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\vspace*{-40pt} 
\centerline{\smalltt INTEGERS: 
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 5 
(2005), \#A19} 
\vskip 30pt 

\begin{center}
{\uppercase{{\bf Decimal expansion of $\mathbf{1/p}$ and subgroup sums}}}
\vskip 20pt
{\bf Ankit Gupta}\\
{\smallit Department of Mathematics and Statistics, Indian Institute of 
Technology, Kanpur, U.P. 208016, India}\\
{\tt ankitg@iitk.ac.in}
\vskip 8pt

{\bf B. Sury}\\
{\smallit Stat-Math Unit, Indian Statistical Institute, Bangalore,  
560 059, India}\\
{\tt sury@isibang.ac.in}
\end{center}
\vskip 30pt
\centerline{\smallit Received: 10/30/04, Revised: 7/1/05, 
Accepted: 8/9/05, Published: 8/10/05}
\vskip 30pt 
\centerline{\bf Abstract}

\noindent
It is well-known and elementary to show that for any prime $p \neq 2,5$, the decimal expansion of $1/p$ is periodic with period dividing $p-1$. 
In fact, the period is $p-1$ if 
and only if $10$ is a primitive root (mod $p)$. In 1836, Midy proved
that if $1/p$ has even period $2d$, then writing
$$\frac{1}{p} = 0.(UV)(UV) \cdots \cdots$$
where $U,V$ are blocks of $d$ digits each, one has $U+V= 10^d-1$
(that is, it is a block of $d$ $9$s). In January 2004, Brian Ginsberg, a
student from Yale University generalized Midy's theorem
to decimal expansions with period $3d$.
His proof is elementary. The purpose of this note is to solve the problem
in complete generality.  This involves some interesting questions about the cyclic
group of order $p-1$.
\pagestyle{myheadings}
\markright{\smalltt INTEGERS: \smallrm ELECTRONIC JOURNAL 
OF COMBINATORIAL NUMBER THEORY \smalltt 5 (2005), \#A19\hfill}

\thispagestyle{empty} 
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\vskip 20pt 

\section*
{\normalsize 1. Sums in $({\bf Z}/p {\bf Z})^{\ast}$}


\noindent
We start with a simple fact that will be useful for us.


\noindent
{\bf Lemma 1.}
{\it Let $p>2$ be a prime and $l>1$ be a divisor of $p-1$. 
Let $G(p,l) \subset \{1,2, \cdots, p-1 \}$ be the representatives of the unique subgroup of order $l$ in the group $({\bf Z}/p {\bf Z})^{\ast}$. Then, 
the sum $s(p,l):= \sum_{g \in G(p,l)} g = rp$
 for some natural number $r$.}

\ni
{\it Proof.}
If $G$ is a nontrivial subgroup of $({\bf Z}/p {\bf Z})^{\ast}$ and $x \neq e$ in $G$,
then, 
$$x \sum_{g \in G} g = \sum_{h \in G} h$$
so that $\sum_{g \in G} g \equiv 0$ (mod $p)$. \B


 
The connection of Lemma 1 with the decimal expansion of $1/p$ is seen
from Theorem 1 below.


\noindent
{\bf Theorem 1.}
{\it Let $p > 5$ be a prime and suppose $l>1$ is a natural number such that the decimal expansion of $1/p$ is periodic,
 of period $ld$. Write
$$\frac{1}{p} = 0.(U_1 U_2 \cdots U_l)(U_1U_2 \cdots U_l) \cdots \cdots$$
where each $U_i$ consists of $d$ digits. Then, one has
$$U_1+U_2+ \cdots + U_l = r (10^d-1)$$
where $s(p,l) = rp$.}



This immediately gives a (different) proof of Midy's and Ginsberg's
theorems.

