ADDITIVE AND SUBTRACTIVE BASES OF $\mathbb {Z}_m$ IN AVERAGE

Abstract

Given a positive integer m, let $\mathbb {Z}_m$ be the set of residue classes mod m. For $A\subseteq \mathbb {Z}_m$ and $n\in \mathbb {Z}_m$, let $\sigma _A(n)$ be the number of solutions to the equation $n=x+y$ with $x,y\in A$. Let $\mathcal {H}_m$ be the set of subsets $A\subseteq \mathbb {Z}_m$ such that $\sigma _A(n)\geq 1$ for all $n\in \mathbb {Z}_m$. Let

$$ \begin{align*} \ell_m=\min\limits_{A\in \mathcal{H}_m}\bigg\lbrace m^{-1}\sum_{n\in \mathbb{Z}_m}\sigma_A(n)\bigg\rbrace. \end{align*} $$

Ding and Zhao [‘A new upper bound on Ruzsa’s numbers on the Erdős–Turán conjecture’, Int. J. Number Theory 20 (2024), 1515–1523] showed that $\limsup _{m\rightarrow \infty }\ell _m\le 192$. We prove

$$ \begin{align*} \limsup\limits_{m\rightarrow\infty}\ell_m\leq 144 \end{align*} $$

and investigate parallel results on subtractive bases of $ \mathbb {Z}_m$.

1 Introduction

Let $\mathbb {N}$ be the set of natural numbers and A a subset of $\mathbb {N}$ . A remarkable conjecture of Erdős and Turán [6] states that if all sufficiently large numbers n can be written as the sum of two elements of A, then the number of representations of n as the sum of two elements of A cannot be bounded. Progress on this conjecture was made by Grekos et al. [8], who proved that the number of representations cannot be bounded by $5$ , later improved to $7$ by Borwein et al. [1]. For more on the Erdős–Turán conjecture, see the books of Halberstam and Roth [10] and Tao and Vu [17].

A set A is called an asymptotic basis of natural numbers if all sufficiently large numbers n can be written as the sum of two elements of A. Motivated by Erdős’ question, Ruzsa [12] constructed an asymptotic basis A of natural numbers which has a bounded square mean value. Ruzsa also considered a variant on the Erdős–Turán conjecture. Let $\mathbb {Z}_m$ be the set of residue classes mod m and A a subset of $\mathbb {Z}_m$ . For any $n\in \mathbb {Z}_m$ , let

$$ \begin{align*} \sigma_A(n)=\#\{(x,y):n=x+y,~x,y\in \mathbb{Z}_m\}. \end{align*} $$

The Ruzsa number $R_m$ is defined to be the least positive integer r so that there exists a set $A\subseteq \mathbb {Z}_m$ with $ 1\le \sigma _A(n)\le r \mbox { for all } n\in \mathbb {Z}_m. $ In his argument, Ruzsa proved that there is an absolute constant C such that $R_m\le C$ for all positive integers m. Employing Ruzsa’s ideas, Tang and Chen [15] proved that $R_m\le 768$ for all sufficiently large m. Later, in [16], they obtained $R_m\le 5120$ for all positive integers m. In [2], Chen proved that $R_m\le 288$ for all positive integers m, and this was recently improved to $R_m\le 192$ by Ding and Zhao [5]. However, Sándor and Yang [13] showed that $R_m\ge 6$ for all $m\ge 36$ .

Ding and Zhao [5] provided an average version of Ruzsa’s number. Precisely, let $\mathcal {H}_m$ be the set of subsets $A\subseteq \mathbb {Z}_m$ such that $\sigma _A(n)\geq 1$ for all $n\in \mathbb {Z}_m$ . Ding and Zhao defined the minimal mean value as

$$ \begin{align*} \ell_m=\min\limits_{A\in \mathcal{H}_m}\bigg\lbrace m^{-1}\sum_{n\in \mathbb{Z}_m}\sigma_A(n)\bigg\rbrace. \end{align*} $$

As they pointed out, their result on $R_m\le 192$ clearly implies

(1.1) $$ \begin{align} \limsup_{m\rightarrow\infty}\ell_m\le 192. \end{align} $$

Ding and Zhao [5, Section 3] thought that ‘any improvement of the bound (1.1) would be of interest’. In this note, we shall make some progress on this problem.

