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Solutions to Problems; Chapter 3 295

This is clearly nonnegative if 0 5 r + s < d. If d 5 r + s < 2d, then at least
one of 2r and 2s is no less than d and the expression is again nonnegative.
8.24. First solution. (x, y, z real). Let x = tanu, y = tan v, z = tan w.
Then (1 - xy - yz - ZX) tan(u + v + w) = x + y + z - xyx. Observe first
thatxy+yz+zx-landx+y+z- xyz cannot vanish simultaneously.
Otherwise, x+y+z = ~(1 -XZ-yz) so that 0 = (x+y)(l+x2). Thus, we
can see that the sum of any two variables is 0, which is impossible. Hence
x+y+~=xy~ifftan(u+v+w)=Oiffu+v+wisamultipleofa.
Suppose x + y + z = xyz. Then u + v + w is a multiple of 7r, so
that 2u + 2v + 2w is a multiple of rr. Then tan 2u + tan 2v + tan2w =
(tan 2u)(tanPv)(tan 2w), and the result follows.
Second solution. Putting the left side over a common denominator yields
the numerator

2(x + y + %) + 2xyz(xy + yz + zx) - 2[(x + y + z)(xy + yz + zx) - 3+yz]

which, under the stated conditions, reduces to 8xy~.

8.25.

dl = (x2!/3 - x3Y2)(x3xl + ?/3&)(21x2 + YlY2)

= xlx2x3(~1x2Y3 - x1?/2x3) + Y1!/2!/3(Ylx2!/3 - YlY223)

+ XlYl(4Y,2 - 4Y,2).

Obtain a similar expression for d2 and d3, and check the required identity
directly.

8.26. (a) a = b + 1 iff 1 is a zero of the quadratic; a + b + 1 = 0 iff -1 is a
zero. Suppose the quadratic has a zero u which is a nonreal root of unity.
Then ii = u-l is also a zero of the quadratic. Since the product of the zeros
is 1, b = 1. Since Iu+u-‘1 < 2, it follows that (a, b) = (-1, l), (0, l), (1,l);
these possibilities yield quadratics whose zeros are primitive cube, 4th and
6th roots of unity respectively.
(b) Let f(t) be the first quadratic and g(t) be the second. If f(1) = 0,
then (b + 1)” = a2 and either g(1) or g(-1) vanishes. If f(-1) = 0, then
a2 + (1 - b)2 = 0 so that a = 0, b = 1 and g(i) = 0. Otherwise b2 = 1
and a2 - 2b = -1, 0, 1. We must have (a, b) = (-1, l), (1,1) and the result
follows.

8.27. We have that au = -bv - c and aTi = -bT- c, whence a2 = b2 + c2 +
2bccos 8, where 0 = Pkr/n for some k, n E N and v = cos 0 + isin 8. Thus,
case is a rational number; write it as p/q in lowest terms with q positive.
Suppose, if possible, that q 1 3. Since cos20 = (2p2 - q2)/q2 and the
greatest common divisor of 2p2 - q2 and q2 is either 1 or 2, cos20 has
a denominator in lowest terms at least equal to iq2 > q. We find that
cos 8, c~s 28, cos 48, cos 88,... are all rationals, each of whose denominators
in lowest terms exceeds that of its predecessor. Hence, these cosines are