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Show that $$f_{n+1}f_{n-1}-f_n^2=(-1)^n$$ when $n$ is a positive integer and $f_n$ is the $n$th Fibonacci number.

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closed as off-topic by Bruno Joyal, T. Bongers, Daryl, Serkan, Stefan Hansen Dec 14 '13 at 8:01

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Have you attempted it? – Zhoe Dec 14 '13 at 4:35
Give the good old induction a try. – user112167 Dec 14 '13 at 4:35
I don't know how to do it. – Out Of Bounds Dec 14 '13 at 4:53
up vote 2 down vote accepted
  1. For a basis $n=1$ the equality holds, as $$f_{k+1}f_{k-1}-f_k^2=f_2f_0-f_1^2=1 \cdot 0 - 1^2=-1=(-1)^1.$$

  2. Assume the equality holds for $n=k$. Then we may assume that $$f_{k+1}f_{k-1}-f_k^2=(-1)^k.$$

    For the final inductive step, we wish to prove that $$f_{k+2}f_k-f_{k+1}^2=(-1)^{k+1}.$$

  3. We begin with the left side.

$$ \begin{align*} f_{k+2}f_k-f_{k+1}^2&= \left( f_k-f_{k+1} \right)f_k-f_{k+1}^2 \\ &=f_k^2-f_{k+1}f_k-f_{k+1}^2 \\ &=-f_{k+1}\left( f_{k+1}+f_k \right)+f_k^2 \\ &=-f_{k+1}f_{k-1}+f_k^2 \\ &= \left( f_{k+1}f_{k-1}-f_k^2 \right)(-1)^1 \\ &=(-1)^k(-1)^1 \\ &=(-1)^{k+1}. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \blacksquare \end{align*} $$

Notes: On line 4 by definition $f_{k+1}+f_k=f_{k-1}$. On line 6 we substituted our assumption.

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First, how should i know that $ f_{2} is 1 $ and $ { f }_{ 1 }^{ 2 } = 1^2 $ ? – Out Of Bounds Dec 14 '13 at 6:46
Well that's the sequence. It goes $\{0,1,1,2,3,5,8,13,21,\ldots\}$. We are just taking two prior numbers and adding them together to get the next number. Every number is the sum of the two prior numbers. Here we let the first term $f_0=0, f_1=1, \ldots$ – J. W. Perry Dec 14 '13 at 6:47
@Tennisman You should know that there is supposed to be a bit of pencil work here. It is not supposed to come out this pretty on your first pass. You actually have to assemble the proof and refine it to something decent. Getting from step 2 to 3 and presenting it all nice is not something you are necessarily be expected to do in a nanosecond. – J. W. Perry Dec 14 '13 at 6:56
Yeah you are right. That's why i am gonna ask you how do you got in step 3 the first line from the LFS to RHS. I mean is that algebra or what :D. Can we do algebra on subscripts ? Iam confused. I can't get what you did in step 3 at all. – Out Of Bounds Dec 14 '13 at 7:00
Ahh yeah got it. Thank you very much appreciate your help :) – Out Of Bounds Dec 14 '13 at 7:23

To start you off, you can prove this by using Mathematical Induction (I suggest you read up for further understanding - more reading). To do so you have three steps, namely: proving true for $n=1$, assuming true for $n=k$, and finally proving true for $n=k+1$ and thus for all $N$.

$\circ$ Prove true for $n=1$

Often called the base case, here you substitute the value $1$ on both sides of the equation, and show that $f_{n+1}f_{n-1} - f_n^2 = (-1)^n$ is true for $n=1$.

$$\implies f_{1+1}f_{1-1} - f_1^2 = (-1)^1 ...$$

$\circ$ Assume true for $n=k$

This is just as it sounds, substitute $k$ where you see $n$ on both sides of the equation.

$$\implies f_{k+1}f_{k-1} - f_k^2 = (-1)^k$$

$\circ$ Prove true for $n=k+1$

Here is where the bulk of the proof lies. Substitute $k+1$ where you see $n$ on both sides of the equation and then perform some gymnastics to prove that it is true for $n=k+1$.