\noindent
{\bf Corollary 1.}
{\it For a prime $p \neq 2,5$, and with notations as above, we have
$s(p,2)=s(p,3) = p$. In particular, Midy's theorem and
 Ginsberg's theorem follow.}

\ni
{\it Proof.}
Note that $G(p,2) = \{1, p-1 \}$ and $G(p,3) = \{1,x,y \}$ for some $x,y < p-1$. Since $1+x+y \equiv 0$
 (mod $p)$ and is less than $1+2(p-1)$, it follows that $1+x+y=p$.
\B

\noindent
{\it Proof of Theorem 1.}
Note that since $10$ has order $ld$ (mod $p)$, the elements of $G(p,l)$
are the images of 
$10^{id}; 1 \leq i \leq l$ modulo $p$. Thus, if $r_i$ is the fractional part $\{ 10^{id}/p \}$, then, 
$$\sum_{i=1}^l r_i = r.$$
Now,  
$$\frac{1}{p} = 0.(U_1 U_2 \cdots U_l)(U_1U_2 \cdots U_l) \cdots \cdots$$
$$\frac{10^d}{p} = U_1.(U_2 U_3 \cdots U_l U_1)(U_2U_3 \cdots U_lU_1) \cdots \cdots$$
$$\frac{10^{2d}}{p} = U_1U_2.(U_3 U_4 \cdots U_1U_2)(U_3U_4 \cdots U_1U_2) \cdots \cdots$$
$$\vdots$$
$$\frac{10^{(l-1)d}}{p} = U_1U_2 \cdots U_{l-1}.(U_l U_1 \cdots U_{l-1})(U_lU_1 \cdots U_{l-1}) \cdots \cdots$$
Thus, we have $U_1U_2 \cdots U_i = [10^{id}/p]$ for all $i < l$. 
Hence, the sum of the numbers to the left of the decimal points on the
right-hand sides of the above equations is $\sum_{i=1}^{l-1}
[10^{id}/p]$. Therefore, the sum of the decimals on the right-hand side
of the above equations is
$\sum_{i=0}^{l-1} \{10^{id}/p \} = r$.
But this sum of decimals is clearly $\frac{U_1+U_2 + \cdots +
U_l}{10^d-1}$. This proves that $U_1+ \cdots + U_l = r(10^d-1)$. \B



In view of this Theorem 1, when one looks for generalizations of Midy's
theorem etc., it is sufficient to consider the more general problem of
determining the value of $s(p,l)$ for various primes $p$ and divisors $l$ of $p-1$. Note that the latter problem is more general because the former one addresses only the cases when $l$ divides the order of $10$ (mod $p)$.
The computation of $s(p,l)$ for any prime $p$ and any divisor $l$ of $p-1$ is equivalent to
the computation of the sum $U_1 + \cdots + U_l$ where $1/p$ is expressed in base $b$ for a primitive root $b$ (mod $p)$.
In particular, the question arises as to whether $s(p,l)$ equals $p$ for any $l>3$ at 
all? We shall now show that there are some cases when it does and some
cases when it does not.


\section*
{\normalsize 2. Mersenne, Sophie Germain, and Dirichlet}


\noindent
Mersenne primes are prime numbers of the form $2^n-1$, in which case
$n$ must also be a prime. We then have two primes $p,n$ with $p$ much
larger than $n$. Another class of primes is the set of those primes $q$ for which $2q+1$ is also prime.
They came up in the proof of the first case of Fermat's last theorem due to Sophie Germain
for such primes $q$. In contrast with the Mersenne primes, here the two
primes $q,2q+1$ are comparable in size. Neither of these classes
of primes is known to be infinite. The behaviour of $s(p,l)$ is different for these two classes
as we show now.
\vskip 3mm

\noindent
{\bf Lemma 2.}
{\it Let $p=2^l-1$ be a (Mersenne) prime. Then, $s(p,l) = p$.}

\ni
{\it Proof.} Clearly, $2^l =1$ in $({\bf Z}/p{\bf Z})^{\ast}$. Therefore,
$2$ has order $l$ in this group. This implies that 
$G(p,l) = \{1,2,2^2, \cdots, 2^{l-1} \}.$
Hence $s(p,l) = 2^l-1=p$. \B


