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388 Answers to Exercises and Solutions to Problems

w2 + w + 1 is a positive real multiple of -(w2 - w + 1). If there is equality,

then lw2 + w + 11 = I w2 - w + 11 which occurs when

2(wz+1)=(wz+w+1)+(wz-w+1)=0.

Hence, except when w2 = -1, it must happen that IwI is strictly less than
the maximum of lw2 + w + 1 I and lw2 - w + 1 I, so at least one of these three
must differ from 1. It follows from this, since f(t) is a polynomial, that the
only zeros of f(t) can be i and -i, and so f(t) = c(t2 + 1)” for c = 1 and
nonnegative integer k. It is readily checked that this works.

46. There is no solution. Otherwise, by the AGM inequality, 2&x 5
    u2 + 3x2, etc., so that 6 < 4fi = 2fi(ux + vy + wt) 5 6, which is a
    contradiction.


47. x3+y3+%3-Qxy% = (x+y+%)(x2+y2+%2-xy-x%-y%) =o*
    x3 + ys + t3 = 3xyz. Similarly, xv3 + yV3 + %-3 = ~x-‘Y-‘%-~. Thus

x6 + y6 + %6 = (x3 + y3 + %3)2 - 2t3y323(x-3 + y-3 + %-“)

= 9x2y2z2 - 6~~~~2~ = 3x2y2t2,

and this yields the result.

48. Let w be a zero of % + 1. Then w is simple, so it suffices to show that
    % = w makes the right side vanish. This is clear for w = -1. Otherwise,
    l-w+w2-.. .+uF’-l = 0 and the right side is equal to (-w-‘-l)(~-‘)~+
    (w + l)~(*-‘)~-’ w h ic h is 0. See Problem 3.7.17 and its solution for an
    alternative solution to a special case.


49. Suppose that (x + 1)3 - x3 = y2. This implies that

3(x2 + x) + 1 = y2 + 3(2x + 1)2 = 4y2 - 1 = (2y - 1)(2y + 1).

Since 2y - 1 and 2y + 1 are relatively prime, either

(1) 2y - 1 = u2, 2y + 1 = 3b2 for some a, b, or

(2) 2y - 1 = 3c2, 2y + 1 = d2 for some c, d.

Case (2) cannot occur, since it would imply that d2 = 3c2 + 2, an im-
possibility. Hence, we must have Case (1) and 4y = u2 + 3b2 = 2(u2 + 1).
Since Q is odd, we can write u = 2u + 1, whence y = u2 + (u + 1)2.

Remark. In Amer. Math. Monthly 57 (1950), 190, it is noted that solu-
tions of the equation are given by (x, y) = (xn, y,,) where (xe, ye) = (0, l),

(xl, ~1) = (7,13) and (xn+l,yn+l) = (142, - G-I+ 6, 14~~ - ~~-1) for
7a> 1.

50. Prom the difference of the first two equations, we obtain that

x(1 - x2) + y(y - 1) + %2(Z - 1) = 0.... (1)