\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.5in} \setlength{\textheight}{8.9in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \def\amod#1 #2{#1\ ({\rm mod}\ #2)} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \begin{center} \vskip 1cm{\LARGE\bf A Further Generalization of a Congruence of \\ \vskip .1in Wolstenholme } \vskip 1cm \large Christian Ballot \\ L.M.N.O., CNRS UMR 6139\\ Universit\'e de Caen \\ F-14032 Caen Cedex \\ France \\ \href{mailto:christian.ballot@unicaen.fr}{\tt christian.ballot@unicaen.fr}\\ \end{center} \vskip .2 in \begin{abstract} Given a pair $(U_t)$ and $(V_t)$ of Lucas sequences, Kimball and Webb showed that $\sum_{01$, we have \begin{equation}\label{eq:hao} \sum_{t=1}^{m-1}\frac{V_t}{U_t}\equiv\ \frac{(m^2-1)D}{6}\cdot\frac{U_m}{V_m}\pmod{w_m^2}, \end{equation} where $m$ is an integer $\ge5$, $D$ is the discriminant of the characteristic polynomial associated with the pair of Lucas sequences $U$ and $V$, and $w_m$ is the largest factor of $U_m$ which is prime to the product $U_2U_3\cdots U_{m-1}$. \footnote{The only case of the Kimball and Webb congruence not implied by (\ref{eq:hao}) is when $p$ is $5$ of rank $4$. However, defining $w_4$ as the largest factor of $U_4$ prime to $3U_2U_3$ would have been sufficient to cover this missing case.} \medskip We recall that given a pair of integers $P$ and $Q$, $Q$ nonzero, the two Lucas sequences $U=(U_t)_{t\ge0}$ and $V=(V_t)_{t\ge0}$ are second order integral linear recurrences $(X_t)$ that both satisfy the recursion \begin{equation}\label{eq:recursion} X_{t+2}=PX_{t+1}-QX_t,\quad\text{ for all }t. \end{equation} Their two initial terms then fully defines them. We have $U_0=0$ and $U_1=1$, whereas $V_0=2$ and $V_1=P$. We write $U(P,Q)$ and $V(P,Q)$ when the dependence on $P$ and $Q$ needs to be reminded. If the zeros of their characteristic polynomial $x^2-Px+Q$ are distinct, say $\alpha$ and $\beta$, then the $t$-th terms of the $U$ and $V$ sequences are given by the so-called Binet formulas, i.e., \begin{equation}\label{eq:Binet1} U_t=\frac{\alpha^t-\beta^t}{\alpha-\beta}\;\text{ and }\;V_t=\alpha^t+\beta^t. \end{equation} In case $x^2-Px+Q$ has a double zero $\alpha$, then the Binet formulas become \begin{equation}\label{eq:Binet2} U_t=t\alpha^{t-1}\;\text{ and }\;V_t=2\alpha^t. \end{equation} The rank of appearance $\rho(m)$, or $\rho_U(m)$, of an integer $m$ is the least positive index $t$ such that $m$ divides $U_t$. It is known to exist for all $m$ prime to $Q$. The rank of a prime $p$ not dividing $Q$ is either equal to $p$, or is a divisor of $p\pm1$. For $(P,Q)=(2,1)$, $1$ is a double zero of $x^2-Px+Q$. Thus, $U_t=t$ and $V_t=2$. Applying the theorem of Kimball and Webb to this particular pair of Lucas sequences yields congruence (\ref{eq:1862}). \medskip We restate the main theorem of Kimball and Webb \cite{KW2}, but in a slightly generalized form, as it appeared in Chapter 3 of the paper \cite{Ba}. \begin{theorem} \label{thm:KW} Let $p$ be a prime at least $5$ which is not a factor of $Q$. Assume the rank $\rho$ of $p$ is maximal. Let $k$ be an integer. Then the sum $$ \sum_{t\in I_{p,k}}\frac{V_t}{U_t}\text{ is congruent to } \amod{0} {p^2},$$ where $I_{p,k}$ is the set of integers in the interval $\big(k\rho,(k+1)\rho\big)$. \end{theorem} The purpose of the present paper is to prove a common generalization, Theorem \ref{thm:wol} stated hereunder, to both the Leudesdorf and the Kimball-Webb congruences, i.e., a common generalization to Theorems \ref{thm:Leud} and \ref{thm:KW}. \begin{theorem} \label{thm:wol} Let $U(P,Q)$ and $V(P,Q)$ be a pair of Lucas sequences. Let $m$ be an integer of maximal rank with respect to $U(P,Q)$. Then, for all integers $k$, we have that $$ S_{m,k}(P,Q)=\sum_{t\in I_{m,k}}\frac{V_t}{U_t}\;\text{ is congruent to }\;0\pmod {h_m\cdot m^2/\gcd(m,3)}, $$ where $I_{m,k}$ is the set of integers in the interval $\big(k\rho(m),(k+1)\rho(m)\big)$ that are not multiples of any rank $\rho(p)$ for all primes $p$ dividing $m$, and where $$ h_m=\begin{cases} 1/2,& \text{ if }m=2^a,\,a\ge1,\text{ and }\nu_2(P)=1;\\ \;2^\sigma,& \text{ if }2||m\;\text{ and }4\mid P;\\ \;2^\tau,& \text{ if }2||m,\;P\text{ is odd and }Q\equiv\amod{1} {4};\\ \;8,& \text{ if }4||m,\;P\text{ is odd and }Q\equiv\amod{3} {4};\\ \;1,& \text{ otherwise,}\end{cases} $$ with $\sigma=\nu_2(P)-2\ge0$ and $\tau=\nu_2(P^2-Q)-1\ge1$, $\nu_2(P)$ and $\nu_2(P^2-Q)$ being respectively the $2$-adic valuations of $P$ and $P^2-Q$. In particular, $S_{m,k}\equiv \amod{0} {m^2/\gcd(m,3)}$ if $m$ is not a power of $2$ or if $\nu_2(P)\not=1$. \end{theorem} \begin{example}\label{exa:only}\quad {\rm Say $(P,Q)=(1,-1)$, i.e., $U=F$ and $V=L$ are the sequences of Fibonacci and Lucas numbers. We will see that $35$ and $500$ are two integers of maximal rank with respect to $F$. Thus, putting $k=0$ in Theorem \ref{thm:wol}, we find that $\bullet$ If $m=35$, then $\rho(35)=40$. Hence, $$ \sum_{\substack{t=1\\5\,\nmid\, t,\;8\,\nmid\, t}}^{40}\frac{L_t}{F_t}\equiv\ 0\pmod{35^2}, $$ where the above sum, $S_{m,0}(1,-1)$, contains $40-(8+5)+1=28$ terms. \medskip $\bullet$ If $m=500=4\cdot5^3$, then $$\sum \frac{L_t}{F_t}\equiv\ 0\pmod{8m^2=2,000,000},$$ where the sum is over all $t$'s between $1$ and $\rho(m)=6\cdot5^3=750$ that are prime to $15$, and, thus, contains $750-(250+150)+50=400$ terms.