Z Z Large zero-free subsets of /p Jean-Marc Deshouillers, Gyan Prakash 9 0 0 Abstract 2 A finite subset of an abelian group is said to be zero-free if the identity n A G a element of cannot be written as a sum of distinct elements from . In this article J we study thGe structure of zero-free subsets of Z/pZ the cardinalityAof which is close 3 tolargestpossible. Inparticular,wedeterminethecardinalityofthelargestzero-free 2 subset of Z/pZ, when p is a sufficiently large prime. ] T For a finite abelian group ( , +) and a subset of , we set ♯ = b : N , = . We say isGzero-free if 0 / ♯; inAotherGwords isAzero-fr{ePe bif∈B0 can h. nBo⊂t bAe exBpre6 sse∅d}as a sum oAf distinct element∈s oAf . A t A a m In 1964, Erd˝os and Heilbronn [5] made the following conjecture, supported by exam- [ ples showing that the upper bound they conjectured is, if correct, very close to being 1 best possible. v 3 Conjecture 1. Let be a subset of Z/pZ. If is zero-free, we have Card( ) √2p. 1 A A A ≤ 7 Upto recently, thebestresultconcerningzero-free subsetsof Z/pZ was that of Hami- 3 . doune and Zémor [3] who proved in 1996 that their cardinality is at most √2p+5lnp, 1 0 thus showing that the constant √2 in the above conjecture is sharp. 9 0 The study of this question has been revived more recently. Freiman and the first : v named author introduced a method based on trigonometrical sums which led to the de- i X scription of large incomplete subsets [2] as well as that of large zero-free subsets [1] of r Z/pZ. Recall that a subset of G is said to be incomplete if ♯ 0 is not equal to a A A ∪{ } G. Szemerédi and Van Vu [6], as a consequence of their result on long arithmetic pro- gressions in sumsets, gave structure results for zero-free subsets leading to the optimal bound for the total number of such subsets of Z/pZ. As it was noticed independently by Nguyen, Szemerédi and Van Vu [4] on one side and us on the other one, both methods readily lead to a proof of the Erd˝os-Heilbronn conjecture for zero-free subsets1. The aim of the present paper is to study the description of rather large zero-free subsets of Z/pZ. We start by reviewing the present knowledge on zero-free subsets of Z/pZ. Notation 2. We denote by σ the canonical homomorphism from Z onto Z/pZ; for an p element a in Z/pZ, we denote by a¯ be the integer in ( p,p] such that a= σ (a¯) and let −2 2 p a = a¯ . Given a set Z/pZ, we denote by ¯ the set a¯ : a . For d Z/pZ, p | | | | A ⊂ A { ∈ A} ∈ we write d := da : a . Given any real numbers x,y with x y, we write [x,y] p ·A { ∈ A} ≤ 1VanH.Vuandthefirstnamedauthorexchangedthisinformationduringaprivateconversationheld in Spring2006. 1 to denote the set σ ([x,y] Z). Given a set Z and non negative real numbers x,y, p ∩ B ⊂ we write (x,y) to denote the set b : x b y and simply write (x) to denote B { ∈ B ≤ | | ≤ } B the set (0,x). B It is evident that Z/pZ is zero-free if and only if the set (A¯)♯ does not contain any A ⊂ multiple of p. This leads to the following examples of zero-free subsets of Z/pZ. Examples 3. (i) Any subset of Z/pZ which satisfy the properties that A¯ is a subset A of [1, p] and a¯ p 1 is a zero-free subset of Z/pZ. 2 Pa¯∈A¯| | ≤ − (ii) Given any integer k with k(k + 1)/2 p + 1, the subset of Z/pZ with = ≤ A A 2,1 [3,k] is a zero-free subset of Z/pZ which has cardinality