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# soln1 - Solutions to Assignment 1 Question 1[Exercises 1.1...

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Solutions to Assignment 1 Question 1. [Exercises 1.1, # 6] Use the division algorithm to prove that every odd integer is either of the form 4 k + 1 or of the form 4 k + 3 for some integer k. Solution: For each positive integer n, the only possible remainders when n is divided by 4 are r = 0 , 1 , 2 , 3 , so that n has exactly one of the forms 4 k, 4 k + 1 , 4 k + 2 , 4 k + 3 for some integer k. Since 4 k and 4 k + 2 are even, then every odd integer has the form 4 k + 1 or 4 k + 3 for some integer k. Question 2. [Exercises 1.1, # 8] (a) Divide 5 2 , 7 2 , 11 2 , 15 2 , and 27 2 by 8 and note the remainder in each case. (b) Make a conjecture about the remainder when the square of an odd integer is divided by 8 . (c) Prove your conjecture. Solution: (a) We have 5 2 = 25 = 24 + 1 = 3 · 8 + 1 7 2 = 49 = 48 + 1 = 6 · 8 + 1 11 2 = 121 = 120 + 1 = 15 · 8 + 1 15 2 = 225 = 224 + 1 = 28 · 8 + 1 27 2 = 729 = 728 + 1 = 91 · 8 + 1 , and the remainder is 1 in each case. (b) The conjecture is that for any odd integer n, we have n 2 = 8 k + 1 for some integer k. (c) If n is an odd integer. then n = 4 m + 1 or n = 4 m + 3 for some integer m, so that either n 2 = (4 m + 1) 2 = 16 m 2 + 8 m + 1 = 8(2 m + 1) + 1 or n 2 = (4 m + 3) 2 = 16 m 2 + 24 m + 1 = 8(2 m + 3) + 1 and in both cases n 2 = 8 k + 1 .

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Question 3. [Exercises 1.2, # 8]. If r Z and r is a nonzero solution of x 2 + ax + b = 0 (where a, b Z ), prove that r | b. Solution: If r is a nonzero solution of x 2 + ax + b = 0 , then r 2 + ar + b = 0 , so that b = - r 2 - ar = - r ( r + a ) , and r | b. Question 4. [Exercises 1.2, # 10]. Prove that ( n, n + 1) = 1 for every positive integer n. Solution: Note that 1 = 1 · ( n + 1) + ( - 1) · n, so that 1 is a linear combination of n + 1 and n, and is the smallest such positive integer. Therefore ( n, n + 1) = 1 .
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