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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?

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  • $\begingroup$ 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
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    $\begingroup$ Usually Fibonacci numbers are defined with $F_0=0$. $\endgroup$ – Max Alekseyev Aug 5 '16 at 15:20
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    $\begingroup$ Squarefree Fibonacci numbers are tabulated at oeis.org/A061305 $\endgroup$ – Gerry Myerson Aug 6 '16 at 6:41
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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).

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ $M(p)$ 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

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