# A note on additive complements of the squares

Yuchen Ding, Yu–Chen Sun, Li–Yuan Wang and Yutong Xia  (Yuchen Ding) School of
Mathematical Science, Yangzhou University, Yangzhou 225002, People’s Republic
of China<EMAIL_ADDRESS>(Yu–Chen Sun) Department of Mathematics and
Statistics, University of Turku, Turku 20014 , Finland<EMAIL_ADDRESS>(Li-Yuan Wang) School of Physical and Mathematical Sciences, Nanjing Tech
University, Nanjing 211816, People’s Republic of China<EMAIL_ADDRESS>(Yutong Xia) School of Mathematical Science, Yangzhou University, Yangzhou
225002, People’s Republic of China<EMAIL_ADDRESS>
###### Abstract.

Let $\mathcal{S}=\\{1^{2},2^{2},3^{2},...\\}$ be the set of squares and
$\mathcal{W}\subset\mathbb{N}$ be an additive complement of $\mathcal{S}$ so
that $\mathcal{S}+\mathcal{W}\supset\\{n\in\mathbb{N}:n\geq N_{0}\\}$ for some
$N_{0}$. Let also
$\mathcal{R}_{\mathcal{S},\mathcal{W}}(n)=\\#\\{(s,w):n=s+w,s\in\mathcal{S},w\in\mathcal{W}\\}$.
Chen and Fang [5] proved that

$\sum_{n=1}^{N}R_{\mathcal{S},\mathcal{W}}(n)-N\gg N^{1/4}\log N.$

In this note, we improve their result and obtain that

$\sum_{n=1}^{N}R_{\mathcal{S},\mathcal{W}}(n)-N\gg N^{1/3}.$

2020 Mathematics Subject Classification. Primary 11B13; Secondary 11B75.
Keywords. Additive complements, Squares.

##  1\. Introduction

Let $\mathbb{N}$ be the set of all non-negative integers. We say that
$A,B\subset\mathbb{N}$ are additive complements if
$A+B\supset\\{n\in\mathbb{N}:n\geq N_{0}\\}$ for some fixed $N_{0}$. If $A,B$
are additive complements, we say that $A$ is an additive complement of $B$.
Let $\mathcal{S}=\\{1^{2},2^{2},3^{2},...\\}$ the set of squares and
$\mathcal{W}\subset\mathbb{N}$ be an additive complement of $\mathcal{S}$. Let
$N$ be a large integer and $\mathcal{W}(N)$ be the number of elements of
$\mathcal{W}$ not exceeding $N$. Clearly, we have $\mathcal{W}(N)\sqrt{N}\geq
N-N_{0}$ for some given integer $N_{0}$, from which we deduce that

$\liminf_{N\rightarrow\infty}\mathcal{W}(N)/\sqrt{N}\geq 1.$

It is of great interest to ask whether
$\liminf_{N\rightarrow\infty}\mathcal{W}(N)/\sqrt{N}$ is strictly larger than
$1$.

In 1956, Erdős [8] posed this problem, which was later settled affirmatively
by Moser [10] with an accurate lower bound $1.06$. The number was then
improved in a few articles [1, 2, 4, 7, 11, 12]. Up to now the best result is

$\displaystyle\liminf_{N\rightarrow\infty}\mathcal{W}(N)/\sqrt{N}\geq 4/\pi,$
(1.1)

given by Cilleruelo [6], Habsieger [9], Balasubramanian and Ramana [3].

Note that $\mathcal{W}(N)/\sqrt{N}>1$ implies that there are some integers
$n\in\mathbb{N}$ which do not have an unique representation. Thus it is
worthwhile to study the number of representations for writing $n$ as a sum
$s+w$ with $s\in\mathcal{S}$ and $w\in\mathcal{W}$.

