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 4k + 1 or of the form 4k + 3 for some integer k. 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 4k, 4k + 1, 4k + 2, 4k + 3 for some integer k. Since 4k and 4k + 2 are even, then every odd integer has the form 4k + 1 or 4k + 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. (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 = 8k + 1 for some integer k. (c) If n is an odd integer. then n = 4m + 1 or n = 4m + 3 for some integer m, so that either n 2 = (4m + 1) 2 = 16m 2 + 8m + 1 = 8(2m + 1) + 1 or n 2 = (4m + 3) 2 = 16m 2 + 24m + 1 = 8(2m + 3) + 1 and in both cases n 2 = 8k + 1.
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. If r is a nonzero solution of x 2 + ax + b = 0, then r 2 + ar + b = 0, so that and r b. b = r 2 ar = r(r + a), Question 4. [Exercises 1.2, # 10]. Prove that (n, n + 1) = 1 for every positive integer n. 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. (n, n + 1) = 1. Therefore Question 5. [Exercises 1.2, # 16]. If (a, b) = d, prove that ( a d, b ) = 1. d If (a, b) = d, then there exist integers x and y such that d = ax + by, and since d 1 > 0, dividing both sides of this equation by d, we have 1 = a d x + b d y, so that 1 is a linear combination of a d and b ( a d, and is the smallest such positive integer. Therefore d, b ) = 1. d Question 6. [Exercises 1.2, # 26]. Let a, b, c Z. Prove that the equation ax + by = c has integer solutions if and only if (a, b) c. Suppose that (a, b) c, then c = k (a, b) for some integer k. Also, (a, b) = ax 0 + by 0 for some integers x 0 and y 0, so that c = k(a, b) = akx 0 + bky 0 = ax + by where x = kx 0 and y = ky 0 are in Z, and the equation ax + by = c has integer solutions. Conversely, suppose that there exist integer solutions to the equation ax + by = c, say, since (a, b) a and (a, b) b, then (a, b) c. ax 0 + by 0 = c,
Question 7. [Exercises 1.2, # 32]. Prove that a positive integer is divisible by 3 if and only if the sum of its digits is divisible by 3. [Hint : 10 3 = 999 + 1 and similarly for other powers of 10.] Every positive integer n has a unique representation as where 0 a i 9 for i = 0, 1, 2,... k. n = a 0 + a 1 10 + a 2 10 2 + + a k 10 k Now, an easy proof by induction shows that for each 1 i k, 10 i 1 = (10 1) (10 i 1 + + 10 + 1) = 9 (10 i 1 + + 10 + 1), which is divisible by 9, and therefore is also divisible by 3. Thus, n (a 0 + a 1 + + a k ) = a 1 (10 1 1) + a 2 (10 2 1) + + a k (10 k 1) is also divisible by 9, and hence also by 3, and therefore the positive integer n is divisible by 3 if and only if the sum of its decimal digits is also divisible by 3. Question 8. [Exercises 1.3, # 2]. Let p be an integer other than 0, ±1. Prove that p is prime if and only if for each a Z either (a, p) = 1 or p a. Suppose that p is a prime, and a Z. Let d = (a, p), since d p, then either d = 1 or d = p, that is, either (a, p) = 1 or p a. Let p be an integer other than 0, ±1, and suppose that p is not a prime, then (assuming p > 1) there exist integers a and b with 1 < a, b < p such that p = a b. Since a p, then (a, p) = a > 1. Also, since 1 < a < p, then p a. Question 9. [Exercises 1.3, # 8]. Prove that (a, b) = 1 if and only if there is no prime p such that p a and p b. Suppose (a, b) = 1, if p is a prime such that p a and p b, then p (a, b) = 1, which is a contradiction. Therefore there is no prime such that p a and p b. Conversely, if d = (a, b) > 1, then there exists a prime p such that p d, and since d a and d b, then p a and p b.
Question 10. [Exercises 1.3, # 12]. (a) If 3 (a 2 + b 2 ), prove that 3 a and 3 b. [Hint : If 3 a, then a = 3k + 1 or a = 3k + 2.] (b) If 5 (a 2 + b 2 + c 2 ), prove that 5 a or 5 b or 5 c. (a) Note that if 3 a, then a = 3k + 1 or a = 3k + 2 for some integer k, and a 2 = (3k + 1) 2 = 3(3k 2 + 2k) + 1 or a 2 = 3(3k 2 + 4k + 1) + 1 so that a 2 leaves a remainder of 1 when divided by 3. Therefore, a 2 + b 2 leaves a remainder of 1 + 0 = 1 or 1 + 1 = 2 depending on whether b is divisible by 3 or not. Thus, if 3 a, then 3 a 2 + b 2. Therefore, if 3 a 2 + b 2, then 3 a. Similarly, if 3 a 2 + b 2, then 3 b. (b) Suppose that a, b, c are integers such that 5 a, 5 b, and 5 c. Since 5 a, then a and a 2 must have one of the forms a = 5k + 1 and a 2 = (5k + 1) 2 = 25k 2 + 10k + 1 a = 5k + 2 and a 2 = (5k + 2) 2 = 25k 2 + 20k + 4 a = 5k + 3 and a 2 = (5k + 3) 2 = 25k 2 + 30k + 5 + 4 a = 5k + 4 and a 2 = (5k + 4) 2 = 25k 2 + 40k + 15 + 1 so that a 2 can only leave a remainder of 1 or 4 when divided by 5, similarly for b 2 and c 2. Therefore, when a 2 + b 2 is divided by 5, the only possible remainders are 2, 0, or 3, since 1 + 1 = 2, 1 + 4 = 5, 4 + 4 = 8, and the only possible remainders when (a 2 + b 2 ) + c 2 is divided by 5 are 1, 2, 3, or 4, since and 5 a 2 + b 2 + c 2. 2 + 1 = 3 and 2 + 4 = 6 0 + 1 = 1 and 0 + 4 = 4 3 + 1 = 4 and 3 + 4 = 7,
Question 11. [Exercises 1.3, # 20]. (a) Prove that there are no nonzero integers a, b such that a 2 = 2b 2. [Hint : Use the Fundamental Theorem of Arithmetic or Theorem 1.8.] (b) Prove that 2 is irrational. [Hint : Use proof by contradiction (Appendix A). Assume that 2 = a/b (with a, b Z) and use part (a) to reach a contradiction.] (a) Suppose there exist positive integers a and b such that a 2 = 2b 2. From the Fundamental Theorem of Arithmetic, we may assume that a and b have no prime factors in common (if they do, just cancel them from both sides of the equation a 2 = 2b 2 ). Since 2 is prime, and 2 a 2 = a a, then either 2 a or 2 a, and hence 2 a, and a = 2k for some integer k. Therefore a 2 = (2k) 2 = 4k 2 = 2b 2, so that b 2 = 2k 2, and the same argument as above shows that 2 b. However, this contradicts the fact that a and b have no prime factors in common. Therefore there are no nonzero integers a and b such that a 2 = 2b 2. (b) Suppose that 2 is rational, then there exist positive integers a and b such that 2 = a b, so that a = 2 b. Squaring both sides, we get a 2 = 2b 2, which is a contradiction. Therefore 2 is irrational.