Divisibility of Fibonacci Number

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

Then:
 * $\forall m, n \in \Z_{> 2} : m \mathrel \backslash n \iff F_m \mathrel \backslash F_n$

where $\mathrel \backslash$ denotes divisibility.

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

Let $n = k m - r$ where $0 \le r < m$

We have:
 * $m \mathrel \backslash n \iff r = 0$

Proof by induction:

For all $m \in \N_{>0}$, let $P \left({m}\right)$ be the proposition:
 * $r = 0 \iff F_m \mathrel \backslash F_{k m - r}$

Basis for the Induction
$P(1)$ is the case $r=0 \iff F_m \mathrel \backslash F_{m-r}$, which holds because $F_{m-r} < F_m$ unless $r=0$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k > 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $r = 0 \iff F_m \mathrel \backslash F_{k m - r}$

Then we need to show:
 * $r = 0 \iff F_m \mathrel \backslash F_{k m + m - r}$

Induction Step
This is our induction step:

Let $F_{k m - r} = a F_m + b$ where $0 \le b < F_m$.

We have:
 * $b = 0 \iff F_m \mathrel \backslash F_{k m - r} \iff r = 0$

by the induction hypothesis.

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

Let $F_m \mathrel \backslash b F_{m -1}$.

Then there exists an integer $k$ such that $k F_m \mathrel \backslash b F_{m-1}$, by the definition of divisibility.

Then, we have $\dfrac k b = \dfrac {F_{m - 1}} {F_m}$.

Since $F_{m-1}$ and $F_m$ are coprime, $\displaystyle \frac {F_{m-1}} {F_m}$ is in the lowest form, by Coprime Numbers form Fraction in Lowest Terms.

Then, we have $F_m \mathrel \backslash b$, by Ratios of Fractions in Lowest Terms.

Since $0 \le b < F_m$, the only case is when $b = 0$.

Therefore, $F_m \mathrel \backslash b F_{m-1} \iff b=0$.

Therefore, $F_m \mathrel \backslash F_{km+m-r} \iff F_m \mathrel \backslash b F_{m-1} \iff b=0 \iff r=0$.

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall m,n > 2 : m \mathrel \backslash n \iff F_m \mathrel \backslash F_n$