TITLE: Fibonacci and Lucas identity
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: This answer only addresses the problem (added in an edit) of showing that the sum of any $10$ consecutive Fibonacci numbers is divisible by $11$. The other problems have been addressed by Martin Sleziak.
We take the Fibonacci numbers to be $0$, $1$, $1$, $2$, $3$, $5$, and so on.  Calculate the remainder when the Fibonacci numbers are divisible by $11$.  Note that the remainder when $F_{m+2}$ is divided by a positive integer $d$ is completely determined by the remainders when $F_{m+1}$ and $F_m$ are divided by $d$.
So modulo $11$ we get
$$0,1,1,2,3,5,8,2,10,1,0,1,1,2,3,5,8,2,10,1,0,1,1,2,\dots.$$
There is obvious apparent cycling with period $10$.  How do we know this goes on forever? Once two consecutive entries are the same as two previous consecutive entries, there must be cycling forever, since the Fibonacci sequence reduced with respect to any modulus $d$ satisfies the relation $F_{n+2}\equiv F_{n+1}+F_n \pmod{d}$.  
So the sum of any $10$ consecutive Fibonacci numbers is, modulo $11$, the same as the sum of the first $10$ entries. This sum, modulo $11$, is $0$.