Acta Arith., 140(2009), no. 4, ON BIALOSTOCKI S CONJECTURE FOR ZERO-SUM SEQUENCES. 1. Introduction

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1 Acta Arith., 140(009), no. 4, ON BIALOSTOCKI S CONJECTURE FOR ZERO-SUM SEQUENCES SONG GUO AND ZHI-WEI SUN* arxiv: v3 [math.co] 16 Dec 009 Abstract. Let n be a positive even integer, and let a 1,...,a n and w 1,...,w n be integers satisfying n a k n w k 0 (mod n). A conjecture of Bialostocki states that there is a permutation σ on {1,...,n} such that n w ka σ(k) 0 (mod n). In this paper we confirmthe conjecturewhenw 1,...,w n formanarithmeticprogressionwith even common difference. 1. Introduction A finite sequence S of terms from an (additive) abelian group is said to have zero-sum if the sum of the terms of S is zero. In 1961 P. Erdős, A. Ginzburg and A. Ziv [3] proved that any sequence of n 1 terms from an abelian group of order n contains an n-term zero-sum subsequence. This celebrated EGZ theorem is is an important result in combinatorial number theory and it has many different generalizations [5, 6, 7, 8] including Sun s recent extension involving covering systems. The following theorem is called the weighted EGZ theorem. It was conjectured by Y. Caro [] and proved by D. J. Grykiewicz [4]. Theorem 1.1 (Weighted EGZ Theorem). Let n be a positive integer and let w 1,...,w n Z n = Z/nZ with n w k = 0. If a 1,a,...,a n 1 is a sequence of elements from Z n, then n w ka jk = 0 for some distinct j 1,...,j n {1,...n 1}. Recently Bialostocki raised the following challenging conjecture. Conjecture 1.1 (Bialostocki [1, Conjecture 14]). Let n be a positive even integer. Suppose that a 1,...,a n and w 1,...,w n are zero-sum sequences with terms from Z n. Then there exists a permutation σ S n such that n w ka σ(k) = 0, where S n denotes the symmetric group of all permutations on {1,...,n}. The conjecture has been verified for n =, 4, 6, 8. It fails for n = 3,5,7,... For example, {a 1,a,a 3 } = {w 1,w,w 3 } = Z 3 gives a counterexample for n = 3. Key words and phrases. Zero-sum sequence, Bialostocki s conjecture, Erdős-Ginzburg- Ziv theorem 000 Mathematics Subject Classification. Primary 11B75; Secondary 05A05, 0D60. *Supported by the National Natural Science Foundation (grant ) of China.

2 SONG GUO AND ZHI-WEI SUN In this paper we mainly establish the following result. Theorem 1.. Let n be a positive even integer, and let a 1,...,a n Z with n a k 0 (mod n). Then there exists a permutation σ S n such that n ka σ(k) 0 (mod n/). Consequently, if w 1,...,w n Z form an arithmetic progression with even common difference, then n w ka σ(k) 0 (mod n) for some σ S n. We are going to present two lemmas in the next section and then give our proof of Theorem 1. in Section 3.. Two Lemmas Lemma.1. Let n = mq with m,q Z + = {1,,3,...} and m. Let d Z + be a divisor of q, and let a 1,...,a n Z. Then, there is a partition I 1,,I m of [1,n] = {1,...,n} such that for each s = 1,...,m we have I s = q and d i I s a i = {a i mod d : i I s } = 1. Proof. By induction on m, it suffices to show that there exists an I [1,n] with I = q such that for each J {I, [1,n] \ I} we have {a j mod d : j J} = 1 or j J a j 0 (mod d). To achieve this we distinguish three cases. Case 1. {a i mod d : i [1,n]} = 1. In this case, I = [1,q] works for our purpose. Case. {a i mod d : i [1,n]} =. Suppose that {a i mod d : i [1,n]} = {rmod d, r mod d}, where r,r [0,d 1], r r (mod d), and a i r (mod d) for at least n/ values of i [1,n]. Choose I 0 {i [1,n] : a i r (mod d)} with I 0 = q n/. Let i 0 I 0 and j 0 Ī0 = [1,n]\I 0 with a j0 r (mod d). When j Ī 0 a j 0 (mod d), we have both a i 0 r +r 0 (mod d) and i (I 0 \{i 0 }) {j 0 } j (Ī0\{j 0 }) {i 0 } a j 0 r +r 0 (mod d). Thus, there always exists an I [1,n] with I = q such that {a i mod d : i I} = 1 or i I and also j Ī a j 0 (mod d). a i 0 (mod d),

