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\begin{center}
\vskip 1cm{\LARGE\bf 
Identities for the Classical Polynomials \\
\vskip .1in
Through Sums of Liouville Type
}
\vskip 1cm
\large
Adama Diene and Mohamed El Bachraoui\\
Department of Mathematical Sciences\\
United Arab Emirates University \\
P.O. Box 17551 \\
Al-Ain \\
United Arab Emirates \\
\href{mailto:adiene@uaeu.ac.ae}{\tt adiene@uaeu.ac.ae} \\
\href{mailto:melbachraoui@uaeu.ac.ae}{\tt melbachraoui@uaeu.ac.ae}\\
\end{center}

\vskip .2 in

\begin{abstract}
We prove identities involving Bernoulli, Euler, Fibonacci, Lucas, Chebychev,
and Dickson polynomials by using sums of Liouville type.
\end{abstract}

\section{Introduction}
Polynomials defined recursively over the integers such as Dickson
polynomials, Chebychev polynomials, Fibonacci polynomials, Lucas
polynomials, Bernoulli polynomials, Euler polynomials, and many others have
been extensively studied in the past. Most of these polynomials have some
type of relationship between them and share a large number of interesting
properties. They have been also found to be topics of interest in many
different areas of pure and applied sciences. Most recently, some of these
families of polynomials have been found to be useful in cryptography and
related topics, which keep making them a very interesting area of research
for many people in this era of communication. In this paper, we use a sum of
Liouiville type to prove new properties concerning many of these families of
polynomials. The following notations and definitions will be adopted
throughout this paper.

The sets of positive integers, nonnegative integers, integers, real numbers,
and complex numbers are respectively denoted by $\mathbb{N}$, $\mathbb{N}_0$%
, $\mathbb{Z}$, $\mathbb{R}$, and $\mathbb{C}$. The sets of even positive
integers and the set of odd positive integers will be written $2\mathbb{N}$
and $2\mathbb{N}-1$ respectively and the sets of even integers and odd
integers will be written $2\mathbb{Z}$ and $2\mathbb{Z}-1$ respectively. The Euler phi function will be denoted by $\phi(n)$ and
the sum of the $k$th powers of positive divisors of $n$ will denoted by $\sigma_k (n)$. 
\begin{definition}
For $n \in \mathbb{N}$ such that $n&gt;1$, let the sets $B(n)$ and $%
B^{\prime}(n)$ be defined as follows:
\begin{equation*}
B(n) = \{(a,b,u,v)\in\mathbb{N}^4:\ au+bv=n \},
\end{equation*}
\begin{equation*}
B^{\prime }(n) = \{(a,b,u,v)\in\mathbb{N}^4:\ au+bv=n,\
\gcd(a,b)=\gcd(u,v)=1 \},
\end{equation*}
and for $n \in 2\mathbb{N}$ let the sets $O(n)$ and $O^{\prime }(n)$ be
defined as follows:
\begin{equation*}
O(n) = \{(a,b,u,v)\in(2\mathbb{N}-1)^4:\ au+bv=n \},
\end{equation*}
\begin{equation*}
O^{\prime }(n) = \{(a,b,u,v)\in(2\mathbb{N}-1)^4:\ au+bv=n,\ \gcd(a,b)=
\gcd(u,v)=1 \}.
\end{equation*}
\end{definition}
Arithmetic sums of the types
\begin{equation}  \label{general-sum}
\sum_{\substack{ (a,b,u,v)\in B(n)}} \bigl(f(a-b)-f(a+b) \bigr)\ \text{and\ }
\sum_{\substack{ (a,b,u,v)\in O(n)}} \bigl(f(a-b)-f(a+b) \bigr)
\end{equation}
were first investigated by Liouville in his seminal work on elementary
methods in number theory, see for instance Liouville \cite{Liouville2,
Liouville4}. A comprehensive treatment of Liouville's methods has been given
in a new book by Williams~\cite{Williams}. Adapting the proof of Williams~%
\cite{Williams} for the sums (\ref{general-sum}), El~Bachraoui~\cite%
{Elbachraoui} found the following related identities for the sums over the
sets $B^{\prime }(n)$ and $O^{\prime }(n)$.
\begin{theorem}\label{main1}
Let $n &gt;1$ be a positive integer.

(a) If $f:\mathbb{Z}\to \mathbb{C}$ is an even function, then
\begin{equation*}
\sum_{(a,b,u,v)\in B^{\prime }(n)} \big( f(a-b)-f(a+b) \big) = \big(f(0) +
2f(1) -f(n) \big) \phi(n) - 2 \sum_{\substack{ 1\leq l &lt; n  \\ (l,n)=1}}f(l).
\end{equation*}
(b) If moreover $n \in 2\mathbb{N}$, then
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)} \big( f(a-b)-f(a+b) \big) = \big(f(0)
-f(n) \big) \phi(n).
