ACTA ARITHMETICA 1102 (2003) On some developments of the Erdős Gnzburg Zv Theorem II by Are Balostock (Moscow, ID), Paul Derker (Moscow, ID), Davd Grynkewcz (Pasadena, CA) and Mark Lotspech (Caldwell, ID) 1 Introducton Let S be a sequence of elements from the cyclc group Z m We say S s zsf (zero-sum free) f there does not exst an m-term subsequence of S whose sum s zero Let g(m, k) (resp g (m, k)) denote the least nteger such that every sequence S wth at least (resp wth exactly) k dstnct elements and length g(m, k) (resp g (m, k)) must contan an m-term subsequence whose sum s zero By an affne transformaton n Z m we mean a map of the form x ax + b, wth a, b Z m and gcd(a, m) = 1 Furthermore, let E(m, s) denote the set of all equvalence classes of zsf sequences S of length s, up to order and affne transformaton, that are not a proper subsequence of another zsf sequence Usng the above notaton, the renowned Erdős Gnzburg Zv Theorem ([1], [11]) states that g(m, 1) = g(m, 2) = 2m 1 for m 2 The functon g(m, k) was ntroduced n [4], where t was shown that g(m, 4) = 2m 3 for m 4 Furthermore, based on a lower bound constructon the authors conectured the value of g(m, k) for fxed k and suffcently large m Concernng the upper bound, they establshed an upper bound for m prme modulo the affrmaton of the Erdős Helbronn conecture (EHC) Snce then, the EHC has been affrmed [9], [2], moreover, the bound gven n [4] was extended for nonprmes n [19] As wll later be seen, t s worthwhle to menton that the affrmaton of the EHC has resulted n several attempted generalzatons and related results [6], [15], [22],[25] Other relevant developments concernng g(m, k) appear n [3], [5], [7], [13], [14], [16], [23] For example, the value g(m, 3) = 2m 2 was determned n [3], and the closely related functon g (m, k), ntroduced n [5] and further nvestgated n [4], [14], [16], was determned for all m and k satsfyng k > m/2 + 1 2000 Mathematcs Subect Classfcaton: Prmary 11B75; Secondary 11B50, 05D10 [173]
174 A Balostock et al Ths paper started under the authorshp of the frst, second and fourth authors, and was cted n [4] as n preparaton Actually, a rough draft was ready determnng g(m, 5) usng the methods of [4] The motvaton for the current paper s twofold Frst, the thrd author was able to determne exactly g(m, k) for fxed k and large m by mprovng the known lower bound constructon and adaptng the proof used by Gao and Hamdoune [19] to obtan a better upper bound than the one conectured n [4] Consequently, the followng conecture of [4] has been affrmed Conecture 11 For every k 2, there exsts an nteger m 0 = m 0 (k) such that f m > m 0, then (a) g(m, k) = 2m c, where c = c(k) s ndependent of m, (b) g (m, k) = g(m, k) Thus, the g(m, k) problem for fxed k and large m has been put to rest Second, as several recent works, eg [21], [27], use the known values of g(m, k) and E(m, s) for k 4 and s 2m 3, and as t s lkely that g(m, 5) wll be needed for further zero-sum applcatons, we determne g(m, 5) for every m 5 The paper s organzed as follows In Secton 2, defntons and notaton are ntroduced and known results needed later n the paper are lsted In Secton 3, frst the upper bound proof of [19] s adapted to fnd, for a sequence S wth S 2m m/4 2, necessary and suffcent condtons n terms of a system of nequaltes over the ntegers for S to be zsf Ths result and a lower bound constructon mply the value of g(m, k) for fxed k and large m Followng s the affrmaton of Conecture 11 Secton 4 contans the evaluaton of g(m, 5) for every m 5 The paper concludes wth an