Solutions to Section 1.3 Homework Problems S. F. Ellermeyer

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1 Solutions to Section 1.3 Homework Problems S. F. Ellermeyer 1. (a) The prime factorization of 38 is 38 = 2 19 so the set of positive divisors of 38 is D (38) = f1; 2; 19; 38g. The prime factorization of 108 is 108 = so the set of positive divisors of 108 is D (108) = f1; 2; 3; 4; 6; 9; 12; 18; 27; 36; 54; 108g. Since D (38) \ D (108) = f1; 2g, we see that gcd (38; 108) = b. It is clear that jnj is the largest positive divisor of n. That is, max (D (n)) = jnj. Therefore gcd (n; n) = max (D (n) \ D (n)) = max (D (n)) = jnj. 3. We want to prove that every common divisor of a and b divides gcd (a; b): Let d = gcd (a; b) and let let c be a common divisor of a and b. Then there exist integers m and n such that d = ma + nb and there exist integers s and t such that sc = a and tc = b. This gives us d = msc + ntc = (ms + nt) c. Since ms + nt is an integer, we see that c divides d. 4. Given that a and c both divide b and that gcd (a; c) = 1, we want to prove that ac divides b. First we give some examples: Example 1: Observe that a = 2 and c = 3 both divide b = 24 and also that gcd (a; c) = gcd (2; 3) = 1. As promised by what we are about to prove, we also observe that ac = 6 divides b = 24. Example 2: Observe that a = 5 and c = 8 both divide b = 400 and also that gcd (a; c) = gcd (5; 8) = 1. As promised by what we are about to prove, we also observe that ac = 40 divides b =

2 Now here is the proof: Suppose that a and c both divide b and that gcd (a; c) = 1. By Proposition 7 (page 22) in the textbook, we know that ac = lcm (a; c) gcd (a; c). Thus ac = lcm (a; c). Also, since a and c both divide b, there exist integers s and t such that sa = b and tc = b. Suppose, for the sake of obtaining a contradiction, that ac does not divide b. Then by the Division Algorithm, there exist integers q and r such that b = q (ac) + r and 0 < r < ac. This implies that r = b qac, which implies that r = sa qac = (s qc) a, which implies that r is a multiple of a. Likewise which implies that r = b qac, r = tc qac = (t qa) c, which implies that r is a multiple of c. However, we have now arrived at a contradiction because we have that r is a common multiple of a and c and that r > 0 and that r < ac = lcm (a; c). (It is not possible that some positive common multiple of a and c is less than lcm (a; c).) This proves that ac divides b. 5. Suppose that x, y, and m are integers and that gcd (x; m) = 1 and that gcd (y; m) = 1. We will prove that this implies that gcd (xy; m) = 1. Since gcd (x; m) = 1, there exist integers s and t such that sx + tm = 1. Likewise, since gcd (y; m) = 1, there exist integers w and z such that wy + zm = 1. 2

3 This gives us (by multiplying each side of the above two equations by each other) (sx + tm) (wy + zm) = 1. By expansion, we obtain which can be written as swxy + szxm + twmy + tzmm = 1 (sw) (xy) + (szx + twy + tzm) m = 1. Since sw and szx + twy + tzm are both integers, we conclude that gcd (xy; m) = Suppose that a > 0 and that a divides b. We want to prove that gcd (a; b) = a. Since a divides b and a > 0, we know that a belongs to the set of positive divisors of b. (That is, a 2 D (b).) Also, it is clear that a is the largest member of the set D (a). Thus a 2 D (a) \ D (b) and a must also be the largest member of this set. Therefore gcd (a; b) = a. 7. c. 8. (a) 442 = 1 (289)+153 tells us that gcd (442; 289) = gcd (289; 153) 289 = 1 (153)+136 tells us that gcd (289; 153) = gcd (153; 136) 153 = 1 (136) + 17 tells us that gcd (153; 136) = gcd (136; 17) 137 = 8 (17) + 0 tells us that gcd (137; 17) = gcd (17; 0) = 17. We conclude that gcd (442; 289) = = 1 (442) = 22 (20) = 10 (2) + 0 tells us that gcd (462; 442) = 2. Thus (462) (442) lcm (462; 442) = 2 = 102;

