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I am given $$ F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}} $$ where, $\varphi = \frac{1 + \sqrt{5}}{2}$ and $\psi = \frac{1 - \sqrt{5}}{2}$.

The textbook states that it's equal to the $n$-th Fibonacci number $F_n$. it is stated that since the Fibonacci numbers are integers, the number in $$ F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}} $$ is an integer as well. can you guys clarify this please?

How do I go on proving that the number $$F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}$$ is an integer using Newton's Binomial Theorem?

Bart Michels
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user2551612
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    It is not clear what you are asking. It looks at one stage that you are asking why $\varphi-\psi=\sqrt{5}$. Surely you are not asking why $\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}$. – André Nicolas Feb 13 '14 at 03:12
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    http://math.stackexchange.com/questions/665604/how-to-prove-this-is-a-rational-number – lab bhattacharjee Feb 13 '14 at 03:32
  • no I wanted to ask, how is it that $$F_n = \frac{\varphi^n - \psi^n}{\varphi-\psi} = \sum_{k=0}^{n-1} \varphi^{n-1-k}\psi^k.$$ because we want a integer and square root five is not an integer.. – user2551612 Feb 13 '14 at 03:38
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    $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y +\cdots+y^{n-1})$. Just multiply and observe the cancellation. Or if you really want to, sum the finite geometric series $x^{n-1}+x^{n-2}y +\cdots+y^{n-1}$. – André Nicolas Feb 13 '14 at 04:34

2 Answers2

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How do I go on proving that the number in is an interger using Newton's Binomial Theorem?

$$\newcommand{\b}[1]{\left(#1\right)} \newcommand{\d}{{\rm d}} \newcommand{\f}{\frac} \newcommand{\s}{\sqrt} \newcommand{\t}{\text} \newcommand{\u}{\underbrace}\bf Answer$$

$$S=\f1{\s5}\b{\b{\f{1+\s5}2}^n-\b{\f{1-\s5}2}^n}=\f1{2^n\s5}\b{\sum_{k=0}^n\binom5k5^{k/2}-\sum_{k=0}^n(-1)^k\binom5k5^{k/2} }=\f1{2^n\s5}\sum_{k=0}^n[1-(-1)^k]\binom5k5^{k/2}$$

For when k is even the terms cancel so we get only odd terms let wherin we let $k=2t-1;k\in\{1,3,5,..n\},t\in\{1,2,..(n+1)/2\}$ $$S=\f1{2^{n-1}\s5}\sum_{t=1}^{(n+1)/2}\binom5{2t-1}5^{t-1/2}=\f1{2^{n-1}}\sum_{t=1}^{(n+1)/2}\binom5{2t-1}5^{t-1}$$ Now it's not irrational anymore and you may use induction.

RE60K
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I think this question was asked before.

First, using induction prove,

$$\Large{\phi^n = F_{n-1} + F_n \phi} \; \; \& \; \; \Large{\psi^n = F_{n-1} + F_n \psi}$$

Where $\phi = \frac{1 + \sqrt{5}}{2}$, and $\psi = 1 - \phi$.

Therefore,

$$\Large{\phi^n - \psi^n = F_{n-1} + F_n \phi - F_{n-1} - F_n \psi} \\ \Downarrow \\ \Large{\phi^n - \psi^n = F_ n ( \phi - \psi ) = F_n \sqrt{5}} \\ \Downarrow \\ \Large{\frac{1}{\sqrt{5}} ( \phi^n - \psi^n ) = F_n }$$

$F_n $ is integer, therefore $\frac{1}{\sqrt{5}} ( \phi^n - \psi^n ) \; $ is also integer.

Tahir Imanov
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