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\topmatter
\leftheadtext{\hskip 8pt \smalltt INTEGERS: \smallrm Electronic Journal
of Combinatorial Number Theory \smalltt 4 (2004), \#A22 }
\rightheadtext{\smalltt INTEGERS:
 \smallrm Electronic Journal of Combinatorial Number Theory \smalltt 4
(2004), \#A22 \hskip 8pt}
\endtopmatter 

%\topmatter\title The square of the Fermat quotient\endtitle
%\author Andrew Granville\endauthor

%\address{D\' epartement de Math\' ematiques et statistique,
%Universit\' e de Montr\'eal, CP 6128 succ.\\Centre-Ville, Montr\' eal
%QC  H3C 3J7, Canada } \endaddress
%\email{andrew{\@}dms.umontreal.ca} \endemail

%\endtopmatter

%\parindent=20pt


\document
\vskip -60pt
\centerline{\smalltt INTEGERS:
 \smallrm ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY \smalltt 4
(2004), \#A22}
\vskip 50pt


\centerline{\bf THE SQUARE OF THE FERMAT QUOTIENT}
\vskip 20pt
\centerline{\bf Andrew Granville}
\centerline{\smallit D\' epartement de Math\' ematiques
et statistique, Universit\' e de Montr\'eal, CP 6128 succ.}
\centerline{\smallit Centre-Ville,
Montr\' eal QC  H3C 3J7, Canada }
\centerline{\tt andrew{\@}dms.umontreal.ca} 
\vskip 30pt
\centerline{\smallit Received: 4/13/04, Revised: 10/20/04,
Accepted: 11/15/04,
Published: 11/30/04}
\vskip 30pt

 
\baselineskip=15pt 


\subhead \nofrills{1. Introduction} \endsubhead

 Fermat quotients, numbers of the
form $(a^{p-1}-1)/p$, played an important r\^ole in the study of
cyclotomic fields and Fermat's Last Theorem [2]. They seem to
appear in many surprising identities, one of the most delightful of which is 
Glaisher's observation that
$$
  \frac{2^{p-1}-1}{p} \equiv - \frac 12 \left( \frac{2^1}{1}+\frac{2^2}{2}+...
+\frac{2^{p-1}}{(p-1)} \right)
  \pmod p . \tag{1}
$$
Recently Skula conjectured that
$$
 \left(\frac{2^{p-1}-1}{p}\right)^2 \equiv 
-\left(\frac{2^1}{1^2}+\frac{2^2}{2^2}+...
+\frac{2^{p-1}}{(p-1)^2}\right)
 \pmod p .\tag{2}
$$
It is stunning that such a simple but elegant generalization of
(1) should have remained unnoticed for so long. In this note we
prove (2), and indeed a further generalization.

One might hazard a guess that the ratio
$$
 \left(\frac{2^{p-1}-1}{p}\right)^k \bigg/ 
\left(\frac{2^1}{1^k}+\frac{2^2}{2^k}+...
+\frac{2^{p-1}}{(p-1)^k}\right)
 \pmod p \tag{3}
$$
should also be a simple fixed rational number for other values of
$k$, but calculations reveal that this is probably not the case.

We will present two proofs of (2), one a substantial simplification of our 
original proof due
to the anonymous referee, the other  a different simplification, but both of 
which contain formulas that are perhaps of independent interest.


\vskip 30pt
\subhead \nofrills{2. The main results} \endsubhead


Let $p$ be a fixed prime $>3$. Define
$$
q(x)= \frac{x^{p}-(x-1)^p-1}{p}, \ \text{with} \ \ g(x)=\sum\Sb{i=1}\endSb^{p-1}
\frac{x^{i}}i\ \ \text{and} \ G(x)=\sum\Sb{i=1}\endSb^{p-1}
\frac{x^{i}}{i^2} .
$$
(Here, and throughout, $x$ is a variable, and the results below are proved 
for polynomials in $x$; of course one may substitute in integers for $x$ to 
obtain integer congruences.) Standard
arguments give that $G(1) \equiv 0 \pmod p$. Since $1/r+1/(p-r)=p/r(p-r)$   
thus $2g(1) \equiv -pG(1)\equiv 0 \pmod {p^2}$. 
Also $G(-1) = \sum_{1\leq j\leq (p-1)/2} ( 1/(2j)^2 -  1/(p-2j)^2) \equiv 0 
\pmod p$.



 We will prove the functional equation
$$
G(x)\equiv G(1-x)+x^pG(1-1/x) \pmod p ,\tag{4}
$$
as well as the two ``mod $p$--identities''
$$
q(x)^2\equiv -2x^p G(x)-2(1-x^p)G(1-x) \pmod p  \tag{5}
$$
and
$$
-G(x) \equiv \frac 1p (q(x)+g(1-x)) \pmod p \tag{6}
$$ 
which lead to two different proofs of (2): Substituting $x=2$ into (5) and then 
into (6) we obtain
$$
q(2)^2 \equiv - 2^{p+1} G(2) - 2 (1-2^p)G(-1)\equiv -4 G(2) \pmod p
$$
which is (2), and then
$$
-G(2) \equiv \frac 1p (q(2)+g(-1)) \pmod p 
$$ 
which gives (2) from Glaisher's result [1] that  
$g(-1) \equiv -  q(2)+ pq(2)^2/4 \pmod {p^2}$.