equal to k. p p {− } ∪ Moreover, one readily sees that if a subset of Z/pZ is zero-free, then it is also the A case for the set s , for any s coprime with p. ·A Building on [2], the first named author proved in [1] the following result Theorem 4. Let c > 1, p a sufficiently large prime and a zero-free subset of Z/pZ A with cardinality larger than c√p. Then , there exists d coprime with p such that da < p+O(p3/4lnp) and da = O(p3/4lnp), (1) X| |p X | |p a a ,da<0 ∈A ∈A where the constants implied in the O symbol depend upon c, and built examples showing moreover that none of the above error-terms can bereplaced by o(p1/2). The error-terms in (1) were reduced to the best possible O(p1/2) by Nguyen, Sze- merédi and Van Vu in [4, Theorem 1.9]. The above mentioned paper of Szemerédi and Van Vu [6] implicitly contains the following result, formally stated in [4] as Theorem 2.1. Theorem 5. Let be a zero-free subset of Z/pZ. Then for some non zero element A d Z/pZ the set d can be partitioned into two disjoint sets and , where ′ ′′ ∈ ·A A A (i) has negligible cardinality: = O(p1/2/log2p). ′ ′ A |A| (ii) We have [1,p/2] and a p 1. A′′ ⊂ p Pa′′∈A′′| ′′|p ≤ − We firstconsider the maximal zero-free subsets of Z/pZ. Thedescription given in the following theorem is a synthesis of the results established in Sections 1 and 2. Theorem 6. Let p be a sufficiently large prime and a zero-free subset of Z/pZ with A maximal cardinality. Then card( ) is the largest integer k such that k(k+1)/2 p+1, (2) A ≤ and one may thus write card( ) = 2p+9/4 1/2 = √2p δ(p), with δ(p) A hp − i (cid:2) (cid:3) − ∈ 0,1 . { } Furthermore, there exists a non-zero element d in Z/pZ such that the set d. is the A union of two sets and , with ′ ′′ A A 2 (i) [ 2(1 + δ(p)), 1] , [1,p/2] , ( ) = and card( ) ′ p ′′ p ′′ ′ ′ A ⊂ − − A ⊂ A ∩ −A ∅ A ≤ 1+δ(p), (ii) a 2(1+δ(p)) and a p 1+3δ(p). Pa′∈A′| ′|p ≤ Pa′′∈A′′| ′′|p ≤ − The Reader will find a more detailed description of extremal zero-free sets in Section 2. In this Introduction, we limit ourselves to a few remarks and examples. Writing 2p+9/4 1/2 = √2p+αp 1/2, we have αp = O(1/√p). Onereadily sees p − − that δ(p) takes the values 1 or 0 according as the fractional part of √2p is smaller than 1/2 α or larger. Thus the density of the primes p for which the maximal zero-free of p − Z/pZ subset has cardinality √2p is 1/2. (cid:2) (cid:3) The sum a can take the values p+1 or p+2 only in very special cases, namely whenPoan′e′∈Ao′f′|p′+′|p2,p + 3,p + 4,p + 5,p + 6, or p + 7 is a value of the poly- nomial x(x + 1)/2 at some integral point x. The number of such primes p up to P is O(√P); the existence of infinitely many such primes is not known and would result from the validity of some standard conjectures, like Schinzel’s hypothesis. The set = 3,1,4,5,6, 14,15 is an example of a zero-free subset of Z/113Z which sat- 113 A {− ··· } isfies Theorem6with card( )= √2p 1, a = p+2and p+7= x(x+1)/2. A ⌊ ⌋− Pa′′∈A′′| ′′|p We now turn our attention to very large zero-free subsets of Z/pZ, i.e. subsets A such that √2p card( )= o(√p). From now on, we fix a function ψ from [2, ) to R+ − A ∞ which tends to 0 at and assume that ∞ e( ) := 2p card( ) ψ(p)√p and p is sufficiently large, (3) A |p − A |≤ the term sufficiently large implicitly refereing to the function ψ. The following