Fibonacci proof question: $f_{n+1}f_{n-1}-f_n^2=(-1)^n$ [closed]

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.

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.

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

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:

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

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?