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Solutions to Problems; Chapter 1 263

for n 2 3. This is a straightforward manipulative exercise for the reader,

9.16. Let z be the sum, with x and y its first and second terms; thus,

z = x + y. If u = 0 or u 1 3/8, then x3 and y3 are real, so that x and y

have uniquely determined real values, as well as nonreal values. For other

values of u, x3 and y3 are nonreal complex conjugates. We will assume that

for each nonreal determination of z and y, they are complex conjugates.

Then xy = 1 - 27~ and z satisfies the equation

z3 = x3 + y3 + 3xyz = (6u - 2) + (3 - 6u)z

or 0 = z3 + (6u - 3)~ - (6u - 2) = (z - 1)(z2 + z + 6u - 2). Hence z = 1

or2z=-lf&KXK.

When u 2 318 and the real determinations of x and y are taken, then

z = 1; test this out on your pocket calculator. The other two values of z are

given by the nonreal xw + p2 and xw2 + yw where w is an imaginary cube

root of unity. Observe that these numbers satisfy the quadratic equation

z2 + z + 6u - 2 = 0 since their sum is -(x + y) = -1 and their product is

x2+ y2 - xy=x3+y3=6u-2.

When u = 0, then z = -2 corresponds to the real determination of x

and y. The nonreal determinations of (x, y) are (-w, -w2) and (-w2, -w)

and these both yield z = 1.

When u # 0, u < 318, then z is always real and takes each of the three

possible values corresponding to the determination of x and y.

9.17. m = (1/2)t(t-l), n = (1/2)t(t+l)satisfy4mn-m-n+1 = (t2-1)2.

9.18. The sum of the zeros, u(l+ u + u2), and the sum of products of pairs

of the zeros, u3(1 + u + u2) must be rational. Either 1 + u + u2 = 0 or u2 is

rational. If u2 is rational, then u(1 + u2) = (u + u2 + u”) - u2 is rational.

Since u is nonrational, 1 + u2 = 0. Thus, the possible values of u are

i, -i, (-1 + id?)/2, (-1 - i&)/2.

9.19. First, observe that, if

u(t) = (at + b)(c,t’ + c,-ltr-l +... + qt + co)

v(t) = (bt + u)(c,.t’ + c,.wltr-’ +... + clt + co),
then l?(u(t)) = u2(Ecq) + Pab(Ccici+l) + b2(Ccf) = l?(v(t)).
We have f(t) = (t + 2)(3t + 1). Let g(t) = (2t + 1)(3t + 1) = 6t2 + 5t + 1.
Then, for n = 1, 2,.. .,

r(f(t)y = r((t + 2)“(3t + 1)“)

= r((2t + i)(t + 2)“-‘(3t + I)“)

= r((2t + ij2(t + 2y2(3t + i)n)

=. ..= r((2t + iy(3t + 1)“) = r(g(t)n).