[seqfan] Re: 1^5+2^5+...+n^5 is a square

[Ignacio Larrosa Cañestro]

El 22/10/2012 16:23, Charles Greathouse escribió:
> It was recently asked (on MathOverflow, I think) whether the formula on A031138:
>
> a(n) =11*(a(n-1)-a(n-2)) + a(n-3)
>
> was proved or merely conjectural. Of course it should be proved to be
> included as it is, but would someone verify this?
>
> This is of course a 6th-degree Diophantine equation:
>
> 12m^2 = n^2 (n+1)^2 (2n^2 + 2n - 1)
>
> Charles Greathouse
> Analyst/Programmer
> Case Western Reserve University
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/

12m'^2 = n^2 (n+1)^2 (2n^2 + 2n - 1)  ===>

3m^2 = 2n^2 + 2n - 1   (#1)

with m = 2m'/(n(n+1))

6m^2 = 4n^2 + 4n - 2 = (2n + 1)^2 - 3

Let q = 2n + 1. Then

q^2 - 6m^2 = 3   (#2)

The first solutions of that equation are (q, m) = (3, 1), (27, 11), (267, 109), ...

The associated regular Pell equation is

q^2 - 6m^2 = 1  (#3)

with minimal non trivial solution solution (q, m) = (5, 2).

It is well known that all the solutions of #3 are then

q(i) + sqrt(6)m(i) = (5 + 2*sqrt(2))^i

The solutions of #2 let be obtained from a finite set of solutions combined with the solutions of #3.

But

(5 + 2*sqrt(6))(3 + sqrt(6)) = 27 + 11*sqrt(6)

the second solution of #2. This say us that all the solutions of #2 are

q(i) + sqrt(6)m(i) = (3 + sqrt(6))(5 + 2*sqrt(6))^i

Then

q(i+1) + sqrt(6)m(i+1) = (5 + 2*sqrt(6))(q(i) + sqrt(6)m(i))

q(i+1) = 5q(i) + 12m(i)
m(i+1) = 2q(i) + 5m(i)

q(i+2) = 5q(i+1) + 12m(i+1) = 5q(i+1) + 24q(i) + 60m(i)
            = 5q(i+1) + 24q(i) + 5q(i+1) - 25q(i) = 10q(i+1) - q(i)

But q(i) = 2n(i) + 1. Then

2n(i+2) + 1 = 20n(i+1) + 10 - 2n(i) - 1
n(i+2) = 10n(i+1) - n(i) + 4

the non-homogeneus linear requrrence ecuation for n(i).  From that

n(i+3) = 10n(i+2) - n(i+1) + 4

n(i+3) - n(i+2) = 10n(i+2) - 10n(i+1) - n(i+1) + n(i)

n(i+3) = 11(n(i+2) - n(i+1)) + n(i),    n(1) = 1, n(2) = 13, n(3) = 133

the homogeneus linear requrrence ecuation for n(i) in 
http://oeis.org/A031138.

Excuse me for my poor English expression ...

-- 
Saludos,

Ignacio Larrosa Cañestro
A Coruña (España)
ilarrosa at mundo-r.com
http://www.xente.mundo-r.com/ilarrosa/GeoGebra/