<s>[INST] Show $F_{n+1} \cdot F_{n-1} = F_n^2 + (-1)^n$ for all $n \in \mathbb{N}$ By calculating for $n\in \{1,2,3,4,5,6,7\}$, I've formulated the rule 
\begin{equation}
F_{n+1} \cdot F_{n-1} = F_n^2 + (-1)^n,
\end{equation} 
where $F_n$ is the $n$th fibonacci number. I want to show that this is true for all $n \in \mathbb{N}$. 
I tried using induction, with $n=1$ as the basis step, but didn't get very far:
For the induction step, we assume the formula holds for a $n = k$, and checks for $n=k+1$:
\begin{align*}
F_{k} \cdot F_{k+2} &= F_{k} \cdot (F_{k+1} + F_{k})  \\
&= F_k \cdot F_{k+1} + F_k^2 \\
\end{align*}
If somehow $F_k \cdot F_{k+1} = (-1)^k$, then I would be done. But I don't see how that's possible. 
Is there a better way of proving this, maybe without using induction? Or am I just going about it the wrong way? 
 [/INST] We can prove the rule without induction using Binet's formula
$$F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}$$
where
$$\varphi = \frac{1 + \sqrt{5}}{2} \qquad \psi = \frac{1 - \sqrt{5}}{2} = -\frac{1}{\varphi}$$
so that
$$\varphi\psi = -1 \qquad \frac{\varphi}{\psi} + \frac{\psi}{\varphi} = -3 = 2 - 5$$
Then
\begin{align}
F_{n+1} F_{n-1} & = \frac{\varphi^{n+1} - \psi^{n+1}}{\sqrt{5}} \cdot \frac{\varphi^{n-1} - \psi^{n-1}}{\sqrt{5}}\\
& = \frac{1}{5}\left(\varphi^{2n} - \varphi^{n+1}\psi^{n-1} - \varphi^{n-1}\psi^{n+1} + \psi^{2n}\right)\\
& = \frac{1}{5}\left(\varphi^{2n} - \left(\frac{\varphi}{\psi} + \frac{\psi}{\varphi}\right)\varphi^n\psi^n + \psi^{2n}\right)\\
& = \frac{1}{5}\left(\varphi^{2n} - (2-5)\varphi^n\psi^n + \psi^{2n}\right)\\
& = \frac{1}{5}\left(\varphi^{2n} - 2\varphi^n\psi^n + \psi^{2n} + 5\varphi^n\psi^n\right)\\
& = \left(\frac{\varphi^n - \psi^n}{\sqrt{5}}\right)^2 + \frac{1}{5}\cdot5(\varphi\psi)^n\\
& = F_n^2 + (-1)^n
\end{align}
</s>