LET n 3 be an integer, recall that (Z/nZ) denotes

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1 Group of Square Roots of Unity Modulo n Rochdi Omami, Mohamed Omami and Raouf Ouni Digital Open Science Index, Mathematical and Computational Sciences waset.org/publication/8554 Abstract Let n 3 be an integer and G (n) be the subgroup of square roots of 1 in (Z/nZ). In this paper, we give an algorithm that computes a generating set of this subgroup. Keywords Group, modulo, square roots, unity. I. INTRODUCTION LET n 3 be an integer, recall that (Z/nZ) denotes the group of units of the ring (Z/nZ). Let n = p α1 1 pα...pαm m the primary decomposition of n, then (Z/nZ) = (Z/p αi i Z) for more details on the structure of (Z/nZ) see [1] and []. The group (Z/nZ) has several applications, the most important is cryptography, that is RSA cryptosystem (see [5]). The security of the RSA cryptosystem is based on the problem of factoring large numbers and the task of finding e th roots modulo a composite number n whose factors are not known. In [8], D.Shanks gives a probabilistic algorithm that computes a square root of an integer modulo an odd prime p. There are other algorithms that compute a square root of an integer modulo an integer n (see [7]) and more generally in a finite fields (see [6]). We denote by G (n) the subgroup of (Z/nZ) which is formed by the integers x that satisfies x =1, such integers are called square roots of unity modulo n. More precisely G (n) contains the unity and elements of order. Recall that elements of order exists always in (Z/nZ) (-1 has for order ), therefore G (n) is not a trivial group. Finally remark that all elements of G (n) except the unity has for order, so G (n) has an order a power of, so we obtain the following result : Proposition Let n 3 be an integer, then there exists an integer t 1 such that : Ord(G (n)) = t. In this article, we will give an algorithm that computes a generating set of G (n) and gives its decomposition into product of cyclic subgroups. Finally this algorithm will be written in MAPLE language. II. SQUARE ROOTS OF UNITY MODULO N Let n 3 be an integer and n = α p α1 1 pα...pαm m its primary decomposition. In this study, we shall distinguish the Rochdi Omami, Mohamed Omami and Raouf Ouni are doctoral students at the Faculty of Science of Tunis : University El Manar, Tunis 09 cases α =0, α =1, α =and α 3. Case 1 : α =0 Let n 3 be an integer and n = p α1 1 pα...pαm m its primary decomposition. Let x be an element of (Z/nZ) such that x =1, that is n divides x 1=(x 1)(x 1).Wehave (x 1) (x 1) =, therefore GCD(x 1,x1) {1, }, so if p i divides x 1 then p αi i divides x 1. If we note, for example, p 1, p,..., p s the primes among the p i which divide x 1, then x is a solution of this system : 1 pα...pαs s K. x 1=p αs1 s1 pαs...pαm It s clear that x is the unique solution of this system modulo n. Conversely, any system of the previous form gives a square root of unity modulo n. Note that a two different systems of this form give two different solutions, indeed let the systems : y 1=p α τ(1) ρ(1) pα ρ() ρ()...pα ρ(r) ρ(r) K 1 y 1=p α ρ(r1) ρ(r1) pρ(r) ρ(r)...pα ρ(m) ρ(m) K where σ and ρ are two permutations of the set {1,,.., m}, if x = y, then the set of prime divisors of x 1 among the p i is the same of y 1. Therefore the set of prime divisors of x 1 among the p i is {p σ(1),p,..., p σ(s) } because p σ(s1),p σ(s),...and p σ(m) does not divide K 1, indeed : p α σ(s1) p α σ(1) =. Thus GCD(K 1,p α σ(s1) σ(m) ) {1, }, so {p σ(1),p,..., p σ(s) } = {p ρ(1),p ρ(),..., p ρ(r) }, it follows that the two systems are identical. We conclude that the number of square roots of unity modulo n is equal to the number of partitions of the set {1,,.., m}, that is m. Note that the empty subset corresponds to 1 and if all p i divide x 1, then x =1.Sowehaveproved: Proposition.1: Let n 3 be an integer, then Ord(G (n)) = ω(n) where ω(n) denote the number of distinct prime factors of n. Now we study the structure of the group G (n). For simplicity throughout this section, we take n 3 to be an odd integer 505

