[seqfan] Re: Primes p = nk-1 dividing Fibonacci( k )

[Max Alekseyev]

On Wed, Nov 25, 2009 at 6:33 PM, Maximilian Hasler
<maximilian.hasler at gmail.com> wrote:

> Least prime p = -1 (mod n) which divides Fibonacci((p+1)/n), or 0 if
> there is no such prime < 99999
> I get
>
> 2, 13, 47, 0, 0, 113, 307, 0, 233, 0, 967, 0, 2417, 797, 0, 0, 1087,
> 233, 5737, 0, 5417, 5653, 1103, 0, 0, 2417, 4373, 0, 6263, ...
>
> Has anyone a simple and/or universal explanation for the 0's ?

Only multiples of 4 and 5 seems to be excluded.

For multiples of 5, there is a simple explanation:
If n is a mulpliple of 5 and  p==-1 (mod n), then p == -1 (mod 5) and
thus p divides F(p-1). But if p also divides F((p+1)/n) then it
divides F(gcd(p-1,(p+1)/n))=F(1)=1, a contradiction.

Regards,
Max