\noindent
{\bf Lemma 3.}
{\it Let $l>3$ be a (Sophie Germain) prime so that $p=2l+1$ is also prime.
 Then, $s(p,l) >p$.}

\ni
{\it Proof.}
Evidently, $s(p,l) \geq 1+2+3+ \cdots + (l-1) = l(l-1)/2 > 2l+1$ if $l>5$.
For $l=5$, it is directly checked that $s(11,5) = 1+3+4+5+9 = 22$.
\B


The question as to whether either of the cases $s(p,l)=p$ and $s(p,l)>p$ can occur infinitely often seems to be difficult to answer. The next result we prove below indicates that if $p$ is comparable in size to $l$, then $s(p,l) > p$ for large $l$. Let us note that the hypothesis of this proposition is conjecturally satisfied for large enough $l$ in the following sense. 
First, by Dirichlet's theorem on primes in progression, given any $l$, there
is a prime $p$ so that $p \equiv 1$ (mod $l)$. The prime number theorem gives the
lower bound for the smallest such $p$ to be at least of the order $l~ \log l$
([R], p.282). Wagstaff noted in 1979 ([R], p.283) that, for heuristic
reasons, the smallest such prime is of the order of $l (\log l)^2$ for
large $l$ except for a set of density zero. Kumar Murty showed in his
Bachelor's thesis ([R], p.281) of 1977 that except for a set of positive
integers $l$ not belonging to a
sequence of density zero, for each $\epsilon > 0$, the least $p \equiv 1$
(mod $l)$ satisfies $p < l^{2 + \epsilon}$. The pair correlation conjecture
-- a deep conjecture of analytic number theory about the zeroes of the
Riemann zeta function -- would imply that for any large $l$, there is a
prime
 $p \equiv 1$ (mod $l)$ such that $p < l^{1+ \epsilon}$.
The smallest exponent $k$ such that $p < Cl^k$ for some $C$ and all
large enough $l$, is known as Linnik's constant; the best unconditional
result in analytic number theory available at present is due to
Heath-Brown ([H]) and gives us $k \leq 5.5$. Even the existence of Linnik's
constant is a very deep theorem due to Linnik.


\noindent
{\bf Proposition 1.}
{\it For any prime $p \geq 11$ and any prime divisor $l$ of $p-1$ such that
 $p < l^2/2$, 
one has $s(p,l)>p$}.

\ni
{\it Proof.}
For any $p \equiv 1$ (mod $l)$, let the unique subgroup of order $l$ of $({\bf Z}/p{\bf Z})^{\ast}$ be generated by $x$. If $G(p,l) = \{1, x_1, \cdots, x_{l-1} \}$ with $x_i$ the residue of $x^i$, then at least one of $x_i$ and
$x_{l-i}$ is greater than $\sqrt{p}$, for each $1 \leq i <l$. The reason
is as follows. If both $x_i,x_{l-i}$ are at most $\sqrt{p}$, then we have
a contradiction since 
$1 \equiv x_ix_{l-i}$ (mod $p)$.
Therefore, at least half of the $x_i$'s for $i \geq 1$ are more
than $\sqrt{p}$. Thus, the largest $(l-1)/2$ of them are bigger than the
numbers
$\sqrt{p}, \sqrt{p}+1, \cdots, \sqrt{p}+(l-3)/2$. The others (including
$1$) are bigger than or equal to the numbers $1,2, \cdots, (l+1)/2$. Hence
$$s(p,l) > \sum_{i=1}^{(l+1)/2} i + \frac{\sqrt{p}(l-1)}{2} +
\sum_{j=1}^{(l-3)/2} j = \frac{l^2+3}{4} + \frac{\sqrt{p}(l-1)}{2}.$$
Since $\sqrt{p} < l/\sqrt{2}$, we can see that $s(p,l) > p$.
This completes the proof.
\B


Given a prime $p$ and any divisor $n$ of $p-1$, it is possible to give
an expression for the natural number $\frac{s(p,n)}{p}$. We do this below
using an element $b$ of order $n$ (mod $p)$ (knowing $b$ is essentially
equivalent to knowing a primitive root $a$ (mod $p)$ because one may take
$b=a^{(p-1)/n}$). In the formula below, we write
$\log_b$ to denote the logarithm to the base $b$.
In other words, $[\log_b(d)] = r$ if $b^r \leq d < b^{r+1}$.


