TAIWANESE JOURNAL OF MATHEMATICS Vol. 1, No. 5, pp. 1191-1199, August 008 This paper is available online at http://www.tjm.nsysu.edu.tw/ THE RESTRICTED SOLUTIONS OF ax + by =gcd(a, b) Ju-Si Lee Abstract. Let D denote a principle ideal domain with identity element 1. Fix three elements a, b, d in D with gcd(a, b) = d. We show there exist two elements x, y in D with either gcd(a, y) =1or gcd(b, x) =1such that ax+by = d. Moreover we show there exist x, y in D such that gcd(a, y) = 1, gcd(b, x) =1and ax + by = d if and only if for each prime divisor p of d with a complete set of residues modulo p containing exactly elements, the power of p appearing in the factorization of a is different to that of b. We apply our results to the study of double-loop networks. 1. INTRODUCTION Let D denote a principle ideal domain (PID) with identity element 1. A special example is when D = Z, the set of integers. Fix three elements a, b, d in D with gcd(a, b)=d. It is well known that there exist x, y D such that (1.1) ax + by = d. For the converse, if ax + by = d for some elements x,y,d D, then d is a divisor of d. Observe each of gcd(a, y), gcd(b, x) and gcd(x, y) is a divisor of d. In fact gcd(x, y) =1otherwise we can divide each of x, y and d in (1.1) by gcd(x, y) to obtain that gcd(a, b) is a divisor of d/gcd(x, y) for a contradiction. Now it is natural to ask if gcd(a, y)=1or gcd(b, x)=1? Our first theorem will show. Theorem A. ax + by = d. There exist two elements x, y in D with gcd(b, x) =1and Moreover our second theorem will show Received July 18, 006, accepted December 15, 006. Communicated by Xuding Zhu. 000 Mathematics Subject Classification: 11A05. Key words and phrases: Principal ideal domain, Double-loop network. 1191
119 Ju-Si Lee Theorem B. There exist x, y in D such that gcd(a, y)=1, gcd(b, x)=1and ax + by = d if and only if for each prime divisor p of d with a complete set of residues modulo p containing exactly elements, the power of p appearing in the factorization of a is different to that of b. See Theorem 3.3 for more detailed description of Theorem B. Note that if D is a unique factorization domain (UFD), Theorem A fails. For example, with D = Z[x], a=and b = x, we have gcd(a, b)=1, but af(x)+bg(x) 1for all f(x),g(x) Z[x]. The necessary condition appearing in our Theorem B can not be omitted. For example for D = Z, a =and b =6, we have x +6y =, where x = +3k and y =1 k for some integer k. Since x and y always have different parity, it can not have gcd(a, y)=1and gcd(b, x)=1. Double-loop networks on integers were introduced by Wang and Coppersmith for the purpose of giving efficient and secure communication [3]. We will generalize the concept of double-loop network on integers to on arbitrary principle ideal domain D with identity element. We apply the above mentioned results to the double-loop network on Z N. The paper is organized as follows. In section we review a few definitions and properties on the principle ideal domain D. The main theorems are proved in section 3. In section 4, we give a few examples of principle ideal domains, and show how our main results are applied to these examples. The double-loop network on D was introduced in section 5.. PRELIMINARIES Let D be a UFD with identity element 1. Fix a nonzero nonunit element n D. For a, b D, we say a and b are congruent modulo n, or in the notation a b (mod n), ifn (a b). Obviously, the congruent is an equivalent relation on D. A nonempty subset S of D is said to be a set of residues modulo n if any two elements of S are not congruent modulo n. A maximum set R of residues modulo n is called a complete set of residues modulo n. In fact, a complete set of residues modulo n is made by choosing one representative from each congruent class. We set r n the number of elements of a complete set R of residues modulo n, that is, r n = R. Note that ar + b := {ar + b r R} is also a complete set of residues modulo n, where a, b are elements in D with gcd(a, n) =1. The following two basic theorems will be used in our proofs. See [1] (p. 355 and 363) for reference. Theorem.1. Let D denote a PID with identity element 1 and fix a, b D, not both zero. If gcd(a, b)=d then there exist x, y D such that ax + by = d.
