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\title{\bf Partitions of $\mathbf{Z}_m$ with same weighted representation functions}

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\author{Zhenhua Qu\thanks{This work was supported by the National Natural Science Foundation of China, Grant No. 11101152.}\\
\small Department of Mathematics\\[-0.8ex]
\small Shanghai Key Laboratory of PMMP\\[-0.8ex]
\small East China Normal University\\[-0.8ex]
\small Shanghai 200241, P.R. China\\
\small\tt zhqu@math.ecnu.edu.cn\\
%\and
%Forgotten Second Author \qquad  Forgotten Third Author\\
%\small School of Hard Knocks\\[-0.8ex]
%\small University of Western Nowhere\\[-0.8ex]
%\small Nowhere, Australasiaopia\\
%\small\tt \{fsa,fta\}@uwn.edu.ao
}

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% \small Mathematics Subject Classifications: comma separated list of
% MSC codes available from http://www.ams.org/mathscinet/freeTools.html}

\date{\dateline{Jan 1, 2012}{Jan 2, 2012}\\
\small Mathematics Subject Classifications: 11B34, 05A15.}

\begin{document}

\maketitle

% E-JC papers must include an abstract. The abstract should consist of a
% succinct statement of background followed by a listing of the
% principal new results that are to be found in the paper. The abstract
% should be informative, clear, and as complete as possible. Phrases
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\begin{abstract}
 Let $\mathbf{k}=(k_1,k_2,\cdots,k_t)$ be a $t$-tuple of integers, and $m$ be a positive integer. For a subset $A\subset\mathbf{Z}_m$ and any $n\in\mathbf{Z}_m$, let
$r_A^{\mathbf{k}}(n)$ denote the number of solutions of the equation $k_1a_1+\cdots+k_ta_t=n$ with $a_1,\cdots,a_t\in A$. In this paper,
we give necessary and sufficient condition on $(\mathbf{k},m)$ such that there exists a subset $A\subset \mathbf{Z}_m$ satisifying
$r_{A}^{\mathbf{k}}=r_{\mathbf{Z}_m\backslash A}^{\mathbf{k}}$. This settles a problem of Yang and Chen.

  % keywords are optional
  \bigskip\noindent \textbf{Keywords:} Representation function, Partition, S\'ark\"ozy problem.
\end{abstract}

\section{Introduction}

We use $\bN$ to denote the set of nonnegative integers. For any subset $A\subset \bN$ and $n\in\bN$,
define the representation functions $R_1(A,n)$, $R_2(A,n)$ and $R_3(A,n)$ to
 be the number of solutions of the equations
 $$ n=a+a',\quad a,a'\in A,$$
$$ n=a+a', \quad a,a'\in A,\quad  a<a',$$
and
$$ n=a+a', \quad a,a'\in A,\quad a\leq a',$$
respectively. Representation functions first appeared in the celebrated paper of Erd\H os and Tur\'an \cite{ET1941}, and were extensively studied by Erd\H os, S\'ark\"ozy and S\'os (see \cite{sequenceI,sequenceII,sequenceIII,sequenceIV,sequenceV}).

S\'ark\"ozy asked for each $i=1,2,3$, whether there exist sets $A$ and $B$ with infinite symmetric difference such that $R_i(A,n)=R_i(B,n)$ for all sufficiently large integers $n$. There have been quite some work around S\'ark\"ozy's problem. Dombi \cite{Dombi2002} observed that the answer is negative for $i=1$,
and constructed a subset $A\subset \bN$ such that $R_2(A,n)=R_2(\bN\bks A,n)$ for all $n\in \bN$. An analogous example for $R_3(A,n)$ was constructed by Chen and Wang \cite{CW2003}. For $i=2,3$, Lev \cite{Lev2004}, S\'andor \cite{Sandor2004} and Tang \cite{Tang2008} determined all subsets $A\subset \bN$ such that $R_i(A,n)=R_i(\bN\bks A,n)$
for all $n\geq 2N-1$. The asymptotic behavior of the representation functions of these special sequences was studied by Chen and Tang (see \cite{Chen2011,CT2009}).

Analogously, for any two positive integers $k_1,k_2$, any subset $A\subset \bN$, one can define the weighted representation function  $r_{k_1,k_2}(A,n)$ as the number of solutions of the equation $n=k_1a_1+k_2a_2$ with $a_1,a_2\in A$. Cilleruelo and Ru\'e \cite{CR2009} proved that $r_{k_1,k_2}(A,n)$ can not be eventually constant. Yang and Chen \cite{YC2012} proved that there exists a set $A\subset \bN$ such that
$r_{k_1,k_2}(A,n)=r_{k_1,k_2}(\bN\bks A,n)$ for all sufficiently large $n$ if and only if $k_1\mid k_2$ and $k_1<k_2$.


