Given two integers a, b with a 0. We say that a divides b, written

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1 1 Divisibility Given two integers a, b with a 0. We say that a divides b, written a b, if there exists an integer q such that b = qa. When this is true, we say that a is a factor (or divisor) of b, and b is a multiple of a. If a is not a factor of b, we write a b. Obviously, any integer n has divisors ±1 and ±n, called the trivial divisors of n. If a is a divisor of n, so is a. Positive divisors of n other than the trivial divisors are called nontrivial divisors. Note that every integer is a divisor of 0. A positive integer p ( 1) is called a prime if it has no nontrivial divisors, i.e., its positive divisors are only the trivial divisors 1 and p. A positive integer is called composite if it is not a prime. The first few primes are listed as follows: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59. Proposition 1.1. Every composite number n has a prime factor p n. 1

2 Proof. Since n is composite, there are primes p and q such that n = pqk, where k P. Then, either p q or q p. Assume p q. We have p 2 pqk = n. Thus p n. Example has the prime factor 2 6; 9 has the prime factor 3 = 9; 35 has the prime factor Is 143 a prime? We find 143 < 12. For i = 2, 3, 5, 7, 11, check whether i divides 143. We find So 143 is a composite number. Is 157 a prime? Since 157 < 13 and for each i = 2, 3, 5, 7, 11, i is not a factor of 157, we see that 157 has no prime factor less or equal to 157. So 157 is not a composite, i.e., 157 is a prime. Proposition 1.2. Let a, b, c be nonzero integers. (a) If a b and b a, then a = ±b. (b) If a b and b c, then a c. (c) If a b and a c, then a (bx + cy) for x, y Z. Proof. (a) Let b = q 1 a and a = q 2 b for some integers q 1 and q 2. Then b = q 1 q 2 b. 2

3 Dividing both sides by b, we have q 1 q 2 = 1. It follows that q 1 = q 2 = ±1. Thus b = ±a. (b) Let b = q 1 a and c = q 2 b for some integers q 1 and q 2. Then c = q 1 q 2 a, that is, a c. (c) Let b = q 1 a and c = q 2 a for q 1, q 2 Z. Then for x, y Z, bx + cy = (q 1 x + q 2 y)a, that is, a (bx + cy). Theorem 1.3. There are infinitely many prime numbers. Proof. Suppose there are finitely many primes, say, Then the integer p 1, p 2,..., p k. a = p 1 p 2 p k + 1 is not divisible by any of the primes p 1, p 2,..., p k because the remainders of a dividing by p 1, p 2,..., p k respectively are always 1. This means that a has no prime factors. By definition of primes, the integer a must be a prime, and this prime is larger than all primes p 1, p 2,..., p k, a contradiction. Theorem 1.4 (Division Algorithm). For any integers a and b, where a > 0, there are unique integers q and r such that b = qa + r, 0 r < a. 3

4 Proof. Consider the set S = {b ta 0 t Z}. Obviously, S is nonempty and is bounded below. Then S has the unique minimum integer r, i.e., there is an unique integer q such that b qa = r. We claim that r < a. Suppose r a, then b (q + 1)a = r a 0 shows that r a is an element of S. This is contrary to that r is the minimum element of S. Example 1.2. For integers a = 21 and b = 358, we have 368 = , q = 18, r = Greatest Common Divisor For integers a and b, not simultaneously 0, a common divisor of a and b is an integer c such that c a and c b. Clearly, there are finitely many common divisors for a and b; the very greatest one is called the greatest common divisor of a and b, and is denoted by gcd(a, b). For convenience, we assume gcd(0, 0) = 0. Two integers a and b are called coprime (or relatively prime) if gcd(a, b) = 1. Theorem 2.1. Let d be the greatest common divisor of integers a and b, i.e., d = gcd(a, b). Then there exist integers x and y such that d = ax + by. Proof. It is obviously true when a = b = 0. Assume that a and b are not simultaneously zero. We consider the set S = {au + bv u, v Z} 4