} \end{example} Section 2 of this paper is a short preliminary section where the relevant definitions, and in particular that of a general integer having maximal rank with respect to a Lucas sequence $U(P,Q)$, followed by a few remarks and comments, are given. Section 3 contains the main theorems and their proofs. Theorem \ref{thm:wol1}, generalizes the theorem of Kimball and Webb to prime powers. Theorem \ref{thm:wol2}, which may be viewed as our chief result, is proved via induction on the number of distinct prime factors of an integer $m$ of maximal rank and so Theorem \ref{thm:wol1} serves as a basis for that proof by induction. Theorem \ref{thm:wol2} is already a generalization of both Theorems \ref{thm:Leud} and \ref{thm:KW}. So we could have ended the paper there. However, it would not have been complete. We wanted a theorem that fully respected the definition we took of an integer of maximal rank. The remaining cases are thus treated in Section 4, but their proofs hinge heavily on the proof of Theorem \ref{thm:wol2} and are established through a series of lemmas, which, combined with Theorem \ref{thm:wol2}, yield Theorem \ref{thm:wol}. \medskip The proofs of the paper are all elementary, but assume familiarity with classical properties of Lucas sequences. We refer to Lucas' original work \cite{Lu1} and Chapters 17 and 18 of \cite{Lu2}, to \cite{Car}, \cite{Rib}, to Chapter 4 of \cite{Wi}, and to Chapter 2 of \cite{Ba}, for properties of Lucas sequences used herein. \medskip The letter $p$ invariably denotes a prime number. If $m$ is an integer, then $\rho(m)$, or $\rho_U(m)$, denotes its rank in $U$. The letter $D$ stands for the discriminant $P^2-4Q$ of $x^2-Px+Q$. If $t$ is a rational number, then $\nu_p(t)$ denotes its $p$-adic valuation. Alternately, as we did in Theorem \ref{thm:wol}, we write $p^a||t$ to mean that $p^a$ divides $t$, but $p^{a+1}$ does not divide $t$. These divisions take place in the ring $A_p$. More generally, the congruences in Theorems \ref{thm:Leud}, \ref{thm:KW} and \ref{thm:wol} and congruences that appear in this paper take place in the ring $A_m$, where $A_m$ is the subring of the rationals which, when expressed in lowest terms, have a denominator prime to $m$. That is, $s\equiv \amod{t} {m}$ means that $s$ and $t$ are in $A_m$ and that $s-t\in mA_m$. Note that $A_m=A_{m^2} =A_{p_1\cdots p_r}$, if $p_1$, \dots, $p_r$ are the prime factors of $m$. We will use the well-known fact \cite{Ja} that $\mathbb{Z}/p^a$ is isomorphic to $A_p/p^a$, $a\ge1$, where the isomorphism is derived from the identity embedding of $\mathbb{Z}$ into $A_p$. \section{Preliminaries} Throughout the paper, we assume that $U=U(P,Q)$ and $V=V(P,Q)$ denote an arbitrary pair of Lucas sequences. \medskip Recall that if $m$ is an integer prime to $Q$, then $m\mid U_t$ iff $\rho(m)\mid t$. Also, if $p$ is a prime that does not divide $Q$, then $\rho(p)\mid p-(D|p)$, where $(D|p)$ is $0$, if $p\mid D$; $1$, if $D$ is a nonzero square (mod $p$) and $-1$, otherwise. \medskip We recall a few classical Lucas identities valid for all integers $s$ and $t$. \begin{equation} \label{eq:w1} 2U_{s+t}=U_sV_t+U_tV_s, \end{equation} \begin{equation} \label{eq:w2} 2V_{s+t}=V_sV_t+DU_sU_t, \end{equation} \begin{equation} \label{eq:w3} V_t^2-DU_t^2=4Q^t. \end{equation} \medskip It is easy to deduce from (\ref{eq:w1}) and the relations $Q^tV_{-t}=V_t$, $Q^tU_{-t}=-U_t$ that \begin{equation} \label{eq:w1+} 2Q^tU_{s-t}=U_sV_t-U_tV_s. \end{equation} \begin{remark} \label{rem:rootD} Suppose $p$ is an odd prime not dividing $Q$. If $t$ is not a multiple of $\rho(p)$, then from (\ref{eq:w3}) we have that the square of the ratio $V_t/U_t$ is well-defined and not congruent to $\amod{D} {p}$. \end{remark} Our intention is to first generalize Theorem \ref{thm:KW} to prime powers, possibly including the primes $2$ and $3$. Recall that the rank of a prime $p$ is said to be {\it maximal} if it is $p+1$, $p$ or $p-1$. We wish our generalized theorems to apply to all positive integers $m$ having a certain maximal rank property. \begin{definition} \label{def:maxrk} {\rm (Maximal Rank)\quad Let $p$ be a prime not dividing $Q$ and $a$ be an integer $\ge1$. We say that the rank of $p^a$ is {\it maximal} whenever $\rho(p)$ is maximal and $\rho(p^a)=p^{a-1}\rho(p)$. An integer $m\ge1$ prime to $Q$ is said to have {\it maximal} rank whenever each prime power dividing $m$ has maximal rank and the ranks of any two prime powers dividing $m$ are coprime.