For any $A,B\subset\mathbb{N}$, let

$R_{A,B}(n)=\\#\\{(a,b):n=a+b,a\in A,b\in B\\}.$

Obviously, if $A,B$ are complements, then

$\liminf_{N\rightarrow\infty}\frac{1}{N}\sum_{n\leq N}R_{A,B}(n)\geq 1.$

Ben Green had a nice observation that for $B=\\{b_{1},b_{2},...\\}$ with
$b_{n}=\frac{\pi^{2}}{16}n^{2}+o(n^{2})$,

$\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=1}^{N}R_{S,B}(n)=1.$

Later, he asked Fang (private communication, see [5]) whether there is an
additive complement of $\mathcal{S}$ has the same property as $B$. Namely, can
we find an additive complement $\mathcal{W}=\\{w_{n}\\}_{n=1}^{\infty}$ of
$\mathcal{S}$ satisfying the asymptotic condition
$w_{n}=\frac{\pi^{2}}{16}n^{2}+o(n^{2})$?

Motivated by Ben Green’s problem, Chen and Fang [5, Theorem 1.1] proved the
following result.

Let $N$ be a sufficiently large integer. For any additive complement
$\mathcal{W}$ of $\mathcal{S}$, we have

$\displaystyle\sum_{n=1}^{N}R_{\mathcal{S},\mathcal{W}}(n)-N\gg\mathcal{W}(2\sqrt{N})\log\mathcal{W}(2\sqrt{N})$
(1.2)

From (1.1), one can immediately obtain the following corollary. If
$\mathcal{W}$ is any complement of $\mathcal{S}$, then

$\sum_{n=1}^{N}R_{\mathcal{S},\mathcal{W}}(n)-N\gg N^{1/4}\log N.$ (1.3)

Chen and Fang [5, Remark 1] also gave a counterexample to show that (1.3) does
not always hold, if $\mathcal{S}$ and $\mathcal{W}$ are substituted with
general additive complements $A,B$.

In this note, we will give an improvement of the result of Chen and Fang and
get the following theorem.

######  Theorem 1.1.

Let $N$ be a sufficiently large integer. For any additive complement
$\mathcal{W}$ of $\mathcal{S}$, we have

$\sum_{n=1}^{N}R_{\mathcal{S},\mathcal{W}}(n)-N\gg\mathcal{W}\left(\frac{1}{5}N^{2/3}\right).$

######  Remark 1.1.

The coefficient $1/5$ can be further optimized by carefully calculating.

From Theorem 1.1 and (1.1), one can immediately obtain the following
corollary.

######  Corollary 1.1.

For any additive complement $\mathcal{W}$ of $\mathcal{S}$, we have

$\sum_{n=1}^{N}R_{\mathcal{S},\mathcal{W}}(n)-N\gg N^{1/3}.$

The proof of Theorem 1.1 is based on the structure of the proof in Chen–Fang
[5]. In Chen–Fang’s proof, for $d_{1},d_{2}\in\mathcal{W}$, they considered
the equation $x^{2}+d_{1}=y^{2}+d_{2}<N$ and required that this equation has
$\gg\log N$ solutions (see [5, Lemma 2.1]). Thus the $\log$-factor, in (1.3),
comes from the number of solutions. However, in Chen-Fang’s arguments, the
above equation has some solutions $(x,y)$ such that $|x-y|$ is very small,
namely $O(1)$, and thus $x+y$ is very large, namely $O(|d_{2}-d_{1}|)$. Since
$\max\\{x^{2},y^{2}\\}<N$, we must have $|d_{2}-d_{1}|\ll N^{1/2}$. For this
reason, in their arguments, they restricted
$d_{1},d_{2}\in\mathcal{W}\cap[cN^{1/2}]$ for some $c>0$. Here $[N]$ denotes
the set $\\{n\in\mathbb{N}:n\leq N\\}$.

Our new idea is that we exclude those solutions $(x,y)$ such that $|x-y|$ is
small. Namely, we require solutions $(x,y)$ to satisfy that $|x-y|\gg
N^{1/6}$, so $x+y$ is not too large, namely, $x+y\ll|d_{2}-d_{1}|/N^{1/6}$.
Hence, in our case, $\max\\{x^{2},y^{2}\\}<N$ follows from
$|d_{2}-d_{1}|/N^{1/6}\ll N^{1/2}$. Thus, one can allow $d_{1}$ and $d_{2}$ to
come from the larger set $\mathcal{W}\cap[cN^{1/6+1/2}]$ for some $c>0$ and
$c$ will be determined in our proof. Here the threshold $N^{1/6}$ is chosen to
balance certain terms in the proof.