Let $\bk=(k_1,k_2,\cdots,k_t)$ be a $t$-tuple of integers, and $m$ be a positive integer. For any $A\subset \bZ_m$ and $n\in \bZ_m$, denote
 the number of solutions of the equation $k_1a_1+\cdots+k_ta_t=n$ with $a_1,\cdots,a_t\in A$ by $r^{\bk}_A(n)$. We
 call $r^{\bk}_A$ the weighted representation function on $\bZ_m$ with respect to $A$ and weight $\bk$. For $t=2$, $\bk=(k_1,k_2)$,
 Yang and Chen \cite{YC2013} characterized all subsets $A\subset \bZ_m$ with the property that $r^{\bk}_A=r^{\bk}_{\bZ_m\bks A}$.

 Note that if $A\subset \bZ_m$ satisfying $r_A^{\bk}=r_{\bZ_m\bks A}^{\bk}$, then $m$ is even and $|A|=\frac m 2$. Indeed, this follows
 from the fact that
 $$|A|^t=\sum_{n\in\bZ_m}r^{\bk}_A(n)=\sum_{n\in\bZ_m}r^{\bk}_B(n)=|B|^t.$$
 For any nonzero integer $k$, we use $v_2(k)$ to denote the largest nonnegative integer $l$ such that $2^l\mid k$.
 The following result is also proved in \cite{YC2013}.

\begin{theorem}
  Let $k_1,k_2$ be nonzero integers, and $\bk=(k_1,k_2)$. For a subset $A\subset\bZ_m$ satisfying $r^{\bk}_A=r^{\bk}_{\bZ_m\bks A}$ to exist,
  it is necessary and sufficient that one of the following holds:

  (i) $k_1+k_2$ is even;

  (ii) $k_1+k_2$ is odd and $v_2(k_1k_2)<v_2(m)$.
\end{theorem}

It is natural to consider the following problem suggested by Yang and Chen \cite{YC2013}.

\begin{problem}
  For $t\geq 3$, determine all $\bk=(k_1,k_2,\cdots,k_t)$ and $m$ such that there exists a subset $A\subset\bZ_m$ with the property that
  $r^{\bk}_A=r^{\bk}_{\bZ_m\bks A}$.
\end{problem}

In this paper, we give a complete answer to this problem. Since $k_1,\cdots,k_t$ are only considered modulo $m$, we may assume $k_1,\cdots,k_t$ are all positive integers, and write
$$|\bk|=k_1+k_2+\cdots+k_t.$$
Let $\cA(\bk,m)$ be the set of all subsets $A\subset \bZ_m$ such that
  $r_A^{\bk}=r_{\bZ_m\bks A}^{\bk}$. We also identify an integer with its canonical image in $\bZ_m$. Our main result is the following.

\begin{theorem}\label{thm1}
  The following statements are equivalent:

  (i) $\cA(\bk,m)$ is nonempty;

  (ii) $\{0,1,\cdots,[\frac m 2]- 1\}\in\cA(\bk,m)$;

  (iii) $m$ is even, and either $|\bk|$ is even, or $0<v_2(k_i)<v_2(m)$ for some $i\in [1,t]$.
\end{theorem}

Currently we have no answer for the following problem.

\begin{problem}\label{prob2}
  Determine the set $\cA(\bk,m)$.
\end{problem}

We give an example illustrating the complexity of Problem \ref{prob2}.
For any even divisor $s\mid m$, a subset $A\subset \bZ_{m}$ is said to be \emph{balanced modulo $s$} if for any integer $k$, we have
$$|\{a\in A:a\equiv k\pmod{s}\}|=|\{a\in A:a\equiv k+\frac s 2\pmod{s}\}|.$$

\begin{example}\label{thm2}
  Let $m=2^l$, $l\geq 2$, $k_1=2$, $k_2=\cdots=k_t=1$.

  (i) If $t$ is even, then $A\in\cA(\bk,m)$ if and only if $|A|=m/2$, and $A$ is balanced modulo 2, in other words, $A$ has same number of odd elements and even elements.