appendx lstng the elements of E(m, s) for every m and s satsfyng 2m 2 s max{2m 8, 2m m/4 2} 2 Prelmnares Let G denote an abelan group of order m As we work smultaneously wth the cyclc group Z m of resdue classes modulo m and the addtve group of the ntegers Z, for α Z m we denote by α the least postve nteger that s congruent to α modulo m A sequence S of elements from G s abbrevated as a strng usng exponental notaton (eg the sequence 1, 1, 1, 2, 2, 2, 2 s abbrevated by 1 3 2 4 ) Furthermore, the length of S s denoted by S Let t be a nonnegatve nteger Denote by S the set of all sums over nonempty subsequences of S, and by t S (resp t S) the set of all sums over subsequences of S of length exactly (resp at least) t If A, B G, then ther sumset, A + B, s the set of all possble parwse sums, e {a + b a A, b B} Followng Kemperman [24], a set A G s H-perodc f t s the unon of H-cosets for some nontrval subgroup H G Furthermore, an n-set
Erdős Gnzburg Zv Theorem II 175 partton of S s a sequence of n nonempty subsequences of S, parwse dsont as sequences, such that every term of S belongs to exactly one subsequence, and the terms n each subsequence are dstnct Thus such subsequences can be consdered sets Fnally let ϕ be the map that takes a sequence to ts underlyng set (eg ϕ(1, 1, 1, 2, 2, 4) = {1, 2, 4}) Frst, we state three well known theorems: the frst one s a generalzed verson of what s known as the Caveman Theorem [12], followed by a generalzed form of the Erdős Gnzburg Zv (EGZ) Theorem [1], [11], and the affrmed EHC [2], [9] The orgnal EGZ Theorem s Theorem 22 wth r = 1; and Theorem 22 s obtaned by r applcatons of the EGZ Theorem Theorem 21 s smlarly obtaned, where the orgnal Caveman Theorem s the case S = m Theorem 21 Let S be a sequence of elements from an abelan group G of order m If S m, then there exsts a subsequence of S of length r whose terms sum to zero, where r satsfes S (m 1) r S Theorem 22 Let r be a postve nteger If S s a sequence of (r + 1)m 1 elements from an abelan group G of order m, then S contans an rm-term subsequence whch sums to zero Theorem 23 Let S be a sequence of k dstnct elements from Z m If m s prme, then h S mn{m, hk h2 + 1} The followng two theorems of Gao [17], [18], respectvely, are central to the proof of Theorem 31 Theorem 24 Let l and m be postve ntegers satsfyng 2 l m/4 + 2, and let S be a sequence of elements from Z m satsfyng S = 2m l If 0 m S, then up to order and affne transformaton, S = 0 u 1 v c 1 c w, where m 2l + 3 v u m 1 and w l 2 Theorem 25 Let m, n and h be postve ntegers Suppose G s a fnte abelan group of order m, and g G Furthermore, let S = 0 h a 1 a n be a sequence of elements from G such that the multplcty of every element n the subsequence T = a 1 a n s at most h If g m h T, then g m S Next, we state the Cauchy Davenport Theorem [8], [26], and a recently proved composte analog of t [20] Theorem 26 For postve ntegers n and m, let A 1,, A n Z p If m s prme, then n n } A mn {m, A n + 1 Theorem 27 Let n be an nteger and let S be a sequence of elements from an abelan group G of order m such that S n and every element of
