Upcoming SlideShare
×

# Remodulization of Congruences

698 views

Published on

Proceedings - NCUR VI. (1992), Vol. II, pp. 1036-1041.

Jeffrey F. Gold
Department of Mathematics, Department of Physics
University of Utah

Don H. Tucker
Department of Mathematics
University of Utah

Introduction

Remodulization introduces a new method applied to congruences and systems of congruences. We prove the Chinese Remainder Theorem using the remodulization method and establish an efficient method to solve linear congruences. The following is an excerpt of Remodulization of Congruences and its Applications [2].

0 Likes
Statistics
Notes
• Full Name
Comment goes here.

Are you sure you want to Yes No
• Be the first to comment

• Be the first to like this

Views
Total views
698
On SlideShare
0
From Embeds
0
Number of Embeds
1
Actions
Shares
0
10
0
Likes
0
Embeds 0
No embeds

No notes for slide

### Remodulization of Congruences

1. 1. Chapter 2 Remodulization of Congruences Proceedings|NCUR VI. 1992, Vol. II, pp. 1036 1041. Je rey F. Gold Department of Mathematics, Department of Physics University of Utah Don H. Tucker Department of Mathematics University of Utah Introduction Remodulization introduces a new method applied to congruences and systems of congruences. We prove the Chinese Remainder Theorem using the remoduliza- tion method and establish an e cient method to solve linear congruences. The following is an excerpt of Remodulization of Congruences and its Applications 2. De nition 1 If a and b are integers, then a mod b = fa; a b; a 2b; : : : g. We write x a mod b, meaning that x is an element of the set a mod b. The common terminology is to say that x is congruent to a modulo b. These sets are frequently called residue classes since they consist of those numbers which, upon division by b, leave a remainder residue of a. 1
2. 2. CHAPTER 2. REMODULIZATION OF CONGRUENCES 2 De nition 2 If a1; a2; : : : ; an; b are integers, then a1 ; a2 ; : : : ; an mod b = a1 mod b a2 mod b an mod b : Theorem 1 Suppose a, b, and c are integers and c 0, then a mod b = a; a + b; : : : ; a + c , 1b mod cb : Proof. We write a mod b = f a , 2cb; a , 2c , 1b; ::: a , c + 1b a , cb; a , c , 1b; ::: a,b a; a + b; ::: a + c , 1b; a + cb; a + c + 1b; ::: a + 2c , 1b; a + 2cb; a + 2c + 1b; ::: a + 3c , 1b; g and rewriting the rows a mod b = f a , 2cb; a + b , 2cb; ::: a + c , 1b , 2cb a , cb; a + b , cb; ::: a + c , 1b , cb a; a + b; ::: a + c , 1b; a + cb; a + b + cb; ::: a + c , 1b + cb; a + 2cb; a + b + 2cb; ::: a + c , 1b + 2cb; g Then, forming unions on the extended columns, the result follows. We refer to this process as remodulization by a factor c. Suppose it is desired to express 1 mod 2 in terms of modulo 8, then, 1 mod 2 is remodulized by the factor 4, i.e., 1 mod 2 = 1; 3; 5; 7 mod 8 : It is convenient to create a notation for the expression a; a + b; : : : ; a + c , 1b mod cb : We write it as c,1 a + kb mod cb : k=0 S Reindexing the symbol , c a mod b = a , b + kb mod cb : k=1 The Chinese Remainder Theorem rst appeared in the rst century A.D. The Chinese mathematician Sun-Ts sought a solution to the following problem: u
3. 3. CHAPTER 2. REMODULIZATION OF CONGRUENCES 3 What numbers n, when divided by 3, 5, and 7, have remainders 2, 3, and 2, respectively? This problem also appeared in the Introductio Arithmeticae, written by Nico- machus of Gerasa, a Greek mathematician circa 100 A.D. The problem asks one to nd the solution to a system of congruences 8 x a1 mod b1 x a2 mod b2 y .. . : x an mod bn where 0 aj bj , and the bj are pairwise relatively prime. The idea is to remodulize each congruence in order to obtain a common mod- ulus, thereby making the solution set the intersection of the resulting classes. These can be determined by direct observation of the sets of residues in