Theorem 1.1. We have

$$ \begin{align*} \limsup\limits_{m\rightarrow\infty}\ell_m\leq 144. \end{align*} $$

Parallel to the additive bases of $\mathbb {Z}_m$ , one naturally considers the corresponding results on subtractive bases of $\mathbb {Z}_m$ . Let A be a subset of $\mathbb {Z}_m$ . For any $n\in \mathbb {Z}_m$ , let

$$ \begin{align*} \delta_A(n)=\#\{(x,y):n=x-y,~x,y\in \mathbb{Z}_m\}. \end{align*} $$

In [3], Chen and Sun proved that for any positive integer m, there exists a subset A of $\mathbb {Z}_m$ so that $\delta _A(n)\ge 1$ for any $n\in \mathbb {Z}_m$ and $\delta _A(n)\le 7$ for all $n\in \mathbb {Z}_m$ with three exceptions. Their result was recently improved by Zhang [18] who showed that $\delta _A(n)\le 7$ could be refined to $\delta _A(n)\le 5$ , again with three exceptions. The exceptions cannot be removed by their method. Motivated by the minimal mean value defined by Ding and Zhao, we consider a parallel quantity

$$ \begin{align*} g_m:=\min\limits_{A\in \mathcal{K}_m}\bigg\lbrace m^{-1}\sum_{n\in \mathbb{Z}_m}\delta_A(n)\bigg\rbrace, \end{align*} $$

where $\mathcal {K}_m$ is the set of subsets $A\subseteq \mathbb {Z}_m$ such that $\delta _A(n)\geq 1$ for all $n\in \mathbb {Z}_m$ . Obviously, Zhang’s bound implies that

$$ \begin{align*} \limsup\limits_{m\rightarrow\infty}g_m\leq 5 \end{align*} $$

since the total sums of $\delta _A(n)$ for the three exceptions contribute only $O(\sqrt {m})$ . Our second main result gives a small improvement on this bound.

Theorem 1.2. We have

$$ \begin{align*} \limsup\limits_{m\rightarrow\infty}g_m\leq 2. \end{align*} $$

There is an old conjecture known as the prime power conjecture (see, for example, [7, 9, 11]) which states that if A is a subset of $\mathbb {Z}_m$ with $\delta _A(n)=1$ for any nonzero $ n\in \mathbb {Z}_m$ , then $m=p^{2\alpha }+p^\alpha +1$ , where $p^\alpha $ is a prime power. The reverse direction was proved by Singer [14] as early as 1938.

As mentioned by Ding and Zhao [5], it is clear that $\liminf _{m\rightarrow \infty }\ell _m\ge 2$ from [13, Lemma 2.2]. They conjectured that $\liminf _{m\rightarrow \infty }\ell _m\ge 3$ [5, Conjecture 3.3]. Based on the results of Singer and Theorem 1.2, it seems reasonable to conjecture that

$$ \begin{align*} \lim\limits_{m\rightarrow\infty}g_m=1. \end{align*} $$

If true, these conjectures reflect rather different features between additive bases and subtractive bases.

2 Proof of Theorem 1.1

For any integer k, let

$$ \begin{align*} Q_k=\{(u,ku^2):u\in \mathbb{Z}_{p}\}\subset \mathbb{Z}_{p}^2. \end{align*} $$

We will make use of the following lemmas.

Lemma 2.1 (Chen [2, Lemma 2]).