$$\implies f_{(k+1)+1}f_{(k+1)-1} - f_{k+1}^2 = (-1)^{k+1} ...$$

EDIT: Note: The Fibonacci Sequence has the properties $f_0=0$, $f_1=1$ and $$f_n=f_{n-1}+f_{n-2}$$ And the sequence looks like this:

$$(f_n)_{n\in N} = (0, 1, 1, 2, 3, 5, 8, 13, 21, 34,...)$$

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How you can prove that the RHS equal to the LHS with n = 1 ? – Out Of Bounds Dec 14 '13 at 6:32
The Fibonacci Sequence has the properties $f_0=0$, $f_1=1$ and $f_n=f_{n-1}+f_{n-2}$ The LHS $=f_2f_0 -(f_1)^2 =1\cdot 0-(1)^2=-1$ (Plugging two into the formula you get $f_2=f_{2-1}+f_{2-2}=f_1+f_0=1+0=1$) The RHS $=(-1)^1 = -1$ Since LHS$=$RHS ($-1=-1$),then you have proven true for $n=1$. – Zhoe Dec 14 '13 at 6:44

Using the definition of Fibonacci numbers, $$f_{n+1}f_{n-1} - f_n^2=(f_n+f_{n-1})f_{n-1}-f_n(f_{n-1}+f_{n-2})$$ $$\implies f_{n+1}f_{n-1} - f_n^2=-(f_nf_{n-2}-f_{n-1}^2)$$

If $\displaystyle u_n=f_{n+1}f_{n-1} - f_n^2,$ we have $u_n=-u_{n-1}$

So, we can write $\displaystyle u_n=(-1)^ru_{n-r}=(-1)^{n-s}u_s $ where $0<n-r\le n,0\le s<n$

Set $r=n$ or $s=0$

What is $u_1?$

Lucas Numbers (1,2) should also satisfy a similar relation, right?

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If this is a homework exercise then this is likely not how you're meant to prove it, since you'll first need to prove that the $n$-th Fibonacci number can be written as $$f_n=\frac{\varphi^n-(-\frac{1}{\varphi})^{n}}{\sqrt{5}}$$ where $\varphi=\frac{1+\sqrt{5}}{2}$. But if you have that then

$$\begin{align}f_{n-1}f_{n+1}-f_{n}^{2} &= \left(\frac{\varphi^{n-1}-(-\frac{1}{\varphi})^{-\left(n-1\right)}}{\sqrt{5}}\right)\left(\frac{\varphi^{n+1}-(-\frac{1}{\varphi})^{n+1}}{\sqrt{5}}\right)-\frac{\varphi^{2n}-2\varphi^{n}(-\frac{1}{\varphi})^{-n}+\left(-\frac{1}{\varphi}\right)^{2n}}{5} \\ &= \frac{\varphi^{2n}+\left(-\frac{1}{\varphi}\right)^{2n}-\varphi^{n-1}\left(-\frac{1}{\varphi}\right)^{n+1}-\left(\frac{1}{\varphi}\right)^{n-1}\varphi^{n+1}}{5}-\frac{\varphi^{2n}-2(-1)^{n}+\psi^{2n}}{5} \\ &= \frac{\varphi^{2n}+\left(-\frac{1}{\varphi}\right)^{2n}-\left(\frac{1}{\varphi}\right)^{2}\left(-1\right)^{n+1}+\varphi^{2}\left(-1\right)^{n-1}}{5}-\frac{\varphi^{2n}-2(-1)^{n}+\left(-\frac{1}{\varphi}\right)^{2n}}{5} \\ &= \frac{\left(-1\right)^{n}\left[\left(\frac{1}{\varphi}\right)^{2}+\varphi^{2}\right]+2(-1)^{n}}{5}\end{align}$$ and, miraculously, $$\left(\frac{1}{\varphi}\right)^{2}+\varphi^{2} = \left(\frac{\sqrt{5}-1}{2}\right)^{2}+\left(\frac{1+\sqrt{5}}{2}\right)^{2} = \frac{5-2\sqrt{5}+1+1+2\sqrt{5}+1}{4} = 3 $$

so finally $$\frac{\left(-1\right)^{n}\left[\left(\frac{1}{\varphi}\right)^{2}+\varphi^{2}\right]+2(-1)^{n}}{5}=\frac{3\left(-1\right)^{n}+2(-1)^{n}}{5}=(-1)^{n}$$

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