176 A Balostock et al S appears at most n tmes n S Furthermore, let p be the smallest prme dvsor of m Then ether () there exsts an n-set partton, A 1,, A n, of S such that n n } A mn {m, (n + 1)p, A n + 1, or () there exsts α G and a nontrval proper subgroup H of ndex a, such that all but at most a 2 terms of S are from the coset α+h, and there exsts an n-set partton, A 1,, A n, of the subsequence of S consstng of terms of S from α + H such that n A = nα + H When applyng the above theorem, the followng two basc propostons about n-set parttons and sumsets are often used, and for the sake of clarty, we provde ther proofs Proposton 21 A sequence S has an n-set partton A f and only f the multplcty of each element n S s at most n and S n Furthermore, a sequence S wth an n-set partton has an n-set partton A = A 1 A n such that A A 1 for all and satsfyng 1 n Proof Suppose S has an n-set partton Then from the defnton the multplcty of each element n S s at most n, and snce empty sets are not allowed n the n-set partton, t follows that S n Next suppose that the multplcty of each element n S s at most n and S n Let ϕ(s) = {s 1,, s u }, and rearrange the terms of S so that all the terms that are equal to s 1 come frst, followed by all the terms that are equal to s 2, and so forth, termnatng wth the terms equal to s u Let us denote ths new sequence by S = x 1 x 2 x kn+r, where S = S = kn + r and 0 r < n Consder the followng sequence A of n subsequences of S wrtten vertcally: A = x 1 x n+1 x (k 1)n+1 x kn+1 x r x n+r x (k 1)n+r x kn+r x r+1 x n+r+1 x (k 1)n+r+1 x n x 2n x kn We wll show that A s an n-set partton of S and hence of S Indeed, snce S n, t follows that none of the sets n A are empty Furthermore, n vew of the defnton of S and the fact that the maxmum multplcty of a term n S does not exceed n, t follows that x 1 n+ x 2 n+, for every and every 1 2 Thus A s an n-set partton of S The furthermore part s clear from the defnton of A
Erdős Gnzburg Zv Theorem II 177 Proposton 22 Let S be a sequence of elements from a fnte abelan group G, and let A = A 1 A n be an n-set partton of S, where n A = r, and s s the cardnalty of the largest set n A Furthermore, let a 1 a n be a subsequence of S such that a A for = 1,, n () There exsts a subsequence S of S and an n -set partton A = A 1 A n of S, whch s a subsequence of the n-set partton A = A 1 A n, such that n r s + 1 and n A = r () There exsts a subsequence S of S of length at most n+r 1, and an n-set partton A = A 1 A n that n A = n A Furthermore, a A of S, where A A for = 1,, n, such for = 1,, n Proof We frst prove () Assume wthout loss of generalty that A 1 = s We wll construct the n -set partton A n n steps as follows; and S wll be mpled mplctly Denote by A (k) = A 1 A a k the sequence constructed after k steps, and hence A = A (n) and n = a n Let A (1) = A 1, and for k = 1,, n 1, let A (k+1) = { A (k) A (k) A k+1 f a k A + A k+1 = a k f a k A + A k+1 > a k A A, It s easly seen by the above algorthm that a n A = n A = r Furthermore, snce each kept term ncreases the cardnalty of the sumset of the prevous terms of A by at least one, and snce A 1 = s, t follows that at most r s terms, excludng A 1, were kept, and thus a n = n 1 + r s The proof of () s smlar to that of () Frst, for = 1,, n, let the elements of A be {a () 1,, a() A }, where a() 1 = a We wll construct the n- set partton A n a two loop algorthm The outer loop has n steps, where at the th step the set A s constructed usng the nner loop In turn, the nner loop, at the th step of the outer loop, constructs A n A steps For a gven, where 1 n, let A (k) denote the set constructed after k steps of the nner loop at the th step of the outer loop, and hence A = A ( A 1 ) 1 A ( A n ) n wth S mpled mplctly For a gven, where 1 n, let A (1) = {a }, and for k = 1,, A 1, let A (k+1) = A ( A ) n A( A ) A (k) A (k) f 1 A( A ) {a () k+1 } f 1 A( A ) + A (k) = 1 + A (k) < 1 A( A ) A( A ) + (A (k) + (A (k) {a () k+1 }), {a () k+1 }) It s easly seen by the above algorthm that A ( A ) A and a for = 1,, n, and that n A( A ) = n A = r; and snce n A, t follows that n A( A ) = n A Further-
