[seqfan] Re: Can a repunit be a Fibonacci number?

[Max Alekseyev]

Integer x is a Fibonacci number iff y^2 - 5*x^2 = +- 4 for some integer y.
If x is a repunit, then x = (z^2 - 1)/9 or x = (10*z^2 - 1)/9 for some
integer z (additionally we know that z must be a power of 10, but we'll
ignore this fact).
The systems of equations
{ y^2 - 5*x^2 = +- 4, x = (z^2 - 1)/9 }
is equivalent to a biquadratic curve y^2 = 5*((z^2 - 1)/9)^2 +- 4 and can
be solved in integers via elliptic curves in SAGE or MAGMA, or with a Thue
solver as explained in my paper:
http://www.emis.ams.org/journals/JIS/VOL17/Alekseyev/alekseyev3.html
Similarly we can solve the system { y^2 - 5*x^2 = +- 4, x = (10*z^2 - 1)/9
}.

So, with some computational effort we can likely find all repunit Fibonacci
numbers (in any fixed base).

Regards,
Max


On Fri, May 13, 2016 at 12:45 PM, Alonso Del Arte <alonso.delarte at gmail.com>
wrote:

> Sorting the base 10 digits of the Fibonacci numbers (A272918) does not
> really seem to go against what we know about Benford's law and the
> Fibonacci numbers. But if we sort the digits in descending order, it seems
> that, aside from the trivial initial exceptions, no term will start with
> the digit 1.
>
> Of course a base 10 repunit is not the only kind of number that would allow
> a Fibonacci number to be in this analogue to A272918. A Fibonacci number
> that has only 1s and 0s would then become something like 111111111000 in
> this sequence, but that seems unlikely as well.
>
> Al
>
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