Homework 1, Solutions

Size: px

Start display at page:

Download "Homework 1, Solutions"

Dulcie Logan
7 years ago
Views:

1 Homework 1, Solutions (1.6) Let a, b, c Z. Use the definition of divisibility to directly prove the following. (a) If a b and b c, then a c. (b) If a b and b a, then a = ±b. (c) If a b and a c, then a (b ± c). Proof: (a) 1 a b & b c Assumption 2 u, v Z, b = au & c = bv Definition of a b & b c 3 c = bv = auv substituting b = au into c = bv 4 c = aw, w Z let w = uv. 5 a c Definition of a c Another way to present the reasoning: Since a b and since b c, by the definition of divisibility, there exist integers u and v such that b = au and c = bv. Substitute b = au into c = bv to get c = auv. Let m = uv. Since u, v are integers, m is also an integer, and so c = am. By the definition of divisibility, a c. (b) If a = 0, then a b forces that b = 0. Hence we assume that a 0 (and so b 0). 1 a b & b a Assumption 2 u, v Z, b = au & a = bv Definition of a b & b a 3 a = bv = auv substituting b = au into a = bv. 4 a(1 uv) = 0 Algebraic manipulation. 5 1 = uv Since a 0, we can cancel a both sides of a = a(1 uv). 6 v = ±1 u, v are integers and uv = 1. 7 a = ±1 a = bv and since v = ±1. Another way to present the reasoning: Since a b and b a, by the definition of divisibility, there exist integers u and v such that b = au and a = bv. Substitute b = au into a = bv to get a = auv. Since a 0, cancelling a both sides yields 1 = uv. Since u and v are integers, either u = v = 1 or u = v = 1, and so v = ±1. It follows that a = bv = ±b. (c) 1

2 1 a b & a c Assumption 2 u, v Z, b = au & c = av Definition of a b & a c 3 b ± c = au ± av = a(u ± v) Algebra/Remove common factor a. 4 b ± c = am Let m = u ± v, which is also an integer. 5 a (b ± c) Definition of divisibility. Another way to present the reasoning: Since a b and since a c, by the definition of divisibility, there exist integers u and v such that b = au and c = av. Add (or subtract) these equalities side by side to get b ± c = au ± av. Remove the common factor a leads to b ± c = a(u ± v). Since u and v are integers, m = u ± v is also an integer. and so by the definition of divisibility, a (b ± c). Additional Properties Proved in Class (a) Suppose that a, b and c are integers with gcd(a, b) = 1. If a (bc), then a c. (b) If p is a prime, and if a and b are integers such that p (ab), then either p a or p b. Proof: (a) 1 gcd(a, b) = 1 Assumption 2 u, v Z, au + bv = 1 Euclidean Algorithm 3 auc + bcv = c multiplying c both sides of au + bv = 1 4 a auc, and a bc Definition of divisibility and assumption. 5 a (auc + bcv) Exercise (1.6)(c) or Proposition 1.4(c) 6 a c Substitution of auc + bcv = c By Euclidean Algorithm there exists integers u and v satisfying 1 = gcd(a, b) = au + bv. Multiplying c both sides yields acu + bcv = c. Since a auc, and a bc, it follows by Exercise (1.6)(c) (or Proposition 1.4(c) in the text) that a c. (b) We assume that p a to show that we must have p b. Since p is a prime and since p a, we have gcd(a, p) = 1. Therefore, by (a), gcd(p, a) = 1 and p (ab) implies that p b. (1.11) Let a and b be positive integers. (a) Suppose that there are integers u and v satisfying au + bv = 1. Prove that gcd(a, b) = 1. (b) Suppose that there are integers u and v satisfying au + bv = 6. Is it necessarily true that gcd(a, b) = 6? If not, give a specific counterexample, and describe in general all of the possible values of gcd(a, b)? 2

3 (c) Suppose that (u 1, v 1 ) and (u 2, v 2 ) are two solutions in integers to the equation au + bv = 1. Prove that a (v 2 v 1 ) and b (u 1 u 2 ). (d) More generally, let g = gcd(a, b) and let (u 0, v 0 ) be a solution to au + bv = g. Prove that every other solution has the form u = u 0 + kb/g and v = v 0 ka/g, for some integer k. Proofs and Solutions: (a) Let g = gcd(a, b). We want to prove that g = 1. Then by the definition of gcd(a, b), we have g 1. It suffices to show that g 1. By assumption, there exist integers u and v such that au + bv = 1. By the definition of gcd, g a and g b, and so g au and g bv. By Exercise (1.6)(c) (or Proposition 1.4(c) in the text), g (au + bv). Since au + bv = 1, g 1, and so g 1. Combining g 1 and g 1, we conclude that g = 1. (b) Let a = 3 and b = 4. Then gcd(3, 4) = 1. However, for u = 6 and b = 6, we have au + bv = 3( 6) + 4(6) = 6. In general, if gcd(a, b) = g and if for some integers u and v with au + bv = h, then by By Exercise (1.6)(c) (or Proposition 1.4(c) in the text), we must have g h. Therefore, if for some integers u and v with au + bv = h, then gcd(a, b) must be a positive factor of h. (c) 1 au 1 + bv 1 = 1 and au 2 + bv 2 = 1 Assumption 2 gcd(a, b) = 1, Exercise (1.11)(a) 3 a(u 1 u 2 ) + b(v 1 v 2 ) = 0 subtraction in Statement 2 4 a(u 1 u 2 ) = b(v 2 v 1 ) Algebraic manipulation in Statement 3 5 gcd(a, b) = 1 Statement 2. 6 a (v 1 v 2 ) and b (u 2 u 1 ) Additional Proposition (a) Another way to present the reasoning: Since au 1 + bv 1 = 1 and au 2 + bv 2 = 1, subtracting one equation from the other yields a(u 1 u 2 ) + b(v 1 v 2 ) = 0, or a(u 1 u 2 ) = b(v 2 v 1 ). Since au 1 +bv 1 = 1, it follows by Exercise (1.11)(a) (just shown) that gcd(a, b) = 1. Now by Additional Property (a) with c = (v 2 v 1 ), we have a (v 2 v 1 ). Similarly (by switching a and b), b (u 1 u 2 ), which is the same to say b (u 2 u 1 ). (d) Let g = gcd(a, b) and let (u 0, v 0 ) be a solution to au + bv = g, and let (u, v) denote the generic solution. Since au + bv = g, it follows from (a/g)u + b/g)v = 1 and Exercise (1.11)(a) that gcd(a/g, b/g) = 1. Therefore, both (u 0, v 0 ) and (u, v) are solutions of (a/g)u + (b/g)v = 1. By the conclusion of Exercise (1.11)(c), we have (b/g) (u u 0 ). By the definition of divisibility, there exists some integer k, such that u u 0 = k(b/g), or u = u 0 + k(b/g). Since (u 0, v 0 ) and (u, v) are solutions to au + bv = g, we have au 0 + bv 0 = g, and au + bv = g. Substitute u = u 0 + k(b/g) into au + bv = g to get au 0 + akb/g + bv = au 0 + bv 0. Cancelling au 0 both sides, we have akb/g + bv = bv 0, or bv = bv 0 bak/g,. Hence v = v 0 ka/g. 3