The Restricted Solutions of ax + by = gcd(a, b) 1193 Corollary.. Let D denote a PID with identity element 1 and fix a, b D, not both zero. Then gcd(a, b) =1if and only if there exist x, y D such that ax + by =1; in particular, gcd(a, y) =1and gcd(b, x)=1. Note that Theorem B is a generalization of Corollary.. 3. PROOFS OF MAIN THEOREMS Throughout this section, let D denote a PID with identity element 1 and let a, b D, not both zero. The following theorem proves Theorem A in the introduction. Theorem 3.1. There exist x, y D such that gcd(b, x) =1and ax + by = gcd(a, b). Proof. Suppose a = a 1 d and b = b 1 d, where a 1,b 1,d D and gcd(a, b)=d. Note that gcd(a 1,b 1 )=1. By Theorem.1, there exist x 0,y 0 D such that a 1 x 0 + b 1 y 0 =1. Note that gcd(b 1,x 0 )=1by Corollary.. Set c := p, where p the product is over all primes p d and p b 1. In particular, gcd(b 1,c)=1. Let R be a complete set of residues modulo c. Hence b 1 R + x 0 is also a complete set of residues modulo c. Pick an element x b 1 R + x 0 which is congruent to 1 modulo c. Say x = rb 1 + x 0 1(modc) for some r R. Then gcd(c, x) =1 and gcd(b 1,x)=1. Hence gcd(b, x) =1. By setting y = ra 1 + y 0, we have ax + by = d. To prove Theorem B we need a lemma first. Lemma 3.. Suppose r p 3 for each common prime divisor p of a and b. Then there exist x, y D such that gcd(a, y) =1, gcd(b, x) =1and ax + by = gcd(a, b). Proof. Set d := gcd(a, b) and suppose d = p t 1 1 p t p tn n, where p i are distinct primes in D and t i are positive integers. Set d 1 = p t 1 1 p t p t n 1 n 1. We prove the lemma by induction on n. For n =0, we have d =1, and the lemma holds by Corollary.. Suppose n 1. Without loss of generality, suppose a = p tn+s n a 1 and b = p tn n b 1 where p n a 1, p n b 1, t n is a positive integer and s is a nonnegative integer. Then gcd(p s na 1,b 1 )= p t 1 1 p t p t n 1 n 1 = d 1. By induction assumption, there exist x 1,y 1 D such that gcd(p s na 1,y 1 )=1, gcd(b 1,x 1 )=1and p s na 1 x 1 + b 1 y 1 = d 1. Hence gcd(p i,y 1 )= 1, gcd(p i,x 1 )=1for i =1,,,n 1, p n y 1 for s>0, and ax 1 + by 1 = d. Let R be a complete set of residues modulo p n. Since p n a 1 and p n b 1, b 1 R+x 1 and ( a 1 )R+y 1 are complete sets of residues modulo p n, each containing
1194 Ju-Si Lee at least three elements. Hence there exists r R such that both b 1 r + x 1 0 (mod p n ) and a 1 r +y 1 0(modp n ). Set x = b 1 r +x 1 and y = p n s a 1 r +y 1. Hence gcd(a, y) =1, gcd(b, x)=1and ax+by = a(b 1 r+x 1 )+b( p n s a 1 r+y 1 )= ax 1 + by 1 = d. By Lemma 3., to prove Theorem B, we consider only each prime divisor p of gcd(a, b) with r p =. In fact, {0, 1} is a complete set of residues modulo p for each prime divisor p of gcd(a, b) with r p =. Let S = {p p is a prime factor of gcd(a, b) withr p =}. Observe that if p S and c D, then p is not a factor of c if and only if c 1 (mod p). The following theorem proves Theorem B. Theorem 3.3. There exist x, y D such that gcd(a, y) =1, gcd(b, x) =1, and ax + by =gcd(a, b) if and only if either S =0, or there exists an integer i such that gcd(a, p i ) gcd(b, p i ) for each p S. Proof. (= ) We suppose that S > 0 and there exists a p S such that gcd(a, p i )=gcd(b, p i ) for each positive integer i. Note that gcd(a, p) = gcd(b, p)=p. To get the contrary let m denote the positive integer such that p m a, p m b, p m+1 a, p m+1 b, and suppose ax + by =gcd(a, b) for some x, y D with gcd(a, y)=1and gcd(b, x)=1. Since p is a divisor of gcd(a, b), x y 1 (mod p). Observe a p m x + b gcd(a, b) y = pm p m, where a p m b gcd(a, b) pm p m 1 (mod p). This implies either x 0 or y 0(modp), a contradiction. ( =) We prove by induction on S. It follows from Lemma 3. if S =0. Suppose S > 0. Pick p S. Let m denote the least positive integer such that gcd(a, p m+1 ) gcd(b, p m+1 ). Set a := a/p m, and b := b/p m and S = {p p is a prime factor of gcd(a,b )withr p =}. Observe S = S {p}, gcd(a, b) =gcd(a,b )p m, and exactly one of a 1 (mod p) and b 1(modp). Without loss of generality, say a 1(modp) and b 0(modp). By induction there exists x,y D such that gcd(a,y )= 1, gcd(b,x )=1, and a x + b y =gcd(a,b ). Hence x 1(modp) and ax + by =gcd(a, b). Set { 0, if y 1 (mod p), c := 1, if y 0 (mod p);
