Factorizing the Sum of Two Fibonacci numbers 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.
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.