How to prove that the Fibonacci sequence $7\mid U_m\Longrightarrow 8\mid m$ and $4\mid U_m\Longrightarrow 6\mid m$

Question

How to prove that the Fibonacci sequence $$7\mid U_m\Longrightarrow 8\mid m$$ and $$4\mid U_m\Longrightarrow 6\mid m$$I was confused because there $\{ 4,7 \}$ in Fibonacci sequece

You have asked several questions on divisibility of Fibonacci numbers already, like here and here and here. What have you learnt? — Calvin Lin, Aug 23 '13 at 06:17
Please ask a new question instead of editing the current one into something completely different. — L. F., Sep 09 '13 at 18:11

score 2 · Answer 1 · edited Apr 13 '17 at 12:21

2

HINT:

Using this, $$(F_m,F_n)=F_{(m,n)}\implies F_m|F_n\iff m|n$$

Now, the smallest Fibonacci number that is divisible by $7$ is $F_8=21$

edited Apr 13 '17 at 12:21

Community

1

answered Aug 23 '13 at 06:17

lab bhattacharjee

274,582

help? details? please. – benjamin_ee Sep 09 '13 at 18:20
@marcelolpjunior, have you understood the link? So, if integer $r$ divides $F_n,$ it must divide $F_m$ as the notations used in my answer – lab bhattacharjee Sep 09 '13 at 18:24

score 1 · Answer 2 · answered Aug 23 '13 at 05:19

1

First you show that the Fibonacci sequence $a_n \mod 7$ is periodic and since $a_8 = 21$ is divisible by 7 you get the result.

answered Aug 23 '13 at 05:19

Uwe Stroinski

705

How to prove that the Fibonacci sequence $7\mid U_m\Longrightarrow 8\mid m$ and $4\mid U_m\Longrightarrow 6\mid m$

2 Answers2