278 Answers to Exercises and Solutions to Problems
3.4. The values of the polynomial at -2, -1, 1, 2 are, respectively, 40, 21,
-11, 48. All but l/2 and 4/3 are eliminated.
3.6. (a) 5. (b) 3/8. (c) 2, 2/5. (d) -1, 3, 7. (e) 3/2, -3/2, -5/2. (f) -7/2,
413.
3.7. (3t - 4)(3t + 5)(2t3 - 6t2 + 9t + 3).
3.8. -7, 2/3, -5/8, (l/2)(1 + id?), (l/2)(1 - id?).
3.9. Newton’s Table is Horner’s Table for the polynomial t3q( l/t) evaluated
at l/5. This polynomial has zero l/5 iff q(t) has zero 5.
3.12. (a) Note that 8/2 and 28 are supplementary.
(b) From (a), cos0 = 2~0~~28 - 1 = 2(2cos2 0 - 1)2 - 1.
(c) 8x4 - 8x2 - x + 1 = (X - 1)(2x + 1)(4x2 + 2x - 1). By (b), cose is a
zero of this polynomial. Since case is not equal to 1 nor -l/2, it must be
azeroof4x2+2a:-1.
(d) (A- 1)/4.4.1. (a) 4, 5, 12, 13, 20, 21, 28, 29, 36, 37.
(b) 2, 7, 12, 17, 22, 27, 32, 37.
(c) For any integer t, 40 divides t2 - 9t - 36 iff 5 and 8 do. The congruence
is satisfied by 12 and 37.
4.4. 22, 82.
4.5. 444.
4.6. By Exercise 1.6.6 (d), th ere is a number w such that a + wu E b (mod
v). Choose c such that 0 5 c _< m - 1 and c E a + wu (mod UV). Then c is
the required number.
If c and d satisfy c = d - a (mod u), c E d - b (mod v), then c - d is
divisible by u and v, hence by UV. Hence c G d (mod m). If both 0 5 c 5
m - 1, 0 5 d < m - 1, then c = d.
4.8. 504 = 7.8.9. n” - n2 = n2(n” - 1) is always divisible by 7 and 9, as
well as by 8 when n G 0, 1,3 (mod 4). If n E 2 (mod 4), n6 - 1 is odd and
n2 is divisible by 4 but not by 8. Hence n8 - n2 is not divisible by 504 iff
nE2 (mod 4).
4.9. (a) t 5 3 (mod 4); t E 0, 2 (mod 9). (c) t E 11, 27 (mod 36).
4.10. (a) None. (b) 11, 35. (c) 21, 45.
4.11. (e) 22. (f) 221. (Note that t satisfies this congruence iff -t does.)
4.12. (a) 1000 = 23.53. 2t3 + t + 3 q 0 (mod 5) is satisfied for t E 3, 4
(mod 5). The solution t 3 3 (mod 5) does not lead to a solution modulo
53 (cf. Exploration E.33). H owever, the congruence modulo 125 is satisfied
by t E 124. The solution modulo 8 is 7. Hence, 2t3 + t + 3 E 0 (mod 1000)
has a unique solution t E 999 (mod 1000).