Let p be an odd prime and m a quadratic nonresidue of p with $m+1\not \equiv 0 \pmod {p}, 3m+1\not \equiv 0\pmod {p}$ and $m+3\not \equiv 0 \pmod {p}$ . Put

$$ \begin{align*} B=Q_{m+1}\cup Q_{m(m+1)}\cup Q_{2m}. \end{align*} $$

Then, for any $(c,d)\in \mathbb {Z}_{p}^2$ , we have $1\le \sigma _B(c,d)\le 16$ , where $\sigma _B(c,d)$ is the number of solutions of the equation $(c,d)=x+y,~x,y\in B$ .

Lemma 2.2 (Prime number theorem; see, for example, [4]).

Let $\pi (x)$ be the number of primes p not exceeding x. Then,

$$ \begin{align*} \pi(x)\sim x/\!\log x \quad \text{as~} x\rightarrow\infty. \end{align*} $$

Lemma 2.3. Let m be a positive integer and A a subset of $\mathbb {Z}_m$ . Then,

$$ \begin{align*} \sum_{n\in \mathbb{Z}_m}\sigma_A(n)=|A|^2, \end{align*} $$

where $|A|$ denotes the number of elements of A.

Proof. Clearly,

$$ \begin{align*} \sum\limits_{n\in \mathbb{Z}_m}\sigma_A(n)=\sum\limits_{n\in\mathbb{Z}_m}\sum\limits_{\substack{a_1+a_2=n\\a_1,a_2\in A}}1=\sum\limits_{\substack{a_1,a_2\in A\\a_1+a_2\in \mathbb{Z}_m}}1=\sum\limits_{a_1,a_2\in A}1=|A|^{2}. \end{align*} $$

This completes the proof of Lemma 2.3.

Lemma 2.4. Let p be a prime greater than $11$ . Then there is a subset $A\subset \mathbb {Z}_{2p^2}$ with $|A|\le 12p$ so that $\sigma _A(n)\ge 1$ for any $n\in \mathbb {Z}_{2p^2}$ .

Proof. Let p be a prime greater than $11$ . Then there are at least $(p-1)/2>5$ quadratic nonresidues mod p, which means that there is some quadratic nonresidue m so that

$$ \begin{align*} m+1\not\equiv 0 \pmod{p}, \quad 3m+1\not\equiv 0\pmod{p} \quad \text{and} \quad m+3\not\equiv 0 \pmod{p}. \end{align*} $$

Let $B=Q_{m+1}\cup Q_{m(m+1)}\cup Q_{2m}$ , $A_1=\lbrace u+2pv:(u,v)\in B\rbrace $ and $A=A_1\cup (A_1+p)$ , where $A_1+p:=\{a_1+p:a_1\in A_1\}$ . Obviously, A can be viewed as a subset of $\mathbb {Z}_{2p^2}$ .

We first show that $\sigma _A(n)\ge 1$ for any $n\in \mathbb {Z}_{2p^2}$ , that is, $A\in \mathcal {H}_{2p^2}$ (by the definition of $\mathcal {H}_{m}$ ). We follow the proof of Chen [2, Theorem 1]. For any $(u,v)\in B$ , we have $0\le u,v\le p-1$ . Let n be an element of $\mathbb {Z}_{2p^2}$ with $0\le n\le 2p^2-1$ . Then, we can assume that

$$ \begin{align*} n=c+2pd \end{align*} $$

with $p\le c\le 3p-1$ and $-1\le d\le p-1$ . By Lemma 2.1, there are $(u_1,v_1),(u_2,v_2)\in B$ so that

$$ \begin{align*} (c,d)=(u_1,v_1)+(u_2,v_2) \pmod{p}, \end{align*} $$

or in other words,

$$ \begin{align*} c\equiv u_1+u_2 \pmod{p} \quad \text{and} \quad d\equiv v_1+v_2 \pmod{p}. \end{align*} $$