  (ii) If $t$ is odd, then $A\in\cA(\bk,m)$ if and only if $|A|=m/2$, and for any integer $s\in [2,l]$, $A$ is balanced modulo $2^{s-1}$ or $2^s$, or both.
\end{example}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Proofs}


For a subset $A\subset \bZ_m$, we always use $B$ to denote the complement $\bZ_m\bks A$.
Let
$$f_A(x)=\sum_{a\in A} x^a,$$
and
$$T(x)=\prod_{i=1}^t f_A(x^{k_i})-\prod_{i=1}^t f_B(x^{k_i}).$$
These polynomials are considered in the ring $\bZ[x]/(x^m-1)$.

\begin{lemma}\label{lem2}
  $A\in\cA(\bk,m)$ if and only if $T(x)=0$.
\end{lemma}

\begin{proof}
  Since
  $$\prod_{i=1}^t f_A(x^{k_i})=\sum_{a_1,\cdots,a_t\in A}x^{k_1a_1+\cdots+k_ta_t}=\sum_{n\in\bZ_m}r^{\bk}_A(n)x^n,$$
  and similarly
  $$\prod_{i=1}^t f_B(x^{k_i})=\sum_{n\in\bZ_m}r^{\bk}_B(n)x^n,$$
  we conclude that $A\in\cA(\bk,m)$ if and only if $T(x)=0$.
\end{proof}

Let $d_i=(k_i,m)$, $i\in [1,t]$. For any positive integer $d$, we write $\xi_d=e^{2\pi i/d}$. For $d\mid m$,
it makes sense to write $f(\xi_d)$ for $f(x)\in\bZ[x]/(x^m-1)$, and we use $I(d)$ to denote the set of indices
$i\in [1,t]$ such that $d\nmid d_i$.

\begin{lemma}\label{lem3}
  For $A\subset\bZ_m$ with $|A|=m/2$, $A\in\cA(\bk,m)$ if and only if for any positive divisor $d\mid m$ with
  $|I(d)|$ odd, there exists $i\in I(d)$ such that $f_A(\xi_{d/(d,d_i)})=0$.
\end{lemma}

\begin{proof}
  By Lemma \ref{lem2}, $A\in\cA(\bk,m)$ if and only if $T(x)=0$. This is true if and only if for every positive divisor $d\mid m$, we have
  $T(\xi_d)=0$. For any $d\mid m$, if $d\mid k_i$, then
  \begin{equation}\label{eq1}
  f_A(\xi_d^{k_i})=f_B(\xi_d^{k_i})=m/2.
  \end{equation}
   If $d\nmid k_i$, then
  $$f_A(\xi_d^{k_i})+f_B(\xi_d^{k_i})=\sum_{n=0}^{m-1}\xi_d^{nk_i}=0,$$
  thus
  \begin{equation}\label{eq2}
  f_A(\xi_d^{k_i})=-f_B(\xi_d^{k_i}).
  \end{equation}
  Combining (\ref{eq1}) and (\ref{eq2}), we have
 $$T(\xi_d)=\left(\frac m 2\right)^{t-|I(d)|}\left(1-(-1)^{|I(d)|}\right)\prod_{i\in I(d)}f_A(\xi_d^{k_i}).$$


  If $|I(d)|$ is even, it is always true that $T(\xi_d)=0$. If $|I(d)|$ is odd, then $T(\xi_d)=0$ if and only if $f_A(\xi_d^{k_i})=0$ for some
  $i\in I(d)$. Since $\xi_d^{k_i}$ is a $d/(d,d_i)$-th primitive root of unity and $f$ has rational coefficients, any primitive $d/(d,d_i)$-th root of unity is a root of $f_A$. In particular $f_A(\xi_{d/(d,d_i)})=0$ and vice versa. This completes the proof of Lemma \ref{lem3}.
\end{proof}

We are now ready to prove Theorem \ref{thm1}.

\begin{proof}[Proof of Theorem \ref{thm1}]
  (ii)$\Rightarrow$(i) is trivial.

  We now show that (i)$\Rightarrow$(iii). Assuming $\cA(\bk,m)$ is nonempty, $m$ must be even. Suppose on the contrary that (iii) fails,
  then $|\bk|$ is odd and either $v_2(k_i)=0$ or $v_2(k_i)\geq v_2(m)$ for every $i\in [1,t]$, and it is clear that the number of
  $i\in [1, t]$ with $v_2(k_i)=0$ is odd. For any positive number $s\leq v_2(m)=:l$, consider
  $d=2^s\mid m$. Since $I(d)=\{i\in [1,t]:v_2(k_i)=0\}$, $|I(d)|$ is odd. By Lemma \ref{lem3}, we have $f_A(\xi_{d})=0$. Since this is true for all
  $s\leq l$, we conclude that the product of all $2^s$-th cyclotomic polynomials for $s\in [1,l]$ divides $f_A(x)$, i.e.
  $$1+x+\cdots+x^{2^l-1}\mid f_A(x).$$
  For $i\in [0,2^l-1]$, let $n_i$ denote the number of elements $a\in A$ such that $a\equiv i\pmod {2^l}$. Then
  $$f_A(x)=\sum_{a\in A}x^a\equiv \sum_{i=0}^{2^l-1} n_ix^i\pmod {1+x+\cdots +x^{2^l-1}},$$
  hence
  $$1+x+\cdots+x^{2^l-1}\mid\sum_{i=0}^{2^l-1} n_ix^i.$$
  It follows that $n_0=n_1=\cdots=n_{2^l-1}=:n$, $|A|=2^ln$. However $|A|=m/2$, $v_2(|A|)=v_2(m)-1=l-1$, and this contradicts $|A|=2^ln$,
  therefore (iii) is true.

