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

[Max Alekseyev]

This is similar to the connection between Weiferich primes and Mersenne numbers.
If p^2 divides Mersenne number 2^q - 1, where p and q are primes, then
p must be a Weiferich prime.
Max

On Tue, Jan 28, 2014 at 2:40 PM, Charles Greathouse
<charles.greathouse at case.edu> wrote:
> 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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