4 (1.14) Let m 1 be an integer and suppose that a 1 a 2 (mod m) and b 1 b 2 (mod m). Prove that a 1 ± b 1 a 2 ± b 2 (mod m) and a 1 b 1 a 2 b 2 (mod m). (Remark: These Properties suggest that for addition, subtraction and multiplication, doing these operations modulo m can be done just like those for integers. ) Proof: 1 a 1 a 2 (mod m) and b 1 b 2 (mod m) Assumption 2 u 1 Z, a 1 = a 2 + u 1 m Definition of mod m congruence 3 u 2 Z, b 1 = b 2 + u 2 m Definition of mod m congruence 4 (a 1 ± b 1 ) = (a 2 ± b 2 ) + (u 1 ± u 2 )m Combining Statements 2 and 3 5 (a 1 ± b 1 ) = (a 2 ± b 2 ) + um Set u = u 1 ± u 2 6 (a 1 ± b 1 ) (a 2 ± b 2 ) (mod m) Definition of mod m congruence 1 a 1 a 2 (mod m) and b 1 b 2 (mod m) Assumption 2 u 1 Z, a 1 = a 2 + u 1 m Definition of mod m congruence 3 u 2 Z, b 1 = b 2 + u 2 m Definition of mod m congruence 4 (a 1 b 1 ) = (a 2 b 2 ) + (a 2 u 2 + b 2 u 1 + u 1 u 2 m)m Combining Statements 2 and 3 5 a 1 b 1 = a 2 b 2 + um Set u = a 2 u 2 + b 2 u 1 + u 1 u 2 m 6 a 1 b 1 a 2 b 2 (mod m) Definition of mod m congruence (1.17) Find all values of x between 0 and m 1 that are solutions of the following congruences. (a) x (mod 37). (b) x (mod 51). (c) x 2 3 (mod 11). (d) x 2 2 (mod 13). (e) x 2 1 (mod 8). (f) x 3 x 2 + 2x 2 0 (mod 11). (g) x 1 (mod 5) and x 2 (mod 7), (Find all solutions mod 35). Solution: (a) x (mod 37). (b) x (mod 51). (c) x (mod 11). Since (mod 11), we have x 2 3 x (x 5)(x + 5) 0 (mod 11). 4

5 Since p = 11 is a prime, and by the property that if p is a prime, then p ab implies p a or p b, we have 11 (x 5) or 11 (x + 5), and so x 5 or x 5 6 (mod 11). (d) Since , , 3 2 9, , , (mod 13), x 2 2 (mod 13) has no solution. (e) From (x 1)(x + 1) 1 (mod 8), we have 8 (x 1)(x + 1). By Exercise (1.6)(a) (with a = 2 and b = 8, c = (x 1)(x + 1)), we have 2 (x 1)(x + 1). Since 2 is a prime, we have x 1 (mod 2), or x 1 (mod 2). Therefore in Z 8, we have x 1, 3, 5, 7. (f) Factoring x 3 x 2 + 2x 2 = x 2 (x 1) + 2(x 1) = (x 2 + 2)(x 1) (mod 11). Thus 11 (x 1), whence x 1 (mod 11), or 11 (x 2 + 2). Since x x 2 9 (x 3)(x + 2), and since 11 is a prime, 11 (x 3)(x + 3) implies that x 3 or x 3 8 (mod 11). Therefore, the solutions are x 1, 3, 8, (mod 11). (g) Since x 1 (mod 5), we have 5 (x 1). Among all integers between 0 and 35, we have x {1, 6, 11, 16, 21, 26, 31}. Since x 2 (mod 7), we have 7 (x 2). From the list {1, 6, 11, 16, 21, 26, 31}, we single out x = 16. Therefore, the solution is x = 16. 5

Applications of Fermat s Little Theorem and Congruences

Applications of Fermat s Little Theorem and Congruences Definition: Let m be a positive integer. Then integers a and b are congruent modulo m, denoted by a b mod m, if m (a b). Example: 3 1 mod 2, 6 4