In the Fibonacci sequence, show that $5\mid u_m\Longleftrightarrow 5\mid m$

Question

Knowing in the Fibonacci sequence$$u_n\mid u_m\Longleftrightarrow n\mid m$$

Question 1: In the Fibonacci sequence, show that $$5\mid u_m\Longleftrightarrow 5\mid m$$

Proof:

$\Longrightarrow$

In the Fibonacci sequence $(1,1,2,3,5,8,...)$, $u_5=5$, therefore we have $5\mid u_m\Longrightarrow u_5\mid u_m\Longrightarrow 5\mid m\;\;\;\;\Box$

$\Longleftarrow$

Having that $5\mid m$ then $u_5\mid u_m\Longrightarrow5\mid u_m\;\;\;\;\Box$$$$$ Correct?

@AhaanRungta sequence$$(1,1,2,3,5,8,13,21,34,55,...)$$ soon $u_8=21$ and $7\mid 21$, if $8\mid m\Longrightarrow u_8\mid u_m\Longrightarrow 21\mid u_m$ Also do not know if I can say that $7\mid u_m$ so I need help. — benjamin_ee, Aug 15 '13 at 17:47
Hm, I haven't solved the problem yet, but I'd try it like this. First, assume $ 6 \mid m $. Then, try to show that $ 4 \mid u_m $ using some consecutive term Euclidean Algorithm argument. Then, go for the other direction. Assume $ 4 \mid u_m $. Show that $ 6 \mid m $. — Ahaan S. Rungta, Aug 15 '13 at 17:48
@AhaanRungta And in the case of "$\Longrightarrow$" I do not know how to do, because there is no sequence number that is 7. — benjamin_ee, Aug 15 '13 at 17:49
Oh, hm. I was only looking at the second one: $$ 4 \mid u_m \iff 6 \mid m. $$ For the first one, if I undersatnd your exact question, $ 7 \mid 21 $, so I don't see a problem. — Ahaan S. Rungta, Aug 15 '13 at 17:51
@AhaanRungta The 7's I think I got just use transitivity, right? Show that $21\mid u_m$ and $7\mid 21$ transitivity $7\mid u_m$ — benjamin_ee, Aug 15 '13 at 17:55
@AhaanRungta I mean so ... Answer to the first part:$$$$ $\Longleftarrow$$$$$
Show: $8\mid m\Rightarrow u_8\mid u_m\Rightarrow 21\mid u_m$ as $7\mid 21$ then $7\mid u_m;;;;\Box$ — benjamin_ee, Aug 15 '13 at 17:59
Do you need to use the given fact? Otherwise, working modulo 7, the sequence is $(1, 1, 2, 3, 5, 1, 6, 0, 1, 1, \ldots )$. Hence $7 | u_m \Leftrightarrow 8 | m $. — Calvin Lin, Aug 15 '13 at 18:32

lhf · Answer 1 · 2020-07-02T11:31:59.277

Mod $5$ the Fibonacci sequence is periodic of period $20$: $$ 0,1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1,0,1,1,2,3,0,\dots $$ The zeros occur at $m=0,5,10,15$ and so occur exactly when $m$ is a multiple of $5$.

Actually, $u_m \equiv 2m 3^m \bmod 5$ and then the claim is clear. This is Binet's formula mod $5$. It follows because $x^2-x-1 \equiv (x-3)^2 \bmod 5$ and so $u_m \equiv (am+b)3^m$ for some integers $a$ and $b$, which are found by considering $u_0$ and $u_1$.

score 1 · Answer 2 · answered Aug 15 '13 at 18:06

1

I'm open to any imprvements / suggestions

answered Aug 15 '13 at 18:06

AlexR

24,905

R: I tried $$7\mid u_m\Longleftrightarrow 8\mid m$$ $\Longleftarrow$$$\text{We have};;8\mid m\implies u_8\mid u_m\implies 21\mid u_m;\text{as};7\mid 21\Longrightarrow 7\mid u_m;\Box$$ Propried transitivity. Only, from $7\mid u_m$ I could not solve. – benjamin_ee Aug 15 '13 at 18:12
@marcelolpjunior That's basically what I have written down, as $8|m \Rightarrow 4|m$ – AlexR Aug 15 '13 at 18:14
Yes, of course. But my question is, how to prove "the trip" – benjamin_ee Aug 15 '13 at 18:15
From $7\mid u_m$ and reach the $8\mid m$ – benjamin_ee Aug 15 '13 at 18:16
Consider $$F(n + 1) = \frac{1}{\sqrt{5}} \left( \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right )^n \right )$$ – AlexR Aug 15 '13 at 18:18
@AlexR Consider fixing your $\LaTeX$, please? – Ahaan S. Rungta Aug 15 '13 at 18:19
@AhaanRungta Sure thing, I forgot a \left (was \eft ^^) – AlexR Aug 15 '13 at 18:20
Ah yeah, looks much better now! – Ahaan S. Rungta Aug 15 '13 at 18:21
@AlexR I could not understand. Sorry, do not speak English, use translator. – benjamin_ee Aug 15 '13 at 18:22
@AlexR I'm really sorry, but I honestly do not understand how to do it, how to solve the problem. – benjamin_ee Aug 15 '13 at 19:14
@marcelolpjunior No problem. $F(n+1)$ is an expanded form of $u_{n+1}$, sorry for failing to point that out. – AlexR Aug 15 '13 at 20:24

In the Fibonacci sequence, show that $5\mid u_m\Longleftrightarrow 5\mid m$

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