1.6 Congruence Modulo m

Transcription

1 1.6 Congruence Modulo m Let a, b 2 N and p be a prime. Prove for all natural numbers n 1, if p n (ab) and p - a, then p n b. 6. In the proof of Theorem it was stated that if n is a prime number p, thenn = p is the only prime factorization of n. Explain why this is true. 7. In our proof of Theorem we stated that A similar argument shows that p r >q s is also impossible. Under the assumptions used in our proof of the theorem, present this similar argument. Exercise Notes: For Exercise 5, use induction on n. 1.6 Congruence Modulo m Karl Friedrich Gauss ( ) has been called the Prince of Mathematicians for his many contributions to pure and applied mathematics. One of Gauss s most important contributions to number theory was the introduction of an equivalence relation on the integers called congruence modulo m, wherem 1 is an integer. We will investigate Gauss s congruence relation. We will also show that the operations of addition, subtraction, and multiplication preserve Gauss s relation (see Theorem 1.6.2). Definition (Congruence modulo m). Let m 1 be an integer. For any integers a and b, we define a b (mod m) if and only if m (a b). For example, 10 2 (mod 4) because 4 (10 2), 5 3 (mod 4) since 4 ( 5 3), and 24 0 (mod 4) because 4 (24 0). Remark. When a b (mod m) we say that a is congruent to b modulo m. a 6 b (mod m), when we wish to say that a is not congruent to b modulo m. We also write The notation a b (mod m) given in Definition is just a statement about divisibility and is used mainly to simplify reasoning about the divisibility concept. When m 1 is an integer and a, b are integers, one can easily verify that the following are all equivalent: 1. a b (mod m), 2. m (a b), 3. a b = km for some k 2 Z, 4. a = b + km for some k 2 Z. Example 1. Let n be any integer. The division algorithm (see Theorem 1.4.1) implies that there are integers k and r such that n =4k + r and 0 apple r<4. Thus, we have n =4k + 0, n =4k + 1, n =4k+2 or n =4k+3. We conclude for every integer n that either n 0 (mod 4) or n 1 (mod 4) or n 2 (mod 4) or n 3 (mod 4) Fundamental Properties In this section we shall establish a series of theorems that will allow us to develop a so-called congruence algebra. The following theorem is a fundamental result showing that the congruence modulo m relation is preserved under the operations of addition, subtraction, and multiplication.

2 48 Chapter 1. The Integers Theorem Let m 1. (a + c) (b + d)(mod m), 2. (a c) (b d)(mod m), 3. ac bd (mod m). 1. Ifa b (mod m) and c d (mod m), then Proof. Assume that a b (mod m) and c d (mod m), that is for some integers i and j. a b = mi (1.8) c d = mj (1.9) 1. To prove (a + c) (b + d)(mod m), we add the corresponding sides of equations (1.8) and (1.9) to obtain (a b)+(c d) = mi + mj. So, (a + c) (b + d) =m(i + j). Therefore, (a + c) (b + d)(mod m). 2. To prove (a c) (b d)(mod m), we subtract the corresponding sides of equations (1.8) and (1.9) to obtain (a b) (c d) =mi mj. So, (a c) (b d) =m(i j). Therefore, (a c) (b d)(mod m). 3. To prove ac bd (mod m), we first multiply (both sides of) equation (1.8) by c and multiply equation (1.9) by b obtaining ac bc = mic (1.10) bc bd = mjb. (1.11) Adding equations (1.10) and (1.11) gives ac bd = m(ic + jb). Thus, ac bd (mod m). This completes the proof. Corollary Let m 1. Ifa b (mod m), then for all integers c we have 1. a + c b + c (mod m), 2. a c b c (mod m), 3. ac bc (mod m). Proof. Assume a b (mod m). Since c c (mod m), Theorem implies 1 3. Theorem Let m 1. Ifa b (mod m), then a 2 b 2 (mod m). Proof. Assume a b (mod m). From Theorem 1.6.2(3) we conclude that aa bb (mod m). Therefore, a 2 b 2 (mod m). Using Theorem 1.6.2(3) and mathematical induction on k, one can prove (see Exercise 13) the following theorem. Theorem Let m 1. For every integer k 1, ifa b (mod m) then a k b k (mod m). Theorem Let Z be the set of integers. Let m 1 be an integer. Let the relation on Z be defined by a b if and only if a b (mod m) (1.12) whenever a, b 2 Z. Then is an equivalence relation.