\end{equation*}
\end{theorem}
In this note we shall apply Theorem \ref{main1} to deduce
identities involving Bernoulli, Euler, Lucas, Fibonacci, Chebychev, and
Dickson polynomials.
\section{ Preliminaries}
Let $(B_n)_{n=0}^{\infty}$ be the sequence of Bernoulli numbers for which we
have $B_1 = -\frac{1}{2}$, $B_{2n+1}=0$ for any  $n\in\mathbb{N}$, and the
first few terms for even $n$ are
\begin{equation*}
B_0 =1,\ B_2= \frac{1}{6},\ B_4= -\frac{1}{30}, B_6= \frac{1}{42},\ B_8=
-\frac{1}{30},\ B_{10}=\frac{5}{66}.
\end{equation*}
Further we have the recursive formula
\begin{equation}  \label{bernoulli-numbers}
B_n = \sum_{l=0}^n \binom{n}{l} B_l, \quad (n\geq 2).
\end{equation}
\begin{definition}
If $n\in \mathbb{N}$, then the \emph{Bernoulli polynomial} is given by
\begin{equation*}
B_n (x) = \sum_{l=0}^{n} \binom{n}{l} B_{l} x^{n-l}=  n B_1 x ^{n-1} +
\sum_{l=0}^{\lfloor \frac{n}{2} \rfloor} \binom{n}{2l} B_{2l} x^{n-2l}.
\end{equation*}
\end{definition}
A well-known property of Bernoulli polynomial is
\begin{equation*}
B_n (x+1) - B_n (x) = n x^{n-1},\quad (n\geq 1)
\end{equation*}
which by iteration yields
\begin{equation}  \label{bernoulli-property}
B_n (x + m) - B_n(x) = n \sum_{l=0}^{m-1} (x+l)^{n-1}, \quad (m, n \geq 1).
\end{equation}
Let $(E_n)_{n=0}^{\infty}$ be the sequence of Euler numbers.  We have $%
E_{2n-1} = 0$ for are all $n \in \mathbb{N}_0$ and the first few terms for
even $n$ are
\begin{equation*}
E_0=1,\ E_2= -1,\ E_4= 5,\ E_6= -61,\ E_8= 1385,\ E_{10}= -50521,\ E_{12}=
2702765.
\end{equation*}
\begin{definition}
For $n\in \mathbb{N}_0$ the \emph{Euler polynomial} is given by
\begin{equation*}
E_n (x) = \sum_{l=0}^{n} \binom{n}{l} \frac{E_{l}}{2^l}\cdot \left(x-\frac{1}{2} \right)^{n-l}
=  \sum_{l=0}^{\lfloor\frac{n}{2} \rfloor} \binom{n}{2l}
\frac{E_{2l}}{4^l} \cdot \left(x-\frac{1}{2}\right)^{n-2l}.
\end{equation*}
\end{definition}
A well-known property of $E_n(x)$ is
\begin{equation*}
E_n (x+1) - E_n(x) = 2 (x^n - E_n(x) ),
\end{equation*}
which by iteration yields
\begin{equation}  \label{Euler-property}
E_n (x+m) - E_n (x) = 2 \sum_{l=0}^{m-1} \bigl( (x+l)^n - E_n (x+l) \bigr).
\end{equation}
\begin{definition}
\label{Dickson} For $n\in \mathbb{N}$, the \emph{Dickson polynomials} of the
first kind and of the second kind of degree $n$ with parameter $a$ are
respectively defined as follows:
\begin{eqnarray*}
D_{n}(x,a) &amp;=&amp;\sum_{j=1}^{\lfloor n/2\rfloor }\frac{k}{k-j}\binom{n-j}{j}
(-a)^{j}x^{n-2j},\text{ \ and \ }D_{0}(x,a)=2. \\
\text{\textsc{D}}_{n}^{(2)}(x,a) &amp;=&amp;\sum_{j=1}^{\lfloor n/2\rfloor } \binom{n-j}{j}
(-a)^{j}x^{n-2j},\text{\ and \ \textsc{D}}_{0}^{(2)}(x,a)=1.
\end{eqnarray*}
\end{definition}
The first few Dickson polynomials of the first and of the second kind of
degree $n$ with parameter $a$ are respectively:
\begin{equation*}
D_{0}(x)=2,\ D_{1}(x)=x,\ D_{2}(x)=x^{2}-2a,\ D_{3}(x)=x^{3}-3xa,
\end{equation*}%
\begin{equation*}
D_{4}(x)=x^{4}-4x^{2}a+2a^{2},\ D_{5}(x)=x^{5}-5x^{3}a+5xa^{2},
\end{equation*}%
\begin{equation*}
\text{\textsc{D}}_{0}^{(2)}(x)=1,\ \text{\textsc{D}}_{1}^{(2)}(x)=x,\
\text{\textsc{D}}_{2}^{(2)}(x)=x^{2}-a,\ \text{\textsc{D}}_{3}^{(2)}(x)=x^{3}-2xa,
\end{equation*}%
\begin{equation*}
\text{\textsc{D}}_{4}^{(2)}(x)=x^{4}-3x^{2}a+a^{2},\ \text{\textsc{D}}
_{5}^{(2)}(x)=x^{5}-4x^{3}a+3xa^{2}.