Consequently, we can improve the lower bound $\mathcal{W}(2N^{1/2})$ in (1.2)
to $\mathcal{W}(cN^{2/3})$ for some explicit $c>0$, but the payoff is that
since we just pick solutions $(x,y)$ such that $|x-y|$ is large, we can only
show that the number of available solutions in our case is $\gg 1$ instead of
$\gg\log N$.

######  Acknowledgment.

The authors would like to thank Kaisa Matomäki for her helpful comments.

The first author was supported by National Natural Science Foundation of China
(Grant No. 12201544), Natural Science Foundation of Jiangsu Province of China
(Grant No. BK20210784) and China Postdoctoral Science Foundation (Grant No.
2022M710121). He was also supported by foundation numbers JSSCBS20211023 and
YZLYJF2020PHD051. The second author was supported by UTUGS funding and was
working in the Academy of Finland project No. $333707$. The third author was
supported by the National Natural Science Foundation of China (Grant No.
12201291) and the Natural Science Foundation of the Higher Education
Institutions of Jiangsu Province (21KJB110001).

##  2\. Proof of Theorem 1.1

The proof of our theorem relies on the following lemma, which is a refinement
of the original argument of Chen and Fang [5, Theorem 2.1].

######  Lemma 2.1.

Let $K$ be a positive integer and $\mathcal{D}\subseteq\mathbb{N}$ be such
that $4K^{2}|d-d^{\prime}$ for any $d,d^{\prime}\in\mathcal{D}$. Then for all
positive numbers $x$, we have

$\sum_{\begin{subarray}{c}n\leq x\\\ R_{\mathcal{S},\mathcal{D}}(n)\geq
1\end{subarray}}\left(R_{\mathcal{S},\mathcal{D}}(n)-1\right)\geq\mathcal{D}(2K\sqrt{x})-2.$

###### Proof.

The lemma is trivial if $\mathcal{D}(2K\sqrt{x})\leq 2$. So we only need to
consider the case $\mathcal{D}(2K\sqrt{x})>2$. In this case, we assume
$\mathcal{D}(2K\sqrt{x})=\ell$ and

$\mathcal{D}\cap\left[0,2K\sqrt{x}\right]=\\{d_{1}<d_{2}<\cdot\cdot\cdot<d_{\ell}\\}.$

For any $2<s\leq\ell$, we have $4K^{2}|d_{s}-d_{1}$. So we can assume
$d_{s}-d_{1}=4K^{2}n_{s}$ for some positive integer $n_{s}>1$. Then one of the
solutions of the equation

$x^{2}-y^{2}=d_{s}-d_{1}=4K^{2}n_{s}$

has the form

$\begin{cases}x_{s}=Kn_{s}+K,\\\ y_{s}=Kn_{s}-K.\end{cases}$ (2.1)

Moreover, the solution given by (2.1) satisfies

$\displaystyle x_{s}^{2}+d_{1}=y_{s}^{2}+d_{s}<x,$ (2.2)

because

$x_{s}^{2}+d_{1}<\left(\frac{d_{s}-d_{1}}{2K}\right)^{2}+2\frac{d_{1}(d_{s}-d_{1})}{(2K)^{2}}<\left(\frac{d_{s}}{2K}\right)^{2}\leq\frac{(2K\sqrt{x})^{2}}{4K^{2}}=x.$

If an integer $n\leq x$ can be written as the sum of $d_{1}$ and a square,
then

$\displaystyle
R_{\mathcal{S},\mathcal{D}}(n)-1\geq\sum_{\begin{subarray}{c}d_{s}>d_{1}\\\
n-d_{s}\in\mathcal{S}\end{subarray}}1,$

from which we deduce that

$\sum_{\begin{subarray}{c}n\leq x\\\ R_{\mathcal{S},\mathcal{D}}(n)\geq
1\end{subarray}}\\!\\!\\!\left(R_{\mathcal{S},\mathcal{D}}(n)-1\right)\geq\\!\\!\\!\sum_{\begin{subarray}{c}n\leq
x\\\
n-d_{1}\in\mathcal{S}\end{subarray}}\\!\\!\\!\left(R_{\mathcal{S},\mathcal{D}}(n)-1\right)\geq\\!\\!\\!\sum_{\begin{subarray}{c}n\leq
x\\\
n-d_{1}\in\mathcal{S}\end{subarray}}\sum_{\begin{subarray}{c}d_{s}>d_{1}\\\
n-d_{s}\in\mathcal{S}\end{subarray}}1=\sum_{\begin{subarray}{c}1<s\leq\ell\end{subarray}}\sum_{\begin{subarray}{c}n\leq
x\\\ n-d_{s}\in\mathcal{S}\\\ n-d_{1}\in\mathcal{S}\end{subarray}}1.$ (2.3)