4 9. We want to prove that if n is an odd integer, then gcd (3n; 3n + 2) = 1. Examples of this are gcd (3; 5) = 1, gcd (15; 17) = 1, and gcd (33; 35) = 1. First observe that if n is an odd integer, then n = 2k + 1 where k is some integer. This means that 3n = 3 (2k + 1) = 6k + 3. Since 6k is an even integer, then 6k + 3 must be odd. Thus 3n is odd. This means that 3n + 2 is also odd. Applying the Division Algorithm to 3n + 2 and 3n, we obtain 3n + 2 = 1 3n + 2 and we know (by Proposition 6 on page 18) that gcd (3n + 2; 3n) = gcd (3n; 2). Since the only positive divisors of 2 are 1 and 2 and since 2 is not a divisor of 3n (because 3n is odd), we see that gcd (3n; 2) = 1. Therefore gcd (3n + 2; 3n) = Let us nd all of the solutions of the Diophantine equation 35x+25y = 15: If we divide both sides of this equation by 5 we obtain 7x + 5y = 3. Any (integer pair) solution of this latter equation is also a solution of the former equation and vice versa. Since gcd (7; 5) = 1 and since 3 is a multiple of 1, we know that our equation does have solutions. We are going to nd a solution of the equation 7x + 5y = 1 (and then multiply our solution by 3 to obtain a solution of 7x + 5y = 3). To nd a solution of 7x + 5y = 1, we use the Euclidean Algorithm: gives us which gives us 7 = 1 (5) = 2 (2) = 2 (2) = 1 (5) = 2 ( 1 (5) + 7) + 5 = 2 (5) 2 (7) + 5 = 2 (7) + 3 (5). 4

5 Thus we see that ( 2; 3) is a solution of the equation 7x+5y = 1. This means that (x 0 ; y 0 ) = ( 6; 9) is a solution of the equation 7x + 5y = 3. Furthermore, all of the solutions of the equation 7x + 5y = 3 are b (x; y) = x 0 + gcd (a; b) t; y a 0 gcd (a; b) t = ( 6 + 5t; 9 7t) where t can be any integer. Thus, for example, another solution of 7x + 5y = 3 (taking t = 2) is (x; y) = ( 16; 23). 11. To solve the Diophantine equation 18x 5y = 15, we note that gcd (18; 5) = 1 and that, since 15 is a multiple of 1, this equation does have solutions. We will rst solve 18x 5y = 1. We will actually solve 18x+5y = 1 (in order for the Division Algorithm to apply) and then note that if (x; y) is a solution of 18x + 5y = 1, then (x; y) is a solution of 18x 5y = 1. gives which gives 18 = 3 (5) = 1 (3) = 1 (2) = 1 (2) = 1 (3) = 3 (5) = 1 (2) + 3 = 1 ( 1 (3) + 5) + 3 = 1 (3) 1 (5) + 3 = 2 (3) 1 (5) = 2 ( 3 (5) + 18) 1 (5) = 6 (5) + 2 (18) 1 (5) = 2 (18) 7 (5). We now see that (2; 7) is a solution of the equation 18x+5y = 1. This means that (2; 7) is a solution of the equation 18x 5y = 1. Therefore 5

6 a solution of the equation 18x 5y = 15 is (x 0 ; y 0 ) = (30; 105). All of the solutions of the equation 18x 5y = 15 are (x; y) = (30 5t; t) where t can be any integer. For example, taking t = 9, we can verify that ( 15; 57) is also a solution. We want to determine which of these solutions are positive integer solutions. Thus we need to determine the values of t for which both 30 5t > 0 and t > 0. The rst inequality tells us that we must have t < 6 and the second inequality tells us that we must have t < 105=18 5:8. Thus, to obtain positive solutions, me must choose t 5. Note that for t = 6, we obtain the solution (x; y) = (0; 3) and that for t = 5 we obtain the solution (x; y) = (5; 15). 6