Suppose that

$$ \begin{align*} c=u_1+u_2+ps \quad \text{and} \quad d=v_1+v_2+ph, \end{align*} $$

with $s,h\in \mathbb {Z}$ . Then, $s=0$ or $1$ or $2$ since $0\le u_1+u_2\le 2p-2$ and $p\le c\le 3p-1$ . Hence,

$$ \begin{align*} n&=c+2pd\\ &= u_1+2pv_1+u_2+2pv_2+ps+2p^2h \\ &\equiv u_1+2pv_1+u_2+2pv_2+ps \pmod{2p^2}. \end{align*} $$

If $s=0$ , then in $\mathbb {Z}_{2p^2}$ ,

$$ \begin{align*} n= (u_1+2pv_1)+(u_2+2pv_2)\in A_1+A_1\subset A+A. \end{align*} $$

If $s=1$ , then in $\mathbb {Z}_{2p^2}$ ,

$$ \begin{align*} n= (u_1+2pv_1+p)+(u_2+2pv_2)\in (A_1+p)+A_1\subset A+A. \end{align*} $$

If $s=2$ , then in $\mathbb {Z}_{2p^2}$ ,

$$ \begin{align*} n= (u_1+2pv_1+p)+(u_2+2pv_2+p)\in (A_1+p)+(A_1+p)\subset A+A. \end{align*} $$

Hence, in all cases, $\sigma _A(n)\ge 1$ for $n\in \mathbb {Z}_{2p^2}$ .

It can be easily seen that $|A_1|\le 2|B|$ from the construction. Therefore, for the set A constructed above,

$$ \begin{align*} |A|\le |A_1|+|A_1+p|=2|A_1|\le 2\times 2|B|=4|B| \end{align*} $$

and

$$ \begin{align*} |B|\leqslant|Q_{m+1}|+|Q_{m(m+1)}|+|Q_{2m}|=3p, \end{align*} $$

from which it follows that

$$ \begin{align*} |A|\le 12p. \end{align*} $$

This completes the proof of Lemma 2.4.

The final lemma gives a relation between the bases of $\mathbb {Z}_{m_1}$ and $\mathbb {Z}_{m_2}$ with certain constraints.

Lemma 2.5. Let $\varepsilon>0$ be an arbitrarily small number. Let $m_1$ and $m_2$ be two positive integers with $(2-\varepsilon )m_1<m_2<2m_1$ . If A is a subset of $\mathbb {Z}_{m_1}$ with $\sigma _A(n)\ge 1$ for any $n\in \mathbb {Z}_{m_1}$ , then there is a subset B of $\mathbb {Z}_{m_2}$ with $|B|\le 2|A|$ such that $\sigma _B(n)\ge 1$ for any $n\in \mathbb {Z}_{m_2}$ .

Proof. Suppose that $m_2=m_1+r$ , so that $(1-\varepsilon )m_1<r< m_1$ . Let

$$ \begin{align*} B=A\cup \{a+r:a\in A\}. \end{align*} $$

Then, $|B|\le 2|A|$ . It remains to prove $\sigma _B(n)\ge 1$ for any $n\in \mathbb {Z}_{m_2}$ .

Without loss of generality, we may assume $0\le a\le m_1-1$ for any $a\in A$ . For $0\le n\le m_1-1$ , there are two integers $a_1,a_2\in A$ so that $ n\equiv a_1+a_2\pmod {m_1}. $ Since $0\le a_1+a_2\le 2m_1-2$ , it follows that

$$ \begin{align*} n=a_1+a_2 \quad \text{or} \quad n=a_1+a_2-m_1. \end{align*} $$

If $n=a_1+a_2$ , then clearly $n\equiv a_1+a_2\pmod {m_2}$ . If $n=a_1+a_2-m_1$ , then

$$ \begin{align*} n+m_2=n+m_1+r=a_1+(a_2+r), \end{align*} $$

which means that $n\equiv a_1+(a_2+r)\pmod {m_2}$ . In both cases, $\sigma _B(n)\ge 1$ for any n with $0\le n\le m_1-1$ . We are left to consider the case $m_1\le n\le m_2-1$ . In this range,

$$ \begin{align*} 0<n-r\le m_2-1-r= m_1-1. \end{align*} $$

Thus, there are two elements $\widetilde {a_1},\widetilde {a_2}$ of A so that

$$ \begin{align*} n-r\equiv \widetilde{a_1}+\widetilde{a_2}\pmod{m_1}. \end{align*} $$

