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.

  • 2
    $\begingroup$ (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. $\endgroup$ May 25, 2019 at 17:24
  • 1
    $\begingroup$ 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, 2019 at 19:12

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.