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\begin{center}
\vskip 1cm{\LARGE\bf 
On a Congruence of Kimball and Webb \\
\vskip .1in
Involving Lucas Sequences
}
\vskip 1cm
\large
Christian Ballot \\
D\'epartement de Math\'ematiques et M\'ecanique\\
Universit\'e de Caen \\
F-14032 Caen Cedex \\
France \\
\href{mailto:christian.ballot@unicaen.fr}{\tt christian.ballot@unicaen.fr} \\
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\begin{abstract} Given a pair $(U_t)$ and $(V_t)$ of Lucas sequences, 
an odd integer $\nu\ge1$, and a prime $p\ge\nu+4$ of maximal rank $\rho_U$, 
i.e., such that $\rho_U$ is $p$ or $p\pm1$, we show that $\sum_{0&lt;t&lt;\rho_U}(V_t/U_t)^\nu  
\equiv0\pmod{p^2}$. This extends a result of Kimball and Webb, who 
proved the case $\nu=1$. Some further generalizations are also established. 
\end{abstract}

\section{Introduction}

  The main purpose of this note is to provide a common generalization to 
Theorems \ref{thm:L} and \ref{thm:KW} below.
 
 \begin{theorem}\label{thm:L} Let $\nu\ge1$ be an odd integer. Suppose $p\ge\nu+4$ is 
a prime number. Then 
$$
H_{p-1}^\nu:=\sum_{t=1}^{p-1}\frac1{t^\nu}\equiv0\pmod{p^2}.
$$ 
\end{theorem}
 
 \begin{theorem} \label{thm:KW} Let $U(P,Q),\;V(P,Q)$ be a pair of Lucas sequences, $p\ge5$, $p\nmid Q$,  
a prime of rank $\rho$ equal to $p-\e_p$, where $\e_p=0$ or $\pm1$. Then   
$$
\sum_{t=1}^{\rho-1}\frac{V_t}{U_t}\equiv0\pmod {p^2}.
$$  
\end{theorem} 
  
  Theorem \ref{thm:L} appears in the book of Hardy and Wright \cite{HW} as Theorem 131 and is one of many  generalizations of the 1862 congruence of Wolstenholme \cite{Wo} which states that  
\begin{equation}\label{eq:1862}
H_{p-1}:=1+\frac12+\frac13+\cdots+\frac1{p-1}\equiv0\pmod{p^2}, 
\end{equation} 
i.e., that the numerator of the rational number $H_{p-1}$ is a multiple of $p^2$, whenever 
$p$ is a prime number at least $5$. Sums of the type $\sum_{t=1}^m1/t^\nu$, where $\nu$ is 
odd and $t$ is prime to $m$, were first studied by Leudesdorf \cite{Leu}. 
Their values modulo $m^2$ have been the subject of many papers and these results were revisited 
in the paper \cite{Sla}.
Historically, Wolstenholme used the congruence (\ref{eq:1862}) 
to prove that for all primes $p\ge5$ we have
\begin{equation}\label{eq:Wol}
\binom{2p-1}{p-1}\equiv1\pmod {p^3},
\end{equation} 
a fact known as Wolstenholme's congruence.
The survey paper \cite{Mes} mentions  
many generalizations of the congruence (\ref{eq:Wol}), but also contains generalizations of the 
related congruence (\ref{eq:1862}), particularly in Sections 4 and 8.   



  Theorem \ref{thm:KW}, as may not readily appear, is yet another (surprising!) generalization of 
 the congruence (\ref{eq:1862}) of Wolstenholme. Contrary to other known generalizations,  
 its discovery by Kimball and Webb \cite{KW2}, who initially proved it for the special 
 pair of Lucas sequences $U(1,-1)$, $V(1,-1)$ in the paper \cite{KW1}, came much later, and more 
 than 130 years after (\ref{eq:1862}) had appeared. For $(P,Q)=(2,1)$, 
 we have $x^2-Px+Q=(x-1)^2$, $U_t=t$ and $V_t=2$ for all integers $t\ge0$ so that every 
 prime $p$ has rank $p$, and we see that the result of Kimball and Webb is indeed a generalization 
 of the congruence of Wolstenholme.  
 
