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

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.

• If you included your proof, we would know what not to offer as an explanation. – Henrik Jul 30 '16 at 15:13
• I'm pretty confident that the identity holds, but I have no proof. – Landei Jul 30 '16 at 15:18
• – lab bhattacharjee Jul 30 '16 at 15:20
• If all else fails expansion of the explicit expression for the $n^{th}$ Fibonacci term given by: $$F_n=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt{5}}{2}\right)^n-\left(\frac{1-\sqrt{5}}{2}\right)^n\right)$$ – Ian Miller Jul 30 '16 at 15:23
• You can devote an entire chapter of a book to Fibonacci identities. A wood nymph wrote such a book in the 12th century, though that one is a little difficult for humans to get ahold of. Then there's Thomas Koshy's Fibonacci and Lucas Numbers with Applications, which you might have a drone deliver straight to your door. And you will write one in 2047. – The Short One Aug 10 '16 at 22:32

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.

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 ).