} \end{definition} If $m$ is a prime $p$, then Definition \ref{def:maxrk} only requires that $\rho(p)$ be $p$ or $p\pm1$. However, if $m=p^a$, with $a\ge2$, has maximal rank, then clearly we also must have $p^1||U_{\rho(p)}$. Note that if $m$ has maximal rank with respect to $U(P,Q)$, then any divisor $n$ of $m$ has maximal rank with respect to $U(P,Q)$. \medskip \begin{remark} \label{rem:simpler} \quad {\rm It may be useful to observe that if $m$ is the product of prime powers which all have maximal rank with respect to some $U(P,Q)$ and $m$ is either odd, or $m$ and $P$ are both even, then $m$ has maximal rank if and only if $\gcd\big(\rho(p),\rho(q)\big)=1$, for any two prime factors $p$ and $q$ of $m$.} \end{remark} \begin{proof} Suppose $\gcd\big(\rho(p),\rho(q)\big)=1$. We only need to verify that $\rho(p^a)$ and $\rho(q^b)$ are coprime, where $p^a||m$ and $q^b||m$. If $p$ and $q$ are odd, one of them, say $q$, must have rank equal to $q$, since $p\pm1$ and $q\pm1$ are both even. Thus, $\rho(q^b)=q^b$. So $q\nmid\rho(p)$ and $q\not= p$ yields that $\gcd\big(\rho(p^a),\rho(q^b)\big)=1$. If $p$ is $2$ and $q$ is odd, then, assuming $P$ is even, we have $\rho(p)=2$ since $U_2=P$, and thence $\rho(q)=q$. Therefore, $\rho(2^a)=2^a$ and $\rho(q^b)=q^b$. The converse is immediate. \end{proof} The upcoming remark essentially says that if $p$ has maximal rank, the condition $p^2\nmid U_{\rho(p)}$ suffices for all $p^a$, $a\ge1$, to have maximal rank. Note that if $U$ is the Fibonacci sequence, then the search for a Wall-Sun-Sun prime, i.e., a prime $p$ such that $p^2$ divides $U_{\rho(p)}$, has gathered attention, but to this date none is known to exist and there does not exist any below $2\times10^{14}$ \cite{MR}. \begin{remark} \label{rem:admiss}\quad {\rm Let $p\nmid Q$ be a prime of maximal rank such that $p^2\nmid U_{\rho(p)}$. Additionally, we assume $P$ is even, if $p$ is $2$. Then $p^a$ has maximal rank for all $a\ge1$.} \end{remark} \begin{proof} By Theorem 9 of \cite{Ba}, the rank of $p^a$ is equal to $p^{a-1}\rho(p)$, if $p$ is odd. Suppose $p=2$. Since $U_1=1$, $U_2=P$ and $P$ is even, we have $\rho(2)=2$. By hypothesis, $4\nmid U_{\rho(2)}$, i.e., $4\nmid P$, and $Q$ is odd. Therefore, $8\mid D= P^2-4Q$. By identity (\ref{eq:w3}), $\nu_2(V_t)=1$ for all integers $t$'s. Thus, the $2$-adic valuation of $U_{2t}=U_tV_t$ is one more than that of $U_t$. So by induction we obtain that $\rho(2^a)=2^{a-1}\rho(2)=2^a$. \end{proof} With the help of Lemma \ref{lem:evencase}, we give the full list of integers of maximal rank with respect to the Fibonacci sequence. \begin{example} \label{example:Fib}\quad {\rm An integer $m\ge1$ has maximal rank with respect to the Fibonacci sequence $F=U(1,-1)$ iff it has the form $$ 2^a\;5^b\;p^c, $$ where $p$ is a prime $\amod{3} {4}$ of rank $\rho(p)=p\pm1$, not a Wall-Sun-Sun prime in case $c\ge2$, and where $a\in\{0,1,2\}$, $b\ge0$, $c\ge0$ with $$ \left.\begin{array}{cl}\bullet\; c=0,\;\;\; & \text{ if }a=2;\\ \bullet\;3\nmid\rho(p),& \text{ if }a=1\text{ and }c\ge1;\\ \bullet\;5\nmid\rho(p),& \text{ if }b\ge1\text{ and }c\ge1.\end{array}\right. $$ Artin's conjecture states that an integer $a$, not a square, with $|a|\ge2$, is a primitive root modulo infinitely many primes. In fact, it has been conditionally proved that such primes have a positive density {\rm\cite{Ho}}. Similarly here, since $Q$ is not a square, it is expected that a positive proportion of the primes $\amod{3} {4}$ have maximal ranks in $F$. In fact, $3$, $7$, $11$, $19$, $23$, $31$ and $43$ all have maximal ranks, $47$ being the smallest prime $\amod{3} {4}$ with rank less than $p-1$. For all primes $p$ that are $\amod{1} {4}$, we have that $p\mid U_{(p-1)/2}$ or $p\mid U_{(p+1)/2}$.} \end{example} \begin{definition} \label{def:Im} {\rm ($I_{m,k}$ and $S_{m,k}$)\quad Given an integer $m\ge1$ and an integer $k$, we define $I_{m,k}$ as the set of integers that lie in the interval $\big[k\rho(m),(k+1)\rho(m)\big]$ and are not a multiple of any $\rho(p)$, for all primes $p$ dividing $m$. Then $S_{m,k}$ will denote the sum of all $V_t/U_t$ in $A_m$ as $t$ varies through $I_{m,k}$. We write $I_{m,k}(P,Q)$ or $S_{m,k}(P,Q)$ if the dependence on $(P,Q)$ needs to be specified. } \end{definition} In coining the two definitions \ref{def:maxrk} and \ref{def:Im}, we had two constraints in our mind. In Leudesdorf's Theorem \ref{thm:Leud}, the sum involved is $S_{m,0}(2,1)$. That sum goes up to $m$, which is viewed as the rank of $m$ in $U_t=t$. Thus, if we wanted a theorem that generalizes Theorem \ref{thm:Leud}, then $I_{m,0}$ had to go up to $\rho(m)$. Actually, considering sums that go up to the least common multiple of the ranks of the prime powers dividing an integer $m$ would also reduce to the right sums when $U_t=t$. However, generally these sums are not $\amod{0} {m^2}$.\footnote{For instance, $3$ and $7$ have maximal ranks in the Fibonacci sequence and the least common multiple of their ranks is $8$. But $\sum_{t=1}^8L_t/F_t$, ($4\nmid t$), is $\amod{0} {21}$, $\amod{6} {9}$ and $\amod{14} {49}$.