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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:

  1. Some computations in Mathematica suggest that $n=4, 5, 7, 11, 13, 17, 47$ are all the integers in $S\cap T$ with $n\leq 10000$.

  2. It follows by the exposition above that an element of $S\cap T$ is either $4$ or prime.

  3. 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.

  4. 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.

  5. I have no strong background on Fibonacci numbers. All remarks are welcome.

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    $\begingroup$ (1). I dk...(2). It is also not known whether $T$ is finite or not...(3). Since $F_{2n}=F_nL_n $ we have $n\in S\cap T$ iff $F_{2n}$ is the product of two primes. $\endgroup$ – DanielWainfleet May 25 at 17:24
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    $\begingroup$ OEIS A080327 is the related sequence and indicates that the next, and conjectured only other, member of $S\cap T$ is $148091$. This in turn links to Prime Curio which remarks on the precision of the related $L_n/F_n$ approximation to $\sqrt{5}$ $\endgroup$ – nickgard Jun 7 at 19:12
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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.

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