In references [1] and [6], Zhang and Ma studied the properties of the Chebyshef polynomials, and obtained some identities involving the Fibonacci numbers and the Lucas numbers. The main results in [1] were the following two identities:

$ \begin{eqnarray} x^{2n}&=& \frac{(2n)!}{4^n(n!)^2}T_0(x)+ \frac{2(2n)!}{4^n}\sum\limits_{k=1}^{n}\frac{1}{(n-k)!(n+k)!}T_{2k}(x)\nonumber\\ &=&\frac{(2n)!}{4^n}\sum\limits_{k=0}^{n}\frac{2k+1}{(n-k)!(n+k+1)!}U_{2k}(x)\end{eqnarray} $

(1.1)

and

$ \begin{eqnarray} x^{2n+1}&=& \frac{(2n+1)!}{4^n}\sum\limits_{k=0}^{n}\frac{1}{(n-k)!(n+k+1)!}T_{2k+1}(x)\nonumber\\ & =&\frac{(2n+1)!}{4^n}\sum\limits_{k=0}^{n}\frac{k+1}{(n-k)!(n+k+2)!}U_{2k+1}(x), \end{eqnarray} $

(1.2)

where $T_n(x)$ is the Chebyshef polynomials of the first kind, and $U_n(x)$ is the Chebyshef polynomials of the second kind, defined respectively by

$ T_n(x)=\frac{1}{2}\left[\left(x+ \sqrt{x^2-1}\right)^n+ \left(x- \sqrt{x^2-1}\right)^n\right] $

and

$ U_n(x)=\frac{1}{2\sqrt{x^2-1}}\left[\left(x+ \sqrt{x^2-1}\right)^{n+1}-\left(x- \sqrt{x^2-1}\right)^{n+1}\right]. $

On the other hand, in a private communication with Cooper, Melham suggested that it would be interesting to discover an explicit expansion for

$\begin{eqnarray*} L_1L_2\cdots L_{2m+1}\sum\limits_{k=1}^nF_{2k}^{2m+1}\end{eqnarray*} $

as a polynomial in $F_{2n+1}$, where $F_n$ and $L_n$ denote the Fibonacci number and Lucas number respectively. Melham [5] also proposed following two conjectures:

Conjecture 1.1  Let $m\geq 1$ be a positive integer. Then the sum

$\begin{eqnarray*} L_1L_3L_5\cdots L_{2m+1}\sum\limits_{k=1}^nF_{2k}^{2m+1}\end{eqnarray*} $

(1.3)

can be expressed as $\left(F_{2n+1}-1\right)^2 P_{2m-1}\left(F_{2n+1}\right)$, where $P_{2m-1}(x)$ is a polynomial of degree $2m-1$ with integer coefficients.

Conjecture 1.2  Let $m\geq 0$ be an integer. Then the sum

$\begin{eqnarray*} L_1L_3L_5\cdots L_{2m+1}\sum\limits_{k=1}^nL_{2k}^{2m+1}\end{eqnarray*} $

(1.4)

can be expressed as $\left(L_{2n+1}-1\right) Q_{2m}\left(L_{2n+1}\right)$, where $Q_{2m}(x)$ is a polynomial of degree $2m$ with integer coefficients.

Wiemann and Cooper [2] obtained some divisibility properties in the study of some conjectures of Melham related to the sum $\sum_{k=1}^nF_{2k}^{2m+1}$. Ozeki [3] proved that

$ \begin{eqnarray*} \sum\limits_{k=1}^nF_{2k}^{2m+1}= \frac{1}{5^m}\sum\limits_{j=0}^{m}\frac{(-1)^j}{L_{2m+1-2j}}\left(^{2m+1}_{\ \ j}\right)\left(F_{(2m+1-2j)(2n+1)}- F_{2m+1-2j}\right). \end{eqnarray*} $

Prodinger [4] studied the more general summation $\sum_{k=0}^nF_{2k+\delta}^{2m+1+\epsilon}$, where $\delta, \ \epsilon \in \{0, \ 1\}$, and obtained many interesting identities, two of which are:

$ \begin{eqnarray*} \sum\limits_{k=0}^nF_{2k+1}^{2m+1}=\sum\limits_{l=0}^{m}F_{2(n+1)}^{2l+1}\sum\limits_{j=0}^{m-l} \frac{5^{l-m}}{L_{2m-2j+1}}\left(^{2m+1}_{\ \ j}\right)\left(^{m-j+l}_{m-j-l}\right)\frac{2m-2j+1}{2l+1} \end{eqnarray*} $

and

$ \begin{eqnarray*} \sum\limits_{k=0}^nF_{2k}^{2m}=\frac{F_{2n+1}}{5^m}\sum\limits_{r=0}^{m-1}L_{2n+1}^{2r+1}\sum\limits_{j=0}^{m-r-1} \left(^{2m}_{ \ j}\right)\left(^{\ m-j+r}_{m-j-r-1}\right)\frac{(-1)^j}{F_{2m-2j}}+ \frac{(-1)^m}{5^m}\left(^{2m}_{\ m}\right)\left(n+\frac{1}{2}\right). \end{eqnarray*} $

For the sum of powers of Lucas numbers, Prodinger [4] also obtained similar conclusions:

$\begin{eqnarray*} \sum\limits_{k=0}^nL_{2k}^{2m+1}=\sum\limits_{r=0}^{m}L_{2n+1}^{2r+1}\sum\limits_{j=0}^{m-r} \left(^{2m+1}_{\ \ j}\right)\left(^{m-j+r}_{m-j-r}\right)\frac{2m+1-2j}{2r+1}\frac{1}{L_{2m+1-2j}}+ 4^m \end{eqnarray*} $

and

$ \begin{eqnarray*} \sum\limits_{k=0}^nL_{2k+1}^{2m+1}&=&\sum\limits_{r=0}^{m}L_{2(n+1)}^{2r+1}\sum\limits_{j=0}^{m-r} \left(^{2m+1}_{ \ \ j}\right)\left(^{m-j+r}_{m-j-r}\right)\frac{2m+1-2j}{2r+1}\frac{(-1)^{m-r}}{L_{2m+1-2j}}\\ && - \sum\limits_{j=0}^{m}\left(^{2m+1}_{ \ \ j}\right)(-1)^j\frac{2}{L_{2m+1-2j}}. \end{eqnarray*} $

The main purpose of this paper is using the combination method to solve Conjecture 1.2 completely, and prove the following:

Theorem 1.3  Let $m\geq 0$ be an integer. Then the sum

$ \begin{eqnarray*} L_1L_3L_5\cdots L_{2m+1}\sum\limits_{k=1}^nL_{2k}^{2m+1} \end{eqnarray*} $

can be expressed as $\left(L_{2n+1}-1\right) Q_{2m}\left(L_{2n+1}\right)$, where $Q_{2m}(x)$ is a polynomial of degree $2m$ with integer coefficients.

In this section, we shall give several simple lemmas which are necessary in the proof of our theorem. First we have the following:

Lemma 2.1  For any nonnegative integers $m$ and $n$, we have

$ \begin{eqnarray*} \sum\limits_{k=1}^mT_{2k}^{2n}(x)= m\frac{(2n)!}{4^n(n!)^2}+\frac{1}{4^n}\sum\limits_{r=1}^{n}\left(^{\ 2n}_{n-r}\right)\frac{U_{4rm+2r-1}(x)-U_{2r-1}(x) }{U_{2r-1}(x)} \end{eqnarray*} $

and

$ \begin{eqnarray*} \sum\limits_{k=0}^mT_{2k+1}^{2n}(x)=(m+1)\frac{(2n)!}{4^n(n!)^2}+ \frac{1}{4^n}\sum\limits_{r=1}^{n}\left(^{\ 2n}_{n-r}\right)\frac{U_{4r(m+1)-1}(x)}{U_{2r-1}(x)}. \end{eqnarray*} $

Proof  First we give a simple proof for (1.1) and (1.2). In fact, for any positive integer $n$ and real number $x\neq 0$, by using the familiar binomial expansion

$\begin{eqnarray*} \left(x+\frac{1}{x}\right)^{n}= \sum\limits_{r=0}^n \left(^n_r\right)x^{n-2r}, \end{eqnarray*} $

we get

$\begin{eqnarray*} \left(x+\frac{1}{x}\right)^{2n}= \frac{(2n)!}{(n!)^2} + \sum\limits_{r=1}^{n} \left(^{\ 2n}_{n-r}\right)\left(x^{2r} +\frac{1}{x^{2r}}\right)\end{eqnarray*} $

(2.1)

and

$\begin{eqnarray*} \left(x+\frac{1}{x}\right)^{2n+1}=\sum\limits_{r=0}^{n}\left(^{ 2n+1}_{\ n-r}\right)\left(x^{2r+1}+ \frac{1}{x^{2r+1}}\right).\end{eqnarray*} $

(2.2)

From the properties of the Chebyshef polynomials we know that for any integer $n$,