5 and the set S + = S P. For u = a, v = b, we have au+bv = a 2 +b 2 > 0. Obviously, S is nonempty and bounded below. Let s be the smallest integer in S +. Write s = au 0 + bv 0 for some u 0, v 0 Z. We claim that d = s. Clearly, d divides every integer in S because d a and d b. In particular, d s. We then have d s. To show s d, we claim that s divides every integer in S. In fact, for any au + bv S with u, v Z, let au + bv = qs + r, 0 r < s. Then r = a(u qv 0 ) + b(v qu 0 ) S. If r was positive, then r S + and s could not be the smallest integer in S +. So r = 0. Thus s (au + bv). In particular, taking (u, v) = (0, 1) and (u, v) = (1, 0), we see that s a and s b. So s is a positive common divisor of a and b. By definition of gcd, s d. Theorem 2.2. Let a, b, d P. Then d = gcd(a, b) if and only if (1) d a, d b; (2) If c a, c b, then c d. Proof. : By definition, gcd is a common divisor of a and b, so we have (1). By Theorem 2.1, there exist integers x, y such that d = ax + by. 5

6 If c a, c b, then c (ax + by), i.e., c d. : (1) indicates that d is a common divisor of a and b. Let c be positive common divisor c of a and b. We have c d by (2). Thus c d. So d is the greatest among all common divisors of a and b. Theorem 2.3. For integers a, b, q, and r, if then b = qa + r, gcd(a, b) = gcd(a, r). Proof. Let d 1 = gcd(a, b), d 2 = gcd(a, r). Since r = b qa, we have d 1 r. This means that d 1 is a common divisor of a and r. Thus d 1 d 2. Since b = qa + r, we have d 2 b. This means that d 2 is a common divisor of a and b. Hence, d 2 d 1. Therefore d 1 = d 2. The above proposition gives rise to a simple constructive method to calculate gcd by repeating the Division Algorithm. Example 2.1. Find gcd(297, 3627) = , 297 = , 63 = , 45 = , 18 = 2 9; gcd(297, 3627) = gcd(63, 297) = gcd(45, 63) = gcd(18, 45) = gcd(9, 18) = 9. 6

7 The procedure to calculate gcd(297, 3627) applies to any pair of positive integers. Let a be a positive integer and b a nonnegative integer. Repeating the Division Algorithm will produce finite sequences of nonnegative integers q i and r i such that b = q 0 a + r 0, 0 r 0 < a, a = q 1 r 0 + r 1, 0 r 1 < r 0, r 0 = q 2 r 1 + r 2, 0 r 2 < r 1, r 1 = q 3 r 2 + r 3,. 0 r 3 < r 2, r k 2 = q k r k 1 + r k, 0 r k < r k 1, r k 1 = q k+1 r k + r k+1, r k+1 = 0. Notice that the sequence {r i } is strictly decreasing; it ends eventually to 0 at some step, say, r k+1 = 0 for the first time. Then r i 0 for all 0 i k. Reverse the sequence {r i } k i=0 and make substitutions as follows: gcd(a, b) = r k, r k = r k 2 q k r k 1, r k 1 = r k 3 q k 1 r k 2,. r 1 = a q 1 r 0, r 0 = b q 0 a. We see that gcd(a, b) can be expressed as an integral linear combination of a and b. This procedure is known as the Euclidean Algorithm. 7

8 Example 2.2. The greatest common divisor of 297 and 3627, written as an integral linear combination of 297 and 3627, can be obtained as follows: gcd(297, 3627) = = 45 2(63 45) = = 3( ) 2 63 = = ( ) = Proposition 2.4. Let a, b, c Z. If a bc, gcd(a, b) = 1, then a c. Proof. By the Euclidean Algorithm, there are integers x, y Z such that ax + by = 1. Then c = 1 c = (ax + by)c = acx + bcy. Since a ac and a bc, thus a c. Theorem 2.5 (Unique Factorization). Every integer a 2 can be uniquely factorized into the form a = p e 1 1 pe 2 2 pe m m, where p 1, p 2,..., p m are distinct primes, e 1, e 2,..., e m are positive integers, and p 1 < p 2 < < p s. Proof. If a has only the trivial divisors, then a itself is a prime, and it obviously has unique factorization. If a has some nontrivial divisors, then a = bc 8

