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

  1. $F_m$|$L_n$ 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:

  1. 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.

  • 1
    For a proof of $F_{m+n} = F_{m-1}F_n + F_m F_{n+1}$, see [this question](http://math.stackexchange.com/questions/11477/showing-that-an-equation-holds-true-with-a-fibonacci-sequence-f-nm-f-n-1).2011-08-27
  • 1
    $L_m=F_{m-1}+F_{m+1}=2F_{m-1}+F_m$ $\Rightarrow$ $(L_m,F_m)=(2F_{m-1}+F_m,F_m)=(2F_{m-1},F_m) \mid 2(F_{m-1},F_m)=2$. See http://math.stackexchange.com/questions/24378/prove-that-two-any-consecutive-terms-of-fibonacci-sequence-are-relatively-prime2011-08-27
  • 0
    Could you give a shortest proof for a) b) and c) of first question.2011-08-27
  • 0
    Thank you so much for present proof. can you provide a shortest proof for a, b and c of first question?2011-08-27

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