NONLINEAR CONGRUENCES IN THE THEORY OF NUMBERS

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1 Bulletin of the Marathwada Mathematical Society Vol. 12, No. 2, December 2011, Pages NONLINEAR CONGRUENCES IN THE THEORY OF NUMBERS S.R.Joshi 8, Karmayog, Tarak Housing Society, Opp to Ramkrishna Mission Ashram, Beed bye pass Road, Aurangabad Abstract The purpose of this paper is to use the concept of isomorphism in solving some nonlinear congruences in the theory of numbers, by considering two finite groups and establishing a suitable isomorphism between them. If m is an integer > 1 and such that m+1 ia a prime number, then we consider two groups A and B, each of order m, such that the binary operations in A and B are addition modulo m and multiplication modulo m+1 respectively. After establishing a suitable isomorphism between A and B, it is shown that the problem of solving a nonlinear congruence related to B can be reduced to a problem on linear congruence related to A. Examples are given to illustrate the results obtained. 1 INTRODUCTION It is well known that Isomorphism is one of the important and useful concepts in Mathematics, as far as the study of algebraic and topological structures is concerned. If A and B are two isomorphic structures and the structure A is easier to study, then many properties of B can be studied with the help of the corresponding properties of A. This is a very useful advantage of isomorphism. In this paper the concept of isomorphic groups is considered and is used to solve a non-linear congruences of the form x n a (mod m+1). If for a given positive integer m > 1, m + 1 is a prime number, then we consider two finite groups A and B each of order m, such that the binary operations in A and B are addition modulo m and multiplication modulo m + 1 respectively and show that they are isomorphic by establishing a suitable isomorphism between them. By using this result, it is then shown that the problem of solving a nonlinear congruence related to B, can be solved by reducing it to a problem on linear congruence related to A. Examples are given to illustrate the theorems proved. The paper ends with some remarks. 24

2 Nonlinear Congruences in the PRELIMINARIES Let Z denote the set of all integers. For a given integer in Z let Z(m) denote the set {0, 1, 2,..., m-1}. Z(m) is also known as the set of all remainders ( or residues ) modulo m. We require the following definitions. Definition 2.1 : If m is a positive integer and x, y are in Z, then we say that x is congruent to y modulo m (written as x y ( mod m) ), if (x-y) is divisible by m. Definition 2.2 : An integer x is said to be co-prime to another integer y or relatively prime to y, if the greatest common divisor of x and y is 1. For example 20 is co-prime to 63 and 77. The integer 1 is co-prime to every integer in Z. Remark 2.1: If m is a prime number, then every non zero member of Z(m) is relatively prime to m. Definition 2.3 : If m is a positive integer, then the total number of members of Z(m), each of which is co-prime to m is denoted by φ(m). The function φ is also known as Euler s phi function. For example φ(10) = φ(12) = 4, φ(6) = 2, φ(7) = 6 etc. Definition 2.4 : If a, b are any two integers and m > 1, then we define a b and a b as remainders obtained when a + b and a b are divided by m respectively. For example if m = 7, a = 24, b = 13, then it can be verified that a b = 2, and a b = 4. The operations and are known as addition modulo m and multiplication modulo m respectively. Definition 2.5 : [1] Let (A, + ) and (B, ) be two groups, where + and are any two binary operations in A and B which need not always represent the usual addition and multiplication of numbers. Then we say that A is isomorphic to B, and write A = B if there exists a one-one and onto mapping f : A B, such that f(a + b) = f(a) f(b), a, b A. (2.1) Definition 2.6 : If (G,.) is a finite cyclic group, then an element g of G is said to be a generator of G, if every element of G can be expressed as a non-negative integral power of g. For example 3 and 5 are the generators for the group G 7 = {1, 2, 3, 4, 5, 6} with binary operation as multiplication modulo 7. Remark 2.2 : If G is a cyclic group of order m then the number of generators of G is φ(m).[1] Notation 2.1:If an integer b is divisible by a nonzero integer a, then we write a b. For example 5 30, , 1 x, x x, where x is any nonzero integer.

