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280 Answers to Exercises and Solutions to Problems

k be the smallest positive exponent for which (<a)k = 1. It can be seen

that the powers of C” cycle through the set { 1, C”, c20,... , c(k-l)o}. Since

(6”)” = 1, n must be a multiple of k so that n = mk. Since ( is a primitive

nth root and C ak = 1, ak must be a multiple of n, say ak = qn. Hence

a = qm, so that m divides n and a. Therefore, m = 1 and C” is a primitive

nth root.

(e) gcd(a, n) = gcd(n - a, n), so that a - n - a is a pairing of positive

integers less than n and coprime with n. Note that a # n - a when n 2 3.

5.9. The zeros of P,(t) consist of all the primitive dth roots of unity, where
d runs through all the positive divisors of n. For each zero, there is a
corresponding linear factor. By collecting together all the linear factors for
the primitive dth roots, we obtain the required representation.
5.11. &$I = t6+t3+ 1;
&lo = t4 -ts+tz-t+1;
f&2 = t4 - t2 + 1;
Q14 = t6 -t5+t4-t3+t2-t+1;
Q15 = ts - t7 + t5 - t4 + t3 - t + 1;
Q1fj = t” + 1.
To find, for example, &Is(t), we look for the complementary factors of
Pls(t) whose zeros are primitive lst, 3rd, 5th roots of unity. Thus

&(t) = (t5 - 1)(t’O + t5 + 1)

= (P-1)(t2+t+1)(t8-P+P-t4+t3-t+1).

5.12. (a) It is straightforward to see that if 6 is a primitive 2kth root of
unity, then C2 is a primitive kth root of unity. Let k be even and C2 be a
primitive kth root of unity. Suppose that C is a primitive rth root of unity.
Then k <_ r 5 2k and either r = 2k or r is odd. Since (C”)” = 1 and Ck # 1,
we must have ck = -1. Hence C’-’ = -1, so that <2(r-k) = 1. Hence
r - k 2 k, so that r 1 2k. Thus, i is a primitive Pkth root of unity.
The primitive kth roots of unity are precisely the numbers of the form
((:k)2, where gcd(a, k) = gcd(a, 2k) = 1.

Qk(t2) = II{(t2 - <a) : 1 _< a < k,gcd(a, k) = 1)

= II{(t - <ik)(t - C,“,‘(l) : 1 5 a < k, gcd(u, k) = 1)

= II{(t - ctk) : 1 _< a < 2k, gcd(a, 28) = 1)

= Qak(t).

(b) Let k be odd and let C be a primitive kth root of unity. Then no
power of C is equal to -1, whence it follows that no odd power of -6 is
equal to 1. Hence, if (-C)’ = 1, then r is even and c’ = 1. But then r is
an even multiple of k, i.e. a multiple of 2k. It follows that -C is a primitive
2kth root of unity.