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\centerline{\smalltt INTEGERS: 
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 6 
(2006), \#A21} 
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%\mbox {}
\centerline{\bf  CONGRUENCES  WITH FACTORIALS MODULO $\mathbf P$}  
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\begin{center}
{\bf Yong-Gao Chen\footnote{Both authors supported by the 
National Natural
Science Foundation  of China, Grant No. 10471064.}}\\
{\smallit Department of Mathematics,
Nanjing Normal University,  Nanjing 210097, P. R. China}
\vskip 10pt
{\bf Li-Xia Dai\footnote{Corresponding author}}\\
{\smallit Department of Mathematics,
Nanjing Normal University,  Nanjing 210097, P. R. China}\\
{\tt lilidainjnu@163.com}
\end{center}
\vskip 20pt \centerline{\smallit Received: 3/20/06,
Revised: 7/18/06,
 Accepted: 8/7/06, Published: 8/28/06}




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\markright{\smalltt INTEGERS: 
\smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 6
(2006), \#A21\hfill}

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\centerline{\bf Abstract}

\noindent It is proved that the number of $a\in \{1,\cdots, p-1\}$
which can be represented as a product of two factorials is at
least $\frac 34\, p +O(p^{1/2}(\log p)^2)$. This improves the
result given by Garaev et.al. [Trans. Amer. Math. Soc., 356
(2004)5089-5102]. Beyond this, we pose several conjectures.

 %\vskip 10pt
%\noindent{\bf Key Words: } factorials, exponential sums,
%congruences.

%\noindent{\bf 2000 Mathematics subject classifications:} 11P05,
%11L40.



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\noindent{ \bf {1.\ \ Introduction}}


\noindent Throughout this paper, $p$ is an odd prime. In [6,F11], 
it is conjectured that about $p/e$ of the residue classes $a\pmod
p$ are missed by the sequence $n!$. If this were so, the sequence
$n!$ modulo $p$ should assume about $(1-1/e)p$ distinct values.
Some results of this spirit have appeared in [1]. The above
conjecture immediately implies that if $p$ is large enough, then
every residue class $a$ modulo $p$ can be represented as a product
of two factorials. Unfortunately, this conjecture appears to be
very hard. Various additive and multiplicative congruences with
factorials have been considered in [2, 3, 4, 5, 7, 8].


We denote by $F_l(a,p-1)$ the number of solutions to the
congruence
$$\prod\limits_{i=1}^ln_i!\equiv a\pmod p,\ 1\leq n_1,\cdots,n_l\leq p-1,$$
where $a\in \{ 1,\ 2,\ldots, p-1\}$. Let $V_l(p-1)$ be the number
of $a\in \{ 1,\ 2,\ldots, p-1\} $ for which $F_l(a,p-1)>0,$ that
is,

$$ V_l(p-1)=\sharp \left\{ \ \prod\limits_{i=1}^ln_i!\pmod p \
 \mid\ 1\leq n_1,\cdots,n_l\leq p-1\ \right\}.$$
Garaev et.al. [3] proved that
$$V_2(p-1)\geq \frac{5}{8}p +O\big (p^{1/2}(\log p)^2 \big).$$

In this paper we prove the following result.



\noindent
{\bf Theorem.}
$$V_2(p-1)\geq \frac{3}{4}p +O\big (p^{1/2}(\log p)^2 \big).$$



We pose the following conjectures.

\noindent
{\bf Conjecture 1.} {\it For any odd prime $p$, any integer $a\in
\{ 1, 2, \cdots , p-1\} $ can be represented as a product of two
factorials except for $p=11$ and $a=7$.}



\noindent
 {\bf Conjecture 2.} {\it If $a$ is a factorial, $a\not= 0$,
 then there are infinitely many primes $p$ for
which there are no integers $n$ with $a\equiv n!\pmod p$.}



\noindent
{\bf Conjecture 3.} {\it Let $a$ be an integer. If for any prime
$p$ there is an integer $n$ with $a\equiv n!\pmod p$, then $a=-1,$ $a=0$, 
or $a$ is a factorial.}