result gives the structure of large zero-free subsets of Z/pZ. It shows that any given large zero-free subset has a dilate, which is a union of sets and , ′ ′′ A A A where is a set closely related to the one given in Example 3 (i) and the cardinality of ′′ A is small. ′ A Theorem 7. When p is sufficiently large, then given any zero-free subset of Z/pZ A with e( ) satisfying (3), there exists a non-zero element d Z/pZ such that d can be A ∈ ·A partitioned into disjoint sets and with the following properties ′ ′′ A A (i) The set ¯ is included in [1, p) and we have a p 1. A′′ 2 Pa′′∈A′′| ′′|p ≤ − (ii) The set ¯ is included in [ ce( ),ce( )] for some absolute constant c and the ′ A − A A cardinality of is O e( )+2 ln(e( )+2) , A′ (cid:16)p A A (cid:17) where e( ) is defined in (3). A To prove Theorems 6 and 7, we prove the following proposition. Proposition 8. Let p be a prime and a zero-free subset of Z/pZ with e( ) satis- A A A fying (3). When p is sufficiently large, there exists a non-zero element d Z/pZ such ∈ that da p+O e( )3/2ln(e( )+2) , (4) X| |p ≤ (cid:16) A A (cid:17) a ∈A da = O e( )3/2ln(e( )+2) (5) X | |p (cid:16) A A (cid:17) a ,d¯a<0 ∈A 3 Remark 9. Noticing that for any zero-free subset of Z/pZ, the corresponding set ¯ Z can contain at most one element from the setAx, x for any integer x we have A ⊂ ¯( ¯+1) { − } Pa¯∈A¯|a¯|≥ |A| |A2| . Usingthis,Conjecture1isanimmediatecorollaryofProposition8. To prove Proposition 8 we use Theorem 4 and the following result from [2]. Theorem 10. ([2, Theorem 2]) Let I > L > 100 and B > 2ClnL be positive integers such that C2 > 500L(lnL)2+2000IlnL. Let be a set of B integers included in [ L,L]. Then there exist d > 0 and a subset B − C of with cardinality C such that B (i) all the elements of are divisible by d, C (ii) contains an arithmetic progression with I terms and common difference d, ∗ C (iii) at most ClnL elements of are not divisible by d. B 1 Proof of Proposition 8 Let p be a sufficiently large prime and Z/pZ be as given in Proposition 8. From A ⊂ Theorem 4, there exists a non-zero element d Z/pZ such that (1) holds. Without loss ∈ of generality, we may indeed assume that d = 1 or, equivalently, replace d by . We ·A A then get a¯ = a p+O(p3/4lnp). (6) X| | X| |p ≤ a¯ ¯ a ∈A ∈A We prove Proposition 8, by showing that if ¯ [ p,p] is as above then we have A ⊂ −2 2 a¯ p+O e( )3/2ln(e( )+2) . (7) X| |≤ (cid:16) A A (cid:17) a¯ ¯ ∈A We shall first show how one can deduce (7) from the following proposition. Proposition 11. Let p be a sufficiently large prime and Z such that ♯ does not K ⊂ K contain any multiple of p. We recall that ψ is a fixed function from [2, ) to R+ which ∞ tends to 0 at . Let us suppose that we have ∞ e( ) := 2p card( ) ψ(p) 2p and k p+s( ), with 0 s( ) p0.9. (8) K |p − K | ≤ p X| | ≤ K ≤ K ≤ k ∈K Then, we have in fact k p+O κ3/2lnκ , (9) X| | ≤ (cid:16) (cid:17) k ∈K where κ = s( )/√p+e( )+2. Moreover we have K K min k , k = O(κ3/2lnκ). { X | | X | |} k ,k>0 k ,k<0 ∈K ∈K 4 The fact that is zero-free and Relations (1) and (3) permit to apply Proposition 11 A with = . When e( ) p1/4, then (9) directly implies (7). But, when e( ) p1/4, K A K ≥ K ≤ we first obtain from (9) the following weaker inequality a¯ p+O(p3/8lnp). X| | ≤ a¯ ¯ ∈A As such, it is weaker than (7) in this case, we may use