Fibonacci identity for $2F_n + F_{n-3}$

Question

For the $n$'th Fibonacci number, I found the following identity:

$$ (2F_n + F_{n-3})^2 = 5\cdot F_n^2 + 4\cdot (-1)^n $$

Now, I don't really expect that this is something new, but I'd like to have some kind of explanation, as this identity with its $\pm$ term seems slightly odd to me.

Answer 1

There's a whole set of identities for the Fibonacci and Lucas numbers which are analogous to trigonometric identities—Fibonacci numbers playing the role of sin, and Lucas numbers playing the role of cos. If you haven't seen Lucas numbers, they satisfy the same recurrence relation as the Fibonacci numbers, but you start with $L_0=2$ and $L_1=1.$

Your identity is the same as $$L_n^{\;2} = 5F_n^{\;2} + 4(-1)^n,$$ which corresponds to $$\cos^2 x = 1 - \sin^2x.$$

$$ $$

The reason for the analogous behavior is the similarity of Binet's formula $$F_n=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-‌​\sqrt{5}}{2}\right)^n\right)$$ and the similar formula for Lucas numbers $$L_n=\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-‌​\sqrt{5}}{2}\right)^n$$ to the formulas $$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$$ and $$\cos x = \frac{e^{ix}+e^{-ix}}{2}.$$

See http://busontime.blogspot.com/2010/07/fibonacci-trigonometry-6-identities.html for more details.

Here's one way to prove this (phrased in terms of hyperbolic functions instead of trig functions because it's perhaps slightly easier this way):

Let $\phi=\frac{1+\sqrt{5}}{2},$ $\phi'=\frac{1-\sqrt{5}}{2},$ and $\psi=\ln(\phi)+\frac{i \pi}{2}.$

Then $e^\psi = i \phi$ and $e^{-\psi} = -i/\phi = i \phi',$ so

\begin{align}\sinh (n \psi) &= \frac{e^{n \psi} - e^{-n \psi}}{2} \\\\ &= i^n \frac{\phi^n - \phi'^n}{2} \\\\ &= \frac{i^n \sqrt{5}}{2} \frac{\phi^n - \phi'^n}{\sqrt{5}} \\\\ &= \frac{i^n \sqrt{5}}{2} F_n, \end{align} so $$F_n=\frac {2 \sinh (n \psi)}{i^n \sqrt{5}}.$$ Similarly, $$L_n=\frac{2 \cosh (n \psi)}{i^n}.$$ It follows that \begin{align} L_n^{\;2}-5F_n^{\;2} &=\frac{4 \cosh^2(n\psi)}{(-1)^n}-5 \cdot \frac{4 \sinh^2(n \psi)}{5 (-1)^n} \\\\ &= 4\cdot(-1)^n (\cosh^2 (n \psi) - \sinh^2 (n \psi)) \\\\ &= 4\cdot(-1)^n. \end{align}

For the OP's equation, just observe that $2F_n+F_{n−3}=F_n+F_{n-1}+F_{n-2}+F_{n-3}=F_{n+1}+F_{n-1}=L_n,$ and you're done.

Answer 2

First notice that $2F_n + F_{n-3}=F_{n+1}+F_{n-1}$ (simply write $F_{n-3}$ as $F_{n-1}-F_{n-2}$ and do the same again for $F_{n-2}$). Then notice that we can also write $F_n=F_{n+1}-F_{n-1}$. So the original equality becomes $$ (F_{n+1}+F_{n-1})^2-(F_{n+1}-F_{n-1})^2=4F_n^2+4(-1)^n.$$ Left side can be expanded and simplified to $4F_{n+1}F_{n-1}$, and so after canceling $4$s, we actually want to prove $$ F_{n+1}F_{n-1}=F_n^2+(-1)^n.\tag{1} $$ But $(1)$ is exactly the Cassini's identity which can be proven for example by comparing determinants of matrices in relation $$\begin{pmatrix} 1 & 1 \\ 1 & 0 \\ \end{pmatrix}^n = \begin{pmatrix} F_{n+1} & F_n \\ F_n & F_{n-1} \\ \end{pmatrix}\tag{2},$$ while $(2)$ is a straightforward implication of recurrence relation $F_{n+2}=F_{n+1}+F_{n}$, just written in a matrix form (see for example Converting recursive equations into matrices ).