Fibonacci divisibilty properties $ F_n\mid F_{kn},\,$ $\, \gcd(F_n,F_m) = F_{\gcd(n,m)}$ Can any one give a generalization of the following properties in a single proof? I have checked the results, which I have given below by trial and error method. I am looking for a general proof, which will cover the all my results below:


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*Every third Fibonacci number is even.

*3 divides every 4th Fibonacci number.

*5 divides every 5th Fibonacci number.

*4 divides every 6th Fibonacci number.

*13 divides every 7th Fibonacci number.

*7 divides every 8th Fibonacci number.

*17 divides every 9th Fibonacci number.

*11 divides every 10th Fibonacci number.

*6, 9, 12 and 16 divides every 12th Fibonacci number.

*29 divides every 14th Fibonacci number.

*10 and 61 divides every 15th Fibonacci number.

*15 divides every 20th Fibonacci number.


The general proof of this is that the fibonacci numbers arise from the expression 
$$F_n \sqrt{-5} = (\frac 12\sqrt{-5}+\frac 12\sqrt{-1})^n - (\frac 12\sqrt{-5}-\frac 12\sqrt{-1})^n$$
Since this is an example of the general $a^n-b^n$, which $a^m-b^m$ divides, if $m \mid n$, it follows that there is a unique number, generally coprime with the rest, for each number.  Some of the smaller ones will be $1$.
The exception is that if $f_n$ is this unique factor, such that $F_n = \prod_{m \mid n} f_n$, then $f_n$ and $f_{np^x}$ share a common divisor $p$, if $p$ divides either.  So for example, $f_8=7$, and $f_{56}=7*14503$, share a common divisor of $7$.  This means that modulo over $49$ must evidently work too.  So $f_{12} = 6$, shares a common divisor with both $f_4=3$ and $f_3 = 4$, is unique in connecting to two different primes.
Gauss's law of quadratic recriprocality applies to the fibonacci numbers, but it's a little more complex than for regular bases.  Relative to the fibonacci series, reduce modulo 20, to 'upper' vs 'lower' and 'long' vs 'short'.  For this section, 2 is as 7, and 5 as 1, modulo 20.  
Primes that reduce to 3, 7, 13 and 17 are 'upper' primes, which means that their period divides $p+1$.  Primes ending in 1, 9, 11, 19 are lower primes, meaning that their periods divide $p-1$.  
The primes in 1, 9, 13, 17 are 'short', which means that the period divides the maximum allowed, an even number of times.  For 3, 7, 11, 19, it divides the period an odd number of times.  This means that all odd fibonacci numbers can be expressed as the sum of two squares, such as $233 = 8^2 + 13^2$, or generally $F_{2n+1} = F^2_n + F^2_{n+1}$
So a prime like $107$, which reduces to $7$, would have an indicated period dividing $108$ an odd number of times.  Its actual period is $36$.  A prime like $109$ divides $108$ an even number of times, so its period is a divisor of $54$.  Its actual period is $27$.  
A prime like $113$ is indicated to be upper and short, which means that it divides $114$ an even number of times.  It actually has a period of $19$.
Artin's constant applies here as well. This means that these rules correctly find some 3/4 of all of the periods exactly.   The next prime in this progression, $127$, actually has the indicated period for an upper long: 128.  So does $131$ (lower long), $137$ (upper short, at 69).  Likewise $101$ (lower short) and $103$ (upper long) show the maximum periods indicated.
No prime under $20*120^4$ exists, where if $p$ divides some $F_n$, so does $p^2$.  This does not preclude the existance of such a number.  
