# Fibonacci and Lucas identity

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.

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For a proof of $F_{m+n} = F_{m-1}F_n + F_m F_{n+1}$, see this question. –  Zev Chonoles Aug 27 '11 at 7:05
$L_m=F_{m-1}+F_{m+1}=2F_{m-1}+F_m$ $\Rightarrow$ $(L_m,F_m)=(2F_{m-1}+F_m,F_m)=(2F_{m-1},F_m) \mid 2(F_{m-1},F_m)=2$. See math.stackexchange.com/questions/24378/… –  Martin Sleziak Aug 27 '11 at 7:28
Could you give a shortest proof for a) b) and c) of first question. –  Gandhi Aug 27 '11 at 8:47
Thank you so much for present proof. can you provide a shortest proof for a, b and c of first question? –  Gandhi Aug 27 '11 at 8:49

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 $0$.

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I believe the pings (notifications) work only in comments. (At least I was not pinged by your answer.) meta.stackexchange.com/questions/43019/… –  Martin Sleziak Aug 27 '11 at 12:16
thank you very much sir. –  Gandhi Aug 28 '11 at 12:56

The result that no Fibonacci number greater than 3 is Lucasian (i.e., a divisor of a Lucas number) can be found in the book Mathematical vistas: from a room with many windows By Peter John Hilton, Derek Allan Holton, Jean Pedersen as Theorem 8 on page 56.

The reference for this result given in this book is the paper Hilton, Pedersen, Somer: On Lucasian Numbers.

This fact is also mentioned in the and of the paper Marjorie Bicknell and Verner E. Hoggatt, Jr.: A primer for the Fibonacci Numbers IX - To Prove: $F_n$ Divides $F_{nk}$, but no reference is given there.

The cases $F_0=F_1=1$ are trivial (every number is divisible by 1). The cases $F_2=2$ and $F_3=3$ can be solved easily by noticing the periodicity of Lucas sequences modulo 2 and 3. Namely, we can notice that residues modulo 2 in the Lucas sequence are
0,1,0,1,0,1,...
hence even terms are precisely the ones, that are divisible by 2. Similarly, if we look at the Lucas sequence modulo 3, we get
2,1,0,1,1,2,0,2,2,1,...
and we see that 2,1,0,1,1,2,0,2 is a period of length 8.

Similar approach was used in these questions:
Fibonacci sequence divisible by 7?
$f_n$ is divisible by $4$ if and only if $n$ is divisible by $6$

Although the question about sum of 10 consecutive Fibonacci numbers was answered by Andre, I'm adding a different possible approach:

$F_n+F_{n+1}+F_{n+2}+F_{n+3}+F_{n+4}+F_{n+5}+F_{n+6}+F_{n+7}+F_{n+8}+F_{n+9}=$
$F_n+F_{n+1}+F_{n+2}+F_{n+3}+F_{n+4}+F_{n+5}+F_{n+6}+2F_{n+7}+2F_{n+8}=$
$F_n+F_{n+1}+F_{n+2}+F_{n+3}+F_{n+4}+F_{n+5}+3F_{n+6}+4F_{n+7}=$
$F_n+F_{n+1}+F_{n+2}+F_{n+3}+F_{n+4}+5F_{n+5}+7F_{n+6}=$
$F_n+F_{n+1}+F_{n+2}+F_{n+3}+8F_{n+4}+12F_{n+5}=$
$F_n+F_{n+1}+F_{n+2}+2F_{n+3}+9F_{n+4}+11F_{n+5}=$
$F_n+F_{n+1}+10F_{n+2}+11F_{n+3}+11F_{n+5}=$
$11F_n+11F_{n+1}+11F_{n+3}+11F_{n+5}=$
$11(F_n+F_{n+1}+F_{n+3}+F_{n+5})$

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Can your provide, shortest proof for a, b and c of first question. Also, Thank you very much for providing some good references. –  Gandhi Aug 27 '11 at 8:48
Thank you. I got it. –  Gandhi Aug 27 '11 at 9:12
Thank you so much and I am extremely heartfelt thanks for your different approach about 11 divisibility. –  Gandhi Aug 27 '11 at 17:03