the remodulized forms. Since the bj are pairwise relatively prime, the smallest common modulus is the product of the bj ; therefore we remodulize the j th congruence by the factor n n 1 Y b = c ; then b c = C = Y b bj k j j j k k=1 k=1 Performing these operations gives: Y n 1 bj bk k=1 n Y xj aj , bj + bj m mod bk m=1 k=1 Simplifying the notation, we nd cj xj aj , bj + bj m mod C : m=1 The solution set to y is the intersection of the sets of initial elements mod C, i.e., n cj z x aj , bj + bj m mod C : j =1 m=1 Thus, for the original problem of Sun-Ts, we have: u 8 x 2 mod 3 : x 3 mod 5 x 2 mod 7
4. 4. CHAPTER 2. REMODULIZATION OF CONGRUENCES 4 Since the bj are prime, the least common modulus is 3 5 7 = 105. The congruences are then remodulized by 35, 21, and 15, respectively. The resulting remodulizations are 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104 mod 105 ; 3, 8, 13, 18, 23, 28, 33, 38, 43, 48, 53, 58, 63, 68, 73, 78, 83, 88, 93, 98, 103 mod 105 ; and 2, 9, 16, 23, 30, 37, 44, 51, 58, 65, 72, 79, 86, 93, 100 mod 105 , where the intersection, 23 mod 105, is the complete solution set among the integers. This ultimately raises the question as to whether n cj z x aj , bj + bj m mod C j =1 m=1 always contains exactly one element, given that 0 aj bj and the bj are pairwise relatively prime. In the same example, if we remodulize to the product 105, we nd that the solution set corresponding to the rst two congruences x 2 mod 3 x 3 mod 5 is characterized as 8; 23; 38; 53; 68; 83; 98 mod 105, which does not appear to be unique; however, this is equivalent to 8 mod 15, which, in the example given, has been remodulized by the factor 7. The solution 8 mod 15 is obtained by solving the rst two congruences directly by the method described. As it hap- pens, if one uses the smallest possible modulus, the answer to our question is yes. Theorem 2 ChineseRemainderTheorem The system y of congruences, where the bj are pairwise relatively prime, has as solution set n cj z x aj , bj + bj m mod C j =1 m=1 Yn C where cj = bj and C = bk . Moreover, the intersection contains only one k=1 element, i.e., one residue class.
5. 5. CHAPTER 2. REMODULIZATION OF CONGRUENCES 5 Proof. In order to show that this element exists, we consider the following: x a1 mod b1 x a2 mod b2 where 0 a1 b1 and 0 a2 b2 and gcdb1 ; b2 = 1. Remodulizing the congruences by the factors b2 and b1 , respectively, x a1 ; a1 + b1 ; : : : ; a1 + b2 , 1 b1 mod b1b2 x a2 ; a2 + b2 ; : : : ; a2 + b1 , 1 b2 mod b1b2 it is required to show that the sets of initial elements intersect, i.e., there ex- ist integers k and h where 1 k b2 , 1 and 1 h b1 , 1, such that a1 + kb1 = a2 + hb2 . Rewriting this equation, we require integers k and h such that kb1 , hb2 = a2 , a1 . Since b1 and b2 are relatively prime, a2 , a1 is divisible by gcdb1 ; b2 . Now notice that if k and h are solutions to kb1 , hb2 = 1, then ka2 , a1 and ha2 , a1 are solutions to the required equation. Euclid's algorithm insures that such integers k and h exist. It follows that the rst pair of congruences have a solution. We wish now to show that the pair has a unique solution modulo b1 b2 . We know that a solution exists, that is, there exists an integer x such that x 2 fa1; a1 + b1 ; : : : ; a1 + b2 , 1b1 g fa2 ; a2 + b2 ; : : : ; a2 + b1 , 1b2 g: Now suppose that the two initial sets intersect in two elements, say x1 = a1 + b1 = a2 + kb2 and x2 = a1 + b1 = a2 + hb2 : Subtracting the second formulation from the rst for each xi , a1 , a2 = kb2 , b1 = hb2 , b1 ; so that k , hb2 = , b1 : Since b1 and b2 are relatively prime it must be that k , h = mb1 and , = nb2 ; for some integers m and n. In other words, k = h + mb1 and = + nb2