$ T_n\left(\frac{1}{2}\left(x+ \frac{1}{x}\right)\right)= \frac{1}{2}\left(x^{n}+\frac{1}{x^n}\right). $

(2.3)

Now substituting $\displaystyle\frac{1}{2}\left(x+\frac{1}{x}\right)$ by $y$ in (2.1) and (2.2), and combining (2.3) we may immediately deduce the identities

$\begin{eqnarray*} y^{2n}= \frac{(2n)!}{4^n(n!)^2}T_0(y)+ \frac{2}{4^n}\sum\limits_{r=1}^{n}\left(^{\ 2n}_{n-r}\right)T_{2r}(y)\end{eqnarray*} $

and

$\begin{eqnarray*} y^{2n+1}= \frac{1}{4^n}\sum\limits_{r=0}^{n}\left(^{ 2n+1}_{\ n-r}\right)T_{2r+1}(y).\end{eqnarray*} $

This proves formulas (1.1) and (1.2).

Let $x=\left(y +\sqrt{y^2-1}\right)^k$ be in (2.1) and (2.2), and hence $\displaystyle\frac{1}{x}= \left(y-\sqrt{y^2-1}\right)^k$, $T_n(T_m(y))= T_{mn}(y)$. Then we have

$ \begin{eqnarray*} T_{k}^{2n}(y)= \frac{(2n)!}{4^n(n!)^2}+ \frac{2}{4^n}\sum\limits_{r=1}^{n}\left(^{\ 2n}_{n-r}\right)T_{2rk}(y)\end{eqnarray*} $

(2.4)

and

$\begin{eqnarray*} T_{k}^{2n+1}(y)= \frac{1}{4^n}\sum\limits_{r=0}^{n}\left(^{ 2n+1}_{\ n-r}\right)T_{k(2r+1)}(y).\end{eqnarray*} $

(2.5)

Now we prove Lemma 2.1. Let $\alpha=x+ \sqrt{x^2-1} $, $\beta=x- \sqrt{x^2-1}$. Then from (2.4) and note that $\alpha \cdot \beta=1$ we may get

$ \begin{eqnarray*} &&\sum\limits_{k=1}^m T_{2k}^{2n}(x)= m\frac{(2n)!}{4^n(n!)^2}+ \frac{2}{4^n}\sum\limits_{r=1}^{n}\left(^{\ 2n}_{n-r}\right)\sum\limits_{k=1}^mT_{4rk}(x)\nonumber\\ &=&m\frac{(2n)!}{4^n(n!)^2}+ \frac{1}{4^n}\sum\limits_{r=1}^{n}\left(^{\ 2n}_{n-r}\right)\sum\limits_{k=1}^m\left(\alpha^{4rk}+ \beta^{4rk}\right)\nonumber\\ &=& m\frac{(2n)!}{4^n(n!)^2}+ \frac{1}{4^n}\sum\limits_{r=1}^{n}\left(^{\ 2n}_{n-r}\right)\left(\frac{\alpha^{4r}(\alpha^{4rm}-1)}{\alpha^{4r}-1}+ \frac{\beta^{4r}(\beta^{4rm}-1)}{\beta^{4r}-1}\right)\nonumber\\ &=&m\frac{(2n)!}{4^n(n!)^2}+ \frac{1}{4^n}\sum\limits_{r=1}^{n}\left(^{\ 2n}_{n-r}\right)\frac{\alpha^{4r}+\beta^{4r}+ \alpha^{4rm}-\alpha^{4r(m+1)}+ \beta^{4rm}-\beta^{4r(m+1)}-2 }{2-\alpha^{4r}-\beta^{4r}}\nonumber\\ &=&m\frac{(2n)!}{4^n(n!)^2}+\frac{1}{4^n} \sum\limits_{r=1}^{n}\left(^{\ 2n}_{n-r}\right)\frac{\left(\alpha^{4rm+2r}-\beta^{4rm+2r}\right)\left(\alpha^{2r}-\beta^{2r}\right)-\left(\alpha^{2r}-\beta^{2r}\right)^2 }{\left(\alpha^{2r}-\beta^{2r}\right)^2} \nonumber\\ &=&m\frac{(2n)!}{4^n(n!)^2}+\frac{1}{4^n} \sum\limits_{r=1}^{n}\left(^{\ 2n}_{n-r}\right)\frac{U_{4rm+2r-1}(x) - U_{2r-1}(x)}{U_{2r-1}(x)}. \end{eqnarray*} $

This proves the first formula of Lemma 2.1. Similarly, we can deduce the second one.

