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Answers to Exercises; Chapter 3 279

(b) 83349 = 35.73. For a solution, we require t F 51, 193 or 242 (mod
243) and t - 41,48, 90, 97, 139,146, 156, 188, 195,237,244,286, 293, 335
or 342 (mod 343). Therefore, there are 3 x 15 = 45 solutions modulo 83349.
For example, t - 51 (mod 243), t E 41 (mod 343) leads to the solution
t E 25080 (mod 83349).

4.13. 675 = 33.52. Any solution x must satisfy x 3 12 or 16 (mod 27) and

x E 2, 7, 9, 12, 17,22 (mod 25). There are 2 x 6 = 12 incongruent solutions

in all modulo 675. An obvious solution is x E 12 (mod 675).

4.14. (a) For any zero r, we must have r 5 = 4r4+411r3+452r2+3322r+828.

If Irl > 2, then

lr15 5 41r14 + 4111r13 + 4521r12 + 3322lrl + 828 5 42001r14 + 828

so that Irl 5 4200 + (828/lr14) < 5000.

(b) Solve the congruence modulo 5000 = 23.54. If r is a root of the

equation, then r E 1, 2 or 3 (mod 8) and r E 7 (mod 25) or r 3 23 (mod

625). The roots are 23 and -18.

4.15. Solving the congruence modulo 32.23 yields t E 18,63 (mod 72). The

only integer root is -9.

5.3. t2 - 1 = (t - 1)(t + 1).

t3 - 1 = (t - l)(t - (-1 + i&)/2)(t - (-1 - ifi)/2).

t4 - 1 = (t - l)(t + l)(t - i)(t + i).

t6 - 1 = (t” - l)(t + l)(t - (1 + ifi)/2) t - (1 - i&)/2).

P-1 = (t”-l)(t-(l+i)/JZ)(t-(l-i)/ J- 2)(t+(l+i)/JZ)(t+(l-i)/JZ).

5.4. Note that

[t - (cos 2kn/n + i sin Pklr/n)][t - (cos 2kr/n - sin Sk?r/n)]

= t2 - (2cos2k?r/n)t + 1

t5 - 1 = (t - l)(? - (2 cos 2?f/5)t + l)(G - (2 cos 47r/5)t + 1).

5.6. Minimum exponents and zeros of &2(t) 1 : 1; 2 : -1; 3 : (-1&i&)/2;

4 : fi; 6 : (1 f i&)/2; 12 : the remaining four zeros.

5.7. (b) Let m be the smallest positive integer for which w”’ = 1. By the

division algorithm, we can write n = qm + r, where q and r are integers

and 0 < r < m. Then w’ = wn(wm)-q = 1. From the minimality of m, it

follows than r = 0. Hence m/n, and, clearly, w is a primitive mth root of

unity.

5.8. (a) By de Moivre’s Theorem, cos(2kr/n) + isin(2k?r/n) = <,k.

(b) Since t” - 1 = (t - l)(t”-l + tnw2 +... + t + l), all nth roots of unity

except 1 itself are zeros of the second factor.

(c) Suppose gcd(a, n) = d. Th en, if C is an nth root of unity, ([“)“ld =

(Cn)a/d = 1. Hence, if C” is primitive, then d = 1. Suppose d = 1 and let