Again, by the constraint $0\le \widetilde {a_1}+\widetilde {a_2}\le 2m_1-2$ ,

$$ \begin{align*} n-r=\widetilde{a_1}+\widetilde{a_2} \quad \text{or} \quad n-r=\widetilde{a_1}+\widetilde{a_2}-m_1. \end{align*} $$

If $n-r=\widetilde {a_1}+\widetilde {a_2}$ , then we clearly have $n-r\equiv \widetilde {a_1}+\widetilde {a_2}\pmod {m_2}$ . Otherwise, we have $n-r=\widetilde {a_1}+\widetilde {a_2}-m_1$ . So, it can now be deduced that

$$ \begin{align*} n+m_2=\widetilde{a_1}+r+\widetilde{a_2}+r, \end{align*} $$

which is equivalent to $n\equiv (\widetilde {a_1}+r)+(\widetilde {a_2}+r)\pmod {m_2}$ .

Proof of Theorem 1.1.

Let $\varepsilon>0$ be an arbitrarily small given number. Then, by Lemma 2.2, there is some prime p so that

(2.1) $$ \begin{align} \sqrt{\frac{m}{4}}<p<\sqrt{\frac{m}{2(2-\varepsilon)}}, \end{align} $$

provided that m is sufficiently large (in terms of $\varepsilon $ ). By Lemma 2.4, there is a subset $A\subset \mathbb {Z}_{2p^2}$ with $|A|\le 12p$ so that $\sigma _{A}(n)\ge 1$ for any $n\in \mathbb {Z}_{2p^2}$ . From (2.1),

(2.2) $$ \begin{align} (2-\varepsilon)2p^2<m<2\times2p^2. \end{align} $$

Thus, by Lemma 2.5, there is a subset B of $\mathbb {Z}_{m}$ with

(2.3) $$ \begin{align} |B|\le 2|A|\le 24p \end{align} $$

such that $\sigma _B(n)\ge 1$ for any $n\in \mathbb {Z}_{m}$ . Hence, by Lemma 2.3,

$$ \begin{align*} \ell_m=\min\limits_{\widetilde{A}\in \mathcal{H}_m}\bigg\lbrace m^{-1}\sum_{n\in \mathbb{Z}_m}\sigma_{\widetilde{A}}(n)\bigg\rbrace\le m^{-1}\sum_{n\in \mathbb{Z}_m}\sigma_{B}(n)=\frac{|B|^2}{m}. \end{align*} $$

Employing (2.2) and (2.3),

$$ \begin{align*} \frac{|B|^2}{m}\le \frac{(24p)^2}{(2-\varepsilon)2p^2}=144\times\frac{2}{2-\varepsilon}. \end{align*} $$

Hence, it follows that

$$ \begin{align*} \limsup_{m\rightarrow\infty}\ell_m\le 144\times\frac{2}{2-\varepsilon} \end{align*} $$

for any $\varepsilon>0$ , which clearly means that

$$ \begin{align*} \limsup_{m\rightarrow\infty}\ell_m\le 144. \end{align*} $$

This completes the proof of Theorem 1.1.

3 Proof of Theorem 1.2

The proof of Theorem 1.2 is based on the following remarkable result of Singer.

Lemma 3.1 (Singer [14]).

Let p be a prime. Then, there exists a subset A of $\mathbb {Z}_{p^{2}+p+1}$ so that $\delta _A(n)=1$ for any $n\in \mathbb {Z}_{p^{2}+p+1}$ with $n\neq \overline {0}$ .