2 Digital Open Science Index, Mathematical and Computational Sciences waset.org/publication/8554 and n = p α1 1 pα...pαm m with this definition : its primary decomposition. we start Definition.1: Let x be a square root of unity modulo n. x is said to be initial if all prime factors of n divide x 1 except only one p i, we said that x is associated with p i. And we note : where K is an integer not divisible by p i and the symbol p αi i means that we remove the factor p αi i. Note that for any i {1,,.., m} there exist only one square root of unity associated with p i which is the solution of this system:. We denote by (n) the set that contains this solution and the unity, so (n) is a cyclic subgroup of G (n) of order. We have the following theorem : Theorem.1: The map ϕ : G p1 (n) Gp (n)... Gpm (n) G (n) (x 1,x,...,x m ) x 1.x,...x m is an isomorphism of groups. It s clear that ϕ is a morphism of groups, we will show first that ϕ is injective. We have ϕ(x 1,x,...,x m ) = 1 x 1.x,...x m = 1. Suppose that there exists an integer i such that x i 1, therefore p i does not divides x i 1. Also, for j i, p i divides x j 1. Then we have: x i =1K i and x j =1p i.k j where p i does not divides K i,so x 1.x,...x m = (1p i.k 1 )..(1 K i )..(1 p i.k m ) = (1p )(1 K i ) = 1(p p K i K i ). Since p i does not divides K i, then p i does not divides x 1.x,...x m 1, that is absurd. Thus x i = 1 for all i {1,,.., m}. Hence ϕ is injective. Finally, we remark that: Ord(G p1 (n) Gp (n)... Gpm (n)) = Ord(G (n)) = m so ϕ is bijective, therefore it s an isomorphism. The fact that ϕ is injective is due to the choice of x i, i.e. the initial square roots of the unity. The previous theorem shows that G (n) is exactly formed by the unity and finished products without the repetition of the initial square roots of the unity. In other words, if x i denote the initial square root of the unity associated with p i, then : G (n) ={ i I x i,avec I {1,,.., m}}. With the convention that the unity is the product over empty set. Remark also that -1 is the product of all x i, Indeed : x i = (1 p α1...p αm i ) = 1 i p α1 i Kn since p α1 i is not divisible by all p i because K i is not divisible by p i, we conclude that x i 1 is not divisible by all p i. It follows x i = 1. Finally, we have the following result : Corollary.1: Let x i be the initial square root of the unity associated with p i, then : G (n) =<x 1,x,...,x m >. Now, we give an algorithm written in MAPLE that computes the x i, i.e. a generating set of G (n). Let us give some explanations. Resuming the system : This system gives the following equation : p αi p α1 = and Bezout algorithm allows us to compute K and K and all x i. Gene :=proc(n) local LB, i, LF act, GEN; GEN := [ ]; LB := [ ]; LF act := if actors(n)[]; Algorithm

3 Digital Open Science Index, Mathematical and Computational Sciences waset.org/publication/8554 An application example : To find the generators of the group of square root of the unity modulo , we can use the previous algorithm with the command Gene ( ); We have the following result [33593, 1319, 3605, 4863], that is the list of generators. The Bezout function which is used in the previous algorithm is not a MAPLE function, but it s a classical algorithm called Extended Euclidean algorithm. Case : α =1 Let n 3 be an integer such that its primary decomposition is n =p α1 1 pα...pαm m. Let x be an element of (Z/nZ) such that x =1, that is n divides x 1=(x 1)(x1).Wehave (x 1) (x 1) =, therefore