On some properties of representation functions related to the Erd˝os-Tur´an conjecture

arXiv:1705.03316v1 [math.NT] 9 May 2017

On some properties of representation functions related to the Erd˝ os-Tur´ an conjecture

Csaba S´ andor

Quan-Hui Yang

1. Institute of Mathematics, Budapest University of Technology and Economics, H-1529 B.O.

Box, Hungary

2. School of Mathematics and Statistics, Nanjing University of Information Science and Technology, Nanjing 210044, China

Abstract

For a setA⊆Nandn∈N, letR^A(n) denote the number of ordered pairs (a, a^′)∈ A×A such that a+a^′ = n. The celebrated Erd˝os-Tur´an conjecture says that, if R^A(n)≥1 for all sufficiently large integersn, then the representation functionR^A(n) cannot be bounded. For any positive integer m, Ruzsa’s number R^mis defined to be the least positive integer r such that there exists a setA ⊆Z^m with 1≤R^A(n)≤r for all n ∈ Z^m. In 2008, Chen proved that R^m ≤ 288 for all positive integers m.

Recently the authors proved thatR^m≥6 for all integersm ≥36. In this paper, for an abelian group G, we prove that if A⊆Gsatisfies R^A(g) ≤5 for allg ∈G, then

|{g:g∈G, R^A(g) = 0}| ≥¹₄m−√

5m. This improves a recent result of Li and Chen.

We also give upper bounds of|{g:g∈G, R^A(g) =i}|fori= 2,4.

2010 Mathematics Subject Classification: Primary 11B34,11B13.

Keywords and phrases: Representation function, Ruzsa’s number, Erd˝os-Tur´an con- jecture

1 Introduction

LetGbe an abelian group. For any setA, B⊆G, let

RA,B(g) =♯{(a, b) : a∈A, b∈B, a+b=g}.

∗Email: csandor@math.bme.hu. This author was supported by the OTKA Grant No. K109789. This paper was supported by the J´anos Bolyai Research Scholarship of the Hungarian Academy of Sciences.

†Email: yangquanhui01@163.com. This author was supported by the National Natural Science Founda- tion for Youth of China, Grant No. 11501299, the Natural Science Foundation of Jiangsu Province, Grant Nos. BK20150889, 15KJB110014 and the Startup Foundation for Introducing Talent of NUIST, Grant No.

2014r029.

Let RA(g) = RA,A(g). If A ⊆N and RA(n)≥1 for all sufficiently large integers n, then we say thatAis a basis ofN. The celebrated Erd˝os-Tur´an conjecture [7] states that if Ais a basis ofN, thenRA(n) cannot be bounded. Erd˝os [6] proved that there exists a basis A and two constantsc1, c2 >0 such thatc1logn≤RA(n)≤c2lognfor all sufficiently large integersn. Recently, Dubickas [5] gave the explicit values ofc1andc2. In 2003, Nathanson [15] proved that the Erd˝os-Tur´an conjecture does not hold on Z. In fact, he proved that there exists a set A ⊆ Z such that 1 ≤ RA(n) ≤ 2 for all integers n. In the same year, Grekos et al. [8] proved that if RA(n) ≥1 for all n, then lim sup_n→∞RA(n) ≥6.Later, Borwein et al. [1] improved 6 to 8. In 2013, Konstantoulas [11] proved that if the upper densityd(N\(A+A)) of the set of numbers not represented as sums of two numbers ofA is less than 1/10, then RA(n)>5 for infinitely many natural numbersn. Chen [3] proved that there exists a basis A of N such that the set of n with RA(n) = 2 has density one.

Later, the second author [20] and Tang [19] generalized Chen’s result. For the analogue of Erd˝os-Tur´an conjecture in groups, one can refer to [9], [10] and [12].

For a positive integerm, let Zmbe the set of residue classes mod m. IfRA(n)≥1 for alln∈Zm, thenA is called an additive basis ofZm.

In 1990, Ruzsa [16] found a basisAofNfor whichRA(n) is bounded in the square mean.

Ruzsa’s method implies that there exists a constantC such that for any positive integerm, there exists an additive basisA of Zm withRA(n)≤C for alln ∈Zm. For each positive integerm, Chen [2] defined Ruzsa’s numberRmto be the least positive integerrsuch that there exists an additive basisAofZm with RA(n)≤rfor all n∈Zm. In the same paper, Chen proved thatRm≤288 for all positive integerm andR2p² ≤48 for all primesp.