For any $2<s\leq\ell$, from (2.1) and (2.2) with $n=x_{s}^{2}+d_{1}$, we have

$\sum_{\begin{subarray}{c}n\leq x\\\ n-d_{s}\in\mathcal{S}\\\
n-d_{1}\in\mathcal{S}\end{subarray}}1\geq 1$

It follows that by (2.3),

$\sum_{\begin{subarray}{c}n\leq x\\\ R_{\mathcal{S},\mathcal{D}}(n)\geq
1\end{subarray}}\left(R_{\mathcal{S},\mathcal{D}}(n)-1\right)\geq\sum_{\begin{subarray}{c}2<s\leq\ell\end{subarray}}1=\ell-2=\mathcal{D}(2K\sqrt{x})-2.$

∎

Now, we turn to prove Theorem 1.1.

###### Proof of Theorem 1.1.

Let $N$ be a sufficiently large integer. It suffices to prove

$\displaystyle\sum_{n=1}^{N}\left(R_{\mathcal{S},\mathcal{W}}(n)-1\right)\gg\mathcal{W}\left(\frac{1}{5}N^{2/3}\right).$

Let $K<N^{1/2}$ be a large number to be chosen later. We follow the proof of
Chen and Fang [5]. For any $1\leq j\leq 4K^{2}$, let

$\mathcal{W}_{j}=\\{w\in\mathcal{W}:w\equiv j\ ({\rm{mod}}\ 4K^{2})\\}.$

Then we have

$\mathcal{W}=\bigcup_{j=1}^{4K^{2}}\mathcal{W}_{j}$

with $\mathcal{W}_{i}\bigcap\mathcal{W}_{j}=\varnothing$ for $i\neq j$. Thus,

$\displaystyle\sum_{n=1}^{N}\left(R_{\mathcal{S},\mathcal{W}}(n)-1\right)=\sum_{n=1}^{N}\left(\sum_{j=1}^{4K^{2}}R_{\mathcal{S},\mathcal{W}_{j}}(n)-1\right)$
$\displaystyle\geq$
$\displaystyle\sum_{n=1}^{N}\sum_{\begin{subarray}{c}j=1\\\
R_{\mathcal{S},\mathcal{W}_{j}}(n)\geq
1\end{subarray}}^{4K^{2}}\left(R_{\mathcal{S},\mathcal{W}_{j}}(n)-1\right)=\sum_{j=1}^{4K^{2}}\sum_{\begin{subarray}{c}n=1\\\
R_{\mathcal{S},\mathcal{W}_{j}}(n)\geq
1\end{subarray}}^{N}\left(R_{\mathcal{S},\mathcal{W}_{j}}(n)-1\right).$ (2.4)

By Lemma 2.1 with $\mathcal{D}=\mathcal{W}_{j}$, we have

$\displaystyle\sum_{\begin{subarray}{c}n=1\\\
R_{\mathcal{S},\mathcal{W}_{j}}(n)\geq
1\end{subarray}}^{N}\left(R_{\mathcal{S},\mathcal{W}_{j}}(n)-1\right)\geq\mathcal{W}_{j}(2K\sqrt{N})-2$
(2.5)

for any $1\leq j\leq 4K^{2}$. Combining (2) and (2.5), we obtain

$\displaystyle\sum_{n=1}^{N}\left(R_{\mathcal{S},\mathcal{W}}(n)-1\right)\geq\sum_{j=1}^{4K^{2}}(\mathcal{W}_{j}(2K\sqrt{N})-2)=\mathcal{W}(2K\sqrt{N})-8K^{2}.$
(2.6)

We choose $K=\frac{1}{10}N^{1/6}$, then our claims follows from (1.1). ∎

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