[seqfan] Re: Is F(p) always squarefree?

[Charles Greathouse]

Suppose F_n is divisible by k^2. Then n is divisible by A001177(k^2) =
A132632(k). So a necessary condition for F_p being squarefree is that
A132632(q) is prime for some prime q. But this can happen only when Wall's
conjecture fails, so if F_p is not squarefree than it is divisible by the
square of a Wall-Sun-Sun prime. (Right?) I think current expectations are
that infinitely many Wall-Sun-Sun primes exist, but they should have only
doubly-logarithmic density and so it seems very hard to find any and
near-impossible to find more than one.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University


On Tue, Jan 28, 2014 at 2:17 PM, Alonso Del Arte
<alonso.delarte at gmail.com>wrote:

> Given a prime p, the number Fibonacci(p) might be composite, but, at least
> for small p, appears to always be squarefree. This seems like something
> that could easily be proven one way or the other with something in Koshy's
> book, but the Library is closed today.
>
> Al
>
> --
> Alonso del Arte
> Author at SmashWords.com<
> https://www.smashwords.com/profile/view/AlonsoDelarte>
> Musician at ReverbNation.com <http://www.reverbnation.com/alonsodelarte>
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