3 ON BIALOSTOCKI S CONJECTURE FOR ZERO-SUM SEQUENCES 3 Case 3. {a i mod d : i [1,n]} >. As n q q 1, by the EGZ theorem there is an I 0 [1,n] with I 0 = q such that i I 0 a i 0 (mod q). For Ī0 = [1,n]\I 0, we clearly have Ī0 = (m 1)q. Set b = a 1 + +a n j Ī 0 a j (mod q). Suppose that a j a i 0 or b (mod d) for any i I 0 and j Ī0. Then {a i mod d : i I 0 )} and {a j mod d : j Ī0)}. If i 1,i I 0, j Ī0 and a j a i1,a i (mod p), then a j a i1 b a j a i (mod d) and hence a i1 a i (mod d). So, if {a i mod d : i I 0 } = then {a j mod d : i Ī0} {a i mod d : i I 0 } which contradicts {a i mod d : i I 0 } >. Similarly, if {a j mod d : j Ī0} = then we also have a contradiction. When {a i mod d : i I 0 } = {a j mod d : j Ī0} = 1, we cannot have {a i mod d : i [1,n]} >. By the above, there are i 0 I 0 and j 0 Ī0 such that Set Then and a j0 a i0 0,b (mod d). I = (I 0 \{i 0 }) {j 0 } and Ī = [1,n]\I = (Ī0 \{j 0 }) {i 0 }. a i = a i a i0 +a j0 = 0 a i0 +a j0 0 (mod d) i I 0 i I a j = j a j0 +a i0 b+a i0 a j0 0 (mod d). j Ī0a j Ī Note that I = q and Ī = (m 1)q. Combining the above and using the induction argument, we see that the desired result holds for any m =,3,4,... Lemma.. Let a 1,...,a n Z with n = p α, where p is an odd prime and α is a positive integer. If n a k 0(mod p) or {a k mod p : k [1,n]} = 1, then there exists a permutation σ S n such that n ka σ(k) 0 (mod n). Proof. If a := n a k 0 (mod p), then there is an l [1,n] such that al + n ka k 0 (mod p) and hence n n n ka σ(k) (k +l)a k ka k +la 0 (mod p α ), where σ(k) is the least positive residue of k l modulo n.

4 4 SONG GUO AND ZHI-WEI SUN In the case a 1 a n (mod p), it is clear that p p ka k a 1 k = a 1 p p+1 0 (mod p). Thus we have the desired result for α = 1. Now let α > 1 and assume the desired result with α replaced by. As mentioned above, the desired result holds if n a k 0 (mod p). Suppose that a 1 a n (mod p) and set b k = (a k a 1 )/p for k = 1,...,n. In light of Lemma.1, there exists a partition I 1 I p of [1,n] with I 1 = = I p = p such that for any s = 1,...,p either {b k mod p : k I s } = 1 or k I s b k 0 (mod p). By the induction hypothesis, there are one-to-one mappings σ s : [1,p ] I s (s = 1,...,p) such that p kb σs(k) 0(mod p ) for all s = 1...,p. For s [1,p] and t [1,p ] define σ(p (s 1)+t) = σ s (t). Then σ S n and n ka σ(k) = n ka 1 +p k = pα (p α +1) a 1 +p p p p kb σ(k) p p (p (s 1)+t)b σs(t) tb σs(t) 0 (mod p α ). This concludes the induction step and we are done. 3. Proof of Theorem 1. Proof of Theorem 1.. We use induction on ν(n), the total number of prime divisors of n. In the case ν(n) = 1, clearly n = and the desired result holds trivially. Now let ν(n) > 1 and assume the desired result for those even positive integers with less than ν(n) prime divisors. Case 1. n = α for some α. By the EGZ theorem, there is an I [1,n] with I = n/ = such that i I a i 0 (mod ). Note that for Ī = [1,n]\I we also have a j = j Ī n a k i I a i 0 (mod ).

