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:


*

*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.

 A: As other posters have already indicated, for every positive integer $N$, there is some $D(N)$ such that every $D$-th Fibonacci is divisible by $N$. 
The next logical question to my mind is how to compute $D(N)$. Observe that, if $a$ and $b$ are relatively prime, then $D(ab)=LCM(D(a), D(b))$ (thanks to hardmath for the correction). In other words, $D$ is determined by its values for prime powers. I'll talk just about computing $D(p)$ for $p$ a prime.
Recall the formula
$$F_n = \frac{1}{\sqrt{5}} \left( \tau^n - (-\tau^{-1})^n \right)$$
where $\tau = (1+\sqrt{5})/2$.
Suppose that the prime $p$ is $\pm 1 \bmod 5$. Then there is a square root of $5$ in $\mathbb{Z}/p$. The above formula is still valid in terms of that square root. For example, if $p=11$, then the square roots of $5$ modulo $11$ are $4$ and $7$. We have $(1+4)/2 \equiv 8 \mod 11$ and $(1+7)/2 \equiv 4 \mod 11$ and, sure enough, $F_n = (1/4) \left( 8^n - 4^n \right) \mod 11$. 
So $p$ divides $F_n$ if and only if $\tau^n = (- \tau^{-1})^n$. In other words, we have to compute the order of $- \tau^2$ in the unit group of $\mathbb{Z}/p$. (In the above example, $- \tau^2 \equiv - 4^2 \equiv -64 \equiv 2 \mod 11$, so the conclusion is that $11$ divides $F_n$ if and only if $2^n \equiv 1 \mod 11$.)
By Lagrange's theorem, we see that $D(p)$ will divide $p-1$ for $p$ which is $\pm 1 \bmod 5$. 
I can say more, but this is really an excellent project for a beginning number theorist to play with for his or herself. What can you say about primes which are $\pm 2 \bmod 5$? What can you say about prime powers? For $p \equiv \pm 1 \mod 5$, when does $D(p)$ divide $(p-1)/2$? There isn't a complete formula here, but there are lots of great things to observe.
A: Many of the divisibility properties of Fibonacci numbers follow from the fact that they comprise a divisibility sequence, i.e. $\rm\,m\,|\,n\ \Rightarrow\ F_m\,|\,F_n.\,$ All of your statements above are special cases of this, e.g. $\rm\,F_{15} = 610,\,$ so $\rm\,15\,|\,n\ \Rightarrow\ F_{15}\,|\,F_n\,\Rightarrow\,610\,|\,F_n,\,$ which is precisely your statement $11,\,$ that $10$ and $61$ divide every $15\,$'th Fibonacci number.
In fact $\rm\,F_n\,$ is strong divisibility sequence, i.e. $\rm\,(F_m,F_n) = F_{(m,n)},\,$ i.e. $\rm\,gcd(F_m,F_n) = F_{\gcd(m,n)}.\,$ This stronger property specializes to the above property if $\rm\,m\mid n\,\ (\!\!\iff \gcd(m,n) = m\,\!).\,$ The proof is not  difficult. Here is a straightforward way to proceed. Recall the Fibonacci addition law $\rm\,F_{n+m} =F_{n+1}\,F_m + F_n\,F_{m-1}.\,$ Applying the shift $\rm\,n\to n-m\ $ this addition law becomes $\rm\,\color{#c00}{F_n}  = F_{n-m+1}\,F_m  + F_{n-m}\,F_{m-1}\!\color{#c00}{\equiv F_{n-m}\,F_{m-1}}\pmod{\!F_m}.\,$ So for $\rm\,k=m-1\,$ we may invoke the Theorem below to conclude that $\rm\,f_n  = F_n\,$ is a strong divisibility sequence.
Theorem $ $  Let $\rm\ f_n\, $ be an integer sequence such that $\rm\ f_{\:\!0} =\, 0,\ f_1 = 1\ $ and such that for all $\rm\,n > m\,$ holds  $\rm\ \, \color{#c00}{f_n\equiv\, f_{\,k}\ f_{n-m}}\,\ (mod\ f_m)\ $ for some $\rm\,k < n,\ (k,m)\, =\, 1.\, $  Then $\rm\ (f_n,f_m) = f_{\,(n,\,m)} $
Proof $\ $ By induction on  $\rm\,n + m.\,$ The theorem is trivially true if $\rm\ n = m\ $ or $\rm\ n = 0\ $ or $\rm\, m = 0.\,$ Assume wlog $\rm\,n > m > 0.\,$  Since $\rm\,k\!+\!m < n\!+\!m,\,$ by induction $\rm\,(f_{\,k},f_m)=f_{\,(k,\,m)}\!=\,f_1 = 1.\,$  So $\rm\ (\color{#c00}{f_n},\,f_m)\overset{\color{#90f}R}=\ (\color{#c00}{f_{\,k}\,f_{n-m}},\,f_m)\, \!\overset{\color{#0a0}E}= (f_{n-m},\,f_m)\, =\, f_{\,(n-m,\,m)} =\, f_{\,(n,\,m)} $ follows by induction  (which applies here since $\rm\,(n-m)+m\,  <\, n+m\,\!),\,$ and by employing  well-known gcd laws, namely $\rm\,\color{#90f}R\!:\ (\color{#c00}a,b) = (\color{#c00}a',\,b)\ \ if\ \ \color{#c00}{a\equiv a'}\pmod{b}\ $ and $\rm\,\color{#0a0}E\!:\ (\color{c00}c\:\!a,b) = (a,b)\,$ if $\rm\,(c,b) = 1.\ \ $ QED
You may find it insightful to simultaneously examine other strong divisibility sequences, e.g. see my this post on $\rm\,f_n = (x^n-1)/(x-1).\,$ In this case $\rm\, \gcd(f_m,f_n) = f_{\,\gcd(m,n)}\,$ may be interpreted as a $\rm\,q$-analog of the integer Bezout identity, for example
$$\begin{align} \rm\displaystyle\ \color{#c00}3 = (\color{#0a0}{15},\color{#90f}{21})\,\  \leadsto\,\ f_{\:\!\large\color{#c00}3\,} &\rm =\, a\, f_{\:\!\large\color{#0a0}{15}}+b\, f_{\:\!\large\color{#90f}{21}},\,\ \text{explicitly:}\\[.3em] 
\rm \frac{x^{\large \color{#c00}3}-1}{x-1} \,&\rm =\, (x^{{15}} + x^9 + 1)\, \frac{x^{\large\color{#0a0}{15}}-1}{x-1} - (x^9+x^3)\, \frac{x^{\large\color{#90f}{21}}-1}{x-1} \end{align}\qquad$$
Generally see this answer for $\rm\, f_n = (x^n-y^n)/(x-y)$.
A: 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.  
A: I guess that the standard way to understand all these divisibility results in one single swoop is to observe that the Fibonacci sequence modulo any number N becomes periodic.
For instance, Fibonacci modulo 2 is 0, 1, 1, 0, 1, 1, 0, ...... proving the even-ness of $F_n$ for $n=0,3,6,9,...$.
Fibonacci modulo $3$ is 0, 1, 1, 2, 0, 2, 2, 1, 0, 1, 1, 2, 0, ..... making obvious that $3$ divides $F_n$ for $n=0, 4, 8, 12, ...$ .
Try yourself the next ones!
NOTE: the same technique can be applied to any linear recursive sequence with constant coefficients.
A: I found that @David's answer here is very clear to understand, but he omitted some minor details. I reproduce @David's answer with full details. All credits are given to @David.