\end{equation*}
Dickson polynomials can also be defined recursively and the
definition extended to all $\mathbb{Z}$ as follow:
\begin{eqnarray*}
D_{0}(x,a) &amp;=&amp;2,\ D_{1}(x,a)=x,\ D_{n+1}(x,a)=xD_{n}(x,a)-aD_{n-1}(x,a) \\
\text{\textsc{D}}_{0}^{(2)}(x,a) &amp;=&amp;1,\ \text{\textsc{D}}_{1}^{(2)}(x,a)=x,\
\text{\textsc{D}}_{n+1}^{(2)}(x,a)=x\text{\textsc{D}}_{n}^{(2)}(x,a)-a\text{\textsc{D}}_{n-1}^{(2)}(x,a)
\end{eqnarray*}
Dickson polynomials are closely related to some well-known polynomials like
the first and the second \emph{Chebychev polynomials} $T_{n}(x)$ and $U_{n}(x)$ defined as follows:
\begin{eqnarray*}
T_{n}(x) &amp;=&amp;
\begin{cases}
1, &amp;\text{if\ }n=0; \\
x, &amp;\text{if\ }n=1; \\
2xT_{n-1}(x)+T_{n-2}(x), &amp; \text{if\ }n\geq 2;
\end{cases}
\quad \text{and}\ T_{-n}(x)=(-1)^{n}T_{n}(x), \\
U_{n}(x) &amp;=&amp;
\begin{cases}
1, &amp; \text{if\ }n=0; \\
2x,  &amp; \text{if\ }n=1; \\
2xU_{n-1}(x)+U_{n-2}(x), &amp; \text{if\ }n\geq 2;
\end{cases}
\quad \text{and\ }U_{-n}(x)=(-1)^{n}U_{n}(x).
\end{eqnarray*}
Dickson polynomials are also related to \emph{Lucas polynomials }$L_{n}(x)$
and the \emph{Fibonacci polynomials }$F_{n}(x)$, see for
instance Piero ~\cite{Piero}. These are defined as follows:
\begin{definition}
\label{Lucas-Fibonacci} For $n\in \mathbb{N}$,
\begin{eqnarray*}
L_{n}(x) &amp;=&amp;\sum_{j=0}^{\lfloor k/2\rfloor }\frac{k}{k-j}\binom{k-j}{j} x^j
 \\
F_{n+1}(x) &amp;=&amp;\sum_{j=1}^{\lfloor k/2\rfloor } \binom{k-j}{j} x^j
\end{eqnarray*}
\end{definition}
Like the Dickson polynomials, these two classes of polynomials can also be
defined recursively in $\mathbb{Z}$ as follow:
\begin{eqnarray*}
L_{n}(x) &amp;=&amp;%
\begin{cases}
2, &amp;  \text{if\ }n=0; \\
x, &amp; \text{if\ }n=1; \\
xL_{n-1}(x)+L_{n-2}(x), &amp; \text{if\ }n\geq 2;%
\end{cases}%
\quad \text{and\ }L_{-n}(x)=(-1)^{n}L_{n}(x),\\
F_{n}(x) &amp;=&amp;
\begin{cases}
0, &amp; \text{if\ }n=0; \\
1, &amp; \text{if\ }n=1; \\
xF_{n-1}(x)+F_{n-2}(x), &amp; \text{if\ }n\geq 2;
\end{cases}
\quad \text{and\ }F_{-n}(x)=(-1)^{n-1}F_{n}(x).
\end{eqnarray*}
The first few Lucas and Fibonacci polynomials are
\begin{equation*}
L_{0}(x)=2,\ L_{1}(x)=x,\ L_{2}(x)=x^{2}+2,\ L_{3}(x)=x^{3}+3x,
\end{equation*}
\begin{equation*}
L_{4}(x)=x^{4}+4x^{2}+2,\ L_{5}(x)=x^{5}+5x^{3}+5x.
\end{equation*}
\begin{equation*}
F_{0}(x)=0,\ F_{1}(x)=1,\ F_{2}(x)=x,\ F_{3}(x)=x^{2}+1,
\end{equation*}
\begin{equation*}
F_{4}(x)=x^{3}+2x,\ F_{5}(x)=x^{4}+3x^{2}+1.