Lemma 2.2  For any nonnegative integers $m$ and $n$, we have

$ \begin{eqnarray*} \sum\limits_{k=1}^mT_{2k}^{2n+1}(x)=\frac{1}{2}\sum\limits_{r=0}^{n}\frac{\left(^{ 2n+1}_{\ n-r}\right)}{4^n} \frac{U_{4mr+2r+2m}(x)- U_{2r}(x)}{U_{2r}(x) } \end{eqnarray*} $

and

$ \begin{eqnarray*} \sum\limits_{k=0}^mT_{2k+1}^{2n+1}(x)=\frac{1}{2\cdot 4^n}\sum\limits_{r=0}^{n}\left(^{ 2n+1}_{\ n-r}\right) \frac{U_{4mr+4r+2m+1}(x)}{U_{2r}(x) }. \end{eqnarray*} $

Proof  From (2.5) we have

$ \begin{eqnarray*} &&\sum\limits_{k=1}^m T_{2k}^{2n+1}(x)= \frac{1}{4^n}\sum\limits_{r=0}^{n}\left(^{ 2n+1}_{\ n-r}\right)\sum\limits_{k=1}^m T_{2k(2r+1)}(x)\nonumber\\ &=&\sum\limits_{r=0}^{n}\frac{\left(^{ 2n+1}_{\ n-r}\right)}{4^n} \sum\limits_{k=1}^m\frac{1}{2}\left(\alpha^{2k(2r+1)} + \beta^{2k(2r+1)}\right)\nonumber\\ &=&\frac{1}{2}\sum\limits_{r=0}^{n}\frac{\left(^{ 2n+1}_{\ n-r}\right)}{4^n} \left(\frac{\alpha^{2(2r+1)}(\alpha^{2m(2r+1)}-1)}{\alpha^{2(2r+1)}-1} + \frac{\beta^{2(2r+1)}(\beta^{2m(2r+1)}-1)}{\beta^{2(2r+1)}-1}\right)\nonumber\\ &=&\frac{1}{2}\sum\limits_{r=0}^{n}\frac{\left(^{ 2n+1}_{\ n-r}\right)}{4^n} \left( \frac{\alpha^{(2m+1)(2r+1)}- \beta^{(2m+1)(2r+1)} -\alpha^{2r+1} + \beta^{2r+1}}{\alpha^{2r+1}-\beta^{2r+1} }\right)\nonumber\\ &=&\frac{1}{2}\sum\limits_{r=0}^{n}\frac{\left(^{ 2n+1}_{\ n-r}\right)}{4^n} \frac{U_{4mr+2r+2m}(x)- U_{2r}(x)}{U_{2r}(x) }, \end{eqnarray*} $

where we have used the identity $\alpha \cdot \beta=1$. This proves the first formula of Lemma 2.2. Similarly, we can deduce the second formula of Lemma 2.2.

Lemma 2.3  For any positive integers $m$ and $n$, we have

$ \begin{eqnarray*} \sum\limits_{r=1}^nL_{2r}^{2m+1}=\sum\limits_{k=0}^{m}\left(^{\ 2m+1}_{\ m-k}\right)\frac{L_{(2n+1)(2k+1)}-L_{2k+1} }{L_{2k+1}}. \end{eqnarray*} $

Proof  Note that $\displaystyle T_{2k}\left(\frac{\sqrt{5}}{2}\right)= \frac{1}{2}L_{2k}$ and $\displaystyle U_{2r}\left(\frac{\sqrt{5}}{2}\right)= L_{2r+1}$, the Lucas number. From the first formula of Lemma 2.2 we may immediately deduce that

$ \begin{eqnarray*} \sum\limits_{r=1}^nL_{2r}^{2m+1}=\sum\limits_{k=0}^{m}\left(^{\ 2m+1}_{\ m-k}\right)\frac{L_{(2n+1)(2k+1)}-L_{2k+1} }{L_{2k+1}}. \end{eqnarray*} $

This proves Lemma 2.3.

