Fibonacci divisibilty properties $ F_n\mid F_{kn},\,$ $\, \gcd(F_n,F_m) = F_{\gcd(n,m)}$

Most 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 $\rm\,F_n = F_{n-m+1}\,F_m + F_{n-m}\,F_{m-1}\!\equiv F_{n-m}\,F_{m-1}\pmod{F_m}.\,$ 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 $\rm\,(f_{\,k},f_m)=f_{\,(k,\,m)}\!=\,f_1 = 1.\,$ Thus $\rm\ (\color{#c00}{f_n},\,f_m)\, =\, (\color{#c00}{f_{\,k}\,f_{n-m}},\,f_m)\, =\, (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\,(a,b) = (a',\,b)\ \ if\ \ a\equiv a'\pmod{b}\ $ and $\rm\,(c\,a,b) = (a,b)\,$ if $\rm\,(c,b) = 1.\quad$ QED

You may find it insightful to simultaneously examine other strong divisibility sequences, e.g. see my post here 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 $$\rm\displaystyle\ 3\ =\ (15,21)\ \ \leadsto\ \ \frac{x^3-1}{x-1}\ =\ (x^{15} + x^9 + 1)\ \frac{x^{15}-1}{x-1}\ -\ (x^9+x^3)\ \frac{x^{21}-1}{x-1}$$

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.

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.