Let $F_n$, $n\geq 0$, be the sequence of Fibonacci numbers, where $F_0=F_1=1$ and $F_{n+1}=F_n+F_{n-1}$ for $n\geq 1$. A number is squarefree if it is is not divisible by the square of a prime number.

Question: Are there infinitely many squarefree Fibonacci numbers?

  • $\begingroup$ I think this an open problem,but from what I just checked in the internet there is no conjecture which says: "There are infinitely many square free Fibonacci numbers".I would be surprised if someone has proved something like this and we did not know. $\endgroup$ – Konstantinos Gaitanas Aug 5 '16 at 14:55
  • $\begingroup$ This is almost surely true, but as with all similar problems about sequences as sparse as the Fibonacci sequence, it's very hard to prove $\endgroup$ – Stanley Yao Xiao Aug 5 '16 at 14:56
  • 4
    $\begingroup$ Usually Fibonacci numbers are defined with $F_0=0$. $\endgroup$ – Max Alekseyev Aug 5 '16 at 15:20
  • 1
    $\begingroup$ Squarefree Fibonacci numbers are tabulated at oeis.org/A061305 $\endgroup$ – Gerry Myerson Aug 6 '16 at 6:41

I assume the traditional definition with $F_0=0$ and $F_1=1$.

Most likely there are infinitely many squarefree Fibonacci numbers. A simple way to construct them is to consider a subsequence $F_p$ for prime $p$. Notice that if $q^2\mid F_p$ for some prime $q$, then $q$ must be a Wall-Sun-Sun prime, whose existence is a big open question (and even if they exist, they would be very rare).

| cite | improve this answer | |
  • $\begingroup$ What do you mean exactly by 'very rare'? Are they conjectured to have density zero among the primes? $\endgroup$ – Sylvain JULIEN Aug 5 '16 at 16:51
  • $\begingroup$ Again, this is most likely the case. At least the current empirical evidence suggests so. $\endgroup$ – Max Alekseyev Aug 5 '16 at 16:59
  • $\begingroup$ Evidently none are known and there are none up to $1.9 \cdot 10^{17}$ $\endgroup$ – Aaron Meyerowitz Aug 7 '16 at 5:11

For prime $p$, let $M(p)$ be the least positive $n$ such that $p^2 \mid F_n$. Then $p^2 \mid F_n$ iff $M(p) \mid n$. Thus at most $N/M(p)$ of the first $N$ Fibonacci numbers are divisible by $p^2$. If we could prove that $\sum_p 1/M(p) < 1$, then at least a positive fraction of all Fibonacci numbers will be squarefree.
Well, it seems to be true numerically; I don't know if it's provable.

| cite | improve this answer | |
  • $\begingroup$ $M(p)$ is tabulated at oeis.org/A065106 $\endgroup$ – Gerry Myerson Aug 6 '16 at 6:38
  • $\begingroup$ Maybe one can try to determine the least $\alpha$ such that $\sum_{p}p^{-\alpha}<1$, and then prove that $M(p)>p^{\alpha}$. $\endgroup$ – Sylvain JULIEN Aug 6 '16 at 6:48

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.