Squarefree Fibonacci Numbers

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?

• 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. – Konstantinos Gaitanas Aug 5 '16 at 14:55
• 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 – Stanley Yao Xiao Aug 5 '16 at 14:56
• Usually Fibonacci numbers are defined with $F_0=0$. – Max Alekseyev Aug 5 '16 at 15:20
• Squarefree Fibonacci numbers are tabulated at oeis.org/A061305 – 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).

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

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