The Restricted Solutions of ax + by = gcd(a, b) 1195 x := x + cb and y := y ca. Then x 1(modp), y 1(modp). Hence we have gcd(a, y)=1, gcd(b, x)=1, and ax + by = ax + by =gcd(a, b). Note that Theorem 3.3 still holds if D is a UFD and assume that there exist x, y D such that ax + by =gcd(a, b). A tittle modification in each of the proofs of Theorem 3.1, Lemma 3. and Theorem 3.3 will do this. 4. EXAMPLES We apply Theorem 3.3 in each of the cases D = Z, D = F [x], and the imaginary quadratic fields D = O( di) where d =1,, 3, 7, 11, 19, 43, 67, 163. I. D = Z. By Theorem 3.1, we have the following theorem. Theorem 4.1. Given two integers a and b, not both zero, there exist integers x and y such that gcd(x, b)=1and ax + by =gcd(a, b). Observe that r p =in Z if and only if p =. By Theorem 3.3, we have the following theorem. Theorem 4.. Given two positive integers a and b. Let a = r a 1 and b = s b 1 where r and s are nonnegative integers and a 1,b 1 are odd. Then there exist integers x and y such that gcd(a, y) =1, gcd(b, x) =1and ax + by = d if and only if either r = s =0or r s. II. D = F [x]. Let D be an integral domain, then D[x] is a PID if and only if D is a field [1] (p. 345). Hence we consider F [x] where F is a field. Obviously, if F has at least three elements, then r p(x) > for each irreducible polynomial p(x) in F [x]. And in Z [x], r p(x) =if and only if either p(x) =x or p(x) =1+x. By Lemma 3. and Theorem 3.3, we have the following two theorems. Theorem 4.3. Let F be a field which has at least three elements. If f(x) and g(x) are polynomials of F [x], not both zero, then there exist polynomials h(x) and k(x) of F [x] such that gcd(f(x),k(x)) = 1, gcd(g(x),h(x)) = 1 and f(x)h(x)+g(x)k(x)=gcd(f(x),g(x)). Theorem 4.4. Suppose f(x), g(x) are two polynomials of Z [x], not both zero, gcd(f(x),g(x)) = d(x). Let f(x) =x r 1 (1 + x) r f 1 (x) and g(x) =x s 1 (1 +
1196 Ju-Si Lee x) s g 1 (x) where r 1,r,s 1,s are nonnegative integers and x and 1+x are not prime divisors of both f 1 (x) and g 1 (x). Then there exist polynomials h(x) and k(x) of Z [x] such that gcd(f(x),k(x)) = 1, gcd(g(x),h(x)) = 1 and f(x)h(x)+ g(x)k(x) =d(x) if and only if either (1) r 1 = r = s 1 = s =0, or () r 1 s 1 and r s. III. In an imaginary quadratic fields with class number 1 (UFD). Let O( D) denote the set of all algebraic integers of Q( D) where D is square-free, then (1) O( D)={l + m D l, m Z} = Z[ D], ifd or 3(mod4). () O( D)={ l + m D l, m Z,l m (mod )}, ifd 1(mod4). There are exactly nine imaginary quadratic fields with class number 1 : d = 1,, 3, 7, 11, 19, 43, 67 and 163 [] where D = d. In the following we discuss the nine imaginary quadratic fields. In this paper, an algebraic integer will be denoted by the form a + b D if D or 3(mod4), α = a + b D if D 1(mod4) where a, b Z. Define the norm of α by a Db if D or 3(mod4), N(α) = a Db if D 1(mod4). 4 Every complete set of residues modulo α has N(α) elements. Hence r p =if and only if N(p)=. The following lemma is clear. Lemma 4.5. (1) p = ±1 ± i in Z(i). () p = ± i in Z( i). r p =in O( di) if and only if (3) ±1 ± 7i in O( 7i). By Lemma 4.5, Lemma 3., and Theorem 3.3, we have 1. d =1.