\vskip 30pt
\subhead \nofrills{3. Proofs} \endsubhead

We begin with the trivial observation that $\binom {p-1}{j}(-1)^{j} \equiv 1 
\pmod p$ for all $0\leq j\leq p-1$. Then
$$
q'(x)= x^{p-1}-(x-1)^{p-1}=-\sum_{j=0}^{p-2} \binom {p-1}{j} (-x)^{j} \equiv 
-\sum_{i=1}^{p-1} x^{i-1}=-g'(x) \pmod p.
$$ 
This, together with the fact that $q(x)$ and $g(x)$ both have degree $<p$, 
implies that $q(x)+g(x)\equiv c_0 \pmod p$ for some constant $c_0$. Substituting 
in $x=0$ we discover that $c_0\equiv 0 \pmod p$ and so 
$$
q(x)+g(x)\equiv 0 \pmod p  . \tag{7}
$$
It is immediate from their definitions that $q(x)=q(1-x)$ and 
$g(x)\equiv -x^pg(1/x) \pmod p$. From these observations and (7) we deduce that
$g(x)\equiv -q(x) = -q(1-x) \equiv g(1-x) \pmod p$ and 
$x^pg(1-1/x)\equiv x^pg(1/x)\equiv -g(x) \pmod p$. Now $G'(x)=g(x)/x$ and so 
$$
\align
\frac d{dx} (G(1-x)+x^pG(1-1/x)) &\equiv  
 -\frac{g(1-x)}{(1-x)} +x^p   \frac{g(1-1/x)}{x^2(1-1/x)} \\ 
 &= \frac{xg(1-x)+x^pg(1-1/x)}{x(x-1)}
\equiv \frac{g(x)}{x} = G'(x) \pmod p ,
\endalign
$$
and therefore $G(x)-G(1-x)-x^pG(1-1/x)\equiv c_1 \pmod p$ for some constant 
$c_1$. Substituting in $x=1$ we discover that $c_1\equiv 0 \pmod p$ and so (4) 
holds.

Similarly, from the above, we have
$$
\align
\frac d{dx} q(x)^2 = 2q(x)q'(x) &\equiv -2g(x) \left(  x^{p-1} -\sum_{j=0}^{p-1} 
x^j\right) \equiv -2x^p \frac{g(x)}{x} +2(1-x^p)\frac{g(1-x)}{(1-x)} \\
& \equiv -2x^pG'(x) - 2(1-x^p)G'(1-x)  \\
&\equiv \frac d{dx} (-2x^pG(x) - 2(1-x^p)G(1-x))  \pmod p. 
\endalign
$$
Therefore $q(x)^2+2x^pG(x)+ 2(1-x^p)G(1-x)\equiv c_2+c_3x^p \pmod p$ since this 
polynomial  has degree $<2p$. Substituting in $x=0$ and $x=1$ we discover that 
$c_2\equiv c_3 \equiv 0 \pmod p$ and we have proved (5).

Finally note that
$$
\align
\sum_{r=1}^{p-1} \frac{(1-x)^r-1}r &= \sum_{j=1}^{p-1}\left( \sum_{r=1}^{p-1} 
\binom {r-1}{j-1} \right) \frac{ (-x)^j}j= \sum_{j=1}^{p-1}  \binom {p-1}{j}  
\frac{ (-x)^j}j\\
&=\sum_{j=1}^{p-1} \left(  \binom {p}{j} -\binom {p-1}{j-1}  \right) \frac{ 
(-x)^j}j \\ &=p \sum_{j=1}^{p-1}   \left\{ (-1)^j\binom {p-1}{j-1} \right\} 
\frac{ x^j}{j^2} +  \frac { (x-1)^p-x^p+1}p , 
\endalign
$$
which implies (6), since each  $(-1)^{j-1}\binom {p-1}{j-1}\equiv 1 \pmod p$ and 
as $g(1)\equiv 0 \pmod p$.
\bigskip





\noindent {\sl Acknowledgements} Thanks to Ladja Skula for finding
this beautiful congruence,  Takashi Agoh for informing me of
Skula's conjecture and the anonymous referee for helping simplify my original 
proof.

\Refs \widestnumber\no{10}

\ref \no 1 \by J.W.L.~Glaisher \paper On the residues of the sums of products of 
the 
first $p-1$ numbers and their powers to modulus $p^2$ or $p^3$
 \jour Quart.~J.~Math.~Oxford \vol 31 \yr 1900
 \pages
 321--353
\endref

\ref \no 2 \by Paulo Ribenboim \book Thirteen lectures on Fermat's
Last Theorem \publ Springer-Verlag \publaddr New York\yr 1979
\endref


\endRefs

\enddocument