GCD(x 1,x1) {1, }. So, if p i divides x 1, then p αi i divides x 1. Also divides (x 1)(x 1), thus divides (x 1) or (x 1). Since (x 1) (x 1) =, then divides (x 1) and (x 1),sox is a solution of a system of this form : where σ is a permutation of the set {1,,.., m}. It s clear that x is the only solution modulo n of this system and every system of this form gives a square root of the unity modulo n. We show in the same way as the previous case, that two different systems gives two distinct solutions. Therefore, the number of square roots of the unity modulo n is the number of partitions of the set {1,,.., m}, that is m. Hence, we have the following result: Proposition.: Let n 3 be an odd integer, then Ord(G (n)) = ω(n) where ω(n) denote the number of distinct prime factors of n. For simplicity throughout this section we take n 3 to be an integer and n =p α1 1 pα...pαm m its primary decomposition. We start the study of G (n) with this definition : Definition.: Let x be a square root of unity modulo n. x is said to be initial if all the prime factors of n divide x 1 except only one p i, we said that x is associated with p i. And we note : x 1=p α1 where K is an integer that does not divisible by p i and the symbol p αi i means that we remove the factor p αi i. We remark that for each i {1,,.., m}, there exists only one square root of unity associated with p i which is the solution of the following system : x 1=p α1. We denote by (n) the set that contains this solution and the unity, so (n) is a cyclic subgroup of G (n) of order. We have the following theorem : Theorem.: The map ϕ : G p1 (n) Gp (n)... Gpm (n) G (n) (x 1,x,...,x m ) x 1.x,...x m is an isomorphism of groups. the previous theorem shows that and we have also G (n) ={ i I x i,avec I {1,,.., m}} x i = 1. Corollary.: Let x i be the initial square root of the unity associated with p i, then G (n) =<x 1,x,...,x m >. We finish this section with the fact that the algorithm 1.1 remains valid with integers of the form n =p α1 1 pα...pαm m, just replacing LF act := if actors(n)[]; by LF act := if actors(n/)[];, it follows the algorithm 1.. Case 3 : α = Let n 3 be an integer such that its primary decomposition is n =4p α1 1 pα...pαm m. If all α i are nuls, then n =4.We know that (Z/4Z) = {1, 1} =< 1 >, therefore, we suppose that at least one of the α i is not null. Let x be an element of (Z/nZ) such that x =1, that is n divides x 1=(x 1)(x1).Wehave(x1) (x 1) =, therefore divides (x 1) and (x 1). But is not an ordinary prime, indeed we have the following equivalence : x 1[] x 1[8]. It follows that 8 divide x 1=(x 1)(x1). Since GCD(x 1,x1)=, therefore 4 divides (x 1) or (x 1),sox is a solution of one of the following systems : x 1=4p α σ(1) σ(s) K 1 x 1=4p α σ(s1) σ(m) K 507

4 Digital Open Science Index, Mathematical and Computational Sciences waset.org/publication/8554 where σ is a permutation of the set {1,,.., m}. It s clear that each one of these systems has a unique solution modulo n and each system of this form gives a square root of the unity modulo n. We shows also that a two different systems gives two distinct solutions. Therefore, the number of square roots of the unity modulo n is twice the number of partitions of the set {1,,.., m}, that is m. Hence, we have the following result: Proposition.3: Let n 3 be an odd integer, then Ord(G (4n)) = ω(n)1 where ω(n) denote the number of distinct prime factors of n. For simplicity throughout this section we take n 3 to be an integer and n =4p α1 1 pα...pαm m its primary decomposition with at least one of the α i as being not null. Now we start studying of G (n). Consider the following systems : x 1=4p α1 1 pα...pαm 1 x 1=K 1 pα...pαm 1 x 1=4K It s clear that 1 is the only solution of the first system. The second system has only