Lemmas:
$f_{s+t}=f_{s-1}f_t+f_sf_{t+1}$ (I presented a proof for this lemma here)
$s\mid t\implies f_s\mid f_t$ (One user presented a proof here)

Let $g=\gcd(m,n)$ and $d$ be any common divisor of $f_m,f_n$. From the definition of Greatest Common Divisor, it's clear that $$\gcd(f_m, f_n) = f_{\gcd(n, m)}=f_g\iff\begin{cases}
f_g\mid f_m\text{ and } f_g\mid f_n\\d\space\space\mid f_g\end{cases}$$


*

*$f_g\mid f_m\text{ and } f_g\mid f_n$


From the second lemma and the fact that $g=\gcd(m,n)$, we have $$g\mid m\Rightarrow f_g\mid f_m\quad\hbox{and}\quad
  g\mid n\Rightarrow f_g\mid f_n$$.


*$d\mid f_g$


From Bézout's identity, there exists $x,y\in\mathbb Z$ such that $g=mx+ny$.
$f_m\mid f_{mx}$ and $d\mid f_m\implies d\mid f_{mx}$. Similarly, $d\mid f_{ny}$. Thus $d\mid f_{mx-1}f_{ny}+f_{mx}f_{ny+1}$, and from the second lemma $d\mid f_{mx+ny}$, or equivalently $d\mid f_g$.
