# HW2 - (6(6(3(3(4(4(5(5(1(1(2(2 p 108#24 We assume that the...

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Unformatted text preview: (6) (6) (3) (3) (4) (4) (5) (5) (1) (1) (2) (2) p. 108 #24 . We assume that the are pairwise relatively prime. Hence . Therefore has a reciprocal mod , which we call . Therefore . Moreover, for , we have , since | . Thus, contains only one nonzero term, which is . p. 105 #8. If , then = . However, if , then , since p is prime, so has a reciprocal mod p, and multiplying both sides of the congruence by that reciprocal, we see that . Hence a solution that is not congruent to 1 mod p must be congruent to mod p. In particular, for p a prime, the integers mod p form a field; and all nonzero elements have a reciprocal. For the ring of integers mod a composite number, the same statement is not true. There will, of necessity, be nonzero zero divisors. Moreover, while it is true that over a field, a quadratic equation has at most two roots, in the more general situation of a quadratic equation over an arbitrary commutative ring, there may be more than two roots. p. 105 #10. We apply the CRT to , , and to discover that the smallest number of people is 58 mod . The next smallest number is 118. restart: 2*3*15+1*2*20+3*3*12 mod (4*3*5); 58 %+60; 118 p. 105 #12. Working mod 101 is like working mod in the exponents. numtheory[phi](101); 100 10203 mod 100; 3 Hence, the remainder is 2^3; 8 Checking: 2&^10203 mod 101; 8 (12) (12) (11) (11) (7) (7) (9) (9) (8) (8) (10) (10) p. 105 # 13 . Working mod 100 is like working mod in the exponents. numtheory[phi](100); 40 562 mod 40; 2 Hence the last two digits are 29. 123^2; 15129 Checking: 123&^562 mod 100; 29 p. 105 #14. a) Working mod 4 is like working mod 4(4) = = 2. with(numtheory): phi(4); 7 mod %; 7^% mod 4; 2 1 3 Hence mod 4 is 7 mod 4 = 3. In particular, = (7^7-3)/4; 205885 b) Working mod 10 is like working mod 4(10) = = 4 in the exponents....
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## This note was uploaded on 02/14/2012 for the course MATH 470 taught by Professor Staff during the Spring '08 term at Texas A&M.

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HW2 - (6(6(3(3(4(4(5(5(1(1(2(2 p 108#24 We assume that the...

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