296 Answers to Exercises and Solutions to Problems
all unequal. However, let n = 2*m with m odd. Since the powers of 2
cannot be all incongruent modulo m, there exist i,j E N with i > j and
2’ G 2j (mod m), whence 23’(2’-j - 1) E 0 (mod m). Let r = i - j. Then
2r+60 - 26C3 = 2J(2r - 1)2k?r/m is a multiple of 2?r and cos 2’+YI = cos 2”0,
yielding a contradiction.
Therefore, the only possible values of case are 0, &f and fl. Since v
is not real, v must be a nonreal fourth or sixth root of unity. Similarly,
so is u. Suppose, if possible, that v = i. Then a2u2 = c2 - b2 + 2bci and
a3u3 = (3b2c - c”) + (b3 - 3bc2)i. Since bc(b2 - 3c2) # 0, u2 and u3 are
nonreal and so u is neither a fourth nor sixth root of unity. Hence v # i;
similarly, v # -i and u # fi. Hence u and v are both nonreal sixth roots
of unity and we have one of u = v, u = -v, u = v2, u = -v2. Since
the first two possibilities would imply that u was rational, we must have
&au2 + bv + c = 0. Since the manic irreducible polynomial with zero v is
t2 f t + 1, the required result follows.
8.28. First solution, Let u = cos(7r/14)+isin(s/l4), so that 2ix = u-u-‘.
Since iu = cos(4n/7)+isin(4r/7), iuisazerooft6+t5+t4+t3+t2+t+1,
whence
(-2 + u4 - u2 + 1) + i(u5 - u3 + 1) = 0
or
-(u” - us3) + i(u2 + uV2) + (u - u-‘) - i = 0.
Since -8ix3 = (u” - u-“) - 3(u - u-l) and -4x2 = (u2 + u-“) - 2, we can
substitute for x, divide by i and obtain the required equation.
Second solution. Applying de Moivre’s Theorem and using cos2(?r/14) =
1 - x2 yields
l=sinn/2 = 7(1 - Z”)“Z - 35(1- .2)%3 + 21(1- x2)x” - x7
= 7x - 56x3 + 112x5 - 64x’.This can be manipulated to
0 = (x + 1)(8x3 - 4x2 - 4x + 1)2.Since z # -1, the result follows.
8.29. Let v = sinr/7. The length of the side of the heptagon is 2v. We
have, by de Moivre’s Theorem,
O=sin7r=7v-56v3+112v5-64~~.Setting x = 2v yields the result. Other roots of the equation are v =
f sin(2n/7), f sin(3s/7), so that the roots of the equation in x are the
lengths of the sides and diagonals and their negatives.
A similar problem posed for a regular undecagon (11-gon) is Problem
2864, Amer. Math. Monthly 27 (1920), 482; 28 (1922), 91.