solution which is x 0 = n/1. This solution is called second trivial square root of the unity, we denote by G 0 (n) the cyclic subgroup which is formed by 1 and x 0. Proposition.4: Let the systems : x 1=4p α σ(1) σ(s) K 1 x 1=4p α σ(s1) σ(m) K if we note by x the solution of the first system and y that of the second. then y = x 0 x (and also x = x 0 y). It s clear thatx 0 x is a square root of the unity. We have : x 0 x = (1p α1 1 pα...pαm 1) (14p α σ(1) ) = 1p α σ(1) σ(s) (4K 1 p α σ(s1) σ(m) K 1)Kn Since K 1 is not divisible by 4 and K 1 is not divisible by p α σ(s1) σ(s1),pσ(s) σ(s)... and pα σ(m) σ(m), therefore x 0x 1 is not divisible by 4, p α σ(s1) σ(s1),pσ(s) σ(s)... and pα σ(m) σ(m). So x 0x is solution of the second system,i.e. x 0 x = y. Definition.3: Let x be a square root of the unity modulo n. We said that x is of the first category if 4 divides x 1, else we said that x is of the second category. From the definition, we see that a square root of the unity of the first category is a solution of a system of the form : x 1=4p α σ(1) also a square root of the unity of the second category is the product of a square root of the unity of the first category by x 0. Definition.4: Let x be a square root of unity modulo n. x is said to be initial if all prime factors of n divide x 1 except only one p i, we said that x is associated with p i. And we note : where K is an integer not divisible by p i. Note that there exist two initial square roots of the unity associated with p i, which are the solutions of the following systems : x 1=4p α1. x 1=4p αi We remark that the solution of the first system is of the first category and that of second is of the second category. If we note by x i the solution of the first system and y i that of second, then y i = x i x 0. So the set {1,x 0,x i,y i } is a subgroup of G (n), which we denote by (n). The set formed by 1 and x i ( the initial square root of the unity of the first category associated with p i ) is a cyclic subgroup of order, which we denote by G pi (n) and we have the following isomorphism : (n) G pi (n) G0 (n). More generally, we have the following result : ϕ : Theorem.3: The map G p1 (n)... G pm (n) G 0 (n) G (n) (x 1,...,x m,y) x 1.x,...x m.y is an isomorphism of groups. It s clear that ϕ is an morphism of groups. For showing that ϕ is an isomorphism, we should prove that ϕ is injective and 508

5 Digital Open Science Index, Mathematical and Computational Sciences waset.org/publication/8554 we conclude by cardinality. We have ϕ(x 1,x,...,x m,y)=1 x 1.x,...x m.y =1, if we suppose that there exists an integer i such that x i 1, then p i does not divides x i 1. Since if j i then p i divides x j 1 and p i divides y. Therefore x 1.x,...x m.y 1 is not divisible by p i, that is absurd. Thus x i =1for all i. Finally we have y =1, therefore ϕ is injective. From the previous theorem, we can see that : G (n) ={ i I x i,avec I {1,,.., m}} {1,x 0 } m and we can also show that x 0 x i = 1. Corollary.3: With the previous notations, we have : G (n) =<x 0,x 1,x,...,x m >. Now we give an algorithm in MAPLE that computes the x i.i.e. a generating set of G (n). x 0 is computed from the relation x 0 = n/ 1. The other x i are computed in the same way as the previous case. Gene :=proc(n) local LB, i, LF act, GEN; GEN := [ ]; LB := [ ]; GEN := [op(gen),n/ 1]; LF act := if actors(n/4)[]; Algorithm 1.3 An application example : To find the generators of the group of square root of the unity modulo , we can use the previous algorithm with the command Gene ( ); We have the following result [4863, 441, 6733, 3433], that is the list of generators. We note that the first value of the given list is the second trivial square root of the unity. Case 4 : α 3 Let n 3 be an integer such that its primary decomposition