s( ) = p3/8lnp, so that K κ = e( )+O(1), and a further application of Proposition 11 leads to Relation (7). K To prove Proposition 11 we need a few lemmas. Lemma 12. Let m Z,ℓ N and let be a subset of [ ℓ, ℓ] Z. We have ∈ ∈ B − ∩ ( m,...,m+ℓ 1 + ) Z = ([m b , m+ℓ 1+ b ]) Z. ∗ { − } B ∩ − X | | − X | | ∩ b ,b<0 b ,b>0 ∈B ∈B Proof. We write k = and = b < b < ... < b < 0 b < ... < b , where 1 2 h h+1 k |B| B { ≤ } h = 0 if all the elements of are non negative. For 0 u k, we define B ≤ ≤ Phi=−1ubi if 0 ≤ u≤ h−1, βu = 0 if u= h, u b if h+1 u k. Pj=h+1 j ≤ ≤ Simply notice that β = min s : s ,β = max s : s and that β < ... < β 0 ∗ k ∗ 0 k { ∈ B } { ∈B } { } is a subset of such that the difference between two consecutive elements of which is ∗ B at most ℓ. Lemma 13. Let Z,c Z,x N,ℓ x+1 be such that (x)♯ contains [c, c+ℓ] Z. B ⊂ ∈ ∈ ≥ B ∩ Then, if there exists an integer y in [x+1, ) such that (y)♯ does not contain ([c ∞ B − b , c+ℓ 1+ b ]) Z, and z is the least such integer, Pthebn∈Bw(xe+h1,ayv),eb<0| | − Pb∈B(x+1,y),b>0| | ∩ z ℓ+ b +1. ≥ X | | b (x+1,z 1) ∈B − Proof. We notice that (z)♯ k (x)♯+ (x+1,z) . Lemma 12 implies that if z has the ∗ B B B required property, then z x+2. Since z x+2, the minimal property of z implies ≥ ≥ that the set (z 1)♯ does contain B − = ([c b , c+ℓ 1+ b ]) Z. I − X | | − X | | ∩ b (x+1,z 1),b<0 b (x+1,z 1),b>0 ∈B − ∈B − By our assumption, the set ( +b) is not an interval. This implies (special case of Lemma 12) that z Iℓ∪+Sb∈B,|b|=z I b +1. ≥ Pb∈B(x+1,z−1)| | Lemma 14. Let be as given in Proposition 11. Then for any k , the element k K ∈ K − does not belong to . K Proof. Ifclaimisnottrue,thenevidently0 ♯ whichiscontrarytotheassumption. ∈ K Lemma 15. We keep the notation of Proposition 11. For x 0.9√2p, the cardinality ≤ of (x) is x+O(e( )+s( )/√p). K K K 5 Proof. Lemma 14 immediately implies that the cardinality of (x) is at most x. Let us K suppose that the cardinality of (x) is x λ(x). Then using Lemma 14 we get K − x λ(x) card( )+λ(x) − K k i+ i X| | ≥ X X k i=1 i=x+1 ∈K Writing each summand in the second sum on the right hand side of the above inequality as (i λ(x))+λ(x) and then noticing that the number of terms in the second sum is − card( ) x, we get the following inequality K − card( ) K k i+λ(x)(card( ) x). (10) X| | ≥ X K − k i=1 ∈K Since x 0.9√2p and card( ) √2p e( ) √2p ψ(p)√p, the second term in the ≤ K ≥ − K ≥ − right hand sideof the above inequality is larger than 0.05√2pλ(x), whereas thefirstterm is p O(e( )√2p). Now comparing the above inequality with (8) we obtain − K λ(x) c(e( )+s( )/√p), ≤ K K for some absolute constant c. The lemma readily follows from this fact. Lemma 16. We keep the notation of Proposition 11. The largest integer y belonging 0 to satisfies y = O e( )√2p+s( ) . o K∪−K (cid:0) K K (cid:1) Proof. Using Lemma 14 we obtain card( ) 1 K − k i+y . (11) X| | ≥ X 0 k i=1 ∈K Now the first term on the right hand side of the above inequality is p O(e( )√2p). − K Therefore comparing the above inequality with (8), the assertion follows. Lemma 17. We keep the notation of Proposition 11 and let x be a sufficiently large integer. Suppose that the cardinality of (x) is at least 0.99x. Then there exists a subset K of A(x) with = O(√xlnx) such that ♯ contains an arithmetic progression of length C |C| C x and common difference