3 26 S.R. Joshi Notation 2.2: If a and b are any two integers not both zero then their g.c.d. will be denoted by (a,b). Lastly we state two theorems from Number Theory related to linear and nonlinear congruences respectively. Theorem 2.1 : (see Theorem 2.13 of [1]). If d = (n,m), then ny b (mod m) has no solution if b is not divisible by d. If d b, then it has d solutions given by, y (b/d)y 0 + t(m/d), t = 0, 1, 2,..., d 1, (2.2) where y 0 is the solution of (n/d)y 1 (mod m/d) and is given by y 0 = (n/d) φ(m/d) 1. (see Cor. 2.9 on page 24 of [2]) Theorem 2.2 : (see Theorem 2.27 of [2]). Let m be an integer > 2 such that m+1 is a prime number. Let d = (n,m), where n is a known positive integer such that (n, m+1) = 1. Then the congruence x n a (mod m+1) has d solutions or no solution according as a m/d 1 ( mod m+1 ) or a m/d 1 ( mod m+1 ). 3 NONLINEAR CONGRUENCES In this section we shall discuss how to solve a nonlinear congruence by reducing it to a corresponding linear congruence. For this purpose we consider two isomorphic groups both being cyclic. We first prove the following two theorems. Theorem 3.1 : Let m be a positive integer > 1 such that (m+1) is a prime number. Let A = Z(m) and B = { 1, 2, 3,..., m } be two groups w.r.t. addition modulo m and multiplication modulo (m+1) respectively. Then A = B. Proof : It is clear that A is a cyclic group of order m with 1 as one of the generators. By Theorem 2.33 on page 59 of [2], the set Z(m+1) is a field. Hence B ( which also coincides with the set of non-zero elements of of Z(m+1)) is a commutative group w.r.t. multiplication modulo (m+1). The order of B is also m. Hence by Problem 3 on page 60 of [2] A = B, and the proof is complete. Remark 3.1 :By Remark 2.2 there are φ(m) generators of the additive group A. Among these 1 is one of the generators. It is easy to verify that any nonzero integer in A which is relatively prime to m is a generator of A. Further B is isomorphic to A and hence the number of generators of B is also φ(m). These are obtained by taking the images of generators of A under some isomorphism f from A onto B. In the Theorem 3.1 the groups A and B are finite each containing m elements.

4 Nonlinear Congruences in the Hence by theory of permutations and the definition of one-one and onto mappings there are m! one-one and onto mappings from A to B. But each of these mappings is not an isomorphism from A to B. The following theorem gives the answer for the number of isomorphisms between A and B. Theorem 3.2 :Let the hypotheses of Theorem 3.1 hold. Then there are k isomorphisms between A and B, where k = φ(m). Proof: By Remark(2.2), there are k generators in each of the groups A and B. Let g 1, g 2, g 3,..., g k be the k generators of B. For a fixed g i define a mapping f i : A B as follows: f i (n) = (g i ) n, 0 n (m 1) (3.1) It is very easy to verify that the mapping f i is an isomorphism between A and B, since the relation (2.1) holds for any two integers a and b in A. Now g i can be selected in k ways only. Hence there are only k isomorphisms between A and B, for each of which A and B are isomorphic. The proof is complete. Theorem 3.3 : Let all the assumptions of Theorem 2.1 hold, and a be any integer in B. Let f be an isomorphism from A onto B, with g as its inverse isomorphism. Then the congruence x n a(mod m + 1) (3.2) has d solutions given by x = f(y), where y is one of the solutions of the linear congruence ny b(mod m), (3.3) and are given by, y (b/d)y 0 + t(m/d), t = 0, 1, 2,..., d 1, (3.4) where y 0 is the solution of (n/d)x 1 (mod m/d) and b = g(a). Proof : Taking the image of the equation (3.2) under g and using the property of group isomorphism we see that the relation (3.2)is reduced to the relation (3.3) which can be solved by applying Theorem 2.1, and the proof is complete. Remark 3.2 : Theorem 3.2 is the same as Theorem 2.2 mentioned in Section 2. The proof given here is clearly easier than that given in [2]. In finding solutions of nonlinear congruences, it is sufficient to determine one generator of B, which can be found by trial method, i.e. by considering different powers of some integer in B, and checking whether they cover the complete set B. If g is one of the generators of B, then it can be shown that g r is also a generator of B, where r is such that (r,m) = 1. (see remarks on page 263 of [1] )