The next lemma is a variant of Lemma 2.3.

Lemma 3.2. Let m be a positive integer and A a subset of $\mathbb {Z}_m$ . Then,

$$ \begin{align*} \sum_{n\in \mathbb{Z}_m}\delta_A(n)=|A|^2, \end{align*} $$

where $|A|$ denotes the number of elements of A.

Proof. It is clear that

$$ \begin{align*} \sum\limits_{n\in \mathbb{Z}_m}\delta_A(n)=\sum\limits_{n\in\mathbb{Z}_m}\sum\limits_{\substack{a_1-a_2=n\\a_1,a_2\in A}}1=\sum\limits_{\substack{a_1,a_2\in A\\a_1-a_2\in \mathbb{Z}_m}}1=\sum\limits_{a_1,a_2\in A}1=|A|^{2}. \end{align*} $$

This completes the proof of Lemma 3.2.

We need another auxiliary lemma.

Lemma 3.3. Let $\varepsilon>0$ be an arbitrarily small number. Let m be a positive integer and p a prime number with

$$ \begin{align*} (2-\varepsilon)(p^{2}+p+1)<m<2(p^{2}+p+1). \end{align*} $$

If A is a subset of $\mathbb {Z}_{p^{2}+p+1}$ with $\delta _A(n)\ge 1$ for any $n\in \mathbb {Z}_{p^{2}+p+1}$ , then there is a subset B of $\mathbb {Z}_{m}$ with $|B|\le 2|A|$ such that $\delta _B(n)\ge 1$ for any $n\in \mathbb {Z}_{m}$ .

Proof. Suppose that $m{\kern-1pt}={\kern-1pt}(p^2{\kern-1pt}+{\kern-1pt}p{\kern-1pt}+{\kern-1pt}1){\kern-1pt}+{\kern-1pt}r$ . Then, $(1-\varepsilon )(p^2{\kern-1pt}+{\kern-1pt}p{\kern-1pt}+{\kern-1pt}1){\kern-1pt}<{\kern-1pt}r{\kern-1pt}<{\kern-1pt} (p^2{\kern-1pt}+{\kern-1pt} p{\kern-1pt}+{\kern-1pt}1)$ . Let

$$ \begin{align*} B=A\cup \{a+r:a\in A\}. \end{align*} $$

Then, $|B|\le 2|A|$ . It remains to prove $\delta _B(n)\ge 1$ for any $n\in \mathbb {Z}_{m}$ .

Without loss of generality, we can assume $0\le a\le p^2+p$ for any $a\in A$ . For $0\le n\le p^2+p$ , there are two integers $a_1,a_2\in A$ so that

$$ \begin{align*} n\equiv a_1-a_2\pmod{p^2+p+1}, \end{align*} $$

which means that

$$ \begin{align*} n= a_1-a_2 \quad \text{or} \quad n=a_1-a_2+(p^2+p+1) \end{align*} $$

since $-p^2-p\le a_1-a_2\le p^2+p$ . If $n=a_1-a_2$ , then we clearly have $n\equiv a_1-a_2 \pmod {m}$ . If $n=a_1-a_2+(p^2+p+1)$ , then

$$ \begin{align*} n-m=n-(p^2+p+1)-r=a_1-(a_2+r), \end{align*} $$

from which it can be deduced that $n\equiv a_1-(a_2+r)\pmod {m}$ . In both cases, we have $\delta _B(n)\ge 1$ for any n with $0\le n\le p^2+p$ . We are left to consider the case $p^2+p+1\le n\le m-1$ . In this case,

$$ \begin{align*} 0<n-r\le m-1-r=p^2+p. \end{align*} $$

Thus, there are two elements $\widetilde {a_1},\widetilde {a_2}$ of A so that

$$ \begin{align*} n-r\equiv \widetilde{a_1}-\widetilde{a_2}\pmod{m}. \end{align*} $$

