3613-s03 - Solutions to Homework Set 3 (Solutions to...

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Solutions to Homework Set 3 (Solutions to Homework Problems from Chapter 2) Problems from § 2.1 2.1.1. Prove that a b (mod n ) if and only if a and b leave the same remainder when divided by n . Proof. Suppose a b (mod n ). Then, by definition, we have a - b = nk for some k Z . Now by the Division Algorithm, a and b can be written uniquely in form (1) a = nq + r b = nq 0 + r 0 with 0 r, r 0 < n . But then (2) a = b + nk = ( nq 0 + r 0 ) + nk = n ( q 0 + k ) + r 0 Comparing (1) and (2) we have a = nq + r , 0 r < n a = n ( q 0 + k ) + r 0 , 0 r 0 < n . By the uniqueness property of the division algorithm, we must therefore have r = r 0 . If a and b leave the same remainder when divided by n then we have a = nq + r b = nq 0 + r . Subtracting these two equations yields a - b = n ( q - q 0 ) , so a b (mod n ) . 2.1.2. If a Z , prove that a 2 is not congruent to 2 modulo 4 or to 3 modulo 4. Proof. By the Division Algorithm any a Z must have one of the following forms a = 4 k 4 k + 1 4 k + 2 4 k + 3 This implies a 2 = 16 k 2 = 4(4 k 2 ) = 4 q 16 k 2 + 8 k + 1 = 4(4 k 2 + 2 k ) + 1 = 4 r + 1 4(4 k 2 + 16 k + 4 = 4(4 k 2 + 8 k + 1) = 4 s 16 k 2 + 24 k + 9 = 4(4 k 2 + 6 k + 2) + 1 = 4 t + 1 1
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2 So a 2 ± 0 (mod 4) 1 (mod 4) . 2.1.3. If a, b are integers such that a b (mod p ) for every positive prime p , prove that a = b . Proof. Since the set of prime numbers in Z is infinite, we can always find a prime number p larger than any given number. In particular we can find a prime number p such that 0 ≤ | a - b | < p . Now by hypothesis, we have, for this prime p , a - b = kp for some k Z (by the definition of congruence modulo p ). Thus, p divides | a - b | . But 0 is the only non-negative number less than p that is also divisible by p . Thus, | a - b | = 0 or a = b . 2.1.4. Which of the following congruences have solutions: (a) x 2 1 (mod 3) We need x 2 - 1 = 3 k By the Division Algorithm, x must have one of three forms x = 3 t 3 t + 1 3 t + 2 x 2 - 1 = 9 t 2 - 1 9 t 2 + 6 t 9 t 2 + 12 t + 3 Thus, if x has the form x = 3 t + 1, then x 2 - 1 = 3(3 t 2 + 2 t ) and so x 2 1 (mod 3). (b) x 2 2 (mod 7) We need x 2 - 2 = 3 k By the Division Algorithm, x must have one of the seven forms x = 7 k 7 k + 1 7 k + 2 7 k + 3 7 k + 4 7 k + 5 7 k + 6 x 2 - 1 = 49 k 2 - 2 = 7 ( 7 k 2 ) + 2 49 k 2 + 14 k - 1 = 7 ( 7 k 2 + 2 k - 1 ) + 6 49 k 2 + 28 k + 2 = 7 ( 7 k 2 + 4 k ) + 2 49 k 2 + 42 k + 7 = 7 ( 7 k 2 + 6 k + 1 ) 49 k 2 + 70 k + 14 = 7 ( 7 k 2 + 8 k + 2 ) 49 k 2 + 70 k + 23 = 7 ( 7 k 2 + 10 k + 3 ) + 2 49 k 2 + 84 + 34 = 7 ( 7 k 2 + 12 k + 4 ) + 6 Thus, if x has the form x = 7 k + 3 or the form x = 7 k + 4, then x 2 - 2 is an integer multiple of 7 and so x 2 2 (mod 7). (c)
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This note was uploaded on 11/03/2010 for the course MATH 262-338-20 taught by Professor Gieseker,d. during the Fall '10 term at UCLA.

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3613-s03 - Solutions to Homework Set 3 (Solutions to...

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