\end{equation*}
Note that $\bigl( L_n(1) \bigr)_{n=0}^{\infty}$ is the Lucas sequence $(L_n)_{n=0}^{\infty}$
and that $\bigl( F_n(1) \bigr)_{n=0}^{\infty}$ is the
Fibonacci sequence $(F_n)_{n=0}^{\infty}$. It is important for the current
purposes to note that Dickson polynomials $D_n (x,a)$ and Lucas polynomials $L_n (x)$
are even functions for even $n$ and odd functions for odd $n$, whereas
Fibonacci polynomials $F_n (x)$ are even functions for odd $n$ and odd
functions for even $n$. For the Lucas and Fibonacci polynomials, it is
well-known that for $m, n\in \mathbb{Z}$,
\begin{eqnarray}
L_{m+n}(x) &amp;=&amp;L_{m}(x)L_{n}(x)-(-1)^{n}L_{m-n}(x),  \label{lucas-prop1} \\
F_{m+n}(x) &amp;=&amp;F_{m}(x)F_{n}(x)+F_{m-1}(x)F_{n-1}(x).  \label{fibonacci-prop1}
\end{eqnarray}
In particular, if $m,n\in 2\mathbb{N}-1$, then the relation (\ref{fibonacci-prop1}) implies
\begin{equation}
F_{m+n+1}(x)-F_{m-n+1}(x)=F_{m+1}(x)\left( F_{n+1}(x)+F_{n-1}(x)\right) .
\label{fibonacci-prop 2}
\end{equation}
Similar properties can be easily proved by induction for the Dickson
polynomials. More precisely, for $n\in \mathbb{Z}$, we have
\begin{eqnarray}
D_{m+n}(x,a) &amp;=&amp;D_{m}(x,a)D_{n}(x,a)-aD_{m-n}(x,a),  \label{Dickson1-prop1}
\\
\text{\textsc{D}}_{m+n}^{(2)}(x,a) &amp;=&amp;\text{\textsc{D}}_{m}^{(2)}(x,a)\text{%
\textsc{D}}_{n}^{(2)}(x,a)-a\text{\textsc{D}}_{m-1}^{(2)^{2}}(x,a)\text{%
\textsc{D}}_{n-1}^{(2)^{2}}(x,a),  \label{Dickson2-prop1}
\end{eqnarray}%
which imply
\begin{eqnarray}
D_{2n}(x,a) &amp;=&amp;D_{n}^{2}(x,a)-2.  \label{Dickson1-prop2} \\
\text{\textsc{D}}_{2n}^{(2)}(x,a) &amp;=&amp;\text{\textsc{D}}_{n}^{(2)}(x,a)-a\text{%
\textsc{D}}_{n-1}^{(2)^{2}}(x,a).  \label{Dickson2-prop2}
\end{eqnarray}
Further for any $n\in \mathbb{Z}$, we have:
\begin{equation}
D_{n}(2xa,a^{2})=2a^{n}T_{n}(x)\ \text{and}\ \text{\textsc{D}}%
_{n}^{(2)}(2xa;a^{2})=a^{n}U_{n}(x),  \label{1}
\end{equation}
in particular
\begin{equation}
D_{n}(x;1)=2T_{n}(x/2)\ \text{and}\ \text{\textsc{D}}
_{n}^{(2)}(x;1)=U_{n}(x/2).  \label{2}
\end{equation}
Many other interesting properties about Dickson polynomials can be
found in Lidl, Mullen, and Turnwald~ \cite{Lidl-Mullen-Turnwald}.
\section{An identity involving Bernoulli polynomials}
\begin{theorem}
\label{bernoulli-application1}  If $n\in 2 \mathbb{N}$, then
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\left( \sum_{l=0}^{2b-1} (a-b +l)^{n-1} +
\frac{(a+b)^{n-1} - (a-b)^{n-1}}{2} \right)
\end{equation*}
\begin{equation*}
= \left( \sum_{l=0}^{n-1} \binom{n}{l} B_l n^{n-l-1} +\frac{1}{2} n^n \right)
\frac{\phi(n)}{n}.
\end{equation*}
\end{theorem}
\begin{proof}
Let $n\in 2\mathbb{N}$ and let $f:\mathbb{Z}\rightarrow \mathbb{C}$ be
defined by:
\begin{equation*}
f(k)=B_{n}(k)-nB_{1}k^{n-1}=B_{n}(k)+\frac{n}{2}k^{n-1}=
\sum_{l=0}^{\frac{n}{2}}\binom{n}{2l}B_{2l}k^{n-2l}.
\end{equation*}
Then clearly $f$ is an even function and thus by Theorem~\ref{main1} we find
\begin{equation}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\left( \bigl(B_{n}(a+b)-B_{n}(a-b)\bigr)+
\frac{n}{2}(a+b)^{n-1}-\frac{n}{2}(a-b)^{n-1}\right)
\label{bernoulli-help1}
\end{equation}
\begin{equation*}
=\bigl(B_{n}(n)+\frac{n}{2}n^{n-1}-B_{n}(0)\bigr)\phi (n)=\bigl(B_{n}(n)+%
\frac{n^{n}}{2}-B_{n}\bigr)\phi (n).