3 1.6 Congruence Modulo m 49 Proof. We shall prove that the relation on Z defined by (1.12) is an equivalence relation; that is, we prove that is reflexive, symmetric, and transitive. Proof that is reflexive. Let x 2 Z be arbitrary. We prove that x x, that is, we prove that x x = km for some k 2 Z. Clearly, for k = 0 we get that x x =0m. Therefore, x x. Proof that is symmetric. Let x, y 2 Z be arbitrary. Assume x y, that is, assume x y = km for some k 2 Z. We prove that y x. Since x y = km, it follows (by multiplying both sides of the equation by 1) that y x =( k)m where k is an integer. Therefore, y x. Proof that is transitive. Let x, y, z 2 Z be arbitrary. Assume x y and y z, that is, assume x y = im and y z = jm for some i, j 2 Z. We prove that x z. Since x y = im and y z = jm, it follows (by adding these latter two equations) that x z = im + jm =(i + j)m where i + j is an integer. Therefore, x z. Theorems 1.6.2, 1.6.5, and allow us to derive congruence relations by using congruence algebra. Example 2. Let m>1 be a whole number. Suppose that a 4(mod m), b 10 (mod m), and c 3(mod m). Show that 3a 2 2b c 3 +4m 1(mod m). Solution. We are given that a 4(mod m), b 10 (mod m), and c 3(mod m). (1.13) We will show that 3a 2 2b c 3 +4m 1(mod m) as follows: 3a 2 2b c 3 +4m 3a 2 2b c 3 (mod m) because 4m 0(mod m) (mod m) by (1.13) 1(mod m) because = 1. Therefore, 3a 2 +2b + c 3 +4m 1(mod m). s Congruence Classes Let m 1 be an integer and let be the equivalence relation on Z defined by (1.12). Recalling Definition 0.5.2, for any a 2 Z the set [a] is the equivalence class of a. Since the relation is so closely connected with m, we shall also use [a] m to also denote [a].thus, [a] m = {x 2 Z : x a (mod m)}. Observe that [a] m = {a + mk : k 2 Z} because x 2 [a] m i x a (mod m) i x = a + mk for some k 2 Z. We shall call [a] m the congruence class of a (mod m). Applying Theorem 0.5.4, we see that a b (mod m) if and only if [a] m =[b] m (1.14) for all a, b 2 Z. We conclude from Theorem and Definition that Z/ = {[a] m : a 2 Z} is the partition of the set of integers Z induced by, where is defined by (1.12). We shall now use the notation Z m to denote this partition Z/. Lemma Let m>1 be an integer. For all a 2 Z and b 2 Z, we have that

4 50 Chapter 1. The Integers 1. [a] m =[b] m if and only if a b (mod m); 2. a 2 [b] m if and only if a b (mod m). Proof. Since congruence modulo m is an equivalence relation on Z, items 1 and 2 follow from Lemma and Corollary 0.5.5, respectively. We now begin to determine the number of distinct equivalence classes that the congruence modulo m relation produces. Theorem Let m 1 be an integer. For any integer a there is exactly one integer r in the list 0, 1,...,m 1 such that a r (mod m). Proof. Let m 1 be an integer and let a be any integer. We will prove that there is a unique integer r in the list 0, 1,...,m 1 such that a r (mod m). Existence: By Theorem 1.4.1, there exists integers q and r such that a = qm + r and 0 apple r<m. Thus, (1) a r = qm and therefore, a r (mod m) and 0 apple r<m. Uniqueness: Let r be as in the existence part of our proof. Let r 0 be any integer also satisfying 0 apple r 0 <mand a r 0 (mod m). We shall prove that r = r 0. Since a r 0 (mod m), there is an integer q 0 such that a r 0 = q 0 m and thus, a = q 0 m + r 0. From (1) we also see that a = qm + r. Since 0 apple r, r 0 <m, Theorem now implies that r = r 0. Problem 3. Show that every perfect square is congruent to 0 or 1 (mod 4). Solution. Let n be a perfect square. So n = k 2 for some integer k. Theorem asserts that k r (mod 4) for some r in the list 0, 1, 2, 3. Thus, either k 0 (mod 4), k 1 (mod 4), k 2 (mod 4), or k 3 (mod 4). (1.15) Thus, (1.15) and Theorem imply that either k (mod 4), k (mod 4), k (mod 4), or k (mod 4). Since (mod 4) and (mod 4), we conclude that in every case either k 2 0 (mod 4) or k 2 1 (mod 4). Therefore, because n = k 2, we conclude that n 0 (mod 4) or n 1 (mod 4). s Given an integer m 1, Theorem asserts that every integer is congruent (mod m) to exactly one of the numbers in the list 0, 1,...,m 1. For this reason, we shall call this list a complete residue system (mod m). Corollary Let m 1 be an integer. For any integer k there is exactly one integer r in the list 0, 1,...,m 1 such that [k] m =[r] m. Proof. Let k be an integer. By Theorem there is exactly one integer r in the list 0, 1,...,m 1 such that k r (mod m). Lemma implies that [k] m =[r] m. Thus, there is exactly one such r where [k] m =[r] m. Corollary Let m 1 be an integer. For integers k 6= r in the list 0, 1,...,m 1, we have that [k] m 6=[r] m and hence, [k] m and [r] m are disjoint. Proof. Let k 6= r both be in the list 0, 1,...,m 1. (1.16) Since k and r are two di erent integers in the list (1.16), Corollary implies [k] m 6=[r] m. Theorem now implies that [k] m \ [r] m = ;.