9 for some positive integers b and c other than 1 and a. Obviously, b < a and c < a. By induction, the positive integers b and c have factorizations into primes. Consequently, a has a factorization into primes. Let a = q f 1 1 qf 2 2 af n n be another factorization, where q 1, q 2,..., q n are distinct primes, f 1, f 2,..., f n are positive integers, and q 1 < q 2 < < q n. We claim that m = n, p i = q i, e i = f i for all 1 i m. Suppose p 1 < q 1. Then p 1 is distinct from the primes q 1, q 2,..., q n. It is clear that gcd(p 1, q i ) = 1, and so gcd(p 1, q f i i ) = 1 for all 1 i n. Note that p 1 q f 1 1 qf 2 2 af n n. Since gcd(p 1, q f 1 1 ) = 1, by Proposition 2.4, we have p 1 q f 2 2 af n n. Since gcd(p 1, q f 2 2 ) = 1, again by Proposition 2.4, we have p 1 q f 2 3 af n n. Repeating the argument, we finally obtain that p 1 q f n n, which is contrary to gcd(p 1, q f n n ) = 1. We thus conclude that p 1 q 1. Similarly, p 1 q 1. Hence p 1 = q 1. Next we claim that e 1 = f 1. Suppose e 1 < f 1. Then p e 2 2 pe m m = p f 1 e 1 1 q f 2 2 qf n n. This implies that p 1 p e 2 2 pe m m. If m = 1, it would imply that p 1 divides 1, which is impossible because p 1 is a prime. If m 2, note that gcd(p 1, p i ) = 1 and so gcd(p 1, p e i i ) = 1 for all 2 i m; by the same token of applying Proposition 2.4 repeatedly, we have p 1 p e m m, which is contrary to gcd(p 1, p e m m ) = 9

10 1. This means that we must have e 1 f 1. Similarly, e 1 f 1. Hence e 1 = f 1. Now we have obtained p e 2 2 pe m m = q f 2 2 qf n n. If m < n, then by induction we have p 1 = q 1,..., p m = q m and e 1 = f 1,..., e m = f m. Thus 1 = q f m+1 m+1 qf n n ; this is impossible because q m+1,..., q n are primes. So m n. Similarly, m n. Hence we have m = n. By induction on m = n, we obtain that e 2 = f 2,..., e m = f m. 3 Least Common Multiple For two integers a and b, a positive integer m is called a common multiple of a and b if a m and b m. The smallest integer among the common multiples of a and b is called the least common multiple of a and b, and is denoted by lcm(a, b). Proposition 3.1. For any nonnegative integers a and b, ab = gcd(a, b) lcm(a, b). Proof. Let a = p e 1 1 pe 2 2 pe n n and b = p f 1 1 pf 2 2 pf n n, where p 1 < p 2 < < p n, e i and f i are nonnegative integers, 1 i n. Then by the Unique Factorization Theorem, gcd(a, b) = p g 1 1 pg 2 2 pg n n, lcm(a, b) = p h 1 1 ph 2 2 ph n n, 10

11 where g i = min(e i, f i ) and h i = max(e i, f i ) for all 1 i n. It is clear that ab = p g 1+h 1 1 p g 2+h 2 2 p g n+h n n, and g i + h i = e i + f i for all 1 i n. Proposition 3.2. An integer m is the least common multiple of integers a and b if and only if (a) a m, b m; (b) if a n, b n, then m n. Proof. By the Unique Factorization Theorem. Example 3.1. Find all integer solutions for the linear equation 25x + 65y = 10. Solution: Applying the Division Algorithm, 65 = , 25 = , 15 = Applying the Euclidean Algorithm, gcd(25, 65) = = 15 (25 15) = = ( ) =

12 Thus the integer solutions are given by { x = 2( 5) + 13k = k y = 2 2 5k = 4 5k, k Z. Exercise 1. Let S c be the solution set of the linear Diophantine equation ax + by = c, i.e., S c = {(x, y) Z 2 ax + by = c}. Show that, if S c, then there is a bijection f : S 0 S c, where S 0 = {(x, y) Z 2 ax + by = 0}. Let (x, y) = (u, v) be a particular solution for ax + by = c. Then f can be given by f(x, y) = (u + x, v + y), (x, y) S 0. The function f has the inverse f 1 : S c S 0, given by f 1 (x, y) = (x u, y v), (x, y) S c. Theorem 3.3. Let a, b Z, not simultaneously zero, d = gcd(a, b). Then the linear Diophantine equation ax + by = c has an integer solution if and only if d c. Moreover, if (x, y) = (u, v) is a particular solution, then all solutions are given by { x = u + bk gcd(a,b) y = v ak gcd(a,b), k Z. 12