d, with d 1,2 . ∈ { } Proof. Applying Theorem 10 with = (x), L = x,I = x+1, C = 100√xlnx , we B K ⌊ ⌋ get that there exists a subset of (x) with = O(√xlnx) such that ♯ contains C K |C| C an arithmetic progression of length x and common difference d dividing at least 0.8x elements of (x). Since (x) is contained in an interval of length 2x, we obtain that K K d 1,2 . ∈ { } Lemma 18. Let x and be as in Lemma 17. Then there exists k (x) such that C ∈ K \C the element k+1 also belongs to (x) . K \C Proof. LetLbethesetconsistingofthoseelementsl [1,x]suchthatoneoftheelements ∈ l or l belongstotheset (x) . ThenLisasetofcardinality atleast0.9xcontained in − K \C anintervaloflengthx. Thereforethereexistsl Lsuchthat l,l+1,l+2,l+3,l+4 L. ∈ { } ⊂ Now by the definitionof L, for any 0 i 4, either l+i (x) or (l+i) (x) . ≤ ≤ ∈ K \C − ∈K \C The lemma follows evidently by showing that there exists i with 0 i 3 for which one ≤ ≤ 6 of thefollowing two sets, l+i,l+i+1 and (l+i), (l+i+1) is includedin (x) . { } {− − } K \C Ifnot,thenreplacing by ifnecessarywehavethat l,l+1,l+3, l 4 (x) . K −K {− − − } ⊂ K \C This would contradict the assumption that 0 does not belong to ♯. Hence the lemma K follows. We are now in a position to prove Proposition 11. Proof of Proposition 11. From Lemma 15, there is an integer x which satisfies the assumption of Lemma 17 and at the same time x = O(e( )+s( )/√p). For this choice K K of x, let be a subset of (x), as provided by Lemma 17. From Lemma 18 we ob- C K tain a subset k,k + 1 of (x) . Then the set = k,k + 1 is a subset of 1 { } K \ C C C ∪ { } (x) with card( ) = O(√xlnx) and ( )♯ contains an interval [y,y + x) of length 1 1 K C C x. With this interval applying Lemma 12 with B = (x) , we obtain that 1 I K \ C (x)♯ contains the interval [y k ,y +x+ k ) of length xK+ k . Then usin−g PLekm∈mK(axs)\C113,ka0c|al|culation, it followPskt∈hKa(tx)f\oCr1|so|me positive absolute constant c , the set (p/c ) contains the interval 0 0 K [y k ,y +x+ k ) of length x+ k . Re−plaPcikn∈gK(p/cb0y)\C1,k<0w|e|may assumPekt∈hKa(pt/cy0)>\C10,k.>0T|h|en since ♯ doesPnok∈tKc(opn/ct0a)i\nC1a|n|y K −K K multiple of p we obtain the following inequalities k p 1 X | | ≤ − k (p/c0) 1 ∈K \C and k c + k c +y. X | | ≤ X | 1| X | | ≤ X | 1| k (p/c0),k<0 c1 1 k (p/c0) 1,k<0 c1 1 ∈K ∈C ∈K \C ∈C From Lemma 16 we have that (p/c ) = . Moreover it is also evident from the con- 0 K K struction of that c = O(x3/2lnx). Since y ♯ we have y c . Therefore theC1assertioPncf1o∈lClo1w| s1.| ∈ C1 ≤ Pc1∈C1| 1| 2 Proof of Theorem 6 Let p be a sufficiently large prime and a zero-free subset of Z/pZ of the largest car- A dinality. From Proposition 8 and Remark 9, we have that card( ) √2p. Moreover, A ≤ since for any prime p the set [1,[√2p] 1] is an example of a zero-free subset of Z/pZ, p − it follows that card( )= [√2p] δ(p) with δ(p) 0,1 . We set A − ∈ { } [√2p] [√2p][√2p+1] s( 2p)= i = . p X 2 i=1 From Example 3 (ii), it follows that when s(√2p) p+1, then δ(p) =0. In this section ≤ we shall show that δ = 0, only when s(√2p) p+1. p ≤ Using Proposition 8, there exists a d (Z/pZ) such that replacing by d. , we ∗ ∈ A A have a p+O(1). (12) X| |p ≤ a ∈A Using (11) with = ¯ and (12), the following lemma is immediate. K A Lemma 19. The largest integer y in ¯ ¯ is O √2p . A∪−A (cid:0) (cid:1) 7 Let G( ) be the collection of all natural numbers g which satisfy the property that none of theAintegers g and g belong to the set ¯, where ¯ is the subset of integers as − A A defined earlier. For the brevity of notation we shall write G to denote the set G( ). Let A G = g < g < g < ..... . 