178 A Balostock et al more, snce each kept element a () k, where k 1 f 1, ncreases the cardnalty of the sumset by at least one, t follows that at most r 1 terms, excludng the a s, were kept, and hence S n + r 1 We wll also need the followng theorem of [16] Theorem 28 Let m and k be ntegers wth m k 2 and m 5 (a) If m/2 + 1 < k m 1, then { g(m, k) = m + 2 m, m odd, (b) If k = m, then g(m, k) = m + 1, m even We conclude the prelmnares wth a theorem of Eggleton and Erdős [10] Theorem 29 Let S be a sequence of k dstnct elements from a fnte abelan group If 0 S and k 4, then S 2k 3 A theorem of Gao and Hamdoune revsted Theorem 31 gves necessary and suffcent condtons for a sequence S of suffcent length to be zsf More precsely, t reduces the problem of determnng extremal zsf sequences of suffcent length to the problem of fndng nteger parttons wth a fxed number of parts and all parts greater than 1 Its proof s an adapton of a proof of Gao and Hamdoune [19] Theorem 31 For ntegers m and l, let S be a sequence of elements from Z m, satsfyng S = 2m l 2m m/4 2 The sequence S does not contan an m-term zero-sum subsequence f and only f there exsts a sequence S = 0 u 1 v a 1 a w1 b 1 b w2, where 1 < a m/2 and 1 b < m/2, that s equvalent to S up to order and affne transformaton, and for whch the followng four nequaltes are satsfed: w 1 w 2 (1) a m v 1 and b m u 1 w 2, (2) m 2l + 3 v u m 1 and w 1 + w 2 l 2 Moreover, equalty holds n both nequaltes of (1) f and only f S belongs to an equvalence class of E(m, 2m l) Proof Frst, suppose S s a sequence of elements from Z m, satsfyng S = 2m l 2m m/4 2, and 0 m S Hence from Theorem 24 t follows that S s equvalent, up to order and affne transformaton, to a sequence S = 0 u 1 v a 1 a w1 b 1 b w2 satsfyng the nequaltes n (2), where 1 < a m/2 and 1 b < m/2 Snce the fact that S s zsf mples that S s zsf, t follows from Theorem 25 that for any gven subsequence T of a 1 a w1 b 1 b w2, (3) ether t m v 1 or t u + 1 + T t T t T
Erdős Gnzburg Zv Theorem II 179 and (4) ether t T t m u 1 T or t T t v + 1 Inducton on r, n vew of (3) and the followng three nequaltes () l m/4 + 2, () m v 1 m/2 (follows from (2) and l m/4 + 2), () 3m 4l + 5 u + 2v (follows from (2)), mples r r (5) a = a m v 1 for every r satsfyng 1 r w 1 Smlarly, nducton on r, n vew of (4) and the nequaltes () (), and the fact that u v, mples r r (6) b = b m u 1 r for every r satsfyng 1 r w 2 Hence (5) and (6) mply (1) Next suppose S s an arbtrary sequence of resdues from Z m that satsfes (1) and (2) Actually, we wll use only the fact that (1) s satsfed and v u m 1 It follows from (1) that any m-term zero-sum modulo m subsequence of S must be zero-sum n Z as well In addton, t follows from (1) that the longest zero-sum n Z subsequence of S that does not contan a zero s of length w 2 + w 2 b m u 1 Hence any m-term zero-sum subsequence must use at least u + 1 zeros, whch exceeds the multplcty of zero n S Thus S s zsf, and as affne transformatons and reorderng preserve m-term zero-sum subsequences, the proof of the man part of the theorem s complete Notce that the two nequaltes n (1) are nterchanged by the affne transformaton whch nterchanges 0 and 1 Hence, the moreover part of the theorem s easly deduced from the man part of the theorem Theorem 32 Let m k 2 be postve ntegers If k s odd and m (k 2 + 4k + 3)/8 + 1 or k s even and m (k 2 + 2k)/8 + 1, then g (m, k) 2m (k 2 2k + 5)/4 Proof If k s even, consder the sequence ( S 0 = k 2 2 ) ( 1)(0) m (k2 +2k)/8 (1) m (k2 +2k)/8 (2) and f k s odd, consder the sequence ( S 1 = k 3 ) ( 1)(0) m (k2 1)/8 (1) m (k2 +4k+3)/8 (2) 2 ( ) k, 2 ( k + 1 It follows from the hypotheses that both strngs are well defned Snce both S 1 and S 2 satsfy (1), and snce v u m 1, where u and v are the 2 )
