TITLE: For $n\ge4,$ prove that $F_n+1$ is not prime, where $F_n$ is $n^{th}$ Fibonacci number.
QUESTION [6 upvotes]: For $n\ge4,$ prove that $F_n+1$ is not prime, where $F_n$ is $n^{th}$ Fibonacci number
What is the idea of the proof? I tried it by contradiction by 
letting $(1+F_n)$  to be prime $\implies$ $F_n$ is not prime $\implies$ WHAT NEXT?

REPLY [5 votes]: As observed by Will Jagy we can always write $F_n+1$ as a product of a Fibonacci number and a Lucas number.
Let $\phi=(1+\sqrt5)/2$. It is well known that then
$$
F_n=\frac1{\sqrt5}(\phi^n-(-\phi)^{-n})
$$
and
$$
L_n=(\phi^n+(-\phi)^{-n}).
$$
Both these sequences of integers satisfy the famous two-step recurrence relation, the difference coming from the initializations $F_0=0, F_1=1$ as opposed to $L_0=2, L_1=1$.
The following factorizations then follow immediately from $F_1=F_2=1$.
$$
\begin{aligned}
F_{2k+1}L_{2k}&=F_{4k+1}+F_1=F_{4k+1}+1,\\
F_{2k-1}L_{2k}&=F_{4k-1}+F_1=F_{4k-1}+1,\\
F_{2k}L_{2k-2}&=F_{4k-2}+F_2=F_{4k-2}+1,\\
F_{2k-1}L_{2k+1}&=F_{4k}+F_2=F_{4k}+1.
\end{aligned}
$$
All the residue classes modulo $4$ were covered, so the claim follows.

As an example:
$$
\begin{aligned}
F_{2k-1}L_{2k+1}&=\frac1{\sqrt5}(\phi^{2k-1}+\phi^{-(2k-1)})(\phi^{2k+1}-\phi^{-(2k+1)})\\
&=\frac1{\sqrt5}(\phi^{4k}+\phi^2-\phi^{-2}-\phi^{-4k})\\
&=F_{4k}+F_2=F_{4k}+1.
\end{aligned}
$$
It's all about polynomials of $\phi$. You do need to be careful with the parities of the exponents due to that $-\phi$ in the base.