# Proof of a Well-Known Fibonacci Identity Involving Cubes of Fibonacci Numbers

The following is due to Lucas in 1876:

$$F_{n + 1}^3 + F_n^3 - F_{n - 1}^3 = F_{3n}$$

I am unable to locate an elementary proof of this identity, and am unable to reproduce it myself. Would anyone mind sharing a proof or a source?

A nice strategy is to use the elementary identity $$F_{n+m} = F_n F_{m-1} + F_{n+1}F_m. \tag1$$ Use it to show that $$\, a_n := F_{3n}\,$$ satisfies the recursion $$a_{n+2} = 4a_{n+1} + a_n. \tag2$$ Now use it to express $$\, F_{n-1} = F_{n+1} - F_n\,$$ and hence express $$b_n := F_{n + 1}^3 + F_n^3 - F_{n - 1}^3 \tag3$$ in terms of $$\, F_n\,$$ and $$\, F_{n+1}.\,$$ Verify that $$\,a_n\,$$ and $$\,b_n\,$$ satisfy the same recursion and initial values. The idea is to express all of $$\,b_n, b_{n+1}, b_{n+2}\,$$ in terms of $$\,F_n, F_{n+1}\,$$ using simple algebra. $$b_n = F_n(2F_n^2 -3F_nF_{n+1} +3F_{n+1}^2), \tag4$$ $$b_{n+1} = F_{n+1}(3F_n^2 +3F_nF_{n+1} +2F_{n+1}^2), \tag5$$ $$b_{n+2} = (F_n+F_{n+1})(2F_n^2 +7F_nF_{n+1} +8F_{n+1}^2). \tag6$$

The OEIS sequence A014445 is defined as $$a_n$$ and its recursion as well as its equivalence to $$b_n$$ is given in the entry.

• Ok, the first relationship (2) can be shown with some playing around. Showing $b_n$ satisfies the same recursion seems to me to be roughly as difficult as the original problem, however. – Johnny Apple Nov 16 '18 at 4:25

This is probably not the most elegant way, but you can use the explicit formula for the Fibonacci numbers. $$F_n = \frac{1}{\sqrt{5}} \tau^n + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^n$$ Here $$\tau = \frac{1}{2}+\frac{\sqrt{5}}{2}, -1/\tau = \frac{1}{2}-\frac{\sqrt{5}}{2}$$. This formula for $$F_n$$ can now be plugged into the left-hand side of the desired identity. \begin{align*} F_{n+1}^3 + F_n^3 - F_{n-1}^3 = & \left(\frac{1}{\sqrt{5}} \tau^{n+1} + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^{n+1}\right)^3\\ & + \left(\frac{1}{\sqrt{5}} \tau^n + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^n\right)^3\\ & - \left(\frac{1}{\sqrt{5}} \tau^{n-1} + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^{n-1}\right)^3 \end{align*}

Expanding out the terms takes a little work but will eventually lead to the desired expression $$\frac{1}{\sqrt{5}} \tau^{3n} + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^{3n}$$ for $$F_{3n}$$.

As an example, considering the leading terms of the three cubes yields $$\frac{1}{5\sqrt{5}}\tau^{3n+3}+\frac{1}{5\sqrt{5}}\tau^{3n} - \frac{1}{5\sqrt{5}}\tau^{3n-3} = \frac{1}{5\sqrt{5}}\left(\tau^3 + 1 - \frac{1}{\tau^3}\right)\tau^{3n}$$ Since $$\tau^3 = 2+ \sqrt{5}$$ and $$-1/\tau^3 = 2-\sqrt{5}$$, we see that this simplifies to $$(1/\sqrt{5})\tau^{3n}$$, which is the first term of the expression for $$F_{3n}$$.

I leave the consideration of the other terms to the reader.

Use Somos' first identity to expand $$F_{2n+n}$$ and then expand again to get a bunch of terms, each the product of three Fibonacci numbers.
Take out the common factor $$F_n$$, and find the factor $$F_n$$ in your left-hand side.
The rest should be fairly easy.