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316 Answers to Exercises and Solutions to Problems

8.22. First solution. Clearly 0 < x < y. If x 5 1, then (x + y)” > 9, so that

y >_ 2. Hence

9(y3 - 1) 5 9(y3 - x3) = 7(x + y)2 _< 7(y + 1)2

* (Y - 2)(9y2 + lly + 8) < 0.

Since 9y2 + lly + 8 has a negative discriminant, it is always positive. Hence

y=2,x= 1. On the other hand, if x > 1, then (x + Y)~ _< 9, so y 5 2.

Then (y - 2)(9ti + lly + 8) > 0 and y = 2, x = 1.

Second solution. Since x > 0, let x = t2. Then 7zW2 + %‘j = y3 =

(32-l - z~)~, whence

0 = 2%’ - 9z6 + 27%s + 7% - 27

= (% - 1)(2%S + 2%7 + 2 z6 - 7%’ - 7%4 - 7z3 + 20z2 + 20% + 27)

= (% - 1)[(%2 + % + 1)(22” - 7%3 + 20) + 71.

The polynomial 2z6 - 7z3 + 20, considered as a polynomial in %3, has
negative discriminant, and so is always positive. Since z2 + z + 1 is also
positive, the term in square brackets never vanishes, and so z = 1. Thus,
(x, y) = (2,l) is the only solution.
8.23. Since x2 = (y + %)2 - 2(1 + a)y% = (y - %)2 + 2(1 - a)yr, it follows
that

whence

2(1+ U)Y% = (x + y + %>(Y + z - x)

2(1- a)yz = (x + y - %)(X + % - y),

4(1 - .2),2y2z2 = (x + y + %>(Y + .% - x)(x + y - %)(X + % - ?/)x2.

Similar equations hold for 1 - b2 and 1 - c2. From these it can be seen
that, when Ial, (bl, ICI are distinct from 1,

x2(1 - u2)-l = y2(1 - b2)--l = ~~(1 - c2)-l.

Ifu=1,thenx2=(y-%)2,whencey=x+zorz=x+y.Ify=x+z,
then 2% = -bx:z and xy = cxy, whence x = 0, or XT # 0, y = 0, b = -1, or
xy # 0, z = 0, c = 1, or xyz # 0, b = -1, c = 1. The cases z = x + y and
a = -1 can be handled similarly.

8.24. Adding the first two equations yields

x2 + y” + %2 = (^30) (*>
whence (x + y + z)~ = 30 + 2(-7) = 16.
Taking the difference of the first two equations yields x2 - %2 = 24.
Case(i):Fromx+z=4-y,x- % = 24(4 - y)-‘, we obtain (x, y, z) =
(U + 621-l, 4 - 2u, u - 6u-l) where 2u = 4 - y. Substituting this into