13 Proof. It is clear that (x, y) = (u, v) + 1 d (kb, ka) are integer solutions. We only need to show that all integer solutions for the equation ax + by = 0. Write the equation as ax = by. We have b ax and a by. This means that ax and by are both common multiples of a and b. Thus lcm(a, b) is a common divisor of ax and by. Let ax = klcm(a, b) for some k Z. Then { ax = k lcm(a, b) k Z. by = k lcm(a, b) Thus { x = k lcm(a,b) bk a = gcd(a,b) y = k lcm(a,b) b = ak gcd(a,b). 13

14 Example x + 668y = 888. Solution. Find gcd(168, 668) = 4 by the Division Algorithm By the Euclidean Algorithm, Dividing = = = = = 168 ( ) = ( 1) 668. = 222, we obtain a special solution (x 0, y 0 ) = 222(4, 1) = (888, 222) Solve 168x + 668y = 0. Dividing both sides by gcd(168, 668), 42x + 167y = 0 i.e. 42x = 167y. The general solutions for 168x + 668y = 0 are given by (x, y) = k(167, 42), k Z. The general solutions for 168x + 668y = 888 are given by (x, y) = (888, 222) + k(167, 42), k Z. { x = k i.e. y = k, k Z. 14

15 4 Modulo Integers Let n be a fixed positive integer. Two integers a and b are said to be congruent modulo n, written a b (mod n) and read a equals b modulo n, if n (b a). Note. For all k, l Z, a b (mod n) a + kn b + ln (mod n). Let a denote set of solutions of the equation x a (mod n); i.e., a = {a + kn k Z}. We call a the residue class of a modulo n. There are n residue classes of modulo n, i.e., 0, 1, 2,..., n 1. Example (mod 2), (mod 8), 8 10 (mod 9), 3 5 (mod 3), (mod 8), 8 18 (mod 9). Proposition 4.1. Let n be a fixed positive integer. If a 1 b 1 (mod n), a 2 b 2 (mod n), then a 1 + a 2 b 1 + b 2 (mod n), a 1 a 2 b 1 b 2 (mod n), 15

16 a 1 a 2 b 1 b 2 (mod n). If a b (mod n), d n, then a b (mod d). Proof. Since a 1 b 1 (mod n), a 2 b 2 (mod n), there are integers k 1, k 2 such that Then Thus b 1 a 1 = k 1 n, b 2 a 2 = k 2 n. (b 1 + b 2 ) (a 1 + a 2 ) = (k 1 + k 2 )n; (b 2 b 1 ) (a 1 a 2 ) = (k 1 k 2 )n; b 1 b 2 a 1 a 2 = b 1 b 2 b 1 a 2 + b 1 a 2 a 1 a 2 = b 1 (b 2 a 2 ) + (b 1 a 1 )a 2 = bk n + kna = (b 1 k 2 + a 2 k 1 )n. a 1 ± a 2 b 1 ± b 2 (mod n); a 1 a 2 b 1 b 2 (mod n). If d n, then n = dl for some l Z. Thus Therefore, a b (mod d). b a = kn = (kl)d. 16

17 Example (mod 8) = (mod 8); 6 14 (mod 8) ( mod 8 ) ; 2 2 However, In fact, (mod 8) 3 7 (mod 8). 3 7 (mod 8). Theorem 4.2. Let c a, c b, and c n. Then a b (mod n) a c b ( mod n ). c c Proof. Write a = ca 1, b = cb 1, n = cn 1. Then a b (mod n) b a = kn for an integer k c(b 1 a 1 ) = kcn 1 b/c a/c = b 1 a 1 = kn 1 a/c b/c (mod n/c). Theorem 4.3. a b (mod m), a b (mod n), a b (mod lcm(m, n)). In particular, gcd(m, n) = 1 a b (mod mn). 17

