Fibonacci(n^2) is a multiple of Fibonacci(n)^2

[Leroy Quet]

We have F(mn) == m F(n) F(n-1)^(m-1) mod F(n)^2, so in particular
F(n^2) is a multiple of F(n)^2 only if n is a multiple of F(n),
and that is true only for n=0,1,2,5.

Cheers,
Robert Israel


On Thu, 5 Apr 2007, Leroy Quet wrote:

> This isn't the biggest math question facing me (or humanity) at the
> moment, but I am wondering...
>
> What is the sequence where n is included if F(n^2) is a multiple of
> F(n)^2, where F(n) is the nth Fibonacci number?
>
> (I say "F(n^2) is a multiple of F(n)^2" instead of "F(n)^2 divides
> F(n^2)" so as to include the n=0 case.)
>
> I get the sequence beginning 0,1,2,5,... (and then my calculator runs out
> of digits).
>
> These terms (0 excluded) are too few to do any meaningful search on the
> EIS. Is this sequence already in the EIS? I bet it is, but under a
> different name.
>
> Or maybe the sequence is finite.
>
> Thanks,
> Leroy Quet
>