For an integer $n\geq 0$ let $F_n$ denote the $n$th Fibonacci number and let $L_n$ denote the $n$th Lucas number.
It is known that $F_n$ is prime only if $n$ is prime or $n=4$.
According to Wikipedia it is known that $L_n$ is prime only if $n$ is $0$, prime or a power of $2$.
The reciprocals are not true. Indeed, $$\begin{array}{ccc} F_2 &= &1,\\ F_{19} &= &37\times 113,\\ L_{23} &= &139\times 461,\\ L_{64} &= &1087 \times 4481. \end{array}$$
Let $S=\{n: F_n\text{ is prime}\}$ and $T=\{n: L_n\text{ is prime}\}$.
Q: Is the set $S\cap T$ finite?
Some remarks:
Some computations in Mathematica suggest that $n=4, 5, 7, 11, 13, 17, 47$ are all the integers in $S\cap T$ with $n\leq 10000$.
It follows by the exposition above that an element of $S\cap T$ is either $4$ or prime.
It is not known whether there are infinitely many Fibonacci prime numbers. So, either my question is an open problem or the answer is yes.
My interest in the set $S\cap T$ arises from a question related to giving approximations of $\sqrt{5}$ as the fraction of two prime numbers. Recall that $\frac{L_n}{F_n}$ tends to $\sqrt 5$ when $n$ tends to infiniy.
I have no strong background on Fibonacci numbers. All remarks are welcome.
Your question may not yet be answerable because we don't know if there exist an infinite number of Fibonacci (or Lucas primes); Moreover, we don't even know if there exist an infinite number of composite Fibonacci (Lucas) numbers.
One thing (among many) that we do know is that $F_{2n} = F_{n}L_{n}$ with the $gcd(F_{n}, L_{n})$ being either 1 or 2.