\vskip 30pt
 \noindent{\bf 2. Proof of the Theorem }



\noindent{\bf Lemma(Zhang [9,\ 10]).} Let $N(p)$ denote the
number of all pairs $(a,\ b)$ with $a,\ b\in \{ 1, 2, \cdots ,
p-1\} $ for which $a$ and $b$ are of opposite parity and $ab\equiv
1\pmod p$. Then
$$N(p)=\frac{1}{2}p+O\big( p^{1/2} (\log p)^2 \big).$$

\noindent
 {\it Proof of  Theorem.} Define
\begin{eqnarray*}
I_1&=&\{\ (a,\ b)\ |\ ab\equiv 1\pmod p,\ 2\ |\ a,\ 2\ \nmid \ b,\
a,\ b=1,2,\cdots , p-1
\},\\
 I_2&=&\{\ (a,\ b)\ |\ ab\equiv 1\pmod p,\ 2\ \nmid\ a,\ 2\  |\
b,\ a,\ b=1,2,\cdots , p-1 \},\\
I_3&=&\{\ (a,\ b)\ |\ ab\equiv 1\pmod p,\ 2\ |\ a,\ 2\ | \ b
,\ a,\ b=1,2,\cdots , p-1\},\\
 I_4&=&\{\ (a,\ b)\ |\ ab\equiv 1\pmod p,\ 2\ \nmid\ a,\ 2\
\nmid \ b,\ a,\ b=1,2,\cdots , p-1\}.
\end{eqnarray*}
It is obvious that
$$|I_2|+|I_4|=|I_1|+|I_4|=\frac{p-1}{2},\eqno(1)$$
$$ |I_1|+|I_3|=|I_1|+|I_4|=\frac{p-1}{2}. \eqno(2) $$
Hence
$|I_1|=|I_2|,\ |I_3|=|I_4|.$
 Thus, by the lemma we have
$$|I_1|=|I_2|=\frac{1}{2}N(p)=\frac{1}{4}p+O \big( p^{1/2} (\log p)^2 \big).\eqno (3)$$
By (1), (2), and (3) we obtain
$$|I_3|=|I_4|=\frac{1}{4}p+O \big( p^{1/2} (\log p)^2 \big).\eqno (4)$$

Let $a,\ b\in \{ 1, 2, \cdots , p-1\} $ with $ab\equiv 1\pmod p$.
Wilson's Theorem implies that (the similar arguments appear in
[1, 2, 3])
$$-1\equiv (p-1)!\equiv (-1)^{a-1}(a-1)!(p-a)!\pmod p$$
and
$$-1\equiv (p-1)!\equiv (-1)^{b}b!(p-b-1)!\pmod p.$$
Hence,
$$a\equiv  (-1)^{a}a!(p-a)!\pmod p, \mbox{ and } $$
$$a\equiv (-1)^{b+1} a \cdot b!(p-b-1)!\equiv (-1)^{b+1}(b-1)!(p-b-1)!\pmod p.$$
Thus, if $a$ is even or if $b$ is odd, then $a$ can be represented
as a product of two factorials. This implies that if $(a,\ b)\in
I_1\cup I_3\cup I_4$, then $a\in V_2(p-1)$.

 Therefore, by (3) and (4) we have
$$ V_2(p-1)\geq |I_1|+|I_3|+|I_4|\geq \frac{3p}{4}+O \big( p^{1/2} (\log p)^2 \big). $$

This completes the proof of the theorem.

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\noindent{\bf {References}}

\begin{description}
\footnotesize

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\item{[2]} \ \ P. Erd\H os and C. Stewart,\ {\it On the greatest
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\item{[3]} \ \ M. Z. Garaev, F. Luca, I. E. Shparlinski,\ {\it
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\item{[4]} \ \ M. Z. Garaev, F. Luca, I. E. Shparlinski,\ {\it
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\item{[5]} \ \ M. Z. Garaev, F. Luca, I. E. Shparlinski,\ {\it
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 \item{[6]} \ \ R. K. Guy,\ {\it Unsolved Problems in Number
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\item{[7]} \ \ F. Luca and P. St$\check{a}$nic$\breve{a}$,\ {\it
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\item{[8]} \ \ C. Stewart,\ {\it On the greatest and least prime
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\item{[9]} \ \ W. P. Zhang,\ {\it On a problem of  D. H. Lehmer
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\item{[10]} \ \ W. P. Zhang,\ {\it On a problem of D. H. Lehmer
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\vspace{-0.2cm}




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