6. 6. CHAPTER 2. REMODULIZATION OF CONGRUENCES 6 so that x1 becomes x1 = a1 + + nb2b1 = a2 + h + mb1b2 = a1 + b1 + nb2 b1 = a2 + hb2 + mb1 b2 : This says that a1 + b1 = a2 + hb2 + m , nb1 b2 ; which is x2 . This implies that m , n = 0 ; i.e., m = n, and hence k = h + mb1 = + mb2 : Therefore, x1 = a1 + b1 = a1 + + mb2 b1 = a1 + b1 + mb1 b2 = x2 + mb1 b2 : This means that x1 and x2 are in the same residue class mod b1 b2 ; i.e., they are congruent mod b1 b2 and the solution set is given as the unique class x d mod b1 b2 ; where d x1 mod b1 b2 x2 mod b1 b2 from above. In the event there exist three congruences, we solve the rst two congruences and combine this result with the third congruence, i.e., x d mod b1 b2 x a3 mod b3 and repeat the argument since b1 b2 and b3 are relatively prime. The induction works and both the existence and the uniqueness are established. Suppose we want to solve cx a mod b for x, where 1 c b and a is divisible by gcdc; b otherwise no solution exists. We consider the case where gcdc; b = 1. By remodulizing a mod b by the factor c, we obtain cx a; a + b; : : : ; a + c , 1 b mod cb: Since the set fa; a + b; : : : ; a +c , 1 bg forms a complete residue system mod c, there exists an element in this set, call it d, which is divisible by c. Since cx a; a + b; : : : ; d; : : : ; a + c , 1b mod cb we nd that the only congruence solvable is cx d mod cb. The remaining congruences, cx a; a + b; : : : ; d , b; d + b; : : : ; a + c , 1 b mod cb
7. 7. CHAPTER 2. REMODULIZATION OF CONGRUENCES 7 are not solvable, since in each case the factor c is pairwise relatively prime with the elements fa; a + b; : : : ; d , b; d + b; : : : ; a +c , 1bg, and thus does not divide them. In the congruence cx d mod cb, dividing through by c, x d mod cb or x d mod b : c c c Note that d c b. To illustrate this procedure, consider the following example. Suppose 5x 3 mod 7. This is solvable since 3 is divisible by gcd5; 7 = 1. Remodulizing 3 mod 7 by the factor 5 gives 5x 3; 10; 17; 24; 31 mod 5 7 so that 5x 10 mod 35 is the only possible solution and, upon dividing all three terms by 5, x 2 mod 7 : Note that 5x 3; 17; 24; 31 mod 35 does not yield any solutions, since in this case gcd5; 35 = 5 does not divide any number in the set f3; 17; 24; 31g. The remodulization method also provides a way of nding solutions to sys- tems of congruences using linear congruences. Suppose we have the following system, x a1 mod b1 x a2 mod b2 where b1 b2 and b1 and b2 are relatively prime. The idea is to multiply the rst congruence by b2 and the second congruence by b1 , i.e., b2 x a1 b2 mod b1b2 b1 x a2 b1 mod b1b2 so that we obtain a common modulus. By subtracting the second linear con- gruence from the rst, we obtain a single linear congruence, b2 , b1 x a1 b2 , a2 b1 mod b1b2 : The unique solution modulo b1 b2 is insured, since gcdb2 , b1 ; b1b2 = 1. If the system consists of more than two congruences, then the solution of the rst two congruences is combined with the third, and so on, to obtain a solution for the entire system.
8. 8. CHAPTER 2. REMODULIZATION OF CONGRUENCES 8 Corollary 1 A system of linear congruences 8 c1 x a1 mod b1 c2 x a2 mod b2 . . . : cn x an mod bn where 1 cj bj , the bj are pairwise relatively prime, and the aj are divisible by gcdcj ; bj , can be reduced to a system 8 x d1 mod b1 x d2 mod b2 . . . : x dn mod bn : By Theorem 2, the solution to this system is n cj x dj , bj + bj m mod C j =1 m=1 n Y Yn where cj = b1j bk and C = bk , and the intersection contains only one k=1 k=1 residue class. References 1 Burton, David M. Elementary Number Theory, Second Edition. Wm. C. Brown Publishers, Dubuque, Iowa, 1989. 2 Gold, J. F. and Don H. Tucker. Remodulization of Congruences and Its Applications. To be submitted. 3 Ore, Oystein. Number Theory and Its History. Dover Publications, Inc., New York, 1988. 4 Stewart, B. M. Theory of Numbers. The MacMillan Co., New York, 1952.