0 1 2 { } From Lemma 15 we obtain that the cardinality of G(x) is O(1) for any x 0.9√2p. ≤ The arguments identical to those used in the proof of Lemma 15 in fact leads to the following lemma. Lemma 20. The set ¯ ( ¯) contains all the integers in [1,√2p/5] with at most δ(p) A∪ −A exception. Proof. The lemma is equivalent to showing that in case card( ) = [√2p], then g > 0 A √2p/5, whereas in case card( )= [√2p] 1, then g > √2p/5. Suppose that this is not 1 A − true. Then if δ(p) = 0, we have g0 1 Card +1 − A [√2p+1][√2p+2] a = a¯ i+ i 2p/5, X| |p X| |≥ X X ≥ 2 −p a a¯ ¯ i=1 i=g0+1 ∈A ∈A whereas in case δ(p) = 1, we have g0 1 g1 1 Card +2 − − A [√2p+1][√2p+2] a i+ i+ i 2 2p/5. X| |p ≥ X X X ≥ 2 − p a i=1 i=g0+1 i=g1+1 ∈A Usingthefactsthat[√2p] √2p 1andforanyintegeri,wehave[√2p+i] = [√2p]+i,it ≥ − followsthateitheroftheseinequalitiesarecontraryto(12). Hencethelemmafollows. Now we determine all the possible structure of ¯(√2p/5), first under the assumption A that g 5. 0 ≥ Lemma 21. When g 5, then replacing by , if necessary, the set ¯ contains 0 ≥ A −A A the whole interval [5,√2p/5) with at most δ(p) exception and ¯(4) is equal to one of the A three sets described in the the first three rows of the second column of Table 1. Proof. Since we have assumed that g 5, replacing by , if necessary, we may 0 ≥ A −A assume that 3 ¯. Then the set ¯(3) is equal to one of the following four sets, ∈ A A 1,2,3 , 1,2,3 , 1, 2,3 , 1, 2,3 . Since is zero-free, among these four pos- { } {− } { − } {− − } A sibilities, the last one cannot occur. We verify that in all the other three possible cases the following always hold 1,2,3,4 ¯(3)♯. { } ⊂ A Thisimplies that the set ¯(4) is equal to one of the three sets described in the second A column of the first three rows of Table 1; that is, the set ¯(4) is equal to one of the A following three sets 1,2,3,4 , 1,2,3,4 , 1, 2,3,4 . We claim that there does not { } {− } { − } exist any integer z [5,√2p/5] with z ¯. The lemma follows immediately using this ∈ − ∈ A claim and Lemma 20. To verify the claim, suppose that the claim is not true and z is 0 the least integer which violates the claim. Then since 1,2,3,4,5,6 is always a subset { } of ¯(4)♯, we have that z is at least 6. Now if z = g +1, then we have z 1 ¯ and 0 0 0 0 A 6 − ∈A thus z ¯(3)♯+z 1 ¯(z 1)♯. Since is zero-free, this implies that z can not 0 0 0 0 ∈ A − ⊂ A − A − belongtotheset ¯whichcontradicts theassumptionthatz is theleast integer violating 0 A the claim. Thus if the claim is not true then z = g +1. But in this case z 2 ¯and 0 0 0 − ∈A thus z ¯(3)♯+z 2 ¯(z 2)♯. This implies that z cannot belong to ¯. Hence 0 0 0 0 ∈A − ⊂ A − − A the claim and thus the lemma hold. 8 Structure of when = √2p δ , with δ 0,1 p p A |A| ⌊ ⌋− ∈ { } Here s = s ( ) = a¯. ♯ a /a 4 δ′′ g′′ A Pa¯∈A¯,a¯>0( ¯)♯ p p 0 { ∈ A | | ≤ } A 1 1,2,3,4 5 [1,s ] ′′ { } ≥ 2 1,2,3,4 5 1 [1,s ] ′′ {− } ≥ {− }∪ 3 1, 2,3,4 5 2, 1 [1,s ] ′′ { − } ≥ {− − }∪ 4 1,2,3 1 4 [1,s ] ′′ { } 5 1,2,3 1 4 1 [1,s ] ′′ {− } {− }∪ 6 1, 2,3 1 4 1, 2 [1,s ] ′′ { − } {− − }∪ 