Divisibility of Fibonacci Number

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Theorem

Let $F_k$ denote the $k$th Fibonacci number.


Then:

$\forall m, n \in \Z_{> 2} : m \divides n \iff F_m \divides F_n$

where $\divides$ denotes divisibility.


Corollary

$\forall m, n \in \Z_{> 0}: F_m \divides F_{m n}$

where $\divides$ denotes divisibility.


Proof

From the initial definition of Fibonacci numbers:

$F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$

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

We have:

$m \divides n \iff r = 0$


The proof proceeds by induction on $k$.


For all $k \in \N_{>0}$, let $\map P k$ be the proposition:

$r = 0 \iff F_m \divides F_{k m - r}$


Basis for the Induction

$\map P 1$ is the case:

$r = 0 \iff F_m \divides F_{m - r}$

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 $\map P k$ is true, where $k > 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is our induction hypothesis:

$r = 0 \iff F_m \divides F_{k m - r}$


Then we need to show:

$r = 0 \iff F_m \divides F_{k m + m - r}$


Induction Step

This is our induction step:

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

We have:

$b = 0 \iff F_m \divides F_{k m - r} \iff r = 0$

by the induction hypothesis.

\(\ds F_{k m + m - r}\) \(=\) \(\ds F_{m - 1} F_{k m - r} + F_m F_{k m - r + 1}\) Honsberger's Identity
\(\ds \) \(=\) \(\ds a F_m F_{m - 1} + b F_{m - 1} + F_m F_{k m - r + 1}\)
\(\ds \) \(=\) \(\ds F_m \paren {a F_{m - 1} + F_{k m - r + 1} } + b F_{m - 1}\)

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

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

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

Then:

$\dfrac k b = \dfrac {F_{m - 1} } {F_m}$

We have that $F_{m - 1}$ and $F_m$ are coprime.

Thus by Coprime Numbers form Fraction in Lowest Terms:

$\dfrac {F_{m - 1} } {F_m}$ is in canonical form.

Then by Ratios of Fractions in Lowest Terms

$F_m \divides b$

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

Therefore:

$F_m \divides b F_{m - 1} \iff b = 0$

Therefore:

$F_m \divides F_{k m + m - r} \iff F_m \divides b F_{m - 1} \iff b = 0 \iff r = 0$

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


Therefore:

$\forall m, n > 2 : m \divides n \iff F_m \divides F_n$

$\blacksquare$


Sources

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