GCD of Fibonacci Numbers

Theorem
Let $F_k$ be the $k$th Fibonacci number.

Then:
 * $\displaystyle \forall m,n > 2 : \gcd \left\{{F_m, F_n}\right\} = F_{\gcd \left\{{m, n}\right\}}$

where $\gcd \left\{{a, b}\right\}$ denotes the greatest common divisor of $a$ and $b$.

Proof
From the initial definition of Fibonacci numbers, we have:
 * $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$

Let $m \le n$. Otherwise, swap $m$ and $n$ because it is symmetrical.

Let $h$ be $\gcd \left\{{m,n}\right\}$, then let $a$ and $b$ be integers such that $m = h a$ and $n = h \left(a+b\right)$.

$a$ and $(a+b)$ are coprime by Divide by GCD for Coprime Integers.

Therefore, $a$ and $b$ are coprime by Integer Combination of Coprime Integers.

Let $u$ and $v$ be integers such that $F_{ha} = u F_h$ and $F_{hb} = v F_h$, whose existence is proved by Divisibility of Fibonacci Numbers.

We have that $F_{ha}$ and $F_{ha-1}$ are coprime by Consecutive Fibonacci Numbers are Coprime.

Therefore, $u$ and $F_{ha-1}$ are coprime by Divisor of One of Coprime Numbers is Coprime to Other.

Therefore:
 * $\displaystyle \forall m,n >2 : \gcd \left\{{F_m, F_n}\right\} = \gcd \left\{{F_m, F_{n-m}}\right\}$

This can be done recurrently to produce the result.

Therefore:
 * $\displaystyle \forall m,n > 2 : \gcd \left\{{F_m, F_n}\right\} = F_{\gcd \left\{{m, n}\right\}}$