is n = α p α1 1 pα...pαm m with α 3. If all α i are null, then n = α with α 3. Recall that (Z/nZ) is not cyclic and its cardinal is n/. Let x be an element of (Z/nZ) such that x = 1, that is α divides x 1=(x 1)(x 1).WehaveGCD(x 1,x1)=, therefore α 1 divides (x 1) or (x 1).Soxis the solution of one of the following systems : x 1= α 1 K 1 x 1=K 1 ; x 1=K x 1= α 1 K The first system has two solutions which are 1 and α 1 1, the second system has two solutions which are -1 and α 1 1. It s clear that all of the previous solutions are square roots of the unity. We have the following result : Proposition.5: Let n = α with α 3, then G (n) ={1,n/ 1,n/1, 1} We remark that (n/ 1)(n/1) = ( α 1 1)( α 1 1) = 1, therefore G (n) =<n/ 1,n/1>. Now we suppose that n = α p α1 1 pα...pαm m with α 3 and at least one of the α i is not null. Let x be an element of (Z/nZ) such that x =1. Since GCD(x 1,x1)=, then x is the solution of one of the following systems : x 1= α 1 p α σ(1) σ(s) K 1 x 1= α 1 p α σ(s1) σ(m) K where σ is a permutation of the set {1,,.., m}. It s clear that each of these systems has two solutions modulo n and each system of this form gives a square root of the unity modulo n, because x is odd. We shows also that a two different systems give distinct solutions. Therefore, the number of square roots of the unity modulo n is four times the number of partitions of the set {1,,.., m}, that is m. Hence, we have the following result: Proposition.6: Let n 3 be an odd integer, then Ord(G ( α n)) = ω(n) with α 3. For simplicity throughout this section we take n 3 to be an integer and n = α p α1 1 pα...pαm m (α 3) its primary decomposition with at least one of the α i is not null. Now we begin to study G (n). Consider the following systems : x 1= α 1 p α1 1 pα...pαm 1 x 1=K ; 1 pα...pαm 1 x 1= α 1 K It s clear that the first system has two solutions modulo n and 1 is one of these solutions, we note by y 0 the other solution. Also the second system has two solutions modulo n, denoted 509

6 Digital Open Science Index, Mathematical and Computational Sciences waset.org/publication/8554 by y 1 and y. We have : y 0 = α 1 p α1 1 pα...pαm m 1=n/1 and y = y 1 α 1 p α1 1 pα...pαm m, therefore y y 1 =1 α 1 p α1 1 pα...pαm m y 1. Since y 1 is odd, then y y 1 = y 0 and y = y 1 y 0. So, the set {1,y 0,y 1,y } is a subgroup of G (n), which is noted by G 0 (n). Finally remark that : G 0 (n) ={1,y 0 } {1,y 1 }. Definition.5: Let x be a square root of the unity modulo n, We said that x is of the first category if α divides x 1, else we said that x is of the second category. Let x G 0 (n), then x is of the first category if and only if x =1. Definition.6: Let x be a square root of unity modulo n. x is said to be initial if all prime factors of n divide x 1 except only one p i, we said that x is associated with p i. And we note :. where K is an integer not divisible by p i. Note that the initial square roots of the unity associated with p i are the solutions of the following systems : x 1= α 1 p α1 x 1= α 1 p αi Since each of these system has two solutions modulo n, therefore there exist 4 initial square roots of the unity associated with p i. Proposition.7: Let the system : x 1= α 1 p α1 If we denote by x 1 and x the solutions of this system, then x 1 = y 0.x. We have x 1 = x α 1 p α1 1 pα...pαm m, therefore x 1.x = 1 α 1 p α1 1 pα...pαm m x. Since x is odd, then x 1.x = y 0 it follows that x 1 = x.y 0. In the same way, we show that the product of the solutions of the following system: is equal to y 0. x 1= α 1 p αi Proposition.8: there exists an only initial square root of the unity associated with p i and of the first category. Indeed, this square root of the unity is the only solution of the system x 1= α p α1 We denote by (n), the cyclic subgroup of order which is formed by 1 and the initial square root of the unity associated with p i and of the first category. Proposition.9: Let us consider these systems : x 1= α 1 p α σ(1) (1) σ(s) K 1 x 1= α 1 p α σ(s1) σ(m) K where σ is a permutation of the set {1,,.., m}, then the product of each solution of (1) by y 1 or y is a solution of (). Let x be a solution of (1). suppose that x is of the first category, that is Therefore x =1 α p α σ(1). () y 1.x = (1p α1 1 pα...pαi ).