5 ON BIALOSTOCKI S CONJECTURE FOR ZERO-SUM SEQUENCES 5 Bytheinductionhypothesis, forsomeone-to-onemappingsσ 0 : [1,n/] I and σ 1 : [1,n/] Ī we have Observe that ka σ0 (k) ka σ1 (k) 0 (mod ). (k 1)a σ1 (k) ka σ1 (k) a j 0 (mod ). j Ī For k [1,n/] and r [0,1] define σ(k r) = σ r (k). Then σ S n and n ja σ(j) = ka σ0 (k) + (k 1)a σ1 (k) 0 (mod ). j=1 Thus we have the desired result for n = α. Case. n has an odd prime divisor p. Write n = p α m with α,m > 0 and p m. With the help of Lemma.1 there is a partition I 1 I m of [1,n] with I 1 = = I m = p α such that for each s = 1,...,m either {a i mod p : i I s } = 1 or i I s a i 0 (mod p). Combining this with Lemma., we see that for each s [1,m] there is a one-to-one mapping σ s : [1,p α ] I s such that p α ta σ s(t) 0 (mod p α ). Set b s = k I s a k for s = 1,...,m. Then m n b s = a k = a k 0 (mod m). k I 1 I m As m and ν(m) < ν(n), by the induction hypothesis, for some τ S m we have m sb τ(s) 0 (mod m) and hence Note also that m Therefore m p α m p α sa στ(s) (t) = ta στ(s) (t) = p α As p α is relatively prime to m, m m sb τ(s) 0 (mod m). p α ta σs(t) 0 (mod p α ). (p α s+mt)a στ(s) (t) 0 (mod p α m). {p α s+mt : s [1,m] and t [1,p α ]}

6 6 SONG GUO AND ZHI-WEI SUN is a complete system of residues modulo n = p α m. For any k [1,n], there are unique s [1,m] and t [1,p α ] such that k p α s+mt (mod n), and we define σ(k) = σ τ(s) (t). Then σ S n and also n ka σ(k) 0(mod n). This concludes the induction step. In view of the above, we have completed the proof of Theorem 1.. References [1] A. Bialostocki, Some problems in view of recent developments of the Erdős-Ginzburg- Ziv Theorem, Integers 7 (007), no., # A07, 10 pp (electronic). [] Y. Caro, Zero-sum problems a survey, Discrete Math. 15 (1996), [3] P. Erdős, A. Ginzburg and A. Ziv, Theorem in additive number theory, Bull. Res. Council Israel 10F (1961), [4] D. J. Grynkiewicz, A weighted Erdős-Ginzburg-Ziv Theorem, Combinatorica 6 (006), [5] D. J. Grynkiewicz, On the number of m-term zero-sum subsequences, Acta Arith. 11 (006), [6] Y. O. Hamidoune, O. Ordaz and A. Ortuño, On a combinatorial theorem of Erdős, Ginzburg and Ziv, Combin. Probab. Comput. 7 (1998), [7] Z. W. Sun, Zero-sum problems for abelian p-groups and covers of the integers by residue classes, Israel J. Math. 170 (009), [8] R. Thangadurai, Non-canonical extensions of Erdős-Ginzburg-Ziv theorem, Integers (00), #A07, 14 pp (electronic). (Song Guo) Department of Mathematics, Huaiyin Normal College, Huaian 3300, People s Republic of China address: guosong77@hytc.edu.cn (Zhi-Wei Sun) Department of Mathematics, Nanjing University, Nanjing 10093, People s Republic of China address: zwsun@nju.edu.cn