In 2016, the authors [17] proved that if m ≥ 36, then Rm ≥ 6. That is, if m ≥ 36 and A⊆Zm satisfiesRA(n)≤5 for all integers n, then there exists an0 ∈Zm such that RA(n0) = 0. Recently, Li and Chen (see [14, Corollary 1.3]) gave a quantitative version of this result.

Li & Chen’s Theorem. Let G be a finite abelian group with |G| = m and A ⊆G. If RA(g)≤5 for allg∈G, then

|{g:g∈G, RA(g) = 0}| ≥ 7 32m−1

2

√10m−1.

In this paper, we improve Li and Chen’s theorem and also give an example on the other hand. For convenience, for a fixed nonnegative integer i, we denote the set{g :g ∈ G, RA(g) =i}bySi.

Theorem 1. (a) Let Gbe a finite abelian group with |G|=m andA ⊆G. If RA(g)≤5 for allg∈G, then|S0| ≥ ¹4m−√

5m.

(b) Let pbe a prime and m= 2(p²+p+ 1). Then there exists a subset A⊆Zm such that RA(n)≤5 for alln∈Zm and|S0|<³₈m.

IfRA(g)≤5 for allg ∈G, then by|S0|+|S2|+|S4| ≤m and Theorem 1 (a), we see that|S2|+|S4| ≤ ³4m+√

5m. In the next two theorems, we give upper bounds for|S2|and

|S4|respectively.

Theorem 2. (a) LetA⊆Gsatisfy RA(g)≤5 for allg∈G. Then|S2| ≤ ¹2m+ 3√ 5m.

(b) Letpbe a prime and m=p²+p+ 1. Then there exists a subsetA⊆Zmsuch that RA(n)≤2for all n∈Zm and|S2|=¹₂m−¹2.

Remark 1. The example in Theorem 2 (b) shows that Theorem 2 (a) is nearly best possible.

If RA(g)≤5 for all g ∈G, by the statement before Theorem 2, we have|S2|+|S4| ≤

3 4m+√

5m, and so|S4| ≤ ³4m+O(√

m). It seems difficult to improve this upper bound. In the following, we will prove this result by a weak conditionRA(g)≤7 for allg∈G.

Theorem 3. (a) LetA⊆Gsatisfy RA(g)≤7 for allg∈G. Then|S4| ≤ ³4m+O(√ m).

(b) Let pbe a prime and m = 2(p²+p+ 1). Then there exists a subset A⊆Zm such that RA(n)≤4 for alln∈Zm and|S4|=¹₂m−1.

2 Preliminary Lemmas

Lemma 1. (See [17, Lemma 3].) Let A⊆Gandc be a positive integer. If RA(g)≤c for allg∈G, then|A| ≤√

cm.

Lemma 2. (See [18, Singer’s Theorem].) If l is a prime power, then there exists A ⊆ Zl²+l+1 such thatRA,−A(n) = 1 for alln∈Zl²+l+1, n6= 0.

Lemma 3. IfA is a subset ofG, then for any positive integerkwe have X

g∈G

(RA(g)−k)²≥km−(2k−1)|A|+k²−k.

Proof. We use Lev and S´ark¨ozy’s argument (see [13]) in the following.

X

g∈G

(RA(g)−k)² = X

g∈G

RA(g)²−2kX

g∈G

RA(g) +k²m

= X

g∈G

RA,−A(g)²−2k|A|²+k²m

= X

g∈G\{0}

RA,−A(g)²−(2k−1)|A|²+k²m

≥ 1

m−1



 X

g∈G\{0}

RA,−A(g)





2

−(2k−1)|A|²+k²m

= (|A|²− |A|)²

m−1 −(2k−1)(|A|²− |A|)−(2k−1)|A|+k²m

= (m−1)

|A|²− |A| m−1 −

k−1 2

2

+k−1 4

!

−(2k−1)|A|+k².

If ^|A|_m−1^2−|A| ≥k or ^|A|_m−1^2−|A| ≤k−1, then we have X

g∈G

(RA(g)−k)²≥k(m−1)−(2k−1)|A|+k²=km−(2k−1)|A|+k²−k, and the result is true.

Ifk−1< ^|A|_m−1^2−|A|< k, then X

g∈G\{0}

RA,−A(g)²≥ min

k₁,k₂,...,k_m−1∈N Pm−1

i=1 ki=|A|²−|A|

m−1

X

i=1

k_i².