(1 α p α σ(1) ) = 1p α σ(1) σ(s) (α K 1 p α σ(s1) σ(m) K)nK. Since α 1 does not divides K and p α σ(s1) σ(s1),pσ(s) σ(s)... and pα σ(m) σ(m) does not divide K 1, then α 1,p α σ(s1) σ(s1),pσ(s) σ(s)...et pα σ(m) σ(m) does not divide α K 1 p α σ(s1) σ(m) K. Hence y 1.x is a solution of (). If z is the other solution of (1), then z = y 0.x. Thus, z.y 1 = y 0.(x.y 1 ). 510

7 Digital Open Science Index, Mathematical and Computational Sciences waset.org/publication/8554 Since (x.y 1 ) is a solution of (), therefore z.y 1 is also a solution of (). Finally, remark that reasoning is also valid to y. If we denote by (n) the set which is formed by the initial square roots of the unity associated with p i and with the elements of G 0 (n), then we have the following result: Corollary.4: (n) is a group and we have : (n) (n) G0 (n). The initial square roots of the unity associated with p i are the solutions of the following systems : x 1= α 1 p α1 i K (1) x 1= α 1 p αi () We deduce that Ord( (n)) = 8. From the previous proposition, we know that the solutions of () are the product of the solutions of (1) by y 1. If we note by x a solution of (1), then the solutions of (1) are x and x.y 0. So, the initial square roots of the unity associated with p i are {x, x.y 0, x.y 1, x.y 0.y 1 }, it follows : (n) ={1,y 0,y 1,y 1.y 0, x, x.y 0, x.y 1, x.y 0.y 1 }. And obviously, we have (n) G pi (n) G0 (n). More generally, we have the following result : ϕ : Theorem.4: The map G p1 (n)... G pm (n) G 0 (n) G (n) (x 1,...,x m,y) x 1,...x m.y is an isomorphism of groups. In the same way as the previous theorem, we show that ϕ is an injective morphism of groups and we conclude by cardinality. The group G 0 (n) is not cyclic, but we have G 0 (n) ={1,y 0 } {1,y 1 }, thus : G (n) G p1 (n) G p Finally we have the following result : (n)... G pm (n) {1,y 0 } {1,y 1 }. Corollary.5: As it is noted above, we have G (n) =<y 0,y 1,x 1,x,...,x m >. Now we give an algorithm in MAPLE that computes x i, y 0 and y 1,i.e. a generating set of G (n). The solution y 0 is computed by the formula y 0 = n/1 and y 1 is a solution of the system : 1 pα...pαm 1 x 1= α 1 K we will choose that satisfied this system 1 pα...pαm 1 x 1= α K Since ( ) implies that α K (n/ α )K 1 =,sowegetk and K 1 with the Bezout algorithm. Therefore y 1 = α K 1n/. The other x i are computed in the same way as the previous case. Gene :=proc(n) local a, LB, i, LF act, GEN; GEN := [ ]; LB := [ ]; a := ifactors(n)[][1][]; GEN := [op(gen),n/ 1]; LB := Bezout( a, n/( a), ); a 1 n/ mod n]; LF act := if actors(n/( a))[]; Algorithm 1.4 An application example : To find the generators of the group of square root of the unity modulo , we can use the previous algorithm with this command : Gene ( ); We have the following result [4863, 441, 6733, 3433], that is the list of generators. We note that the first value of the given list is y 0, and the second is y 1. The choice of y 1 allows us to have : m y 0.y 1 x i = 1. ( ) 511