 
 
  Putting Theorems \ref{thm:L} and \ref{thm:KW} next to each other and computing a few sums  
 such as $\sum_{t=1}^7L_t^3/F_t^3\pmod {49}$ or $\sum_{t=1}^9L_t^5/F_t^5\pmod{121}$, where 
 $L_t$ and $F_t$ are the $t$-th Lucas and the $t$-th Fibonacci numbers, and finding 
 in both cases $0\pmod{p^2}$ for $p=7$ and $p=11$, respectively, led to conjecturing (\^o res 
 mirabile!) that Theorem \ref{thm:B} holds.  
 
  \begin{theorem} \label{thm:B} Let $\nu\ge1$ be an odd integer and $U(P,Q),\;V(P,Q)$ be a pair of Lucas sequences. If $p\ge\nu+4$, $p\nmid Q$, is a prime of maximal rank $\rho$, i.e., of rank $p-\e_p$, where $\e_p=0$ or $\pm1$, then   
$$
\sum_{t=1}^{\rho-1}\frac{V_t^\nu}{U_t^\nu}\equiv0\pmod {p^2}.
$$  
\end{theorem} 
    



 We briefly remark that Theorem \ref{thm:KW} of Kimball and Webb has recently been   
generalized in several directions. One generalization is found in \cite{Pan}, another 
in Chapter 4 of \cite{Ba}, and again another in \cite{Ba1}, where 
congruences involving sums of ratios $V_t/U_t$ of Lucas sequences taken modulo $m^2$, 
where $m$ may be composite, not necessarily prime, were considered.  
   



  Theorem \ref{thm:KW} was stated in a mildly generalized form in both papers 
 \cite{Ba} and \cite{Ba1}, in part because it led to the further generalizations 
 found in these papers. In anticipation of yet further tentative generalizations 
 we wish to prove a more general version of Theorem \ref{thm:B}, namely 
 
 
\begin{theorem} \label{thm:LKW} Let $\nu\ge1$ be an odd integer. Let $U(P,Q)$ and $V(P,Q)$ be a pair 
of Lucas sequences and $p\nmid Q$ be a prime number of maximal rank $\rho$ with respect to $U(P,Q)$. Assume 
further that $p-1\nmid\nu\pm1$ if $\nu\ge3$ and $p-1\nmid\nu+1$ if $\nu=1$. 
Then, for all integers $k$, we have that  
$$
S:=\sum_{t=k\rho+1}^{(k+1)\rho-1}\frac{V_t^\nu}{U_t^\nu}\equiv0\pmod {p^2}. 
$$   
\end{theorem}

 Note that if $p\ge\nu+4$ then $p-1\nmid\nu\pm1$ when $\nu\ge3$ and $p-1\nmid\nu+1$ when $\nu=1$ so that Theorem \ref{thm:LKW} 
 implies Theorem \ref{thm:B} by setting $k=0$.     
  
  A secondary purpose of our note is to indicate how one can obtain further results that 
combine the approach taken in \cite{Ba1}, i.e.,  congruences with composite moduli $m$ 
of maximal ranks, with the general odd exponents $\nu\ge1$ of the present paper. This 
is done in the third and final section. 

  Throughout this note we assume familiarity with Lucas sequences. The introduction 
of the paper \cite{Ba1} contains references to Lucas theory and useful remarks and 
definitions concerning ranks of primes and integers, and congruences of rational numbers modulo 
an integer $m$. 
   