180 A Balostock et al multplctes of 0 and 1 respectvely, t follows from the proof of the second drecton of Theorem 31 that S 1 and S 2 are zsf We conclude the secton wth Theorem 33, whch determnes the value of g(m, k) for fxed k and suffcently large m, dsprovng Conecture 51 of [4], and provng Conectures 11(a) and 11(b) n parts (a) and (b) respectvely Agan, ts proof s an adaptaton of the proof n [19] Theorem 33 Let m k 2 be postve ntegers If k s even and m k 2 2k 4 or k s odd and m k 2 2k 3, then (a) g(m, k) = 2m (k 2 2k + 5)/4, (b) g (m, k) = g(m, k) Proof From Theorem 32, and from the trval fact that g (m, k) g(m, k), t suffces for both parts (a) and (b) to show g(m, k) 2m (k 2 2k + 5)/4 Assume to the contrary that there s a sequence S of elements from Z m wth S = 2m (k 2 2k + 5)/4 and 0 m S From the hypotheses and the fact that k 2 0 or 1 mod 4, t follows that (k 2 2k + 5)/4 m/4 + 2 Hence from Theorem 31 t follows that wthout loss of generalty S satsfes (1) and (2) Let c 1 = {a 1,, a w1 } and c 2 = {b 1,, b w2 } It follows from the frst nequalty n (1) that 2 + 3 + + (c 1 + 1) + 2(w 1 c 1 ) m v 1, mplyng that c 2 1 c 1 (7) + 2w 1 m v 1 2 Lkewse from the second nequalty n (1), t follows that c 2 2 c 2 (8) + 2w 2 m u 1 w 2 2 Inequaltes (7) and (8) mply c 2 1 c 1 + c2 2 c 2 m v 1 w 1 + m u 1 w 2 = l 2, 2 2 whch, n turn, yelds l (c 1 + c 2 ) 2 + c 1 + c 2 (k 2)2 + 2 + k 2 + 2 = k2 2k + 4 + 1 > l, 4 2 4 2 4 whch s a contradcton; and the proof s complete 4 The Erdős Helbronn conecture and g(m, 5) In vew of Theorem 33, g(m, 5) has been determned for m 12 In ths secton, we present an abbrevated proof determnng g(m, 5) for all m 5 We wll make use of the followng conecture, whch can be verfed for k 5 wth some effort by consderng the equatons generated by the 2-sums of a 5-set S wth 2 S < 7
Erdős Gnzburg Zv Theorem II 181 Conecture 41 Let S be a sequence of k 2 dstnct elements from Z m If 2 S < 2k 3, then ether h S s H-perodc, where H > 2, or S s K-perodc, where K = 2 Theorem 41 Let m 5 Then g(6, 5) = 8, and f m 6, then g(m, 5) = 2m 5 Proof For m 6 the result follows from Theorem 28 Suppose S s zsf and S = 2m 5 We may assume that 0 has the greatest multplcty n S Case 1: The multplcty of 0 n S s at most m 2 Applyng Conecture 41 wth k = 5 to all possble 5-sets of ϕ(s) that nclude 0, we can ether fnd a 5-set A ϕ(s) such that 2 A = 3 A 7 and 0 A, or else there exsts a subgroup H of cardnalty h = 5 or h = 6 such that ϕ(s) H In the latter case, m 10, and so from Theorem 22 t follows that any subsequence wth length m + h 1 2m 5 must contan an m-term zero-sum subsequence, a contradcton So 3 A 7 In vew of the assumpton of the case and Proposton 21, there exsts an (m 3)-set partton P of S \ A wth m 7 sets of cardnalty two Applyng Theorem 27 to S \ A, and usng 3 A 7, we obtan an m-term zero-sum subsequence of S, provded concluson () of Theorem 27 holds Hence we are done for m 8 So assume that concluson () of Theorem 27 holds wth coset α + H of ndex a, and wthout loss of generalty assume α = 0 Let P be the (m 3)-set partton mpled by concluson () of Theorem 27 Applyng Proposton 22() followed by Proposton 22() to P we obtan an (m/a 1)-set partton P of a subsequence Q of