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\documentclass[a4paper,11pt]{amsart}
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\usepackage{amsmath,amssymb,amsfonts,latexsym}
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\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newcommand{\fd}{\mathbb{F}}
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\begin{document}
\title[Short title]{Title}
\author{Author1}
\address{Affiliation1}
\email{email1}
\author{Author2}
\address{Affiliation2}
\email{email2}
\author{Author3}
\address{Affiliation3}
\email{email3}
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\begin{abstract}
This is an abstract
\vspace{2mm}
\noindent\textsc{2010 Mathematics Subject Classification.} 11T23,
20G40, 94B05.
\vspace{2mm}
\noindent\textsc{Keywords and phrases.} key1, key2.
\end{abstract}
\thanks{This work was supported by ..}
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\maketitle
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\section {Equation numbering}
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Let $\psi$ be a nontrivial additive character of the finite field
$\fd{_q}$ with $q=p^{r}$ elements ($p$ a prime). Then the
Kloosterman sum $K(\psi;a)$ is defined by
\begin{equation}
K(\psi;a)=\sum_{\alpha \in \fd_{q}^{*}}\psi(\alpha+a \alpha^{-1})\\
(a \in \fd_{q}^{*}).
\end{equation}
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\section {Theorem numbering}
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\begin{theorem}
Let $\psi$ be any nontrivial additive character of $\fd_{q}$. Then
we have
\begin{equation}
\sum_{w \in O(3,q)} \psi(Tr w)=\psi(1)qK(\psi;1).
\end{equation}
\end{theorem}
\begin{proposition}
Let $q=2^r$. Then we have the following.
\begin{enumerate}
\item[(a)] $T_{0}K=1+\frac{1}{2}(-1)^{r}q$.
\item[(b)] $T_{1}K=\frac{1}{2}(-1)^{r+1}q$.
\end{enumerate}
\begin{proof}
\begin{enumerate}
\item[(a)]
\begin{align}
T_{0}K&=\sum_{tr a=0}\sum_{x \in \fd_q^*}\lambda(x^{-1}+ax) \notag\\
&=\frac{1}{2}\sum_{x \in \fd_q^*}\lambda(x^{-1})\sum_{\alpha \in \fd_q \backslash \{0,1\}}\lambda((\alpha^2+\alpha)x) \notag \\
&=1+\frac{1}{2}\sum_{x \in \fd_q^*}\lambda(x^{-1})\sum_{\alpha \in
\fd_q}\lambda(x \alpha^2+x \alpha).
\end{align}
We have
\begin{equation*}
\sum_{\alpha \in \fd_q}\lambda(x \alpha^2+x \alpha)
=\begin{cases}
q, &\textmd{if}\;\;\; x^2+x=0\;\;(\textmd{i.e},\;\; x \in \fd_2),\\
0 ,& \textmd{otherwise}.
\end{cases}
\end{equation*}
Thus (3) equals
\begin{equation*}
1+ \frac{q}{2}\sum_{x \in
\fd_2^{*}}\lambda(x^{-1})=1+\frac{1}{2}(-1)^{r}q.
\end{equation*}
\item[(b)] This follows from (a), since $T_{1}K=MK-T_{0}K=1-T_{0}K$.
\end{enumerate}
\end{proof}
\end{proposition}
\begin{corollary}
Let $\lambda$ be the canonical additive character of $\fd_q$, and let $a \in \fd_{q}^{*}$. Then we have
\begin{equation}
\sum_{ w \in O(3,q)} \lambda(aTr w)=\lambda(a)qK(\lambda ;a ).
\end{equation}
\end{corollary}
\begin{lemma}
Let $c(a)=(tr(aTrg_{1}), \cdots,tr(aTrg_{N})) \in C(O(3,q))^{\bot}$,
for $a \in \fd_q^{*}$. Then the Hamming weight $w(c(a))$ can be
expressed as follows:
\begin{equation}\label{a23}
w(c(a))=\frac{1}{2}q \{(q^2-1)-\lambda(a)K(\lambda;a)\}.
\end{equation}
\end{lemma}
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\section {Table and Figure caption position}
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\begin{table}[h!]
\caption{table title} TABLE HERE
%\begin{tabular}{|c|c|c|} \hline
%A & B & C \\
%\hline
%1 & 2 & 3\\
%\hline
%\end{tabular}
\end{table}
\begin{figure}[h!]
\begin{center}
%\includegraphics[width=7cm]{figure.eps}
FIGURE HERE
\end{center}
\caption{figure title}
\end{figure}
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\begin{thebibliography}{9}
\bibitem{L69}
L. Carlitz, \emph{Gauss sums over finite fields of order
$2^n$}, Acta Arith. 15 (1969), 247-265.
\end{thebibliography}
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\end{document}