\section{Proof of Theorem \ref{thm:LKW}}

 We begin by a few lemmas and remarks. 

\begin{lemma}\label{lem:newton} Let $\nu\ge1$ be an odd integer, $p$ be a prime 
number and $x$ and $y$ be two rational integers whose sum is divisible by $p$.  
Then there is a rational integer $c$, which may depend on $\nu$, such that 
\begin{equation}\label{eq:vi}
x^\nu+y^\nu\equiv c(xy)^{(\nu-1)/2}(x+y)\pmod {p^2}. 
\end{equation}  
\end{lemma}
\begin{proof} Choosing $c=1$ we see that (\ref{eq:vi}) holds for $\nu=1$. We 
proceed by strong induction on $\nu$. So given $\nu\ge3$ we assume that for each 
$i=1,\dots,(\nu-1)/2$, we have 
$$  
x^{\nu-2i}+y^{\nu-2i}\equiv c_i(xy)^{(\nu-2i-1)/2}(x+y)\pmod {p^2}, 
$$ for some $c_i\in\mathbb{Z}$.

Expanding $(x+y)^\nu$ by Newton's formula and pairing terms in $x^ky^{\nu-k}$ 
with terms in $x^{\nu-k}y^k$, we find that 
\begin{eqnarray*}
x^\nu+y^\nu &amp; = &amp; (x+y)^\nu-\binom{\nu}{1}xy(x^{\nu-2}+y^{\nu-2})-\binom{\nu}{2}x^2y^2
(x^{\nu-4}+y^{\nu-4})\\
&amp; &amp; -\cdots-\binom{\nu}{(\nu-1)/2}(xy)^{(\nu-1)/2}(x+y).
\end{eqnarray*}
The term $(x+y)^\nu$ is divisible by $p^2$ and, by the inductive hypothesis, 
each term $(xy)^i(x^{\nu-2i}+y^{\nu-2i})$ is congruent to $c_i(xy)^{(\nu-1)/2}
(x+y)$ modulo $p^2$ for some integer $c_i$. Hence, putting $c=-\sum c_i\binom{\nu}
{i}$, $i=1,\dots,(\nu-1)/2$, the lemma follows.  
\end{proof}

\begin{lemma}\label{lem:char} Let $p$ be an odd prime and $e\ge0$ be an integer such 
that $p-1\nmid e$ if $e\ge1$. Then the sum $\sum_{t=1}^pt^e$ is divisible by $p$. 
\end{lemma}
\begin{proof} If $e$ is $0$, then the result is trivial. Assume $e\ge1$. 
If $g$ is a primitive root modulo $p$, then $g^e\not\equiv1\pmod p$. 
But $\sum_{t=1}^pt^e\equiv\sum_{t=1}^p(gt)^e\pmod p$. Hence, $p$ divides $(g^e-1)\sum_{t=1}^pt^e$.
\end{proof}

 Given a pair $U$, $V$ of Lucas sequences with parameters $(P,Q)$ we denote the discriminant $P^2-4Q$ by $D$ and 
recall a few relevant identities. For all integers $s$ and $t$, we have 
\begin{eqnarray}
2U_{s+t} &amp; = &amp; U_sV_t+U_tV_s,\label{summ1}\\
4Q^t &amp; = &amp; V_t^2-DU_t^2,\label{Pyt}\\
2Q^tU_{s-t} &amp; = &amp; U_sV_t-U_tV_s,\label{subt}\\
2Q^tV_{s-t} &amp; = &amp; V_sV_t-DU_sU_t\label{subt1}.
\end{eqnarray} 


 We are ready for a proof of Theorem \ref{thm:LKW}. 

