, it follows that uy - vz # 0 (otherwise z = Ay and
āu = kv). Hence, we can solve the system bu + cz = bv + cy = 1 for b and c.
Multiplying the equations (*) respectively by 1, -b, -c and adding yields
0 = (1 - bw - cz)(x -y), whence bw+cz= 1.
From (-CZ) 3 = (bu - 1)3 and -x3 = u3 - 03, etc., we find that the
equation
(cā - b3)t3 + 3b2t2 - 3bt + (1 - a3c3) = 0
has (unequal) roots u, v, w, so that (b3 - c3)uvw = 1 - a3c3. Similarly,
(b3-c3)t3+3c2t2-3ct+(l-a3b3) = 0 has roots x, y, z, so that (b3-c3)xyz =
a3b3 - 1. Hence
(b3 - c3)(uvw + xyr) = a3(b3 - c3).
Since each of the two cubic equations has three roots and since b and c
cannot both be zero, at least one of the cubic equations is nontrivial and
b3 - c3 # 0. The result follows.
Suppose CC = y = z. Then u = v = w and the result is trivial. If 2 = y,
then u = v and the linear condition holds automatically. However, the
result may fail. Let (u,v, w) = (12, 12, lo), (z,y,z) = (1,1,9), so that the
cubic condition is satisfied with a3 = 1729. But uvw + zyz = 1449. If we
put these numbers into the above solution, we find that (*) consists of
p+q+r=lQ+q)+lOr=(p+q)+9r=O
and that b = 4/49, c = l/49. Thus, u = 12, w = 10 are the roots of the
equation
0 = 63t3 - 2352t2 + 2881% - 115920 = 21(t - lO)(t - 12)(3t - 46)
and x = 1, z = 9 are the roots of
0 = 21(3t3 + 7t2 - 3432 + 333) = 21(t - 1)(t - 9)(3t + 37).
Thus, if we take rather (u, v, w) = (12,46/3, lo), (x, y, z) = (1, -37/3,9),
then the hypotheses and the conclusion of the problem are satisfied.
- The given inequality has no solution when x < 1 since all summands on
the left side are negative and no solutions for large x when all summands
assume very small positive values. The inequality will be satisfied when 2