S \ A of length at most 2m/a 2, whose sumset s also H Then there exsts a subsequence R of S \ A of length a 1 whose terms are from H and are not used n P We can apply Theorem 21 to a subsequence of S \ {Q R} of length m m/a + 1 wth ts terms consdered as elements from Z m /H to obtan a subsequence T of S \ {Q R} whose sum s an element of H and of length r, where r satsfes m m/a a + 2 r m m/a + 1 Snce the sumset of P s H, we can fnd m/a 1 terms from P whch along wth T and an approprate number of terms from R gve an m-term subsequence wth sum zero Case 2: The multplcty of 0 n S s m 1 Let T be a subsequence of S that conssts of 4 dstnct nonzero resdue classes and 3 zeros In vew of Proposton 21, t follows that there exsts an (m 4)-set partton P of S\T wth m 8 cardnalty two sets Applyng Theorem 27 to P and Theorem 29 to ϕ(t )\{0}, we fnd an m-term zero-sum subsequence provded concluson () of Theorem 27 holds If concluson () of Theorem 27 holds nstead, then snce m 4 > a 2 mples 0 α+h, the arguments from the end of Case 1 complete the proof
182 A Balostock et al Appendx In the followng table, Theorem 31 s used to lst the values of E(m, s) for all m and s satsfyng 2m 2 s max{2m 8, 2m m/4 2} m s E(m, s) m 2 2m 2 0 m 1 1 m 1 m 4 2m 3 0 m 1 1 m 3 2 m 8 2m 4 0 m 1 1 m 5 2 2 ( 1)0 m 3 1 m 3 2 0 m 1 1 m 4 3 m 12 2m 5 0 m 1 1 m 7 2 3 ( 1)0 m 3 1 m 5 2 2 0 m 1 1 m 6 23 ( 1)0 m 3 1 m 4 3 0 m 1 1 m 5 4 m 16 2m 6 0 m 1 1 m 9 2 4 ( 1)0 m 3 1 m 7 2 3 ( 1) 2 0 m 5 1 m 5 2 2 0 m 1 1 m 8 2 2 3 ( 1)0 m 3 1 m 6 23 ( 2)0 m 4 1 m 5 2 2 0 m 1 1 m 7 24 0 m 1 1 m 7 3 2 ( 1)0 m 3 1 m 5 4 ( 2)0 m 4 1 m 4 3 0 m 1 1 m 6 5 m 20 2m 7 0 m 1 1 m 11 2 5 ( 1)0 m 3 1 m 9 2 4 ( 1) 2 0 m 5 1 m 7 2 3 0 m 1 1 m 10 2 3 3 ( 1)0 m 3 1 m 8 2 2 3 ( 2)0 m 4 1 m 7 2 3 ( 1) 2 0 m 5 1 m 6 23 0 m 1 1 m 9 23 2 0 m 1 1 m 9 2 2 4 ( 1)0 m 3 1 m 7 24 ( 1)0 m 3 1 m 7 3 2 ( 2)0 m 4 1 m 6 23 ( 1) 2 0 m 5 1 m 5 4 0 m 1 1 m 8 25 0 m 1 1 m 8 34 ( 1)0 m 3 1 m 6 5 ( 2)0 m 4 1 m 5 4 0 m 1 1 m 7 6 m 24 2m 8 0 m 1 1 m 13 2 6 ( 1)0 m 3 1 m 11 2 5 ( 1) 2 0 m 5 1 m 9 2 4 ( 1) 3 0 m 7 1 m 7 2 3 0 m 1 1 m 12 2 4 3 ( 1)0 m 3 1 m 10 2 3 3 ( 2)0 m 4 1 m 9 2 4 ( 1) 2 0 m 5 1 m 8 2 2 3 ( 2)( 1)0 m 6 1 m 7 2 3 0 m 1 1 m 11 2 3 4 0 m 1 1 m 11 2 2 3 2 ( 1)0 m 3 1 m 9 2 2 4 ( 1)0 m 3 1 m 9 23 2 ( 2)0 m 4 1 m 8 2 2 3 ( 1) 2 0 m 5 1 m 7 24 ( 1) 2 0 m 5 1 m 7 3 2 ( 3)0 m 5 1 m 7 2 3 ( 2)( 1)0 m 6 1 m 6 23 0 m 1 1 m 10 2 2 5 0 m 1 1 m 10 234 0 m 1 1 m 10 3 3 ( 1)0 m 3 1 m 8 25 ( 1)0 m 3 1 m 8 34 ( 2)0 m 4 1 m 7 24 ( 2)0 m 4 1 m 7 3 2 ( 1) 2 0 m 5 1 m 6 5 ( 3)0 m 5 1 m 6 23 0 m 1 1 m 9 26 0 m 1 1 m 9 35 0 m 1 1 m 9 4 2 ( 1)0 m 3 1 m 7 6 ( 2)0 m 4 1 m 6 5 ( 3)0 m 5 16 m 5 4 0 m 1 1 m 8 7 References [1] N Alon and M Dubner, Zero-sum sets of prescrbed sze, n: Combnatorcs, Paul Erdős s Eghty, Vol 1, Bolya Soc Math Stud, János Bolya Math Soc, Budapest, 1993, 33 50 [2] N Alon, M B Nathanson and I Ruzsa, The polynomal method and restrcted sums of congruence classes, J Number Theory 56 (1996), 404 417
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184 A Balostock et al [26] M B Nathanson, Addtve Number Theory Inverse Problems and the Geometry of Sumsets, Grad Texts n Math 165, Sprnger, New York, 1996 [27] R Sabar, On a famly of nequaltes and the Erdős Gnzburg Zv Theorem, preprnt Department of Mathematcs Unversty of Idaho Moscow, ID 83844, USA E-mal: math@udahoedu Department of Mathematcs Albertson College Caldwell, ID 83605, USA E-mal: mlotspech@albertsonedu Mathematcs 253-37 Caltech Pasadena, CA 91125, USA E-mal: dambr@hotmalcom Receved on 392002 and n revsed form on 1422003 (4359)