\begin{proof}  All unmarked sums are taken with indices running 
from $k\rho+1$ to $(k+1)\rho-1$. 
Also we put $r:=(2k+1)\rho$. Then we see that 
$$
S=\frac12\sum\bigg(\frac{V_t^\nu}{U_t^\nu}+\frac{V_{r-t}^\nu}{U_{r-t}^\nu}\bigg)
=\frac12\sum\frac{V_t^\nu U_{r-t}^\nu+U_t^\nu V_{r-t}^\nu}{U_t^\nu U_{r-t}^\nu}.
$$
Put $x=V_tU_{r-t}$ and $y=U_tV_{r-t}$. Then, by equation (\ref{summ1}), $x+y=2U_r$, which is 
divisible by $p$. Thus, by Lemma \ref{lem:newton}, we have for some integer $c$
$$
V_t^\nu U_{r-t}^\nu+U_t^\nu V_{r-t}^\nu\equiv 2cU_r(V_tU_{r-t}U_tV_{r-t})^{(\nu-1)/2}
\pmod {p^2}.
$$ Therefore, if $p^2\mid cU_r$, then $S\equiv0\pmod {p^2}$. So we assume $p^2\nmid cU_r$. 
Then $S\equiv0\pmod {p^2}$ if and only if $p$ divides 
\begin{equation}\label{eq:med}
\sum\frac{(V_tU_{r-t}U_tV_{r-t})^{(\nu-1)/2}}{U_t^\nu U_{r-t}^\nu}.
\end{equation}
By the subtraction formulas (\ref{subt}) and (\ref{subt1}), we find that 
\begin{eqnarray*}
(2U_{r-t})^{(\nu-1)/2}&amp;\equiv&amp; (-1)^{(\nu-1)/2}Q^{-(\nu-1)t/2}U_t^{(\nu-1)/2}V_r^{(\nu-1)/2}\pmod p,
\text{ and}\\  
(2V_{r-t})^{(\nu-1)/2}&amp;\equiv&amp; Q^{-(\nu-1)t/2}V_t^{(\nu-1)/2}V_r^{(\nu-1)/2}\pmod p.
\end{eqnarray*} 
Thus, there exist integers $\lambda_1$ and $\lambda_2$, prime to $p$ and not dependent on $t$,   
such that modulo $p$ the numerator of the $t$-th term in (\ref{eq:med}) is equal 
to $\lambda_1Q^{-(\nu-1)t}U_t^{\nu-1}V_t^{\nu-1}$, while its denominator is equal to 
$\lambda_2Q^{-\nu t}U_t^{2\nu}$ since $(2U_{r-t})^\nu\equiv -Q^{-\nu t}U_t^\nu V_r^\nu\pmod p$.  
Indeed, by (\ref{Pyt}), $p\nmid V_r$. Thus, $p^2$ divides $S$ iff 
$$
p\;\text{ divides }\; \sum\frac{Q^tV_t^{\nu-1}}{U_t^{\nu+1}}.
$$
By (\ref{Pyt}) we see that $4Q^tV_t^{\nu-1}=V_t^{\nu+1}-DU_t^2V_t^{\nu-1}$. 
Hence, $S$ is zero modulo $p^2$ iff $p$ divides 
\begin{equation}\label{eq:lar}
\sum\frac{V_t^{\nu+1}}{U_t^{\nu+1}}-D\sum\frac{V_t^{\nu-1}}{U_t^{\nu-1}}.
\end{equation}

 Note that since $p$ is an odd prime and $p\nmid Q$, if $s$ and $t$ are distinct integers that belong to the 
interval $(k\rho,(k+1)\rho)$, then $V_s/U_s$ and $V_t/U_t$ are distinct modulo $p$ by (\ref{subt}). Also, if $D$  
is a quadratic residue modulo $p$, then $V_t/U_t$ and $\pm\sqrt{D}$ are not congruent modulo $p$ 
by (\ref{Pyt}).  

 If $\rho$ is $p+1$, then, as $t$ varies from $k(p+1)+1$ to $k(p+1)+p$, $V_t/U_t\pmod p$ runs 
through all of $\mathbb{Z}/p\mathbb{Z}$. 
Thus, by Lemma \ref{lem:char}, as $\nu\not\equiv\pm1\pmod{p-1}$, we find that both sums appearing in (\ref{eq:lar}) 
are divisible by $p$. The same divisibilities hold if $\rho$ is $p$, since, then, only the 
residue $0\pmod p$ is absent from each of the two sums. If $\rho$ is $p-1$, then, as $t$ varies 
from $k\rho+1$ to $k\rho+p-2$, $V_t/U_t\pmod p$ runs through all of $\Z/p\Z$ but $\pm\sqrt{D}\pmod p$. 
Hence we deduce respectively that
$$
\sum\bigg(\frac{V_t}{U_t}\bigg)^{\nu\pm1}+(\sqrt{D})^{\nu\pm1}+(-\sqrt{D}\,)^{\nu\pm1} 
\equiv0\pmod p.
$$ So we find that 
$$\sum\frac{V_t^{\nu+1}}{U_t^{\nu+1}}-D
\sum\frac{V_t^{\nu-1}}{U_t^{\nu-1}}
\equiv(0-2D^{(\nu+1)/2})-D(0-2D^{(\nu-1)/2})\equiv0\pmod p.$$ 
\end{proof}