It is known that ifPm−1

i=1 ki is fixed, whereki ∈N, thenPm−1

i=1 k_i² gets the minimal value when |ki−kj| ≤1 for all 1≤i, j ≤m−1. Let|A|²− |A|=q(m−1) +r, where q, r are nonnegative integers and 0≤r < m−1. Thenq=j_|A|2−|A|

m−1

k

and r=n_|A|2−|A|

m−1

o

(m−1).

Hence X

g∈G\{0}

RA,−A(g)² ≥ min

k₁,k₂,...,km−1∈N Pm−1

i=1 ki=|A|²−|A|

m−1

X

i=1

k_i²=rk+ (m−1−r)(k−1)

=

|A|²− |A| m−1

(m−1)k²+

1−

|A|²− |A| m−1

(m−1)(k−1)²

= (k−1)²(m−1) + (2k−1)

|A|²− |A| m−1

(m−1).

Therefore, X

g∈G

(RA(g)−k)² = X

g∈G\{0}

RA,−A(g)²−(2k−1)|A|²+k²m

≥ (k−1)²(m−1) + (2k−1)

|A|²− |A| m−1

(m−1)

−(2k−1)(|A|²− |A|)−(2k−1)|A|+k²m

= (k−1)²(m−1)−(2k−1)(m−1)(k−1)−(2k−1)|A|+k²m

= km−(2k−1)|A|+k²−k.

3 Proofs

Proof of Theorem 1. LetAbe a given subset ofGsuch thatRA(g)≤5 for allg∈G. Then X

g∈G

(RA(g)−3)² = 9|S0|+ 4|S1|+|S2|+|S4|+ 4|S5|

≤ 8|S0|+ 3(|S1|+|S3|+|S5|) + (|S0|+|S1|+|S2|+|S3|+|S4|+|S5|).

It is clear that

|S0|+|S1|+|S2|+|S3|+|S4|+|S5|=|{g: g∈G,0≤RA(g)≤5}|=m,

|S1|+|S3|+|S5|=|{g:g∈G,2∤RA(g)}|=|{2a:a∈A}| ≤ |A|. Hence we have

X

g∈G

(RA(g)−3)²≤8|S0|+ 3|A|+m.

(1)

On the other hand, by Lemma 3, takingk= 3, we have X

g∈G

(RA(g)−3)²≥3m−5|A|+ 6.

(2)

Therefore, by (1),(2) and Lemma 1, it follows that

|S0| ≥ 1

4m− |A|+3 4 ≥1

4m−√ 5m.

Now we prove part (b). Letpbe a prime number and m= 2(p²+p+ 1). By Lemma 2, there is a setB ⊆Z_p2+p+1 such thatRB,−B(n) = 1 for alln∈Z_p2+p+1 and n6= 0. Then for any integerlwith 0≤l≤p²+p, we define

Al= 2B∪(2B+ 2l+ 1) modm, where 2B={2b: b∈B}. Now we first prove thatRAl(n)≤4 for alln∈Zm.

If 2 | n, then n =a1+a2 with a1, a2 ∈2B or a1, a2 ∈ 2B+ 2l+ 1. Hence RAl(n) = RB(ⁿ₂) +RB(ⁿ₂−(2l+ 1))≤2 + 2 = 4.

If 2∤n, then n=a1+a2 with a1 ∈2B, a2 ∈2B+ 2l+ 1 ora1∈2B+ 2l+ 1, a2∈2B.

Hence,RAl(n) =RB(^n−2l−1₂ )×2≤4.

Therefore,RAl(n)≤4 for alln∈Zm. LetP be a statement and we define

I(P) =











1, if the statementP is true;

0, if the statementP is false.

Let

X_odd^l ={2k+ 1 : 2k+ 1∈Zm andRAl(2k+ 1) = 0}, X_even^l ={2k: 2k∈Zm andRAl(2k) = 0}.

Then S0=X_odd^l ∪X_even^l . It is clear thatRAl(2n+ 1) = 0 if and only ifRB(n−l) = 0.