18 Proof. Write l = lcm(m, n). If a b (mod m), a b (mod n), then m (b a) and n (b a). Thus l (b a), i.e., a b (mod l). Conversely, if a b (mod l), then l (b a). Since m l, n l, we have m (b a), n (b a). Thus a b (mod m), a b (mod n). In particular, if gcd(m, n) = 1, then l = mn. Definition 4.4. An integer a is called invertible modulo n if there exists an integer b such that ab 1 (mod n). If so, b is called the inverse of a modulo n. Proposition 4.5. An integer a is invertible modulo n if and only if gcd(a, n) = 1 Proof. : If a is invertible modulo n, say its inverse is b, then exists an integer k such that ab = 1 + kn, i.e., 1 = ab kn. Thus gcd(a, n) divides 1. Hence gcd(a, n) = 1. : By the Euclidean Algorithm, there exist integers u, v such that 1 = au + nv. Then au 1 (mod n). Example 4.3. The invertible integers modulo 12 are the following numbers 1, 5, 7, 11. Numbers 0, 2, 3, 4, 6, 8, 9, 10 are not invertible modulo

19 Theorem 4.6. Let gcd(c, n) = 1. Then a b (mod n) ca cb (mod n) Proof. By the Euclidean Algorithm, there are integers u, v such that 1 = cu + nv. Then 1 cu (mod n). : c c (mod n), a b (mod n) ca cb (mod n). : ca cb (mod n) and u u (mod n) imply that uca ucb (mod n). Applying 1 cu (mod n), we have a b (mod n). Example 4.4. Find the inverses of 2, 4, 7, 8, 11, 13 modulo 15 respectively. Solution (mod 15), (mod 15). Write 15 = Then = 1. Thus (mod 15). Since 2 13 (mod 15), then (mod 15). Similarly, 15 = , 11 = , 4 = 3 + 1, then 1 = 4 3 = 4 (11 2 4) = = 3 (15 11) 11 = Thus the inverse of 11 is 4 11 (mod 15), i.e., (mod 15). Theorem 4.7. Let n P. The congruence equation ax b (mod n) has a solution if and only if gcd(a, n) divides b. 19

20 Proof. Let d = gcd(a, n). The congruence equation has a solution if and only if there exist integers x and k such that b = ax + kn. This is equivalent to d b. Exercise 2. Let S b be the set of solutions of the congruence equation ax b (mod n). If x = u is a particular solution of S b. Show that the function f : S 0 S b, defined by is a bijection. Note. For all k, l Z, f(x) = x + u, ax b (mod n) (a + kn)x b + ln (mod n). Example 4.5. Find all integers x for 9x 27 (mod 15). Solution. Find gcd(9, 15) = 3. Dividing both sides by 3, 3x 9 (mod 5) 3x 4 (mod 5). Since gcd(3, 5) = 1, the integer 3 is invertible and its inverse is 2. Multiplying 2 to both sides, 6x 8 (mod 5). Since 6 1 (mod 5), 8 3 (mod 5), then Thus x 3 (mod 5). x = 3 + 5k, k Z. 20

21 Example 4.6. Find all integers x such that 668x 888 (mod 168). Solution. Find gcd(668, 168) = 4, then 167x 222 (mod 42). By the Division Algorithm, 167 = ; 42 = By the Euclidean Algorithm, 1 = = 42 ( ) = Then (mod 42); the inverse of 167 is 1. Multiplying 1 to both sides, we have x 222 (mod 42). Thus x 12 (mod 42) or x 30 (mod 42); i.e. x = k, k Z. Exercise 3. Let S be the solution set of the system { a1 x b 1 (mod n 1 ) a 2 x b 2 (mod n 2 ) Let S 0 be the solution set of the same system with (b 1 = b 2 = 0. If S has a particular solution x = u, then the function f : S 0 S, defined by is a bijection. f(x) = x + u, x S 0 Example 4.7. How to solve the system of modular equations { a1 x b 1 (mod n 1 ) (2) a 2 x b 2 (mod n 2 ) (1) 21