5 28 S.R. Joshi 4 EXAMPLES ON NONLINEAR CONGRUENCES Example 4.1 Solve the following congruences. (i) x 3 6x x 15 0 (mod 19), (ii) x 3 2 (mod 7), (iii)x 15 7 (mod 19). Solution :(i): The given congruence can be expressed in the form t 3 7(mod19), where t = (x 2). (4.1) Here m = 18, n = 3, d = (n,m) = 3. Consider the groups (A, ), and (B, ) where A = {0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,1516,17}, B = {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18} Note that A and B are groups under the operations of addition modulo 18 and multiplication modulo 19 respectively. Since φ (18) = 6, there are 6 generators for each of the groups A and B. Further it can be verified that 2 is one of the 6 generators of B. By considering powers of 2 (starting with 0 ) mod 19, the group B can be expressed as B = {1,2,4,8,16,13,7,14,9,18,17,15,11,3,6,12,5,10} With this representation of B the groups A and B are isomorphic and if f is the isomorphism we may express f in the form of a table given below : T able 4.1 x A f(x) B Here f(i) 2 i (mod 19), i = 0,1,2..., 18. By Theorem 3.3, the congruence (4.1) is reduced to 3y 6, (mod 18) because f(6) = 7. Note that b = 6, and d b. Hence by Theorem 2.1, it has 3 solutions since (3,18)= 3, and they are given by y = 2,8,14. The images of these numbers under f are 4, 9, and 6 respectively. But x = t + 2, and hence x = 6, 11, 8. It can be verified that these three integers are the solutions of the given congruence. Instead of 2 we may take any other generator and establish an isomorphism

6 Nonlinear Congruences in the between A and B. For example by Remark 3.2, 2 5 is also a generator of B, since (5,2) = 1. But ( mod 19 ). Hence 13 is a generator of B. Taking powers of 13 (starting with 0 ), and with h as the corresponding isomorphism we may express h in the form of a table given below : T able 4.2 x A h(x) B Here h(i) 13 i (mod 19), i = 0,1,2..., 18. Now by Theorem 3.3 the congruence (4.1) reduces to 3y 12 (mod 18), because h(12) = 7. Again by Theorem 2.1, this congruence has 3 solutions viz. y = 4,10,16. By considering the images of these elements under h we see that t = 4,6,9, and hence x = t+2 = 6, 8, 11, the same answer. (ii): For this congruence, we have m = 6, n = 3, d = 3, m/d = 2, and a = 2. Here 2 2 = 4, and 4 is not congruent to 1 modulo 7. Hence the condition mentioned in Theorem 2.2 is not satisfied, and therefore the given congruence has no solution. (iii): Here the power 15 of x is greater compared to 3 in Example (i). Hence even if we reduce it to a corresponding linear congruence, it will require more time to solve it. Now the power 15 is more than the integer part of 19/2. Hence we can adopt the method given in [3]. By multiplying both sides of the given congruence by x 3 (the power 3 = ) we get the following congruence x 18 7x 3 (mod 19) (4.2) By Fermat s Theorem x 18 1 (mod 19). Hence we get the following congruence 7 x 3 1 (mod 19) (4.3) By considering the isomorphism f used in Example(i) we observe that the above congruence can be reduced to the following linear congruence y 0 (mod 18) (4.4) The solutions of this congruence are 2, 8 and 14. Hence the corresponding values of x are 17, 16 and 5 respectively,which can be verified to be the solutions. Remark 4.1 : In the following problem on nonlinear congruence m+1 is not a prime number but a product of two different primes m 1 and m 2. To solve such a problem we first solve two congruences related to m 1 and m 2 separately and then obtain final solution by applying Chinese Remainder Theorem (Theorem 2.14 of [1] ). We state below this theorem in the modified form for two prime factors only.