\end{equation*}%
By the relations (\ref{bernoulli-numbers}) we have
\begin{equation*}
B_{n}(n)-B_{n}=\sum_{l=0}^{n}\binom{n}{l}B_{l} n^{n-l}-B_{n}=n\sum_{l=0}^{n-1}\binom{n}{l}B_{l} n^{n-l-1}
\end{equation*}
and by the relation (\ref{bernoulli-property}) we have
\begin{equation*}
B_{n}(a+b)-B_{n}(a-b)=B_{n}(a-b+2b)-B_{n}(a-b)=%
n \sum_{l=0}^{2b-1}(a-b+l)^{n-1}.
\end{equation*}
Now put in (\ref{bernoulli-help1}) and divide by $n$ to obtain the desired
identity.
\end{proof}
\begin{remark}
If $n\in 2\mathbb{N}$, then by Theorem~\ref{main1} applied to the
function $f(k)=k^{n}$ we have
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl((a+b)^{n}-(a-b)^{n}\bigr)=n^{n}\phi(n)\quad (n\in 2\mathbb{N}).
\end{equation*}
On the other hand, if $n\in 2\mathbb{N}$, we do not know an evaluation for the sum
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl((a+b)^{n-1}-(a-b)^{n-1}\bigr)
\end{equation*}
which appears in the left hand side of the formula in Theorem~\ref{bernoulli-application1}.
\end{remark}
\section{An identity for Euler polynomials}
\begin{theorem}
\label{euler-application}  If $n \in 2\mathbb{N}$, then
\begin{equation*}
\sum_{(a,b,u,v) \in O^{\prime }(n)} \left(\sum_{l=0}^{2b-1} (a-b +\frac{1}{2}
+l)^n - E_n(a-b+\frac{1}{2} +l) \right)
\end{equation*}
\begin{equation*}
=  \phi(n) \sum_{l=0}^{n-1} \bigl( (\frac{1}{2} +l)^n -E_n(\frac{1}{2} + l) \bigr).
\end{equation*}
\end{theorem}
\begin{proof}
Let $n\in 2\mathbb{N}$ and let $f:\mathbb{Z}\to \mathbb{C}$ be defined as
follows:
\begin{equation*}
f(k) = E_n (k + \frac{1}{2} ) = \sum_{l=0}^{\frac{n}{2}} \binom{n}{2l} \frac{E_{2l}}{4^l} k^{n-2l}.
\end{equation*}
Then clearly $f$ is an even function and so by Theorem~ \ref{main1} we get
\begin{equation}  \label{euler-help1}
\sum_{(a,b,u,v)\in O^{\prime }(n)} \left( E_n (a+b +\frac{1}{2} ) - E_n (a-b + \frac{1}{2} ) \right)
\end{equation}
\begin{equation*}
= \left(E_n(n+\frac{1}{2}) - E_n(\frac{1}{2}) \right) \phi(n).
\end{equation*}
By the property (\ref{Euler-property}) we have
\begin{multline*}
E_n (a+b + 1/2) - E_n (a-b+1/2) = E_n(a-b +1/2 + 2b) - E_n(a-b +1/2) \\
= 2 \left(\sum_{l=0}^{2b-1} (a-b + 1/2 +l)^n - E_n(a-b+1/2 +l) \right)
\end{multline*}
and
\begin{equation*}
E_n(n+ 1/2) - E_n(1/2) = 2 \sum_{l=0}^{n-1} \bigl( (1/2 +l)^n -E_n(1/2 + l) \bigr),
\end{equation*}
which put in (\ref{euler-help1}) yield the desired formula.
\end{proof}
\section{Identities for Lucas polynomials}
\begin{theorem}
\label{lucas-application1} If $n\in 2 \mathbb{N}$ and $x\in \mathbb{R}$,
then
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)} L_{a} (x) L_{b} (x)= \phi(n) (L_{n} (x)-2 ).
\end{equation*}
\end{theorem}
\begin{proof}
Let $x\in \mathbb{R}$ and let the function $f(k)$ be defined as follows:
\begin{equation*}
f(k)=
\begin{cases}
0, &amp;  \text{if\ }k\in 2\mathbb{Z}-1; \\
L_{k}(x), &amp; \text{if\ }k\in 2\mathbb{Z}.
\end{cases}
\end{equation*}
Clearly $f$ is an even function. Then application of Theorem~\ref{main1}
to the function $f$ yields
\begin{equation}  \label{lucas-help1}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\left( L_{a+b}(x)-L_{a-b}(x)\right)
=\left( L_{n}(x)-L_{0}(x)\right) \phi (n)
\end{equation}
\[
= \bigl( L_{n}(x)- 2 \bigr) \phi(n),
\]
which by virtue of the relation (\ref{lucas-prop1}) and the fact that $L_{0}(x)=2$ gives
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\left(
L_{a}(x)L_{b}(x)-(-1)^{b}L_{a-b}(x)-L_{a-b}(x)\right) =\left(
L_{n}(x)-2\right) \phi (n),
\end{equation*}
and the desired formula follows since $(-1)^{b}=-1$.