Again, by the constraint $-p^2-p\le \widetilde {a_1}-\widetilde {a_2}\le p^2+p$ , we have

$$ \begin{align*} n-r=\widetilde{a_1}-\widetilde{a_2} \quad \text{or} \quad n-r=\widetilde{a_1}-\widetilde{a_2}+(p^2+p+1). \end{align*} $$

If $n-r=\widetilde {a_1}-\widetilde {a_2}$ , then we clearly have $n-r\equiv \widetilde {a_1}-\widetilde {a_2}\pmod {m}$ . Otherwise, we have $n-r=\widetilde {a_1}-\widetilde {a_2}+(p^2+p+1)$ , from which it clearly follows that

$$ \begin{align*} n-m=\widetilde{a_1}-\widetilde{a_2}. \end{align*} $$

So we also deduce $n\equiv \widetilde {a_1}-\widetilde {a_2}\pmod {m}$ .

We now turn to the proof of Theorem 1.2.

Proof of Theorem 1.2.

Let $\varepsilon>0$ be an arbitrarily small given number. By Lemma 2.2, there is some prime p so that

$$ \begin{align*} \frac{\sqrt{2m-3}-1}{2}<p<\frac{\sqrt{\frac{4}{2-\varepsilon}m-3}-1}{2} \end{align*} $$

providing that m is sufficiently large (in terms of $\varepsilon $ ). Equivalently,

(3.1) $$ \begin{align} (2-\varepsilon)(p^{2}+p+1)<m<2(p^{2}+p+1). \end{align} $$

By Lemma 3.1, there is a subset A of $\mathbb {Z}_{p^{2}+p+1}$ so that $\delta _A(n)=1$ for any $n\in \mathbb {Z}_{p^{2}+p+1}$ with $n\neq \overline {0}$ . Employing Lemma 3.2,

$$ \begin{align*} |A|^2=\sum\limits_{n\in \mathbb{Z}_{p^{2}+p+1}}\delta_A(n)=\sum\limits_{n\in \mathbb{Z}_{p^{2}+p+1},~n\neq \overline{0}}\delta_A(n)+\delta_A(0)=p^{2}+p+|A|, \end{align*} $$

from which it follows clearly that

$$ \begin{align*} |A|=p+1. \end{align*} $$

By Lemma 3.3 and (3.1), there is a subset B of $\mathbb {Z}_{m}$ with

(3.2) $$ \begin{align} |B|\le 2|A|\le 2(p+1) \end{align} $$

such that $\delta _B(n)\ge 1$ for any $n\in \mathbb {Z}_{m}$ . Thus, by the definition of $g_m$ and Lemma 3.2 again,

$$ \begin{align*} g_m=\min\limits_{\widetilde{A}\in \mathcal{K}_m}\bigg\lbrace m^{-1}\sum_{n\in \mathbb{Z}_m}\delta_{\widetilde{A}}(n)\bigg\rbrace\le m^{-1}\sum_{n\in \mathbb{Z}_m}\delta_{B}(n)=\frac{|B|^2}{m}. \end{align*} $$

From (3.1) and (3.2),

$$ \begin{align*} \frac{|B|^2}{m}\le \frac{4(p+1)^2}{(2-\varepsilon)(p^{2}+p+1)}\le \frac{4}{2-\varepsilon/2}, \end{align*} $$

provided that m (hence p) is sufficiently large (in terms of $\varepsilon $ ). Hence, we conclude that

$$ \begin{align*} \limsup_{m\rightarrow\infty}g_m\le \frac{4}{2-\varepsilon/2} \end{align*} $$

for any $\varepsilon>0$ , which clearly means that

$$ \begin{align*} \limsup_{m\rightarrow\infty}g_m\le 2. \end{align*} $$

This completes the proof of Theorem 1.2.