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

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



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


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

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@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
@AhaanRungta And in the case of 7? – marcelolpjunior Aug 15 '13 at 17:51

(partial Answer)
$u_8 = 21$, 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.

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

I'm open to any imprvements / suggestions

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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

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