8 Digital Open Science Index, Mathematical and Computational Sciences waset.org/publication/8554 Indeed, y 0.y 1 is the solution of ( ). Therefore y 0.y 1 m x i = (1p α1 1 pα...pαm 1 ) (1 α p α1 i ) = (1p α1 1 pα...pαm 1 )(1 α p α1 i Kn) = 1[p α1 1 pα...pαm 1 α p α1 i ] Kn It s clear that the term between the brackets is not divisible by α 1,p α1 1,pα...,pαm m. So, y 0.y 1 x i is a solution of this system x 1=K 1 x 1= α 1 p α1 1 pα...pαm Since the solutions of this system are -1 and (n/ 1). To m conclude, just shows that α divides y 0.y 1 x i 1. We have m y 0.y 1 x i 1=(y 0.y 1 1) x i ( x i 1) so it s clear that (y 0.y 1 1) is divisible by α because y 0.y 1 is solution of ( ), and x i 1= α p α1 i Kn, thus x i 1 is divisible by α it follow that α divides m y 0.y 1 x i 1. Now we give an explicit formula for y 1 in special cases. Proposition.10: Let n be an integer of the form 8b, with b is an odd positive integer, then : y 1 = n/41 if b 1[4]. y 1 =3n/41 if b 3[4]. On the first hand, we have (n/4 1) = (p 1) = 14p(p 1), and since divides p 1, then n divides 4p(p 1). Hence (n/41) =1. On the other hand, (n/4 1) 1=n/4 is divisible by all the prime factors of n. Since (n/41)1=(p 1) and b 1[4], then p 1 is divisible by and not by 4. Thus (n/41)1 is divisible by 4 and not by 8, hence the result. We will show this point in the same way. Proposition.11: Let n be an integer of the form α b with b is an odd positive integer and α 3. ifb 1[ α 1 ], the solution of ( ) is : Therefore We have y = (α 1 1)n α 1 1. y 1 = (α 1)n α 1 1. y = (b( α 1 1)1) = 14b ( α 1 1) 4b( α 1 1) = 14b( α b( α 1) α 1 b 1). Since α 1 divides b 1, then n divides 4b( α b( α 1) α 1 b 1), therefore y =1. It s clear that all the prime factors of n divide y 1. Onthe other hand, y 1=b( α 1 1) = α b (b 1), then α divides y 1. So, y is solution of ( ). We know that y 1 = y n/, it follows the expression of y 1. III. CONCLUSION For the cardinal of G (n), we have the following theorem : Theorem 3.1: Let n 3 be an odd integer, then : Ord(G (n)) = ω(n) Ord(G (n)) = ω(n) Ord(G (4n)) = ω(n)1 Ord(G ( α n))= ω(n) with α 3 where ω(n) is the number of distinct prime factors of n. Now we give an algorithm that computes a generating set for G (n), where n is an integer. Gene :=proc(n) local a, LB, i, LF act, GEN; GEN := [ ]; LB := [ ]; if(n mod =1)then LF act := if actors(n)[]; else a := ifactors(n)[][1][]; if a =1then LF act := if actors(n)[]; elifa =then GEN := [op(gen),n/ 1]; 51

9 Digital Open Science Index, Mathematical and Computational Sciences waset.org/publication/8554 LF act := if actors(n/4)[]; else GEN := [op(gen),n/ 1]; LB := Bezout( a, n/( a), ); a 1 n/ mod n]; LF act := if actors(n/( a))[]; Algorithm 1.5 Complexity of the algorithm : It s clear that the complexity of the Algorithm 1.5 is the same as the Algorithm 1.1. Recall that the number of distinct prime factors of a number n is denoted ω(n). We know that ω(n) =O(ln(ln n)) (see [9] and [10]), and the complexity of the Extended Euclidean algorithm is O(ln n) (see [3] and [4]). Therefore the complexity of Algorithm 1.1 without the factorization is O(ln(ln n) ln n). REFERENCES [1] J-P. Serre, A Course in Arithmetic. Graduate Texts in Mathematics, Springer, 1996 [] S. Lang, Undergraduate Algebra, nd ed. UTM. Springer Verlag,1990 [3] H. Cohen, A course in computational algebraic number theory. Springer-Verlag, [4] V. Shoup, A Computational Introduction to Number Theory and Algebra. Cambridge University Press, 005. [5] David M. Bressoud, Factorization and Primality Testing. Undergraduate Texts in Mathematics, Springer-Verlag, New York, [6] E. Bach, A note on square roots in finite fields. IEEE Trans. Inform. Theory, 36(6): , Eric [7] E. Bach and K. Huber, Note on taking square-roots modulo N. IEEE Transactions on Information Theory, 45():807809, [8] D. Shanks, Five number-theoretic algorithms. In Proc. Second Munitoba Conf. Numerical Math , 197. [9] Hardy, G. H, Ramanujan: Twelve Lectures on Subjects Suggested by His Life and Work, 3rd ed. New York: Chelsea, G. H. [10] Hardy and E. M.Wright, An introduction to the theory of numbers, 4th ed. Oxford University Press,