Show $F_{n+1} \cdot F_{n-1} = F_n^2 + (-1)^n$ for all $n \in \mathbb{N}$

Question

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?

Answer 1

Use the induction hypothesis to replace $F_k^2$ by $F_{k-1} \cdot F_{k+1} - (-1)^k$.

Answer 2

This is Cassini's identity. It has a nice proof using determinants: $$ f_{n-1}f_{n+1} - f_n^2 =\det\left[\begin{matrix}f_{n+1}&f_n\\f_n&f_{n-1}\end{matrix}\right] =\det\left[\begin{matrix}1&1\\1&0\end{matrix}\right]^n =\left(\det\left[\begin{matrix}1&1\\1&0\end{matrix}\right]\right)^n =(-1)^n $$ This matrix formulation of Fibonacci numbers is well worth knowing and easily proved by induction: $$ \left[\begin{matrix}f_{n+1}&f_n\\f_n&f_{n-1}\end{matrix}\right] = \left[\begin{matrix}1&1\\1&0\end{matrix}\right]^n $$

Answer 3

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}

Answer 4

The Catalan's identity tell us that $$F_{n}^{2}-F_{n+r}F_{n-r}=\left(-1\right)^{n-r}F_{r}^{2} $$ so taking $r=1 $ we have $$F_{n}^{2}-F_{n+1}F_{n-1}=\left(-1\right)^{n-1} $$ as wanted. It is also called the Cassini's identity.