22 Solution. Apply the Division Algorithm to find d i = gcd(a i, n i ), i = 1, 2. Reduce the system (2) to the following by dividing both side by d i respectively. { a 1 x b 1 (mod n 1) a 2x b 2 (mod n (3) 2) where a i = a i/d i, b i = b i/d i, n i = n i/d i, and gcd(a i, n i) = 1, i = 1, 2. Apply the Euclidean Algorithm to find u i, v i such that a iu i + n iv i = 1. Then a i u i 1 (mod n i ), i = 1, 2. Thus (3) can be written as { x u1 b 1 (mod n 1) x u 2 b 2 (mod n (4) 2) Set c i = u i b i, m i = n i, i = 1, 2. We have { x c1 (mod m 1 ) x c 2 (mod m 2 ) Let x = c 1 + ym 1 = c 2 + zm 2 for y, z Z. Solve (y, z) for m 1 y m 2 z = c 2 c 1 to get a special solution (y, z) = (y 0, z 0 ). Hence x 0 = c 1 + m 1 y 0 = c 2 + m 2 z 0. Finally, set m = lcm(m 1, m 2 ); the system (2) becomes x x 0 (mod m). 22 (5)

23 Example 4.8. Find all integral solutions for { 12x 96 (mod 20) 20x 70 (mod 30) Solution. Applying the Division Algorithm to find gcd(12, 20) = 4, gcd(20, 30) = 10. Then { 3x 24 (mod 5) 2x 7 (mod 3) Applying the Euclidean Algorithm to find gcd(3, 5) = 1 = so that (mod 5); and to find gcd(2, 3) = 1 = so that 1 2 = 1 (mod 3). (Equivalently, (mod 3).) Then, 2 is the inverse of 3 modulo 5; 1 or 2 is the inverse of 2 modulo 3. Thus { 2 3x 2 24 (mod 5) 1 2x 1 7 (mod 3) { x 48 (mod 5) x 7 (mod 3) { x 3 (mod 5) x 2 (mod 3) 23

24 Set x = 3 + 5y = 2 + 3z, where y, z Z, i.e., 5y 3z = 1. We find a special solution (y 0, z 0 ) = (1, 2); so x 0 = 8. Thus the original system is equivalent to and all solutions are given by x 8 (mod 15) x = k, k Z. Example 4.9. Find all integer solutions for the system { x 486 (mod 186) x 386 (mod 286) Solution. The system can be reduced to { x 114 (mod 186) x 100 (mod 286) Set x = y = z, i.e., 186y 286z = 14. Applying the Division Algorithm, 286 = , 186 = , 100 = , 86 =

25 Then gcd(186, 286) = 2. Applying the Euclidean Algorithm, 2 = = 86 6(100 86) = = 7( ) = = ( ) = Note that 14 2 = 7. So we get a special solution (y 0, z 0 ) = 7(20, 13) = ( 140, 91). Thus x 0 = y 0 = z 0 = Note that lcm(186, 286) = The general solutions for { x 0 (mod 186) x 0 (mod 286) are given by x 0 (mod lcm(168, 268)), i.e., x 0 (mod 26598). Hence x (mod 26598); i.e. x = k, k Z. Theorem 4.8 (Chinese Remainder Theorem). Let n 1, n 2,..., n k P. If gcd(n i, n j ) = 1 for all i j, then the system of congruence equations x b 1 (mod n 1 ) x b 2 (mod n 2 ). x b k (mod n k ). has a unique solution modulo n 1 n 2 n k. 25

26 Exercise 4. In the Chinese Remainder Theorem, if gcd(n i, n j ) = 1, is not satisfied, does the system have solutions? If yes, are the solutions unique modulo some integers? 26

27 5 Important Facts 1. Let a denote set of solutions of the equation x a (mod n); i.e., a = {a + kn k Z}. We call a the residue class of a modulo n. There are n residue classes of modulo n, i.e., 0, 1, 2,..., n If c a, c b, c n, then a b (mod n) a c b c ( mod n c). 3. An integer a is called invertible modulo n if there exists an integer b such that ab 1 (mod n). If so, b is called the inverse of a modulo n. 4. An integer a is invertible modulo n gcd(a, n) = If gcd(c, n) = 1, then a b (mod n) ca cb (mod n). 6. Equation ax b (mod n) has solution gcd(a, n) b. 7. For all k, l Z, ax b (mod n) (a + kn)x b + ln (mod n). 27