} Also we wanted the arguments used in proving Theorem \ref{thm:KW} to remain valid, which meant that $S_{m,k}$ should have about $m$ terms. These observations guided us in defining the notion of maximal rank. \begin{lemma} \label{lem:distinct} Let $k$ be an integer. Let $m$ be an integer $\ge1$ prime to $Q$. If $m$ is odd, then all ratios $V_t/U_t$ are distinct (mod $m$) as $t$ varies through $I_{m,k}$. If $m$ is even and $2||P$, then all ratios $V_t/(2U_t)$ are distinct (mod $m$) as $t$ varies through $I_{m,k}$. \end{lemma} \begin{proof} Assume first $m$ is odd. If $s$ and $t$ are distinct integers in $I_{m,k}$, then $U_{s-t}$ is not divisible by $m$. Thus, dividing (\ref{eq:w1+}) through by $U_sU_t$, we get that \begin{equation} \label{eq:distinct} \frac{V_t}{U_t}-\frac{V_s}{U_s}=2Q^t\frac{U_{s-t}}{U_sU_t}. \end{equation} Since the right-hand term of (\ref{eq:distinct}) is not $\amod{0} {m}$, all ratios $V_t/U_t$ are pairwise incongruent modulo $m$ as $t$ varies through $I_{m,k}$. For the case $m$ even, note that $P=2P'$ with $P'$ odd. Since $D=4(P'^2-Q)$, $8$ divides $D$. We have $V_t^2-DU_t^2=4Q^t$, so each $V_t$ is even and the ratios $V_t/(2U_t)$ do belong to the ring $A_m$ for $t\in I_{m,k}$. The reasoning is then similar to that of $m$ odd. \end{proof} \section{Main Theorems and Proofs.} \begin{theorem} \label{thm:wol1} Let $U(P,Q)$ and $V(P,Q)$ be a pair of Lucas sequences. Let $p$ be a prime number. Suppose $p^a$ has maximal rank with respect to $U(P,Q)$, where $P$ is even and not divisible by $4$ in case $p$ is $2$. Then for all integers $k$ $$ S_{p^a,k}=\sum_{t\in I_{p^a,k}}\frac{V_t}{U_t}\;\text{ is congruent to } \amod{0} {p^{c}}, $$ where $$ c=\begin{cases} 2a,& \text{ if }p\ge5;\\ 2a-1,& \text{ if }p=2\text{ or }3.\end{cases} $$ \end{theorem} \begin{proof} Fix an integer $k$. Note that $I_{p^a,k}$ contains $p^{a-1}\rho(p)-p^{a-1}=p^{a-1}\big(\rho(p)-1\big)$ integers. By convention, all unmarked sums appearing in the proof are taken over the set $I_{p^a,k}$, and thus contain $p^{a-1}\big(\rho(p)-1\big)$ terms. Denote $(2k+1)p^{a-1}\rho(p)$ by $b$. Since the rank of $p^a$ is $p^{a-1}\rho(p)$, we have that $p^a$ divides $U_{p^{a-1}\rho(p)}$ and that $U_b$ is at least divisible by $p^a$. By (\ref{eq:w1}), $$ 2S_{p^a,k}=\sum\bigg(\frac{V_t}{U_t}+\frac{V_{b-t}}{U_{b-t}}\bigg)= 2\sum\frac{U_b}{U_tU_{b-t}}.$$ {\bf Suppose $p$ is odd}. Hence, it suffices to show that $\sum2/(U_tU_{b-t})$ is divisible by $p^d$, where $d=a$ if $p>3$, and $d=a-1$ if $p=3$. By (\ref{eq:w3}) with $t=b$, $\gcd(U_b,V_b)$ divides $2Q^{\lfloor b/2\rfloor}$, so that $p$ does not divide $V_b$. Thus, it also suffices to show that $$ S_1:=\sum\frac{2V_b}{U_tU_{b-t}} $$ is divisible by $p^d$. Now, by (\ref{eq:w2}), $S_1$ is equal to $$ \sum\frac{V_tV_{b-t}}{U_tU_{b-t}}\;+\;Dp^{a-1}\big(\rho(p)-1\big). $$ But $$\sum\frac{V_tV_{b-t}}{U_tU_{b-t}}= \sum\frac{U_tU_{b-t}V_tV_{b-t}}{(U_tU_{b-t})^2}= \frac12\sum\frac{(U_tV_{b-t}+U_{b-t}V_t)^2}{(U_tU_{b-t})^2} -\sum\bigg(\frac{V_t}{U_t}\bigg)^2. $$ In the above final expression, the numerators of the terms in the first sum are all equal to $4U_b^2$ by (\ref{eq:w1}). Since $p^a$ divides $U_b$, we have that $p^d$ divides $S_1$, if and only if, $p^d$ divides $$S_2:= Dp^{a-1}\big(\rho(p)-1\big)-S_3,$$ where $S_3:=\sum(V_t/U_t)^2$. If $\rho(p)$ is $p+1$, then $S_2$ is congruent to $\amod{-S_3} {p^a}$. But since by Lemma \ref{lem:distinct} the $p^a$ rational numbers $V_t/U_t$ that appear in $S_3$ are all distinct $\amod{0} {p^a}$, we have that $$ S_3\equiv \ \sum_{t=1}^{p^a}t^2=\frac{p^a(p^a+1)(2p^a+1)}{2\cdot3}\pmod{p^a}, $$ which yields that $$ S_3\equiv \amod{0} {p^d}. $$ If $\rho(p)$ is $p$, then $p$ divides $D$. Thus, again, $S_2\equiv \amod{-S_3} {p^a}$. Since $p$ has odd rank, say by Theorem 10 in \cite{Ba}, no term of the $V$ sequence is divisible by $p$. Thus, the $p^a-p^{a-1}$ ratios $V_t/U_t$ that appear in $S_3$, being all distinct (mod $p^a$), are all the invertible elements of the ring $A_p/p^a$. As they form a cyclic multiplicative group generated by, say, $g$, we have $$ S_3\equiv\sum_{k=0}^{\varphi(p^a)-1}g^{2k}=\frac{g^{2\varphi(p^a)}-1}{g^2-1}\pmod{p^a}, $$ where $\varphi$ denotes Euler's totient function. For odd primes $p$, any primitive root (mod $p^a$) reduces to a primitive root (mod $p$). Therefore, the order of $\amod{g} {p}$ is $p-1$ and, thus $p\mid g^2-1$ iff $p-1\mid 2$, i.e., iff $p=3$. But if $p=3$ and $a\ge2$, then $g$ being a primitive root (mod $9$), the quantity $g^2-1$ is divisible by $3$, but not by $9$. If $p=3$ and $a=1$, then $d=0$. Therefore, for all cases, we have $$ S_3\equiv\ 0\pmod {p^d}. $$ If $\rho(p)$ is $p-1$, then the set $I_{p^a,k}$ is made up of $p^{a-1}$ integer intervals of length $p-2$. Indeed, integers in $I_{p^a,k}$ are not divisible by $\rho(p)$, i.e., by $p-1$. As $t$ varies through each such interval all $V_t/U_t$ are distinct (mod $p$), and are not equal to any of the two square roots of $\amod{D} {p}$ by Remark \ref{rem:rootD}. Thus, the summands $(V_t/U_t)^2$ in $S_3$ take the value $0$ once and every nonzero quadratic residue (mod $p$) twice, except $D$, on each interval of length $p-2$ of $I_{p^a,k}$. Thus, as $t$ runs through $I_{p^a,k}$ and since all $V_t/U_t$ are distinct (mod $p^a$), none of the summands $(V_t/U_t)^2$ of $S_3$ is equal to any of the residues $D$, $D+p$, $D+2p, \hdots, D+(p^{a-1}-1)p$ (mod $p^a$). Thus, $$ S_3\equiv \sum_{i=1}^{p^a}i^2-2\sum_{j=1}^{p^{a-1}}(D+jp)\equiv\ 0-2p^{a-1}D\pmod{p^d}. $$ Therefore, $S_2=Dp^{a-1}(p-2)-S_3\equiv\ -2Dp^{a-1}+2p^{a-1}D=\amod{0} {p^d}$. \medskip {\bf Suppose $p=2$}. The symbols $S_1$, $S_2$ and $S_3$ refer to the same sums we had defined in the case where $p$ was odd. Since $P$ is even and $U_2=P$, $\rho(2)=2$ and $b=(2k+1)2^a$. As seen in proving Remark \ref{rem:admiss}, $\nu_2(U_{2^a})=a$ and so we also have $\nu_2(U_b)=a$. Therefore, as $S_{2^a,k}=\sum\frac{U_b}{U_tU_{b-t}}$, we have $$ \nu_2(S_{2^a,k})=a+\nu_2\bigg(\sum\frac1{U_tU_{b-t}}\bigg) =a+\nu_2\bigg(\sum\frac{2V_b}{U_tU_{b-t}}\bigg)-2, $$ since, as shown in Remark \ref{rem:admiss}, $\nu_2(V_b)=1$. Thus, we need to show that $\nu_2(S_1)\ge a+1$. In fact, we will see that $\nu_2(S_1)=a+1$ so that $\nu_2(S_{2^a,k})$ is exactly $2a-1$ for all integers $k$. Because $\frac12(4U_b^2)$ is divisible by $2^{2a+1}$, which is $>2^{a+1}$, $\nu_2(S_1)=a+1$ iff $\nu_2(S_2)=a+1$. Since $D=4(P'^2-Q)$, where $P=2P'$ and $P'$ is odd, $8\mid D$. Therefore, $\nu_2(S_2)$ is $a+1$ iff $\nu_2(S_3)=a+1$. But $S_3=4S_4$, where $S_4=\sum\frac{V_t^2}{4U_t^2}$. The $2^{a-1}$ terms $V_t/2U_t$ are all distinct (mod ${2^a}$), by Lemma \ref{lem:distinct}, and each such term is a unit of the ring $A_2$, by the proof of Remark \ref{rem:admiss}. Therefore, $$ S_4\equiv\sum_{t=1}^{2^{a-1}}(2t-1)^2\pmod{2^a}. $$ But $\sum_{t=1}^N(2t-1)^2=4(N-1)N(N+1)/3-N$ so $2^{a-1}|| S_4$. Since $S_3=4S_4$, we have $\nu_2(S_3)=a+1$ as required. \end{proof} \begin{remark} \label{rem:exact} {\rm Under the hypotheses of Theorem \ref{thm:wol1}, we have actually shown that the sums $S_{2^a,k}$ have $2$-adic valuation exactly equal to $2a-1$, for all integers $k$.} \end{remark} We are ready to move on to our main result. \begin{theorem} \label{thm:wol2} Let $U(P,Q)$ and $V(P,Q)$ be a pair of Lucas sequences. Let $m$ be an integer of maximal rank with respect to $U(P,Q)$. If $m$ is even, then we make the additional hypothesis that $P$ is even and not divisible by $4$. Then for all integers $k$ $$ S_{m,k}(P,Q)=\sum_{t\in I_{m,k}}\frac{V_t}{U_t}\;\text{ is congruent to }\;0\pmod {m^2/d_m}, $$ where $I_{m,k}$ is the set of integers in the interval $\big(k\rho(m),(k+1)\rho(m)\big)$ that are not multiples of any rank $\rho(p)$ for all primes $p$ dividing $m$, and where $$ d_m=\begin{cases}\; 2,& \text{ if }m=2^a,\,a\ge1;\\ \;3,& \text{ if }3\mid m;\\ \;1,& \text{ otherwise.}\end{cases} $$ \end{theorem} \begin{proof} We proceed by induction on $\omega(m)$, where $\omega$ denotes the distinct prime factors counting function. More precisely, our inductive hypothesis at level $i$, $i\ge1$, says that for all integers $k$, all integers $m\ge1$ with $\omega(m)=i$ and all Lucas sequences $U=U(P,Q)$ with respect to which $m$ satisfies the hypotheses of Theorem \ref{thm:wol2}, the sum $S_{m,k}(P,Q)$ is congruent to $\amod{0} {m^2/d_m}$. \medskip The case $i=1$ was established in Theorem \ref{thm:wol1}, so we assume that $i\ge2$ and that the result holds for integers having $i-1$ distinct prime factors. To prove the inductive hypothesis at level $i$, we establish two properties (\ref{eq:one}) and (\ref{eq:two}) which together clearly suffice to prove the inductive step. The first property says that \begin{equation}\label{eq:one} \text{if }m=p^an, \text{ then }n^2/d_n\text{ divides }S_{m,k}, \end{equation} where $p$ is a prime that does not divide $n$ and $a\ge1$. Note that (\ref{eq:one}) is sufficient to prove the inductive step if $m$ is odd, but not if $m$ is even, unless we know the theorem holds for even integers at level $i=2$. Thus, in addition, we prove that \begin{equation}\label{eq:two} \text{ if }m=2^aq^b, \text{ then }2^{2a} \text{ divides }S_{m,k}, \end{equation} where $q$ is an odd prime, $a$ and $b\ge1$. \medskip We begin by establishing (\ref{eq:one}) inductively. To this end, we note that $$ S_{m,k}=S^*-S^{**}, $$ where \begin{equation}\label{eq:fund} S^*=\sum_{j=0}^{\rho(p^a)-1}\quad\sum_{\substack{t=k\rho(m)+j\rho(n)\\ \rho(q)\,\nmid\, t\text{ if }q\mid n}}^ {k\rho(m)+(j+1)\rho(n)}\frac{V_t}{U_t}\qquad\text{ and }\qquad S^{**}=\sum_{\substack{t=k\rho(m)\\\rho(p)\,\mid\, t\\\rho(q)\,\nmid\, t,\, \forall q\mid n}}^{(k+1)\rho(m)}\frac{V_t}{U_t}, \end{equation} and the letter $q$ stands for a prime. Moreover, we have \begin{equation}\label{eq:imp} S^*=\sum_{j=0}^{\rho(p^a)-1}\;S_{n, j+k\rho(p^a)}. \end{equation} Each inner sum $S_{n, j+k\rho(p^a)}$ in $S^*$ is, by the inductive hypothesis, divisible by $n^2/d_n$, so we are left with showing that $S^{**}$ is also divisible by $n^2/d_n$. But \begin{equation}\label{eq:s**} S^{**}=\sum_{\substack{t=k\rho(n)p^{a-1}\\\rho(q)\,\nmid\, t,\,\forall q\mid n}}^{(k+1)\rho(n)p^{a-1}} \frac{V_{\rho(p)t}}{U_{\rho(p)t}}=U_{\rho(p)}^{-1}\sum_{j=kp^{a-1}}^{(k+1)p^{a-1}-1}S_{n,j} \big(V_{\rho(p)},Q^{\rho(p)}\big). \end{equation} In (\ref{eq:s**}), the sum $S_{n,j}\big(V_{\rho(p)},Q^{\rho(p)}\big)$ is associated with the pair of Lucas sequences attached to the recursion $x^2-V_{\rho(p)}x+Q^{\rho(p)}$. Indeed, we have \begin{equation}\label{eq:change}V_{\rho(p)t}(P,Q)=V_t(P',Q') \;\text{ and }\; U_{\rho(p)t}(P,Q)=U_t(P',Q')\cdot U_{\rho(p)}(P,Q), \end{equation} where $P'=V_{\rho(p)}(P,Q)$ and $Q'=Q^{\rho(p)}$. Identities (\ref{eq:change}) may be derived from the Binet formulas for $U_t$ and $V_t$ whether the discriminant $P^2-4Q$ is zero or nonzero. For instance, suppose $x^2-Px+Q=(x-\alpha)^2$. Then $U_t(P,Q)=t\alpha^{t-1}$ and denoting $\rho(p)$ by $\rho$ we have \begin{eqnarray*} U_{\rho t}(P,Q) & = & \rho t\alpha^{\rho t-1}=\rho t\alpha^{\rho(t-1)+(\rho-1)}\\ & = & \rho\alpha^{\rho-1}\cdot t(\alpha^\rho)^{t-1}\\ & = & U_{\rho}(P,Q)\cdot U_t(P',Q'). \end{eqnarray*} By the inductive hypothesis $n^2/d_n$ divides each sum $S_{n,j}(P',Q')$, for all $j$'s, provided we check, on the one hand, that $n$ has maximal rank with respect to $U'=U(P',Q')$ and, on the other, in case $n$ is even, that $\nu_2(P')=1$. \medskip Since $m$ is prime to $Q$, $n$ is prime to $Q'$. Let $q$ be a prime factor of $n$, $\rho(q)$ be its rank in $U=U(P,Q)$ and $\rho'(q)$ its rank in $U'=U(P',Q')$, where as above $\rho(p)$ is the rank of $p$ in $U$. By hypothesis, $\rho(q)$ and $\rho(p)$ are coprime so that $q\nmid U_{\rho(p)}$. Thus, as $U_{\rho(p)t}=U_{\rho(p)} \cdot U'_t$, we have that $q\mid U'_t$ iff $q\mid U_{\rho(p)t}$ iff $\rho(q)\mid\rho(p)t$ iff $\rho(q)\mid t$. Thus, $\rho(q)=\rho'(q)$. We conclude that the rank of $q$ in $U'$ is maximal. Similarly, since $\rho(q^s)=q^{s-1}\rho(q)$ is also prime to $\rho(p)$, the ranks of $q^s$ in $U$ and in $U'$ are identical for all $s\ge2$. Therefore, if $q^\alpha||m$, then, as $\rho(q^\alpha)$ exists and is equal to $q^{\alpha-1}\rho(q)$, we have $\rho'(q^\alpha)=q^{\alpha-1}\rho'(q)$. That is, $q^\alpha$ has maximal rank in $U'$. Moreover, distinct prime factors of $n$ have coprime ranks in $U'$. Suppose $n$ is even. Then $m$ is even. Hence, $Q$ is odd and $\nu_2(P)=1$, which, as proved in Remark \ref{rem:admiss}, implies that $\nu_2(V_t)=1$ for all integers $t$. But $P'=V_{\rho(p)}$. Thus, since each prime factor $q$ of $n$ is prime to $U_{\rho(p)}$, we deduce that $S^{**}$ is divisible by $n^2/d_n$. We have proved (\ref{eq:one}). To prove (\ref{eq:two}), we assume $i=2$ and `recycle' the proof of (\ref{eq:one}). So given $m=2^aq^b$ we have, by (\ref{eq:one}), that $2^{2a-1}q^{2b}/d_q= m^2/(2d_m)$ divides $S_{m,k}$. Reexamining the proof of (\ref{eq:one}) with $p=q$, we will show that in fact $2^{2a}$ divides $S_{m,k}$. By Remark \ref{rem:exact}, each of the $\rho(q^b)$ sums $S_{2^a,j+k\rho(q^b)}$ in (\ref{eq:imp}) is divisible by $2^{2a-1}$, but not by $2^{2a}$. But, by (\ref{eq:s**}), we see that $S^{**}$ is also the sum of $q^{b-1}$ sums each of which is exactly divisible by $2^{2a-1}$. Therefore, $S_{m,k}$ is the sum of $\rho(q^b)+q^{b-1}$ terms each divisible exactly by $2^{2a-1}$. Thus, it suffices to check that $\rho(q^b)+q^{b-1}$ is even. But $\rho(2)=2$, and so $\rho(q)$ must be odd. As the rank of $q$ is maximal, it must be that $\rho(q)=q$. Hence, $\rho(q^b)+q^{b-1}=q^{b-1}(q+1)$. \end{proof} \section{The Remaining Cases} Theorem \ref{thm:wol2} clearly contains Theorem \ref{thm:KW}, but it also implies Theorem \ref{thm:Leud}. Indeed, for $(P,Q)=(2,1)$, we have $U_t=t$ and $V_t=2$, for all $t$'s. In particular, $\nu_2(P)=1$ and all integers $m\ge1$ have maximal rank. Say, for instance, that $3\mid m$. Then, by Theorem \ref{thm:wol2}, $\sum_{t\in I_{m,0}}2/t\equiv\ \amod{0} {m^2/3}$, which implies $$\sum_{\substack{t=1\\\gcd(t,m)=1}}^m1/t\equiv\ 0 \pmod {m^2/\gcd(m,6)},$$ as $\gcd(t,m)=1$ iff $\rho(p)=p\nmid m$ for all primes $p$ dividing $m$. \medskip But Theorem \ref{thm:wol2} does not apply to even integers having maximal rank in $U(P,Q)$, if either $4\mid P$, or $P$ is odd. The next lemmas treat those missing cases, and, combined with Theorem \ref{thm:wol2}, they yield Theorem \ref{thm:wol} stated in the introduction. The first lemma fully describes when powers of $2$ have maximal rank. \begin{lemma}\label{lem:evencase} Assume $Q$ is odd. Then $2^a$, $a\ge1$, has maximal rank in $U(P,Q)$ if and only if $$ \begin{cases}\; a=1,& \text{ if }4\mid P,\text{ or if }P\text{ is odd and } Q\equiv \amod{1} {4};\\ \;a=1\text{ or }2,& \text{ if }P\text{ is odd and }Q\equiv\amod{3} {4};\\ \;a\ge1,& \text{ if }P\equiv \amod{2} {4}.\end{cases} $$ \end{lemma} \begin{proof} Since $U_2=P$ and $U_3=P^2-Q$, either $P$ is even and $\rho(2)=2$, or $P$ is odd and $\rho(2)=3$. Thus, $2^a$ has maximal rank for $a=1$. If $4\mid P$, or if $P$ is odd and $Q\equiv\amod{1} {4}$, then $\rho(4)=\rho(2)$. By (\ref{eq:w3}), if $U_t$ is even, then $V_t$ is even. Since $U_{2t}=U_tV_t$, we have $\nu_2(U_{2t})\ge1+\nu_2(U_t)$. Therefore, since $\rho(4)=\rho(2)$, $\rho(2^a)<2^{a-1}\rho(2)$ for all $a\ge2$. If $P$ is odd and $Q\equiv \amod{3} {4}$, then $\rho(2)=3$ and $\rho(4)=6$. Thus, $4$ has maximal rank. However, $U_6=U_2U_3(P^2-3Q)$ and $4\mid P^2-3Q$, so $8\mid U_6$. Hence, $\rho(8)=\rho(4)$ and $\rho(2^a)<2^{a-1}\rho(2)$, for all $a\ge3$. The case $P\equiv \amod{2} {4}$ was taken care of in Remark \ref{rem:admiss}. \end{proof} \begin{lemma} \label{lem:wol3} Let $U(P,Q)$ and $V(P,Q)$ be a pair of Lucas sequences, where $P$ is divisible by $4$ and $Q$ is odd. Let $m$ be an even integer of maximal rank with respect to $U(P,Q)$. Then for all integers $k$ $$ S_{m,k}(P,Q)\equiv\ 0\pmod {2^\sigma m^2/\gcd(m,3)}, $$ where $\sigma=\nu_2(P)-2$. \end{lemma} \begin{proof} By Lemma \ref{lem:evencase}, $m$ is not divisible by $4$. So we write $m=2n$ with $n$ odd. Since $\rho(2)=2$, all odd prime factors $p$ of $m$ satisfy $\rho(p)=p$. As in proving Theorem \ref{thm:wol2}, we carry an induction on $i=\omega(m)$, where the inductive hypothesis at level $i$ assumes the lemma to hold for all integers $m$ with $\omega(m)=i$, all pairs $(P',Q')$, $4\mid P'$ and $Q'$ odd, with respect to which $m$ has maximal rank in $U(P',Q')$ and all integers $k$. If $i=1$, that is, if $m=2$, then $S_{2,k}=V_{2k+1}/U_{2k+1}$. An easy induction using (\ref{eq:recursion}) yields that $\nu_2(V_{2t})=1$ and $\nu_2(V_{2t+1})=\nu_2(P)$, for all integers $t$. Hence, $2^{\nu_2(P)}||S_{2,k}$. But $2^{\nu_2(P)}=2^\sigma m^2$. Assume that $i\ge2$ and that the inductive hypothesis holds at level $i-1$. We decompose $S_{m,k}$ into the difference of $S^*$ and $S^{**}$ as we did in the proof of Theorem \ref{thm:wol2} by choosing a prime $p$ that divides $m$. If $p$ is $2$, then, by Theorem \ref{thm:wol2}, both the inner sums $S_{n,j+2k}(P,Q)$ of $S^*$ in (\ref{eq:imp}) and the inner sums $S_{n,j}(P',Q')$ of $S^{**}$ in (\ref{eq:s**}) are divisible by $n^2/d_n$, where $d_n=\gcd(n,3)$. But $\gcd(n,3)=\gcd(m,3)$ so $n^2/\gcd(m,3)$ divides $S_{m,k}$. It remains to see that $2^{\nu_2(P)}$ divides $S_{m,k}$. If $p$ is odd, then put $\ell=m/p^a$, where $a=\nu_p(m)$. The inner sums $S_{\ell,j+k\rho(p)}(P,Q)$ of $S^*$ are now divisible by $2^\sigma\ell^2/\gcd(\ell,3)$, and by $2^{\nu_2(P)}$ in particular, by the inductive hypothesis. The inner sums $S_{\ell,j}(P',Q')$ are divisible by $2^{\nu_2(P')}$ by the inductive hypothesis, which is applicable since $4\mid P'$ and $Q'$ is odd. But $P'=V_{\rho(p)}=V_p$ and $\nu_2(V_p)=\nu_2(P)$, since $p$ is odd. Hence, $S_{m,k}$ is divisible by $2^\sigma m^2/\gcd(m,3)$. \end{proof} \begin{lemma} \label{lem:wol4} Let $U(P,Q)$ and $V(P,Q)$ be a pair of Lucas sequences, where $P$ is odd and $Q$ is congruent to $\amod{1} {4}$. Let $m$ be an even integer of maximal rank with respect to $U(P,Q)$. Then for all integers $k$ $$ S_{m,k}(P,Q)\equiv\ 0\pmod {2^\tau m^2/\gcd(m,3)}, $$ where $\tau=\nu_2(P^2-Q)-1\ge1$. \end{lemma} \begin{proof} One may carry a successful induction that closely follows that of the proof of Lemma \ref{lem:wol3}. Note that $4\nmid m$ and that $\rho(2)=3$. If $m=p^a\ell$, $\ell$ even, then by induction the sums $S_{\ell,j}(P',Q')$ in $S^{**}$ are divisible by $2^\tau\ell^2/\gcd(\ell,3)$. Indeed, $2\mid V_t$ iff $3\mid t$. Thus, since $P'=V_{\rho(p)}$ and $\rho(2)\nmid\rho(p)$, $P'$ is odd. Moreover, $Q'=Q^{\rho(p)}\equiv \amod{1} {4}$. If $m=2n$, then, by Theorem \ref{thm:wol2} applied to the inner sums in both $S^*$ and $S^{**}$, $n^2/d_n$ divides $S_{m,k}$, where $d_n$ was defined in Theorem \ref{thm:wol2}. For the base step of the induction, note that if $\omega(m)=1$, then $m=2$. So $I_{2,k}=\{3k+1,3k+2\}$ and $S_{2,k}= 2U_{6k+3}/U_{3k+1}U_{3k+2}$. Therefore, we have $\nu_2(S_{2,k})=\nu_2\big(2U_3\big)=2+\tau$. Indeed, $U_{3(2k+1)}=U_3\cdot U'_{2k+1}$, where $U'=U(V_3,Q^3)$, and $U'_t\pmod2=0,1,0,1,0,1,\cdots$ ($t\ge0$), since $V_3$ is even. \end{proof} \begin{lemma} \label{lem:wol5} Let $U(P,Q)$ and $V(P,Q)$ be a pair of Lucas sequences, where $P$ is odd and $Q$ is congruent to $\amod{3} {4}$. Let $m$ be an even integer not divisible by $4$ of maximal rank with respect to $U(P,Q)$. Then for all integers $k$ $$ S_{m,k}(P,Q)\equiv\ 0\pmod {m^2/\gcd(m,3)}. $$ \end{lemma} \begin{proof} An induction very similar to those used in proving Lemmas \ref{lem:wol3} and \ref{lem:wol4} works fine. Note however that besides Theorem \ref{thm:wol2}, Lemma \ref{lem:wol4} also comes into play. Indeed, if $m=p^a\ell$, $p$ an odd prime not dividing $\ell$, then the sums $S_{\ell,j}$ in $S^{**}$ are associated with a Lucas sequence $U(P',Q')$, where $P'$ is odd, but $Q'=Q^{\rho(p)}$ is congruent to $\amod{1} {4}$, if $\rho(p)$ is even. The base step of the induction corresponds to $m=2$ and, as in the proof of Lemma \ref{lem:wol4}, we have $\nu_2(S_{2,k})=\nu_2(2U_3)$. However, $U_3=P^2-Q$ is even, but not divisible by $4$. \end{proof} \begin{lemma} \label{lem:wol6} Let $U(P,Q)$ and $V(P,Q)$ be a pair of Lucas sequences, where $P$ is odd and $Q$ is congruent to $\amod{3} {4}$. Let $m$ be an integer divisible by $4$ of maximal rank with respect to $U(P,Q)$. Then for all integers $k$ $$ S_{m,k}(P,Q)\equiv\ 0\pmod {8m^2}.\footnote{Thus, $2^7$ divides $S_{m,k}$. Unless we refine further the hypotheses on $P$ and $Q$, the exponent $7$ is optimal. For instance, $S_{4,0}(1,-1)=\frac{128}{15}$.