\begin{remark} If $D=0$, then the hypothesis $p-1\nmid\nu+1$ suffices. 
\end{remark}

\section{Two more theorems}

 Given a Lucas sequence $U(P,Q)$ and an integer $m$ prime to $Q$ we denote the rank of $m$ in 
$U$ by $\rho(m)$. 

By the definition which was introduced and motivated in \cite{Ba1},
we say that a composite integer 
$m\ge1$ has maximal rank 
with respect to a Lucas sequence $U(P,Q)$ if every prime power dividing $m$ has 
maximal rank and the ranks of any two such prime powers are relatively prime. 
A power $p^a$ has maximal rank in $U(P,Q)$ iff $p$ has maximal rank and the 
rank $\rho(p^a)$ is equal to $p^{a-1}\rho(p)$. 
 
 Here, unlike in the paper \cite{Ba1}, we will not look for an exhaustive theorem  
dealing with sums of terms $V_t^\nu/U_t^\nu$ modulo squares of integers of maximal rank, but the 
partial results we will state 
and sketch will be enough to see how one could obtain such congruences. 
 
  We start by an extension of Theorem \ref{thm:LKW} to prime powers. The case $\nu=1$ 
is in fact a corollary of Theorem 12 of \cite{Ba1}. 
  
\begin{theorem}\label{thm:pp} Let $\nu\ge1$ be an odd integer and $U$, $V$ be a pair 
of Lucas sequences with parameters $P$ and $Q$. Suppose $p\nmid Q$ is a prime satisfying  
$p&gt;(\nu+3)/2$, $p\not=\nu$ and $p\not=\nu+2$. Assume $p^a$ has maximal rank for some integer $a\ge1$. 
Then for all integers $k$ we have 
$$
S:=\sum_{t\in I}\frac{V_t^\nu}{U_t^\nu}\equiv0\pmod{p^{2a}},
$$ where $I$ is the set of integers in $\big(k\rho(p^a),(k+1)\rho(p^a)\big)
\;\backslash\;\rho(p)\Z$. 
\end{theorem}
\begin{proof} The proof is actually an extension of that of Theorem \ref{thm:LKW}.    
It is easy to upgrade Lemma \ref{lem:newton}.  
 Suppose $p^a\mid x+y$. Then for each odd integer $\nu\ge1$ there is a rational integer $c$   
such that  
 \begin{equation}
x^\nu+y^\nu\equiv c(xy)^{(\nu-1)/2}(x+y)\pmod {p^{2a}}. 
\end{equation}   
Note that the sum $S$ contains $p^{a-1}\big(\rho(p)-1\big)$ terms. 
We now have $r=(2k+1)\rho(p^a)$. So $2cU_r$ is at least divisible by $p^a$. So we will be 
through if we can show that the sum corresponding to the sum in (\ref{eq:med}) is $0\pmod{p^a}$. 
But that will be true if the expression in (\ref{eq:lar}) is also $0\pmod{p^a}$, where now 
both sums are taken over all integers $t$ in $I$. At this point note that Lemma \ref{lem:char} 
can also be upgraded so that $\sum_{t=1}^{p^a}t^e\equiv0\pmod{p^a}$ provided, again, $p-1\nmid e$. 
Indeed, if $g$ is a primitive root modulo $p^a$, then it reduces to a primitive root modulo $p$, 
so $p\nmid g^e-1$ if $p-1\nmid e$ and this gives the lemma. Note that 
as $t$ varies through $I$ all $V_t/U_t$ are distinct modulo $p^a$. 
If $\rho(p)$ is $p+1$, then since the cardinality of $I$ is $p^a$ we deduce that  
$$
\sum_{t\in I}\bigg(\frac{V_t}{U_t}\bigg)^{\nu\pm1}\equiv\sum_{t\in\Z/p^a}t^{\nu\pm1}\equiv0\pmod{p^a}.
$$ If $\rho(p)$ is $p$ then $V_t$ is never $0\pmod p$ and the terms $V_t/U_t\pmod{p^a}$ 
run through all invertible elements of $\Z/p^a\Z$ as $t$ runs through $I$. Thus $g$ being 
a primitive root modulo $p^a$ and $\varphi$ being Euler's totient function, we find that 
$$
\sum_{t\in I}\bigg(\frac{V_t}{U_t}\bigg)^{\nu\pm1}\equiv\sum_{t=0}^{\varphi(p^a)-1}g^{(\nu\pm1)t}=
\frac{g^{(\nu\pm1)\varphi(p^a)}-1}{g^{\nu\pm1}-1}\equiv0\pmod{p^a},
$$ since $g^{(\nu\pm1)}-1$ is not divisible by $p$. This argument is not valid if $\nu=1$ 
and the exponent of $g$ is $\nu-1$. But in that case the result nevertheless holds because 
$$
D\sum_{t\in I}\bigg(\frac{U_t}{V_t}\bigg)^{\nu-1}=D\cdot |I|\equiv0\pmod{p^a}, 
$$ so the expression in (\ref{eq:lar}) is $0\pmod{p^a}$. 

