Proving that $n|m\implies f_n|f_m$

Question

Question: Let $m,n\in\mathbb{N}$, prove that if $n|m$, $F_n|F_m$.

I've tried to use induction, but I don't really know where to start since there's $2$ numbers: $n$ and $m\ \dots$ I did induction before with just $1$ number, like proving $1+2+\cdots+n=\frac{n(n+1)}{2}$, that's only dealing with one number $n$, but with this I have no clue how to do it.

Can anyone give me a little hint on how to start, I don't want you to do whole problem for me but can I have some hint so I know where to start?

Thanks for the help!

Is $F_n$ the $n^{th}$ Fibonacci number? You should use only $F$ or $f$ if they are the same thing. — Git Gud, May 01 '13 at 11:19
From Lemma# 4 of http://www.fq.math.ca/Scanned/4-3/halton.pdf, $(F_m,F_n)=F_{(m,n)}$ — lab bhattacharjee, May 01 '13 at 11:53

score 7 · Accepted Answer · answered May 01 '13 at 11:19

7

Hint: Try to prove that $f_{n+m}=f_{n-1}\ f_m+f_n\ f_{m+1}$.

Then, let $m=n\times k$ for some $k\in \mathbb{N}$.

Now, prove that $f_m$ is divisible $f_n$ by induction on $k$.

answered May 01 '13 at 11:19

I think I got it, appreciated for help! – Charles G May 01 '13 at 11:38

score 0 · Answer 2 · answered Jan 28 '17 at 17:40

0

You can prove the desired result without induction using Binet's formula, $$F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}$$ where $\varphi$ and $\psi$ are the roots of $x^2 + x + 1$, and the factoring identity $$a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \cdots + ab^{n-2} + b^{n-1}).$$

answered Jan 28 '17 at 17:40

user0

3,247

Does this mean that: $$ψ = \frac{1 - \sqrt{5}}{2} = φ - \frac{\sqrt{20}}{2}$$ – Mr Pie Sep 24 '17 at 04:20
@user477343 You should be able to answer that yourself. – user0 Sep 24 '17 at 15:50

Proving that $n|m\implies f_n|f_m$

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