By the trial and error method I have observed the following identity by taking some numerical values. Those are  


$F_m$|$L_n$ is valid only if one of the following holds.
a) $m = 1$ or $m =2$
b) $m = 3$ or $3|n$
c) $n$ is congruent to $2\pmod 4$ and $m = 4$.  


The other identity is:


a) $F_{m+n}$ = $F_{m-1}$ $F_n$ + $F_m$ $F_{n+1}$
b) Either ($L_m$, $F_m$) = 1 or 2.


Unfortunately, I could not get any proofs for the above stated identities. But, numerically and by trial and error methods, the above stated identities are very correct. I am looking for a proof(s) of the above identities...  

edited The sum of any ten consecutive Fibonacci numbers is always evenly divisible by 11.

You have

\begin{align*}
I &=\int\limits_{-a}^{a}\frac{f(x)}{1+e^{x}} \ dx \qquad\qquad \cdots (1)\\\ I &= \int\limits_{-a}^{a} \frac{f(x)}{1+e^{-x}} \ dx \qquad\qquad \Bigl[ \small\because \int\limits_{a}^{b}f(x)\ dx = \int\limits_{a}^{b}f(a+b-x)\ dx \ \Bigr] \quad \cdots (2) \\\ \Longrightarrow 2I &= \int\limits_{-a}^{a} \biggl[ \frac{f(x)}{1+e^{x}} + \frac{e^{x}\cdot f(x)}{1+e^{x}} \biggr] \ dx  \quad\qquad \cdots (1) + (2)\\\  &=\int\limits_{-a}^{a} f(x) \ dx = 2 \int\limits_{0}^{a} f(x) \ dx \qquad \Bigl[ \small  \text{since}\ f \  \text{is even so} \ \int\limits_{-a}^{a} f(x) = 2\int\limits_{0}^{a} f(x) \Bigr]
\end{align*}



$\textbf{Note.}$ A similar problem, which uses result $(2)$ can be found here:


Integration of a trigonometric function