Then

|X_odd^l | = p²+p+ 1−♯{n:n∈Z_p2+p+1, RB(n) = 2} −♯{n:n∈Z_p2+p+1, RB(n) = 1}

= p²+p+ 1− p+ 1

2

−(p+ 1) = 1 2p²−1

2p < 1 4m.

and

p²+p

X

l=0

|X_even^l | =

p²+p

X

l=0

|{n: n∈Zp²+p+1, RB(n) = 0 andRB(n−2l−1) = 0}|

=

p²+p

X

l=0 p²+p

X

n=0

I(RB(n) = 0 andRB(n−2l−1) = 0)

=

p²+p

X

l=0 p²+p

X

n=0

I(RB(n) = 0)I(RB(n−2l−1) = 0)

=

p²+p

X

n=0

I(RB(n) = 0)

p²+p

X

l=0

I(RB(n−2l−1) = 0)

= p²

2 −p 2

^p2+p X

n=0

I(RB(n) = 0) = p²

2 −p 2

² .

Hence there is an integerl such that

|X_even^l | ≤ 1

4· (p²−p)² p²+p+ 1 < 1

4(p²+p+ 1) = 1 8m.

Therefore, for this integerl,

|S0|=|X_odd^l |+|X_even^l |<1 4m+1

8m= 3 8m.

Proof of Theorem 2. By Lemma 3, takingk= 2, we have X

g∈G

(RA(g)−2)²≥2m−3|A|+ 2.

(3)

On the other hand, X

g∈G

(RA(g)−2)² = 4|S0|+|S1|+|S3|+ 4|S4|+ 9|S5|

≤ 4(|S0|+|S4|) + 9(|S1|+|S3|+|S5|) (4)

≤ 4(|S0|+|S4|) + 9|A|.

Hence, by (3) and (4), we have|S0|+|S4| ≥ ¹₂m−3|A|+¹₂ ≥¹₂m−3√

5m. Since

|S0|+|S1|+|S2|+|S3|+|S4|+|S5|=m, it follows that

|S2| ≤ 1

2m+ 3√ 5m.

Now we prove the part (b). By Lemma 2, there exists a subset A ⊆ Zm such that RA,−A(n) = 1 for alln∈Zm,n6= 0. It is easy to see that|A|=p+ 1 andRA(n)≤2 for alln∈Zm. Hence

|S2|= |A|

2

=1

2(p²+p+ 1)−1 2 = 1

2m−1 2.

Proof of Theorem 3. By Lemma 3, takingk= 4, we have X

g∈G

(RA(g)−4)²≥4m−7|A|+ 12.

On the other hand, by|S1|+|S3|+|S5|+|S7| ≤ |A|, we have X

g∈G

(RA(g)−4)² = 16|S0|+ 9|S1|+ 4|S2|+|S3|+|S5|+ 4|S6|+ 9|S7|

≤ 4(|S0|+|S2|+|S4|+|S6|) + 9|A|+ 12|S0| −4|S4|

≤ 4m+ 9|A|+ 12|S0| −4|S4|. Hence|S4| ≤4|A|+ 3|S0|+ 3. SinceP7

i=0|Si|=m, it follows that m≥ |S0|+|S4| ≥|S4| −4|A| −3

3 +|S4|=4

3|S4| −4

3|A| −1.

By Lemma 1, we have

|S4| ≤ 3

4m+|A|+3 4 ≤3

4m+√ 7m+3

4.

Now we prove the part (b). Let p be a prime and m = 2(p²+p+ 1). By Lemma 2, there exists a subset Ap ⊆ Zp²+p+1 such that RAp,−Ap(n) = 1 for all n 6= 0. Let A= 2Ap∪(p²+p+ 1 + 2Ap)⊆Zm.

If 2 | n, then RA(n) = RAp(ⁿ₂) +RAp(ⁿ₂) ≤ 2 + 2 = 4. If 2 ∤ n, then RA(n) = 2RAp(^n−(p2₂^+p+1))≤4. HenceRA(n)≤4 for alln∈Zm.

|S4| = |{n: 2|n, n∈Zm andRAp(n 2) = 2}|

+ |{n: 2∤n, n∈Zm andRAp(n−(p²+p+ 1) 2 ) = 2}|

= |{n:n∈Zp²+p+1 andRAp(n) = 2}|+|{n: n∈Zp²+p+1 andRAp(n−p²+p 2 ) = 2}|

= 2

p+ 1 2

=p²+p= 1 2m−1.

4 Acknowledgement

This work was done during the second author visiting to Budapest University of Technology and Economics. He would like to thank Dr. S´andor Kiss and Dr. Csaba S´andor for their warm hospitality.

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