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October 9th, 2014, 03:43 AM   #1
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Lightbulb A discovery in the Fibonacci sequence

From a very long time, I have been interested in the mysterious properties of this seemingly simple sequence.
While playing with squares I stumbled upon another two properties which thrilled me:

Let F be the set of Fibonacci numbers such that,
$F = \{F_n = F_{n-1} + F_{n-2}; F_0 = 0; F_1 = 1; n ϵ N\}$
then,
(i) $F_{2n} = (F_{n+1})^2 - (F_{n-1})^2$
(ii) $F_{2n+1} = (F_n)^2 + (F_{n+1})^2$

Thus far, I have never come across any website which lists these along the other properties of the Fibonacci numbers.
At the same time three requests come to my mind for the great gentlemen reading this thread.
(i) Help me with the proof of these properties.
(ii) Enlighten me with the practical use of the Fibonacci sequence (if any)?
(iii) Is there a resource(a website for instance), where we can get a summary of the properties of the Fibonacci sequence?
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October 9th, 2014, 04:16 AM   #2
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Your formulae fail for $n=1$. Which $n$ do they work?
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Last edited by v8archie; October 9th, 2014 at 04:31 AM.
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October 10th, 2014, 04:41 AM   #3
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Thanks for pointing it out,
but I think it works fine for n>1.
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October 10th, 2014, 05:37 AM   #4
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They don't work for $n=2$ either. I think you have you indices wrong.

Perhaps you mean $$F_{2n+1} = F_{n+1}^2 - F_{n-1}^2 \\ F_{2n} = F_{n-1}^2 + F_n^2$$
These do at least work for $n \in \{1, 2\}$.

Here's a trick that might lead to the result.
\begin{align*}F_n &= F_{n-1}+F_{n-2} = F_1 F_{n-1} + F_0F_{n-2} \\ &= F_1(F_{n-2} + F_{n-3}) + F_0 F_{n-2} = (F_0 + F_1) F_{n-2} + F_1 F_{n-3} \\ &= F_2 F_{n-2} + F_1 F_{n-3} \\ &\vdots \end{align*}

Note that this doesn't work for the Brady numbers!
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Last edited by v8archie; October 10th, 2014 at 06:07 AM.
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October 10th, 2014, 11:16 AM   #5
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Note that, in the previous derivation, the indices of the first term on the right sum to the index of the term on the left.

Thus, \begin{align*}F_{2n} &= F_1 F_{2n-1} + F_0 F_{2n-2} \\ &= F_2 F_{2n-2} + F_1 F_{2n-3} \\ &\vdots \\ &= F_n F_n + F_{n-1} F_{n-1} \\ &= F_n^2 + F_{n-1}^2 \\[12pt] F_{2n+1} &= F_1 F_{2n} + F_0 F_{2n-1} \\ &= F_2 F_{2n-1} + F_1 F_{2n-2} \\ &\vdots \\ &= F_n F_{n+1} + F_{n-1} F_n \\ &= (F_{n+1} - F_{n-1})F_{n+1} + F_{n-1}(F_{n+1} - F_{n-1}) \\ &= F_{n+1}^2 - F_{n-1}^2 \end{align*}
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Last edited by v8archie; October 10th, 2014 at 11:18 AM.
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October 14th, 2014, 05:01 AM   #6
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But hey, there must be some misunderstanding.
Both my formulae work fine for natural numbers including 1 and 2.

Given, $F_0=0; F_1=1; F_2=1; F_3=2; F_4=3; F_5=5; F_6=8; F_7=13;$ and so on...

(i)$ F_{2n} = (F_{n+1})^2 - (F_{n-1})^2 $
When $n=1$,
L.H.S. :
$F_{2*1} = F_2 = 1$
R.H.S. :
$(F_{1+1})^2 - (F_{1-1})^2 = {F_2}^2 - {F_0}^2 = 1^2 - 0^2 = 1$
Hence, L.H.S. = R.H.S.

(ii)$ F_{2n+1} = (F_n)^2 + (F_{n+1})^2$
When $n=1$,
L.H.S. :
$ F_{2*1+1} = F_3 = 2$
R.H.S. :
$ (F_1)^2 + (F_{1+1})^2 = (F_1)^2 + (F_2)^2 = 1^2 + 1^2 = 2$
Hence, L.H.S. = R.H.S.

Similarly, other natural number values for n also satisfies the equations.
So, what now?!!!
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October 14th, 2014, 09:07 AM   #7
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My apologies, I misread, assuming the usual $F_0 = F_1 = 1$. The proofs above can be easily adjusted for the change of index.
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