\end{proof}
Taking $x=1$ in Theorem ~\ref{lucas-application1} we have the following result on Lucas numbers.
\begin{corollary}
\label{lucas:cor1}  If $n\in 2 \mathbb{N}$, then
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)} L_a L_b = \phi(n) (L_n -2).
\end{equation*}
\end{corollary}
\begin{theorem}
\label{lucas-application2} If $n\in 2\mathbb{N}$, then
\begin{multline*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(L_{a+b}(a+b)-L_{a-b}(a-b)\bigr)
=\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(L_{n}(a+b)-L_{n}(a-b)\bigr) \\
=\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(L_{a+b}(n)-L_{a-b}(n)\bigr).
\end{multline*}
\end{theorem}
\begin{proof}
Note first that by formula~(\ref{lucas-help1}) the rightmost sum equals
\begin{equation*}
\bigl( L_n (n) - 2 \bigr) \phi(n).
\end{equation*}
Next let $f(k)=L_{k}(-k)$ and $g(k)=L_{n}(k)$. Then it is easily verified
that both $f$ and $g$ are even functions. Thus by Theorem~\ref{main1}
applied to $f$ and $g$ we find respectively
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(L_{a+b}(-a-b)-L_{a-b}(b-a)\bigr)=%
\bigl(L_{n}(n)-L_{0}(0)\bigr)\phi (n)
\end{equation*}
and
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(L_{n}(a+b)-L_{n}(a-b)\bigr)=\bigl(
L_{n}(n)-L_{n}(0)\bigr)\phi (n).
\end{equation*}
Now combine the previous two relations with the facts that $L_{n}(0)=L_{0}(0)=2$, $L_{a+b}(-a-b)=L_{a+b}(a+b)$,
and $L_{a-b}(b-a)=L_{a-b}(a-b)$ to deduce the equality of the three sums.
\end{proof}
\section{Identities for Fibonacci polynomials}
\begin{theorem}
\label{fibonacci-application1} Let $n\in 2 \mathbb{N}$ and $x\in \mathbb{R}$. Then
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)} \bigl(F_{a+1}(x) \left( F_{b+1}(x) +
F_{b-1}(x) \right) \bigr) = \phi(n)(F_{n+1}(x) - 1).
\end{equation*}
\end{theorem}
\begin{proof}
Note first that if $f$ is even, then
\begin{multline}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\left( f(a+b)-f(a-b)\right)  \label{help1}
\\
=\left( f(2)-f(0)\right) \phi (n)+2\sum_{\substack{ (a,b,u,v)\in O^{\prime}(n)  \\ a&gt;b}}\left( f(a+b)-f(a-b)\right) .
\end{multline}%
Let $f$ be the function defined by:
\begin{equation*}
f(k)=
\begin{cases}
0, &amp; \text{if\ }k\in 2\mathbb{Z}-1; \\
F_{k+1}(x), &amp; \text{if\ }0\leq k\in 2\mathbb{Z}; \\
F_{k-1}(x), &amp; \text{otherwise.}
\end{cases}
\end{equation*}
It is easily verified that $f$ is an even function. Then by Theorem~\ref{main1} we obtain
\begin{equation}  \label{fibonacci-help1}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\left( F_{a+b+1}(x) - F_{a-b+1}(x) \right)
\end{equation}
\[
= \phi(n) \bigl( F_{n+1}(x) - F_1 (x) \bigr) = \phi(n) \bigl( F_{n+1}(x) - 1
\bigr).
\]
By virtue of the relation (\ref{help1}) the previous formula is equivalent
to
\begin{equation*}
\left( F_{3}(x)-F_{1}(x)\right) \phi (n)+2\sum_{\substack{ (a,b,u,v)\in
O^{\prime }(n)  \\ a&gt;b}}\left( F_{a+b+1}(x)-F_{a-b+1}(x)\right)
\end{equation*}
\begin{equation*}
 =\left(
F_{n+1}(x)-F_{1}(x)\right) \phi (n).
\end{equation*}
That is,
\begin{equation*}
\sum_{\substack{ (a,b,u,v)\in O^{\prime }(n)  \\ a&gt;b}}\left(
F_{a+b+1}(x)-F_{a-b+1}(x)\right) =\frac{F_{n+1}(x)-F_{3}(x)}{2}\phi (n),
\end{equation*}
which by identity (\ref{fibonacci-prop1}) and the fact that $F_{3}(x)=x^{2}+1 $ gives
\begin{equation*}
\sum_{\substack{ (a,b,u,v)\in O^{\prime }(n)  \\ a&gt;b}}\bigl(
F_{a+1}(x)(F_{b+1}(x)+F_{b-1}(x))\bigr)=\frac{F_{n+1}(x)-x^{2}-1}{2}\phi (n).