If $\rho(p)$ is $p-1$, then $I$ 
contains $p^a-2p^{a-1}$ elements. So $V_t/U_t\pmod{p^a}$ runs through all $\Z/p^a$ except 
$\pm\sqrt{D}+ip\pmod{p^a}$ for $i=1,2,\dots,p^{a-1}$. Hence,  
$(V_t/U_t)^2\pmod{p^a}$ hits each nonzero quadratic 
residue in $\Z/p^a$ twice except all $D+ip$, for $i=1,\dots, p^{a-1}$, as $t$ runs through $I$. 
Therefore, 
$$
\sum_{t\in I}\bigg(\frac{V_t}{U_t}\bigg)^{\nu+1}\equiv\sum_{i=1}^{p^a}(i^2)^{(\nu+1)/2}
-2\sum_{j=1}^{p^{a-1}}(D+jp)^{(\nu+1)/2}\pmod{p^a}.
$$
Now 
$$
\sum_{j=1}^{p^{a-1}}(D+jp)^{(\nu+1)/2}=\sum_{u=0}^{(\nu+1)/2}\binom{(\nu+1)/2}{u}D^{(\nu+1)/2-u}p^u
\sum_{j=1}^{p^{a-1}}j^u.
$$ As $p-1&gt;(\nu+1)/2$, $p^{a-1}\mid\sum_{j=1}^{p^{a-1}}j^u$ and, unless $u=0$, 
$p^a\mid p^u\sum_{j=1}^{p^{a-1}}j^u$. Hence, as $p\not=\nu+2$, we have 
$$
\sum_{t\in I}\bigg(\frac{V_t}{U_t}\bigg)^{\nu+1}\equiv0-2D^{(\nu+1)/2}p^{a-1}\pmod{p^a}. 
$$ Since a similar argument is valid for $\nu-1$ instead of $\nu+1$ we see that 
$\sum_{t\in I}\frac{V_t^{\nu+1}}{U_t^{\nu+1}}-D\sum_{t\in I}\frac{V_t^{\nu-1}}{U_t^{\nu-1}}$ 
is congruent to 
$$ 
(0-2p^{a-1}D^{(\nu+1)/2})-D(0-2p^{a-1}D^{(\nu-1)/2})=0\pmod {p^a}.
$$ \end{proof}
 
