GCD of Fibonacci Numbers

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

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

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 $m \le n$.

Let $h$ be $\gcd \left\{{m, n}\right\}$.

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 Integers Divided by GCD are Coprime.

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

Let $u$ and $v$ be integers such that $F_{h a} = u F_h$ and $F_{h b} = v F_h$, whose existence is proved by Divisibility of Fibonacci Number.

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

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

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

This can be done recurrently to produce the result, in a fashion similar to the Euclidean Algorithm.

Since $a$ and $b$ are co-prime, the result would be $\gcd \left\{{F_h, F_h}\right\}$.

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