\end{equation*}
Now use the previous formula together with the facts that $F_{0}(x)=0$, $F_{2}(x)=x$, and
\begin{multline*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\left(
F_{a+1}(x)(F_{b+1}(x)+F_{b-1}(x))\right) =F_{2}(x)\left(
F_{2}(x)+F_{0}(x)\right) \phi (n) \\
+2\sum_{\substack{ (a,b,u,v)\in O^{\prime }(n)  \\ a&gt;b}}\left(
F_{a+1}(x)(F_{b+1}(x)+F_{b-1}(x))\right)
\end{multline*}
to derive the result.
\end{proof}
Taking $x=1$ in Theorem ~\ref{fibonacci-application1} we have the following identity involving Fibonacci numbers.
\begin{corollary}
\label{fibonacci:cor1}  If $n \in 2 \mathbb{N}$, then
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)} \left( F_{a+1} \bigl( F_{b+1} + F_{b-1} \bigr) \right) = \phi(n) (F_{n+1} -1).
\end{equation*}
\end{corollary}
\begin{theorem}
\label{fibonacci-application2} Let $n\in 2\mathbb{N}$. Then we have
\begin{equation*}
(a)\quad \sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(F_{a+b}(a+b)-F_{a-b}(a-b)
\bigr)=F_{n}(n)\phi (n).
\end{equation*}
\begin{equation*}
(b)\quad \sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(F_{n+1}(a+b)-F_{n+1}(a-b) \bigr)
 = \sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(F_{a+b+1}(n)-F_{a-b+1}(n) \bigr).
\end{equation*}
\end{theorem}
\begin{proof}
To prove part (a) apply Theorem~\ref{main1} to the even functions $f(k)=F_{k}(k)$.
As to part (b) note first that by the relation~ (\ref{fibonacci-help1}) the sum on the right equals
\begin{equation*}
\bigl( F_{n+1} (n) - 1 \bigr) \phi(n).
\end{equation*}
Next by Theorem~\ref{main1} applied to the even function $g(k)=F_{n+1}(k)$
we have
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(F_{n+1}(a+b)-F_{n+1}(a-b) \bigr)
=  \bigl(F_{n+1}(n)- F_{n+1}(0) \bigr)\phi (n)  =  \bigl(F_{n+1}(n)- 1 \bigr)\phi (n).
\end{equation*}
This completes the proof.
\end{proof}
\section{Identities involving Dickson polynomials}
To simplify the notation in this section, we write $D_{n}(x)$ for $D_{n}(x,1)$.
\begin{theorem}
\label{Dickson-application1} Let $n\in \mathbb{N}$ and $x\in \mathbb{R}$.
Then
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(D_{a}(x)D_{b}(x)\bigr)\bigl(
D_{a}(x)D_{b}(x)-2D_{a-b}(x)\bigr)=\phi (n)\bigl(D_{2n}(x)-2\bigl).
\end{equation*}
\end{theorem}
\begin{proof}
Clearly,
\begin{equation*}
D_{2n}(x)=D_{n}^{2}(x)-2
\end{equation*}
is an even function. Therefore if we define $f$ on $\mathbb{Z}$ by
\begin{equation*}
f(k)=
\begin{cases}
0, &amp; \text{if\ }k\leq 0; \\
D_{2k}(x), &amp; \text{if\ }k&gt;0;
\end{cases}
\end{equation*}
then $f$ is also even. By Theorem~\ref{main1} we have
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}f(a+b)-f(a-b)=(f(n)-f(0))\phi (n),
\end{equation*}
or equivalently,
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\left( D_{2(a+b)}(x)-D_{2(a-b)}(x)\right)
=(D_{2n}(x)-D_{0}(x))\phi (n),
\end{equation*}
which by identity (\ref{Dickson1-prop2}) and the fact that $D_{0}(x)=2$
gives
\begin{eqnarray*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(D_{a+b}^{2}(x)-D_{a-b}^{2}(x)\bigr)
&amp;=&amp;(D_{2n}(x)-2)\phi (n). \\
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(D_{a+b}(x)+D_{a-b}(x)\bigr)\bigl(
D_{a+1}(x)-D_{a-b}(x)\bigr) &amp;=&amp;(D_{2n}(x)-2)\phi (n).
\end{eqnarray*}
Now application of the formula~ (\ref{Dickson1-prop1}) yields
\begin{equation*}
\bigl(D_{a+b}(x)+D_{a-b}(x)\bigr)=D_{a}(x)D_{b}(x),
\end{equation*}
and
\[
\bigl(
D_{a+b}(x)-D_{a-b}(x)\bigr)=D_{a}(x)D_{b}(x)-2D_{a-b}(x).
\]
That is,
\begin{eqnarray*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(D_{a+b}(x)+D_{a-b}(x)\bigr)\bigl(
D_{a+1}(x)-D_{a-b}(x)\bigr) &amp;=&amp;(D_{2n}(x)-2)\phi (n)\text{ \ or} \\
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(D_{a}(x)D_{b}(x)\bigr)\bigl(
D_{a}(x)D_{b}(x)-2D_{a-b}(x)\bigl) &amp;=&amp;\phi (n)\bigl(D_{2n}(x)-2\bigr),
\end{eqnarray*}
which completes the proof.