7 1,2, 3 1 4 [ 4, 1] [1,s ] ′′ {− − } − − ∪ 8 1,2,4 1 3 [1,s ] ′′ { } 9 1,2,4 1 3 1 [1,s ] ′′ {− } {− }∪ 10 1, 2,4 1 3 2, 1 [1,s ] ′′ { − } {− − }∪ 11 1,2, 4 1 3 [ 4, 1] [1,s ] ′′ { − } − − ∪ 12 1, 2,4 1 3 [ 3, 1] [1,s ] p ′′ {− − } − − ∪ 13 1,3,4 1 2 [1,s ] 2,s 2 ′′ ′′ { } \{ − } 14 1,3,4 1 2 1 [1,s ] 1,s 2 ′′ ′′ {− } {− }∪ \{ − } 15 1, 3,4 1 2 3, 2 [1,s ] s 2 ′′ ′′ { − } {− − }∪ \{ − } 16 2,3,4 1 1 [2,s ] s 1 ′′ ′′ { } \{ − } 17 2,3,4 1 1 2, 1 [1,s ] s 1 ′′ ′′ {− } {− − }∪ \{ − } 18 2, 3,4 1 1 3, 1 [1,s ] s 1 ′′ ′′ { − } {− − }∪ \{ − } 19 2,3, 4 1 1 4, 2, 1 [1,s ] s 1 ′′ ′′ { − } {− − − }∪ \{ − } Table 1: Subset sum of a largest zero-free set 9 Lemma 22. Let A be as in Lemma 21, A′ = A ∩ [−2p,−1]p and A′′ = A ∩ [1, p2]p. Then we have [ 2, 1] . Moreover the set ( )♯ contains the interval [1,s ] with ′ p ′′ p A ⊂ − − A s = a and is equal to one of the sets described in the fifth column of the first th′′reeProwa′′s∈Aof′′T|a′′b|lpe 1, the three possibilities corresponding to three possible structures for ¯(4). We have A s p 1. ′′ ≤ − Proof. For any integer z we set s (z) = a¯. ′′ X ′′ a¯′′ ¯′′(z) ∈A We claim that there is an absolute constant c such that for any integer z with 5 z p, ≤ ≤ c the set ¯(z) ♯ contains the interval [1,s (z)]. The claim is easily verified with z = 5. ′′ (cid:0)A (cid:1) Suppose the claim is not true and z is the least integer violating the claim. Since using 0 the previous lemma we always have s (5) 5+1 = 6, we apply Lemma 13 with x = 5 ′′ ≥ and obtain the following inequality. z s (z 1)+1. (13) 0 ′′ 0 ≥ − Using the previous lemma, for any integer y with y [6,√2p/5], we have ∈ y(y+1) s (y) = a ǫ, ′′ ′ 2 − X | |− a′ ¯′(4) ∈A where ǫ = 0 if y g and ǫ = g if y > g . Using this it follows that (13) cannot hold 0 0 0 ≤ with z √2p/5. Therefore we have 0 ≤ p z s ( 2p/5) , 0 ′′ ≥ p ≥ c wherecisanabsoluteconstant. Hencetheclaim follows. Usingtheclaim andLemma19, it follows that the set ( )♯ contains the interval [1,s ] . Since is zero-free, it follows ′′ p A A that s p 1. ′′ ≤ − Since ¯(√2p/5) ♯ contains the interval [1,s (√2p/5)], it follows that there is no integer ′′ (cid:0)A (cid:1) y p, √2p with y ¯. Using the previous lemma and Lemma 19, it follows that ∈ h−c − 5 i ∈ A′ ¯ = ¯ [ 4, 1] [ 2, 1]. Using this it may be easily verified that the set ( )♯ is ′ A A∩ − − ⊂ − − A equal to one of the sets described in the fifth column of the first three rows of Table 1. Hence the lemma follows. Theorem 23. Let p be a sufficiently large prime and a zero-free subset of Z/pZ of the A largest cardinality. Then card( )= √2p δ(p), where δ(p) = 0 if s(√2p) p+1 and A (cid:2) (cid:3)− ≤ is equal to 1 otherwise. In other words, card( ) is the largest integer k with the property A k(k+1) that p+1; that is card( )= 2p+9/4 1/2 . 2 ≤ A hp − i Proof. From theremarks madein the beginningof this section, it follows that card( )= A [√2p] δ(p) with δ(p) 0,1 . If s(√2p) p+1, then the set 2, 1 3,[√2p] − ∈ { } ≤ {− − }p ∪(cid:2) (cid:3)p is an example of a zero-free subset of Z/pZ and since is a largest zero-free subset, A we have δ(p) = 0, in this case. Now in case δ(p) = 0, then from the remarks made in the beginning of this section there is a d (Z/pZ) , such that replacing by d. , the ∗ ∈ A A 10

Large zero-free subsets of Z/pZ

Similar Large zero-free subsets of Z/pZ