} $$ \end{lemma} \begin{proof} A proof by induction on $i=\omega(m)$ of the same model as in the three previous lemmas works fine. The rank of $4$ is $6$. Since the rank of $3$ must be $2$, $3$ or $4$, and none of these is prime to $6$, $3\nmid m$. Writing $m$ as $4n$ with $n$ odd, we find that $n^2$ divides $S_{m,k}$ by Theorem \ref{thm:wol2}. Thus, we are left with seeing that $2^7$ divides $S_{m,k}$. Write $m$ as $p^a\ell$, where $p$ is a prime $\ge5$ and $p\nmid\ell$. We have that $4\mid\ell$ and that the inner sums $S_{\ell,j+k\rho(p^a)}$ in $S^*$ are divisible by $8\ell^2$ and by $2^7$ in particular, using the inductive hypothesis. How about the inner sums $S_{\ell,j}(P',Q')$ of $S^{**}$? Note that $\rho(2)$ is $3$. So $m$ being of maximal rank, $3\nmid\rho(p)$. But $2\mid V_t$ iff $3\mid t$. So $P'=V_{\rho(p)}$ is odd. In fact, as $\rho(4)=6$, $\rho(p)$ must also be odd, that is to say, $\rho(p)=p$. Thus, $Q'=Q^{\rho(p)}\equiv \amod{3} {4}$. The inductive hypothesis yields that $S^{**}$ is divisible by $2^7$. We have not checked the initial step of the induction, i.e., the case $m=4$. Note that $I_{4,k}=\{6k+1,6k+2,6k+4,6k+5\}$. Combining, on one hand, the first and fourth terms of $S_{4,k}$, and, on the other hand, the middle ones, yields $$ S_{4,k}=\frac{2U_{12k+6}}{U_{6k+1}U_{6k+5}}+\frac{2U_{12k+6}}{U_{6k+2}U_{6k+4}}. $$ Hence, $\nu_2(S_{4,k})=\nu_2(2U_6)+\nu_2(U_{6k+2}U_{6k+4}+U_{6k+1}U_{6k+5})$. But $U_6=U_2U_3(P^2-3Q)=P(P^2-Q)(P^2-3Q)$, so that \begin{equation}\label{eq:1st} \nu_2(2U_6)=2+\nu_2(P^2-3Q)\ge\begin{cases}\;5,\quad\text{ if }Q\equiv \amod{3} {8};\\ \;4,\quad\text{ if }Q\equiv\amod{7} {8} .\end{cases} \end{equation} Since $\rho(8)=6$, we have $U_{t+6}\equiv \amod{U_7U_t} {8}$ for all $t\ge0$. Indeed, the congruence holds for $t=0$ and $t=1$, and, by (\ref{eq:recursion}), it must hold for all $t$'s. Thus, $$ U_{6k+2}U_{6k+4}+U_{6k+1}U_{6k+5}\equiv \amod{U_7^{2k}(U_2U_4+U_1U_5)} {8}. $$ A direct calculation gives $U_2U_4+U_1U_5=2P^4-5P^2Q+Q^2\equiv \amod{3(Q+1)} {8}$. Therefore, $$ \nu_2(U_{6k+2}U_{6k+4}+U_{6k+1}U_{6k+5})\ge \begin{cases}\; 2,\quad\text{ if }Q\equiv \amod{3} {8};\\ \; 3,\quad\text{ if }Q\equiv \amod{7} {8},\end{cases} $$ which, combined with (\ref{eq:1st}), yields that $\nu_2(S_{4,k})\ge7$ for all integers $k$. \end{proof} \section{Acknowledgments} We thank the two referees for time spent reading this paper and writing up comments and suggestions. One of the referees pointed out reference \cite{Pan}. \begin{thebibliography}{99} \bibitem{Ba} Christian Ballot, Lucas sequences with cyclotomic root field, submitted preprint, 120 pages. \bibitem{Car} R. D. Carmichael, On the numerical factors of the arithmetic forms $\alpha^n\pm\beta^n$, {\it Ann. of Math.\/}, {\bf 15}, (1913--14), 30--48, 49--70. \bibitem{Cho} S. Chowla, Leudesdorf's generalization of Wolstenholme's theorem, {\it J. London Math. Soc.\/}, {\bf 9} (1934), 246. \bibitem{HW} G. H. Hardy and E. M. Wright, {\it An Introduction to the Theory of Numbers}, Clarendon Press, 1960. \bibitem{Ho} C. Hooley, On Artin's conjecture, {\it J. reine angew. Math.\/} {\bf 225} (1967), 209--220. \bibitem{Ja} G. J. Janusz, {\it Algebraic Number Fields}, Academic Press, 1973. \bibitem{KW1} William Kimball and William Webb, A congruence for Fibonomial coefficients modulo $p^3$, {\it Fibonacci Quart.\/}, {\bf 33} (1995), 290--297. \bibitem{KW2} William Kimball and William Webb, Some generalizations of Wolstenholme's theorem, in {\it Applications of Fibonacci Numbers} Vol.\ 8, Kluwer, 1999, pp.\ 213--218. \bibitem{Leu} C. Leudesdorf, Some results in the elementary theory of numbers, {\it Proc. London Math. Soc.\/}, {\bf 20} (1889), 199--212. \bibitem{Lu1} \'Edouard Lucas, Th\'eorie des fonctions simplement p\'eriodiques, {\it Amer. J. Math.\/}, {\bf 1} (1878), 184--240, 289--321. \bibitem{Lu2} \'Edouard Lucas, {\it Th\'eorie des Nombres}, Librairie scientifique et technique Albert Blanchard, 1961. \bibitem{MR} Richard J. McIntosh and Eric L. Roettger, A search for Fibonacci-Wieferich and Wolstenholme primes, {\it Math. Comp.\/}, {\bf 76} (2007), 2087--2094. \bibitem{Pan} Hao Pan, A generalization of Wolstenholme's harmonic series congruence, {\it Rocky Mountain J. Math.\/}, {\bf 38} (2008), 1263--1269. \bibitem{Rib} Paulo Ribenboim, The Fibonacci numbers and the Artic Ocean, Proceedings of the 2nd Gauss Symposium. Conference A: {\it Mathematics and Theoretical Physics\/}, de Gruyter, 1995, pp. 41--83. \bibitem{Wi} Hugh C. Williams, {\it \'Edouard Lucas and Primality Testing}, Wiley, 1998. \bibitem{Wo} Joseph Wolstenholme, On certain properties of prime numbers, {\it Quarterly J. Pure Applied Math.\/}, {\bf 5} (1862), 35--39. \end{thebibliography} \bigskip \hrule \bigskip \noindent 2010 {\it Mathematics Subject Classification}: Primary 11B39; Secondary 11A07. \noindent \emph{Keywords: } Lucas sequence, rank of appearance, congruence, Wolstenholme, Leudesdorf. \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received June 6 2012; revised version received October 8 2012. Published in {\it Journal of Integer Sequences}, October 8 2012. Minor typographical corrections, November 24 2013. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document} .