\end{proof}
\begin{theorem}
\label{Dickson-application2} Let $n\in 2\mathbb{N}$ and $x\in \mathbb{R}$.
Then
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(D_{a}(x)D_{b}(x)-2D_{a-b}(x)\bigr)
\bigl(D_{a}(x)D_{b}(x)-1\bigr))=\phi (n)(D_{2n}(x)-D_{n}(x)).
\end{equation*}
\end{theorem}
\begin{proof}
Using the same argument as in the previous theorem with
\begin{equation*}
g(k)=
\begin{cases}
0, &amp; \text{if\ }k\in 2\mathbb{Z}-1; \\
D_{k}(x), &amp; \text{if\ }k\in 2\mathbb{Z};
\end{cases}
\end{equation*}
we obtain:
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\left( D_{(a+b)}(x)-D_{(a-b)}(x)\right)
=(D_{n}(x)-D_{0}(x))\phi (n),
\end{equation*}
or equivalently \
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(D_{a}(x)D_{b}(x)-2D_{a-b}(x)\bigr)
=\phi (n)\bigl(D_{n}(x)-2\bigr).
\end{equation*}
Combining with the result of Theorem~ \ref{Dickson-application1} we obtain:
\begin{multline*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(D_{a}(x)D_{b}(x)\bigr)\bigl(
D_{a}(x)D_{b}(x)-2D_{a-b}(x)\bigr) \\
+\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(
D_{a}(x)D_{b}(x)-2D_{a-b}(x)\bigr)
\end{multline*}
\[
=\phi (n)\bigl(D_{2n}(x)-2\bigr)+\phi (n)\bigl(D_{n}(x)-2\bigr).
\]
or equivalently
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(D_{a}(x)D_{b}(x)-2D_{a-b}(x)\bigr)
\bigl(D_{a}(x)D_{b}(x)-1\bigr))=\phi (n)(D_{2n}(x)-D_{0}(x)).
\end{equation*}
\end{proof}
As a corollary, we have
\begin{corollary}
\label{Chebychev1-application1} Let $n\in 2\mathbb{N}$ and $x\in \mathbb{R}$. Then
\begin{equation*}
\sum_{(a,b,u,v)\in O^{\prime }(n)}\bigl(T_{a}(x)T_{b}(x)-2T_{a-b}(x)\bigr)
\bigl(T_{a}(x)T_{b}(x)-1\bigr))=\frac{1}{2}\phi (n)(T_{2n}(x)-T_{n}(x)).
\end{equation*}
\end{corollary}
\begin{proof}
The proof is immediate from the previous theorem and relation (\ref{2}).
\end{proof}
\begin{remark}
Similar identities can be proven for the Dickson polynomials of the second
kind.
\end{remark}
\section{Concluding remarks}
It should be noticed that in deducing our formulas involving the classical polynomials, we restricted ourselves
to sums over the set $O'(n)$ as the formulas we obtain when considering sums over the sets $B(n)$, $O(n)$, and $B'(n)$
are not easy to simplify. However, under some restrictions on the positive integer $n$ one can get to nice identities. 
For our example in this matter we need the following particular case of a sum
of Liouville~\cite{Liouville2} (Problem~13 of Chapter~10 in Williams~\cite{Williams}). 
If $N$ is twice an odd positive integer, then
\begin{equation}\label{twice-odd}
\sum_{(a,b,x,y)\in O(N)} ( f(a+b)-f(a-b) ) = -f(0) \sigma(N/2) + \sum_{d\mid N/2} d f(2d).
\end{equation} 
Let $N$ be twice an odd positive integer.
Using relation (\ref{twice-odd}) one can show that the analogue of Theorem~\ref{bernoulli-application1} is
\begin{equation*}
\sum_{(a,b,u,v)\in O(N)}\left( \sum_{l=0}^{2b-1} (a-b +l)^{N-1} +
\frac{(a+b)^{N-1} - (a-b)^{N-1}}{2} \right)
\end{equation*}
\begin{equation*}
= 2^{N-2} \sigma_{N}(N/2) - \frac{B_N}{N} \sigma(N/2) + \frac{1}{N} \sum_{d\mid N/2} d B_N(2d).
\end{equation*}
\section{Acknowledgment} The authors are grateful to the referee
 for valuable comments and interesting suggestions.

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\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B39; Secondary 11B68, 11B83, 11A25, 11B75.

\noindent \emph{Keywords: } 
Liouville sum, Bernoulli polynomial,
Chebychev polynomial, Dickson polynomial, Euler polynomial,
Fibonacci polynomial.

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received June 3 2012;
revised version received  July 25 2012.
Published in {\it Journal of Integer Sequences}, August 5 2012.

\bigskip
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\noindent
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\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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