5 1.6 Congruence Modulo m 51 Let m 1 be an integer. Corollaries and assert that there are exactly m many distinct congruence classes (mod m) for a given integer m 1. We conclude that Z m = {[0] m, [1] m, [2] m,...,[m 1] m }. Example 4. Let Z be the set of integers. Consider the equivalence relation on Z defined by a b if and only if a b (mod 5) and let Z 5 be the partition of Z induced by (see Figure 1.3). Note that: 1. For every a 2 Z, [a] 5 = {a +5k : k 2 Z}. 2. Corollaries and imply that Z = [0] 5 [ [1] 5 [ [2] 5 [ [3] 5 [ [4] 5 and the sets [0] 5, [1] 5, [2] 5, [3] 5, [4] 5 are all mutually disjoint. Thus, Z 5 = {[0] 5, [1] 5, [2] 5, [3] 5, [4] 5 }. 3. The set Z 5 is a partition of Z as represented by the diagram in Figure 1.3. Z = " " " " " [0] 5 [1] 5 [2] 5 [3] 5 [4] 5 Figure 1.3: Partition of Z induced by. Remark Let m 1 be an integer. When the integer m is understood, we shall drop the subscripts; that is, we will just use [a] instead of [a] m and shall write Z m = {[0], [1], [2],...,[m 1]}. Many texts just let Z m = {0, 1, 2,...,m 1}, which clearly simplifies the notation, and at times we will do the same. Exercises Which of the following are valid congruences? 5 13 (mod 4), 18 2 (mod 10) and 4 14 (mod 6). 2. By dividing 97 by 7, find a so that a 97 (mod 7) and 0 apple a<7. Now find b so that b 97 (mod 7) and 0 apple b<7. 3. Let m 1 be an integer. Show that if n 1(mod m), then n 2 + n 2(mod m) for every integer n. 4. Find a counterexample showing that the converse of Theorem does not hold. 5. Prove that for every odd integer k we have k 2 1 (mod 8). 6. Prove that (m 1) 2 1(mod m) for every whole number m>1. 7. Let m 1 be an integer and let a, b, k be integers where k 1. Prove that if a b (mod m) and k m, thena b (mod k).

6 52 Chapter 1. The Integers 8. Let m 1 be an integer and let a, b, k be integers where k 1. Prove that if a b (mod m), then ak bk (mod mk). 9. Let m 1 be an integer and let a, b, k be integers. Suppose that gcd(k, m) = 1. Prove that if ka kb (mod m), then a b (mod m). 10. Show that every perfect square is congruent to 0, 1 or 4 (mod 8). 11. Show that for all integers n, n 3 n (mod 3). 12. For every pair of integers x and y, show that x 2 + y 2 is congruent to 0, 1 or 2 (mod 4). 13. Prove Theorem using mathematical induction. 14. Let m = 6 and define the equivalence relation on Z: a b if and only if a b (mod 6). Let Z 6 = {[0], [1], [2], [3], [4], [5]} be the partition induced by, where[a] =[a] 6 for any a 2 Z. Determine which of the following are true. Justify your answers. (a) [2] = [3]. (b) [2] = [ 4]. (c) 5 2 [8]. (d) 5 2 [11]. (e) [3 + 4] = [1]. (f) [3 4] = [0]. 15. Let a be an integer and let m>1 be a whole number. Prove that [a] m = [0] m if and only if m a. 16. Let a and b be integers and let m>1 be a whole number. Prove that [a] m =[b] m if and only if a b = mi for some integer i. 17. Let Z 6 = {[0], [1], [2], [3], [4], [5]}. Consider the proposed function f : Z 6! Z defined by f([a]) = a for all [a] 2 Z 6. Show that f is not well-defined. Exercise Notes: For Exercise 3, use Theorems and For Exercise 5, see Exercise 4 on page 41. For Exercise 11, we know by Theorem that there are three cases to consider: n 0 (mod 3), n 1 (mod 3), or n 2 (mod 3). For Exercise 12, use Example 3 and congruence algebra. For Exercise 14, review Lemma For Exercise 17, review Lemma and Problem 4 on page Modular Arithmetic Modular arithmetic is a system of arithmetic that is based on the congruence modulo m relation. Carl Friedrich Gauss first introduced modular arithmetic in 1801 and it has become an important tool in number theory. In the next definition, we will first recall the rules of integer arithmetic that we all learned in elementary school. These rules are referred to as the ring axioms. We shall soon see that modular arithmetic also satisfies these same rules. Definition Let Z be the set of integers with the usual binary operations and +. We say that (Z, +, ) is a ring because it satisfies the following seven Ring Axioms: 1. a + b = b + a for all a, b 2 Z. 2. (a + b)+c = a +(b + c) for all a, b, c 2 Z.