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. 84≡(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).
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.
10: k=2, 5, 10.
c. k2−3=qk, so k 3. Thus k=1, 3 so, (as k≥2 by assumption) k=3.
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.
(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 ...
, so , . Since 515=6·85+5 we get . Hence ...Get Introduction to Abstract Algebra, Solutions Manual, 4th Edition now with the O’Reilly learning platform.
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