  Finally we show that (iii)$\Rightarrow$(ii). So $m$ is even, and we put
  $$A=\{0,1,\cdots,\frac m 2-1\}.$$ Then
 $$f_B(x)=x^{m/2}f_A(x),$$
  and
  $$T(x)=\left(1-x^{|\bk|m/2}\right)\prod_{i=1}^t f_A(x^{k_i}).$$
  If $|\bk|$ is even, then $x^m-1$ divides $1-x^{|\bk|m/2}$, thus $T(x)=0$. By Lemma \ref{lem2}, we have $A\in\cA(\bk,m)$.
    Now suppose $|\bk|$ is odd, and there exists $j\in [1,t]$ such
  that $0<v_2(k_{j})<v_2(m)$. Let $d$ be any positive divisor of $m$ such that $|I(d)|$ is odd. If $d\mid m/2$, then for any $i\in I(d)$, letting $d'=d/(d,d_i)$, we have
  $$f_A(\xi_{d'})=\sum_{i=0}^{m/2-1} \xi_{d'}^{i}=\frac{\xi_{d'}^{m/2}-1}{\xi_{d'}-1}=0.$$
  If $d\nmid m/2$, then $v_2(d)=v_2(m)$, and we have $j\in I(d)$. Since $2\mid (d_j,d)$, therefore $d/(d,d_j)\mid m/2$.
  Let $d'=d/(d_j,d)$, then again,
  $$f_A(\xi_{d'})=\sum_{i=0}^{m/2-1} \xi_{d'}^{i}=\frac{\xi_{d'}^{m/2}-1}{\xi_{d'}-1}=0.$$
  By Lemma \ref{lem3}, we conclude that $A\in\cA(\bk,m)$. This completes the proof of Theorem \ref{thm1}.
\end{proof}

We now explain Example \ref{thm2}. Assume therefore that $m=2^l$, $l\geq 2$, $k_1=2$, $k_2=\cdots=k_t=1$, and $A\subset \bZ_m$ with $|A|=m/2$.

\begin{lemma}\label{lem4}
For any integer $s\in [1,l]$, $f_A(\xi_{2^s})=0$ if and only if $A$ is blanced modulo $2^s$.
\end{lemma}

\begin{proof}
  For $k\in [0,2^{s}-1]$, let $n_k$ denote the number of elements $a\in A$ such that $a\equiv k\pmod {2^s}$.
  $f_A(\xi_{2^s})=0$ if and only if $(1+x^{2^{s-1}})\mid f_A(x)$. We have
  $$f_A(x)=\sum_{a\in A}x^a\equiv \sum_{k=0}^{2^{s-1}-1}(n_k-n_{k+2^{s-1}})x^k\pmod{1+x^{2^{s-1}}}.$$
  It follows that $(1+x^{2^{s-1}})\mid f_A(x)$ if and only if $n_k=n_{k+2^{s-1}}$ for any $k\in [0,2^{s-1}-1]$, i.e.
  $A$ is balanced modulo $2^s$.
\end{proof}

\begin{proof}[Explanation of Example \ref{thm2}]
  If $t$ is even, consider $d\mid m$ with $|I(d)|$ odd, it is easy to see that $d=2$. By Lemma \ref{lem3}, $A\in\cA(\bk,m)$
  if and only if $f_A(\xi_2)=0$. By Lemma \ref{lem4}, this is equivalent to $A$ being balanced modulo 2.

  If $t$ is odd, then $d\mid m$ with $|I(d)|$ odd if and only if $d=2^s$ such that $2\leq s\leq l$. By Lemma \ref{lem3}, $A\in \cA(\bk,m)$
  if and only if for any $s\in [2,l]$, we have either $f_A(\xi_{2^s})=0$ or $f_A(\xi_{2^{s-1}})=0$. By Lemma \ref{lem4}, this is equivalent to $A$ being
  balanced modulo $2^{s-1}$ or $2^s$, or both.
\end{proof}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection*{Acknowledgements}
I am grateful to the referee for his/her detailed comments.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
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% If you use BibTeX to create a bibliography
% then copy and past the contents of your .bbl file into your .tex file

\begin{thebibliography}{10}

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\end{thebibliography}

\end{document}
