Can $F_n^2-F_m^2$ be factored as a product of Fibonacci or Lucas numbers when $n-m$ is odd?

Question

The Fibonacci and Lucas numbers are defined for all integers $n$ by the recurrence relations $$F_n=F_{n-1}+F_{n-2}\text{ where }F_1=1\text{ and }F_2=1,$$ $$L_n=L_{n-1}+L_{n-2}\text{ where }L_1=1\text{ and }L_2=3.$$ I would like to simplify expressions of the form $$F_n^2-F_m^2.$$ By "simplify" I mean write as some product of Fibonacci or Lucas numbers, or a quotient of said products. When $n-m=2k$ is even this is easy to do using the formulae: \begin{align*} F_{n+k}+F_{n-k}=F_nL_k\text{ where $k$ is even,}\\ F_{n+k}+F_{n-k}=L_nF_k\text{ where $k$ is odd.}\\ \end{align*} Therefore, \begin{align*} F_n^2-F_m^2&=(F_n-F_m)(F_n+F_m)\\ &=(F_{m+2k}-F_m)(F_{m+2k}+F_m)\\ &=(F_{(m+k)-k}-F_{(m+k)-k})(F_{(m+k)+k}+F_{(m+k)-k})\\ &=L_{m+k}F_{m+k}L_kF_k\\ \end{align*} This trick obviously doesn't work when $n-m$ is odd. Does anyone have any suggestions as to how to find a similar expression in this instance?

Answer 1

I want to point out that, in the case $n-m$ is even, then you can also write $$F_n^2-F_m^2 = F_{n+m}\cdot F_{n-m}.$$

Which probably can be obtained easily from Binet expression for Fibonacci numbers.

For the case $n-m$ odd, I have only the following negative result:

In general, the expression $F_n^2-F_m^2$ when $n-m$ is odd cannot be written as a product of $8$ (or less) Fibonacci and/or Lucas numbers.

Since this is a negative result, I established it simply by computing the expression for various values of $n$ and $m$ and begin unable to decompose it as the product of Fibonacci and/or Lucas number.

The only non trivial result I found empirically (I'm sure it can be proved easily) is the following, which holds only for $n=4k+2$, and $m=n/2=2k+1$ (hence $n-m$ is odd): $$ F_n^2-F_{n/2}^2 = F_5\cdot F_{n/2-1}\cdot F_{n/2}\cdot F_{n/2} \cdot F_{n/2+1} $$