7 30 S.R. Joshi Theorem 4.1: For an integer variable x, let F(x) stand for a polynomial with integer coefficients. For any two different primes m 1 and m 2 let a 1, a 2,...,a k1, be k 1, solutions of the congruences F(x) 0 (mod m 1 ) and let b 1, b 2,...,b k2, be k 2 solutions of the congruences F(x) 0 (mod m 2 ). Let d 1 and d 2 be two integers satisfying the conditions m 2 d 1 1 (mod m 1 ) and m 1 d 2 1 (mod m 2 ) respectively. Then the solutions of the congruence F (x) 0 (mod m 1 m 2 ) are given by u m 2 d 1 a i + m 1 d 2 b j (mod m 1 m 2 ), i = 1, 2,..., k 1, and j = 1, 2,..., k 2. (4.5) Example 4.2 Solve the congruence x 3 34 (mod 91). Solution : Here F(x) = x We may take m 1 = 7 and m 2 = 13. We shall first solve the congruences x 3 34 (mod 7 ), and x 3 34 (mod 13) separately. These congruences can be expressed in the form and x 3 6 (mod 7), (4.6) x 3 8 (mod 13). (4.7) By applying the method used for Example 4.1 these congruence can be reduced to linear congruences given below: 3y 6 (mod 6). (4.8) 3y 9 (mod 12). (4.9) By Theorem 2.1 the solutions of (4.5) and (4.6) are y = 1, 3, 5 and y = 3, 7, 11 respectively. If A and B are the groups with binary operations as addition modulo 6 (or 12) and multiplication modulo 7 (or 13) respectively and if f 1 and f 2 are the isomorphisms between them related to these two congruences, then we may express f 1 and f 2 as follows; T able 4.3 x A f 1 (x) B T able 4.4 x A f 2 (x) B

8 Nonlinear Congruences in the Here f 1 (i) = 5 i (mod 7 ), i = 0,1,2..., 5 and f 2 (i) = 7 i (mod 13), i = 0,1,2..., 12. Note that while considering the isomorphisms f 1 and f 2 we have taken 5 and 7 respectively as generators of B. From Table 4.3 and Table 4.4 we see that the values of x corresponding to the values of y are x = 3,5, 6 and 2,5,6 respectively. In respect to the notations used in Theorem 4.1 we have a 1 = 3, a 2 = 5,a 3 = 6, b 1 = 2, b 2 = 5, b 3 = 6. Further it is easy to show that d 1 = -1, and d 2 = 2. Hence using the formula (4.2) we get 9 different solutions of the given congruence, since k 1 k 2 = 3 3 = 9. One of the 9 solutions is obtained by taking a 1 = 3, b 1 = 2. It is -11 which is congruent to 80 modulo 91. The other 8 solutions are 41, 45, 54, 5, 19, 50, and CONCLUDING REMARKS The author thinks that the method of isomorphism for reducing a given non linear congruence to a linear congruence is easier than the usual method given in almost all text books on Number Theory. A minute observation of the method we have adopted indicates that our method is not far different from the usual method, because the concept of generator we have used, coincides with the concept of primitive root in Number Theory. Of course when the power n in the congruence x n a (mod m + 1) is large both methods require almost equal time to solve such a congruence. When the integer n is greater than m/2, then the degree of the congruence x n a (mod m + 1) can be reduced to a smaller one by adopting the method given in [3], as we have done in solving Example 4.1(iii). When the integer m is large and without calculator it becomes difficult to solve a nonlinear congruence then the only way to solve it quickly is to make use of a computer program and then solve it. References [1] I.N.Herstein, Topics in Algebra,Vikas Publishing House, Pvt. Ltd., New Delhi,7th Indian Edition(1978). [2] Ivan Niven and Herbert S. Zuckerman, An Introduction to the theory of Numbers,Wiley Eastern Limited,(1976). [3] Peta beogradska, Solutions of Some Classes of Congruences,The Teaching of Mathematics, Vol. IX, 1 (2006),

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