\noindent
{\bf Proposition 2.}
{\it Let $p$ be a prime, $n|(p-1)$, and $b < p$ be an element of order $n$
in
$({\bf Z}/p {\bf Z})^{\ast}$. Then, we have}
$$\frac{s(p,n)}{p} = 
\frac{b^n-1}{p(b-1)} - (n-1)[\frac{b^{n-1}}{p}]
+ \sum_{i=1}^{[\frac{b^{n-1}}{p}]} [\log_b (ip)].$$



For example, take $p=11,n=5,b=4$. Then, $s(p,n) = 1+4+5+9+3 = 22.$
Since $[\log_4 (11i)]$ equals $1$ for $i=1$, equals $2$ for $2 \leq i \leq 5$
and, equals $3$ for $6 \leq i \leq 23$,
the expression on the right side of the proposition gives
$31-92+(1+8+54)=2$.
Another class of examples easily seen from the above is that of Mersenne primes $p=2^n-1$.
Then, $b=2$ and the sum is empty and one evidently has $\frac{s(p,n)}{p} = 1$.


\noindent
{\it Proof.}
We separate the powers $1,b,b^2, \dots ,b^{n-1}$ into the various ranges
$((i-1)p,ip)$ for $1 \leq i \leq [\frac{b^{n-1}}{p}]$.
Now, the largest $r$ for which the power $b^r$ is in the range $(0,p)$, equals
$[\log_b p]$. Counting in this manner, we have $b^{r_i+1}, b^{r_i+2},
\dots, b^{r_{i+1}}$ in the range  $(ip,(i+1)p)$ where $r_i = [\log_b
(ip)]$. These powers contribute 
$\sum_{j=r_i+1}^{r_{i+1}} (b_j - ip)$ to the sum $s(p,n)$. If $t$ is the largest number for which $r_t < n-1$, then the interval $(tp,(t+1)p)$ contains the powers 
$b^{r_t+1}, \cdots, b^{n-1}$. Hence, we get
$$s(p,n) = \sum_{j=0}^{n-1} b^j - \sum_{i=1}^{t-1}(r_{i+1}-r_i)ip -
(n-1-r_t)tp,$$ which simplifies to the expression
$$s(p,n) = \frac{b^n-1}{b-1} - p(n-1)[\frac{b^{n-1}}{p}]
+ \sum_{i=1}^{[\frac{b^{n-1}}{p}]} p [{\rm \log}_b (ip)].$$
This completes the proof.
\B


We end with the following question which is interesting because
finding a primitive root (mod $p)$ is far from easy.


\noindent
{\bf Question.}
{\it Given any prime $p$ and any divisor $n>1$ of $p-1$, give an expression for the natural number $s(p,n)/p$ in terms of $p$ and $n$ alone.}


\ni
{\bf Acknowledgement.}
We would like to thank the referee for a careful reading of the paper
and for suggestions to improve the language.


\section*
{\normalsize References}
\footnotesize

\noindent
[G] Brian D. Ginsberg, `Midy's (nearly) secret theorem - an extension
after 165 years', {\it College Math. Journal} {\bf 35} (2004), 26-30.

\noindent
[H] D.R.Heath-Brown, `Zero-free regions for Dirichlet L-functions and
the least prime in an arithmetic progression', {\it Proc. Lond. Math.
Soc.} {\bf 64} (1992), 265-338.

\noindent
[R] P. Ribenboim, The new book of prime number records, 3rd ed., Springer,
1996.



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