# 1.3 Integers Modulo n

1.

a. True. 40−13=3·9

c. True. −29−6=(−5)7

e. True. 8−8=0·n for any n.

g. False. 8

^{4}≡(64)^{2}≡(−1)^{2}≡1 (mod13).2.

a. 2k−4=7q, so q is even. Thus k=2+7x for some integer x; that is k≡2 (mod 7).

c 2k≡0 (mod 9), so 2k=9q. Thus 2 q, so k=9x for some integer x; that is k≡0 (mod 9).

3.

a. 10≡0 (mod k), so k 10: k=2, 5, 10.

c. k

^{2}−3=qk, so k 3. Thus k=1, 3 so, (as k≥2 by assumption) k=3.5.

a. a≡b (mod 0) means a−b=q·0 for some q, that is a=b.

6.

a. a≡a for all a because n (a−a). Hence if n (a−b), then n (b−a). Hence if a−b=xn and b−c=yn, , then a−c=(x+y)n.

7. If n=pm and a≡b(mod n), then a−b=qn=qpm. Thus a≡b(modm).

8.

a. In , so , . Since 515=6·85+5 we get . Hence ...

Get *Introduction to Abstract Algebra, Solutions Manual, 4th Edition* now with O’Reilly online learning.

O’Reilly members experience live online training, plus books, videos, and digital content from 200+ publishers.