The Restricted Solutions of ax + by = gcd(a, b) 1197 Theorem 4.6. Given α, β Z[i],not both zero, let α = (1 + i) r α 1 and β =(1+i) s β 1 where r and s are nonnegative integers and 1+i is not a divisor of both α 1 and β 1. Then there exist integers x and y of Z[i] such that gcd(α, y) =1, gcd(β, x)=1and αx + βy =gcd(α, β) if and only if r = s =0or r s.. d =. Theorem 4.7. Given α, β Z[ i],not both zero, let α =( i) r α 1 and β = ( i) s β 1 where r and s are nonnegative integers and i is not a divisor of both α 1 and β 1. Then there exist integers x and y of Z[ i] such that gcd(α, y) =1, gcd(β, x)=1and αx + βy) =gcd(α, β) if and only if r = s =0or r s. 3. d =7. Theorem 4.8. Given α, β O( 7i), not both zero, let α = γ r 1 δ r α 1 and β = γ s 1δ s β 1 where γ = 1+ 7i, δ = 1 7i, r 1,r,s 1,s are nonnegative integers, and N(α 1 ), N(β 1 ) are odd. Then there exist integers x and y of O( 7i) such that gcd(α, y) =1, gcd(β, x) =1and αx + βy =gcd(α, β) if and only if either (1) r 1 = r = s 1 = s =0, or () r 1 s 1 and r s. 4. d =3, 11, 19, 43, 67 and 163. Theorem 4.9. Given α, β O( di), not both zero, then there exist integers x and y of O( di) such that gcd(α, y) =1, gcd(β, x) =1and αx + βy = gcd(α, β). 5. AN APPLICATION Let D be a principle ideal dommain with identity 1. Fix elements a, b, N D and a complete set D N of residues modulo N. Definition 5.1. Let DL(D N,a,b) be a digraph with vertex set D N and directed edge set {xy y x + a (mod N) ory x + b (mod N), x,y D N }. DL(D N,a,b) is called a double-loop network on D N modulo N. Wong and Coppersmith [3] introduced the network DL(N;1,h) (DL( Z N ;1,h)), where N, h are positive integers. Fiol et al. [4] extended it to DL(N; a, b) for nonnegative integers a, b, N with 0 a<b<n. They proved that DL(N; a, b) is strongly connected if and only if gcd(n, a, b)=1.
1198 Ju-Si Lee The minimum-distance diagram of a double-loop network DL(N; a, b) gives a shortest path from node u to node v for any u, v. Since a double-loop network is node-symmetric, it suffices to give a shortest path from node 0 to any other node. Let 0 occupy cell (0,0). Then, v occupies cell (i, j) if and only if ia + jb v(modn) and i + j is the minimum among all satisfying the congruence. Wong and Coppersmith [3] proved that the diagram is always an L-shape. Let d(k) denote the number of cells (i, j) in a L-shape such that i + j = k. Hwang and Xu [5] defined two double-loop networks,or two L-shapes, to be equivalent if they have the same d(k) for all k. Hwang and Li [6] defined two double-loop networks DL(N; a, b) and DL(N; a,b ) to be isomorphic if there exists a z prime to N such that a az, b bz (mod N). It can be easily seen that two isomorphic double-loop networks are equivalent, but the reverse is not true. Given a strongly connected double-loop network DL(N; a, b), gcd(n, a, b) =1. Theorem 5.. Suppose gcd(n, a) =d and gcd(d, b) =1. Then DL(N; a, b) is isomorphic to DL(N; d, s) for some s with gcd(d, s)=1. Proof. By Theorem 4.1, there exists z such that gcd(n, z) =1and az d (mod N). Let s bz (mod N) and 0 s N 1. Since gcd(n, a, b)=1,we have gcd(d, s)=1and DL(N; a, b) is isomorphic to DL(N; d, s). Hence, when double-loop networks DL(N; a, b) are discussed, it suffices to consider only those DL(N; a, b) with properties that a N, and gcd(a, b)=1. ACKNOWLEDGMENT The author wishes to thank Professor Chih-Wen Weng for providing comments to the paper. REFERENCES 1. D. S. Malik, John N. Mordeson and M. K. Sen, Fundamentals of Abstract Algebra, The McGraw-Hill Companies, Inc. 1997, Chapter 15.. H. M. Stark, A complete determination of the complex quadratic fields of classnumber one, Michigan Math. J., 14 (1967), 1-7. 3. C. K. Wong and D. Coppersmith, A combinatorial Problem Related to Multimodule Organizations, J. ACM, 1 (1974), 39-40. 4. M. A. Fiol, M. Valero, J. L. A. Yebra and T. Lang, Optimization of Double-Loop equivalent Structures for Local Networks, Proc. 19th int l Symp. MIMI 8, 198, pp. 37-41.
The Restricted Solutions of ax + by = gcd(a, b) 1199 5. F. K. Hwang and Y. H. Xu, Ddouble-loop networks with minimum delay, Discr Math., 66 (1987), 109-118. 6. F. K. Hwang and W.-C. W. Li, Reliabilities of double-loop networks, Prob Eng Inf Sci., 5 (1991), 55-7. Ju-Si Lee Department of Mathematics, National Kaohshiung Normal University, Kaoshiung 80, Taiwan E-mail: jusilee@nknucc.nknu.edu.tw