Now we use Lemma 2.3 to prove Theorem 1.3. For any integer $k\geq 0$, applying mathematical induction we can prove that $L_{2n+1}-1$ divide $L_{(2n+1)(2k+1)} - L_{2k+1}$. In fact note that $L_1 = 1$, so $\left(L_{(2n+1)}-1\right)| \left(L_{(2n+1)(2k+1)} - L_{2k+1}\right)$ holds for $k=0$. If $k=1$, then note that $L^3_{2n+1}= L_{3(2n+1)}- 3L_{2n+1}$ (This identity can be deduced directly from the Binet formula) and $L_3=4$, we have

$ L_{3(2n+1)}- L_3= L^3_{2n+1}+ 3L_{2n+1}-4= \left(L_{2n+1}-1\right)\left(L^2_{2n+1} + L_{2n+1} +4\right) . $

So that $\left(L_{(2n+1)}-1\right)$ divide $\left(L_{3(2n+1)} - L_{3}\right)$. Suppose that

$ \left(L_{(2n+1)}-1\right)| \left(L_{(2n+1)(2k+1)} - L_{2k+1}\right) $

for all integers $k\leq m$. Then for $k = m+1$, note that

$ L_{2(2n+1)} L_{(2n+1)(2m+1)}= L_{(2n+1)(2m+3)}+ L_{(2n+1)(2m-1)}, $

we have

$ \begin{eqnarray} &&L_{(2n+1)(2m+3)} - L_{2m+3}= L_{2(2n+1)} L_{(2n+1)(2m+1)}- L_{(2n+1)(2m-1)}-L_{2m+3}\nonumber\\ &=&\left(L_{2n+1}^2+2\right) L_{(2n+1)(2m+1)}- \left(L_{(2n+1)(2m-1)}- L_{2m-1}\right)-3L_{2m}-3L_{2m-1}\nonumber\\ &=&\left(L_{2n+1}^2-1\right) L_{(2n+1)(2m+1)}+3\left( L_{(2n+1)(2m+1)}-L_{2m+1}\right)\nonumber\\ &&- \left(L_{(2n+1)(2m-1)}- L_{2m-1}\right).\end{eqnarray} $

(3.1)

Now our conclusion follows from (3.1) and the inductive hypothesis.

On the other hand, note that

$ \begin{eqnarray}L_{(2n+1)(2m+3)}& =& L_{2(2n+1)}L_{(2n+1)(2m+1)}- L_{(2n+1)(2m-1)}\nonumber\\ &=&(L_{2n+1}^2+2) L_{(2n+1)(2m+1)}-L_{(2n+1)(2m-1)}, \end{eqnarray} $

(3.2)

so by mathematical induction we can also prove that

$ L_{(2n+1)(2m+1)}- L_{2m+1}=f_{2m+1}\left(L_{2n+1}\right), $

where $f_{2m+1}(x)$ is a polynomial of degree $2m+1$ with integer coefficients. In fact if $m=0$, then $L_{(2n+1)(2m+1)}- L_{2m+1}=L_{(2n+1)}-L_1= f_{1}\left(L_{2n+1}\right)$, where $f_{1}(x)$ is a polynomial of degree $1$ with integer coefficients. Suppose that $L_{(2n+1)(2k+1)}- L_{2k+1}=f_{2k+1}\left(L_{2n+1}\right)$ for all integers $k\leq m$. Then for $k=m+1$, note that (3.2), we have

$ \begin{eqnarray*}&&L_{(2n+1)(2m+3)}- L_{2m+3} =(L_{2n+1}^2+2) L_{(2n+1)(2m+1)}- L_{2m+3}-L_{(2n+1)(2m-1)}\\ & =&(L_{2n+1}^2+2)\left(f_{2m+1}(L_{2n+1})+ L_{2m+1}\right)- L_{2m+3}-L_{(2n+1)(2m-1)} =f_{2m+3}(L_{2n+1}), \end{eqnarray*} $

where $f_{2m+3}(x)$ is a polynomial of degree $2m+3$ with integer coefficients. This proves $L_{(2n+1)(2m+1)}- L_{2m+1}=f_{2m+1}\left(L_{2n+1}\right)$ for all integers $m\geq 0$.

Combining Lemma 2.3, $\left(L_{2n+1}-1\right)$ divide $\left(L_{(2n+1)(2k+1)} - L_{2k+1}\right)=f_{2k+1}\left(L_{2n+1}\right) $ and $\left(L_{2n+1}-1, \ L_{2n+1}\right)=1$, we may immediately deduce that

$ \begin{eqnarray*} && L_1L_3L_5\cdots L_{2m+1}\sum\limits_{k=1}^nL_{2k}^{2m+1}\\ &=& L_1L_3L_5\cdots L_{2m+1}\left(\sum\limits_{k=0}^{m}\left(^{\ 2m+1}_{\ m-k}\right)\frac{L_{(2n+1)(2k+1)}-L_{2k+1} }{L_{2k+1}} \right)\\ &=& \left(L_{2n+1}-1\right) Q_{2m}\left(L_{2n+1}\right), \end{eqnarray*} $

where $Q_{2m}(x)$ is a polynomial of degree $2m$ with integer coefficients. This completes the proof of our theorem.