\begin{theorem}\label{thm:comp} Let $\nu\ge1$ be an odd integer and $U$, $V$ be a pair 
of Lucas sequences with parameters $P$ and $Q$. Suppose $m$ is a positive integer prime to $Q$ 
of maximal rank. Assume further that each prime factor $p$ of $m$ satisfies  
$p&gt;(\nu+3)/2$, $p\not=\nu$ and $p\not=\nu+2$.  
Then for all integers $k$ we have 
$$
S:=\sum_{t\in I}\frac{V_t^\nu}{U_t^\nu}\equiv0\pmod{m^2},
$$ where $I$ is the set of integers in $\big(k\rho(m),(k+1)\rho(m)\big)
\;\backslash\;\bigcup_{p\mid m}\rho(p)\Z$. 
\end{theorem}
  
  The proof we sketch is similar to, but simpler than that of Theorem 14 of the paper \cite{Ba1}.
  
\begin{proof} Assume for simplicity that $k=0$. We proceed by induction 
on the number of distinct prime factors of $m$. If $m$ is a prime power, then this is 
Theorem \ref{thm:pp}. Say $m=p^an$, where $p\nmid n$ is prime and $n&gt;1$. 
It suffices to show that $S\equiv0\pmod{n^2}$. Note that since $n$ has maximal rank 
the theorem will hold for $n$ by the inductive hypothesis. Write $S$ as $S^*-S^{**}$, 
where 
$$
S^*:=\sum_{i=0}^{\rho(p^a)-1}\quad\sum_{\substack{t=i\rho(n)+1\\\rho(q)\,\nmid\, t,\text{ if }q\mid n}}^{(i+1)\rho(n)-1}
\bigg(\frac{V_t}{U_t}\bigg)^\nu, 
$$ and 
$$
S^{**}:=\sum_{\substack{t=1\\ \rho(q)\,\nmid\, t, \text{ if }q\mid n}}^{p^{a-1}\rho(n)}\bigg(\frac{V_{\rho(p)t}}{U_{\rho(p)t}}\bigg)^\nu=
\sum_{i=0}^{p^{a-1}-1}U_{\rho(p)}^{-\nu}\sum_{\substack{t=i\rho(n)+1\\ \rho(q)\,\nmid\, t, \text{ if }q\mid n}}^{(i+1)\rho(n)-1}\bigg(\frac{V_t'}{U_t'}\bigg)^\nu,
$$ where the sequences $U_t'=U_{\rho(p)t}\cdot U_{\rho(p)}^{-1}$ and $V_t'=V_{\rho(p)t}$ are the Lucas 
sequences associated with $(P',Q')=(V_{\rho(p)},Q^{\rho(p)})$ and the letter $q$ stands for a prime.
Note that all inner sums in $S^*$ are $0\pmod{n^2}$ and that, as $n$ has maximal 
rank with respect to $U'$, the same is true, by our inductive hypothesis again, of the inner sums of $S^{**}$. 
\end{proof}




\begin{example} Consider the Lucas sequences $U$ and $V$ with parameters $P=1$ and $Q=-9$.  
Then $D=37$, $\rho(7)=6$ and $\rho(37)=37$. Thus, $m=259=7\cdot37$ has maximal rank. 
Hence, by Theorem \ref{thm:comp}, we have for $\nu=1,3$ or $9$
$$
\sum_t\bigg(\frac{V_t}{U_t}\bigg)^\nu\equiv0\pmod{7^2\cdot37^2},
$$ where $t$ runs over all integers between $1$ and $222=6\cdot37$ except for multiples of $6$ or $37$. 
\end{example} 

\section{Acknowledgments} 

 We thank a nearly anonymous referee for writing typed and handwritten 
suggestions meant to improve the clarity of some points in the paper.  


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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B39; Secondary 11A07.

\noindent \emph{Keywords: } 
Lucas sequence, rank of appearance, congruence, Wolstenholme,  Leudesdorf. 

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\hrule
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\vspace*{+.1in}
\noindent
Received  July 22 2013;
revised versions received   November 9 2013; November 28 2013.
Published in {\it Journal of Integer Sequences},
December 15 2013.

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\noindent
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