Large zero-free subsets of Z/pZ

0.21 MB·

by  Jean-Marc Deshouillers

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 Z´emor [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. Szemer´edi 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, Szemer´edi 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-
mer´edi and Van Vu in [4, Theorem 1.9].
The above mentioned paper of Szemer´edi 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 re-
sult 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 as-
sumption 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,ka<n0d| 1|5, after aPn ke∈leKm(xe)n\Ct1a,rky>0c|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