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 know for what values of $k\in\mathbb{N}$ can one write $$F_{n+(2k+1)}\pm F_n=cP(k,n)$$ where $c\in\mathbb{N}$ and $P(k,n)$ is some product of Fibonacci or Lucas numbers. Note that it is easy to show that that: \begin{align*} F_{n+2k}+F_{n}=F_{n+k}L_k\text{ where $k$ is even,}\\ F_{n+2k}+F_{n}=L_{n+k}F_k\text{ where $k$ is odd.}\\ F_{n+2k}-F_{n}=F_{n+k}L_k\text{ where $k$ is even,}\\ F_{n+2k}-F_{n}=L_{n+k}F_k\text{ where $k$ is odd.}\\ \end{align*} It is also easy to see that \begin{align*} F_{n+1}+F_{n}=F_{n+2}\\ F_{n+1}-F_{n}=F_{n-1}\\ F_{n+3}+F_{n}=2F_{n+2}\\ F_{n+3}-F_{n}=2F_{n+1}\\ \end{align*} Are these the only such expressions?

Ideas tried: I've tried the Binet formula to see what insights this might provide, but I can't see anything. I've also tested small values of $n$ numerically but couldn't find any further examples than those four given.


It seems unlikely that there will be a simple factorization for $F_{n+5}\pm F_{n}$ because these numbers are prime for several values of $n$.

Primes of the form $F_{n+5}+F_{n}$ are listed in oeis/A091157.

Primes of the form $F_{n+5}-F_{n}$ are listed in oeis/A153892.

  • $\begingroup$ Right. So you suspect that there will be no such $k$ for which the this will be possible? $\endgroup$ – Auslander Nov 30 '16 at 22:54

Among Lucas numbers $L_{n+1}^2-L_n^2=L_{n+2}L_{n-1}, n\ge 2$. This is just a differeence of squares factorization together with applying the recursive relation.

  • $\begingroup$ I'm not sure how this helps...? This example is the Lucas counterpart to my first example. $\endgroup$ – Auslander Oct 22 '16 at 13:24

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