Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

Show:

$\Longrightarrow$

In 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 to $5\mid m$ then $u_5\mid u_m\Longrightarrow5\mid u_m\;\;\;\;\Box$$$$$ Correct?

share|improve this question
    
What have you tried for Question 2? –  Ahaan S. Rungta Aug 15 '13 at 17:44
    
@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. –  marcelolpjunior 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. –  marcelolpjunior 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

1 Answer 1

(partial Answer)
2.1
$u_8 = 21$, so $8|m \Leftrightarrow 21|u_m \Leftrightarrow 3|u_m \wedge 7|u_m$
now if $3|u_m \Leftrightarrow 4|m$ and thus
$$8|m \Leftrightarrow 4|m \wedge 7|u_m$$
is all you can get by the requirements.

2.2
$$6|m \Leftrightarrow 3|m \wedge 2|m \Leftrightarrow 2|u_m \wedge 1|u_m \Leftrightarrow 2|u_m$$ Same here...

I'm open to any imprvements / suggestions

share|improve this answer
    
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. –  marcelolpjunior 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" –  marcelolpjunior Aug 15 '13 at 18:15
    
From $7\mid u_m$ and reach the $8\mid m$ –  marcelolpjunior 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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.