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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) Claim 1 2 3 4 5 Statement a|b & b|c ∃u, v ∈ Z, b = au & c = bv c = bv = auv c = aw, w ∈ Z a|c Reason Assumption Definition of a|b & b|c substituting b = au into c = bv let w = uv. 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 6= 0 (and so b 6= 0). Claim 1 2 3 4 5 6 7 Statement a|b & b|a ∃u, v ∈ Z, b = au & a = bv a = bv = auv a(1 − uv) = 0 1 = uv v = ±1 a = ±1 Reason Assumption Definition of a|b & b|a substituting b = au into a = bv. Algebraic manipulation. Since a 6= 0, we can cancel a both sides of a = a(1 − uv). u, v are integers and uv = 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 6= 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 Claim 1 2 3 4 5 Statement a|b & a|c ∃u, v ∈ Z, b = au & c = av b ± c = au ± av = a(u ± v) b ± c = am a|(b ± c) Reason Assumption Definition of a|b & a|c Algebra/Remove common factor a. Let m = u ± v, which is also an integer. 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) Claim 1 2 3 4 5 6 Statement gcd(a, b) = 1 ∃u, v ∈ Z, au + bv = 1 auc + bcv = c a|auc, and a|bc a|(auc + bcv) a|c Reason Assumption Euclidean Algorithm multiplying c both sides of au + bv = 1 Definition of divisibility and assumption. Exercise (1.6)(c) or Proposition 1.4(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 6 |a to show that we must have p|b. Since p is a prime and since p 6 |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 (c) Suppose that (u1 , v1 ) and (u2 , v2 ) are two solutions in integers to the equation au + bv = 1. Prove that a|(v2 − v1 ) and b|(u1 − u2 ). (d) More generally, let g = gcd(a, b) and let (u0 , v0 ) be a solution to au + bv = g. Prove that every other solution has the form u = u0 + kb/g and v = v0 − 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) Claim 1 2 3 4 5 6 Statement au1 + bv1 = 1 and au2 + bv2 = 1 gcd(a, b) = 1, a(u1 − u2 ) + b(v1 − v2 ) = 0 a(u1 − u2 ) = b(v2 − v1 ) gcd(a, b) = 1 a|(v1 − v2 ) and b|(u2 − u1 ) Reason Assumption Exercise (1.11)(a) subtraction in Statement 2 Algebraic manipulation in Statement 3 Statement 2. Additional Proposition (a) Another way to present the reasoning: Since au1 + bv1 = 1 and au2 + bv2 = 1, subtracting one equation from the other yields a(u1 − u2 ) + b(v1 − v2 ) = 0, or a(u1 − u2 ) = b(v2 − v1 ). Since au1 + bv1 = 1, it follows by Exercise (1.11)(a) (just shown) that gcd(a, b) = 1. Now by Additional Property (a) with c = (v2 − v1 ), we have a|(v2 − v1 ). Similarly (by switching a and b), b|(u1 − u2 ), which is the same to say b|(u2 − u1 ). (d) Let g = gcd(a, b) and let (u0 , v0 ) 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 (u0 , v0 ) 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 − u0 ). By the definition of divisibility, there exists some integer k, such that u − u0 = k(b/g), or u = u0 + k(b/g). Since (u0 , v0 ) and (u, v) are solutions to au + bv = g, we have au0 + bv0 = g, and au + bv = g. Substitute u = u0 + k(b/g) into au + bv = g to get au0 + akb/g + bv = au0 + bv0 . Cancelling au0 both sides, we have akb/g + bv = bv0 , or bv = bv0 − bak/g,. Hence v = v0 − ka/g. 3 (1.14) Let m ≥ 1 be an integer and suppose that a1 ≡ a2 (mod m) and b1 ≡ b2 (mod m). Prove that a1 ± b1 ≡ a2 ± b2 (mod m) and a1 b1 ≡ a2 b2 (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: Claim 1 2 3 4 5 6 Claim 1 2 3 4 5 6 Statement a1 ≡ a2 (mod m) and b1 ≡ b2 (mod m) ∃u1 ∈ Z, a1 = a2 + u1 m ∃u2 ∈ Z, b1 = b2 + u2 m (a1 ± b1 ) = (a2 ± b2 ) + (u1 ± u2 )m (a1 ± b1 ) = (a2 ± b2 ) + um (a1 ± b1 ) ≡ (a2 ± b2 ) (mod m) Statement a1 ≡ a2 (mod m) and b1 ≡ b2 (mod m) ∃u1 ∈ Z, a1 = a2 + u1 m ∃u2 ∈ Z, b1 = b2 + u2 m (a1 b1 ) = (a2 b2 ) + (a2 u2 + b2 u1 + u1 u2 m)m a1 b1 = a2 b2 + um a1 b1 ≡ a2 b2 (mod m) Reason Assumption Definition of mod m congruence Definition of mod m congruence Combining Statements 2 and 3 Set u = u1 ± u2 Definition of mod m congruence Reason Assumption Definition of mod m congruence Definition of mod m congruence Combining Statements 2 and 3 Set u = a2 u2 + b2 u1 + u1 u2 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 + 17 ≡ 23 (mod 37). (b) x + 42 ≡ 19 (mod 51). (c) x2 ≡ 3 (mod 11). (d) x2 ≡ 2 (mod 13). (e) x2 ≡ 1 (mod 8). (f) x3 − x2 + 2x − 2 ≡ 0 (mod 11). (g) x ≡ 1 (mod 5) and x ≡ 2 (mod 7), (Find all solutions mod 35). Solution: (a) x ≡ 23 − 17 ≡ 6 (mod 37). (b) x ≡ 19 − 42 ≡ 19 + 51 − 42 ≡ 28 (mod 51). (c) x2 − 3 ≡ 0 (mod 11). Since 52 ≡ 3 (mod 11), we have x2 − 3 ≡ x2 − 52 ≡ (x − 5)(x + 5) ≡ 0 (mod 11). 4 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 12 ≡ 122 ≡ 1, 22 ≡ 112 ≡ 4, 32 ≡ 9, 42 ≡ 92 ≡ 3, 52 ≡ 82 ≡ 12, 62 ≡ 72 ≡ 10 (mod 13), x2 ≡ 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 Z8 , we have x ≡ 1, 3, 5, 7. (f) Factoring x3 − x2 + 2x − 2 = x2 (x − 1) + 2(x − 1) = (x2 + 2)(x − 1) (mod 11). Thus 11|(x − 1), whence x ≡ 1 (mod 11), or 11|(x2 + 2). Since x2 + 2 ≡ x2 − 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