[seqfan] Re: Pythagorean triples of triangular numbers?

[Andrew N W Hone]

Hi Brad and Jon, 

The equation 

T(x)^2 + T(y)^2 = T(z)^2

is not homogeneous in x,y,z, so it defines an affine surface in three dimensions. If there is some way to transform 
it to a quartic plane curve of genus 3, then that would be interesting. Please supply details!

Faltings' theorem says that curves of genus g>1 have only finitely many rational points. However, getting
good effective bounds on the size (height) of rational solutions can be very hard, and there can be huge solutions. 
This article by Everest and Ward has a nice discussion about this problem: https://www.tandfonline.com/doi/abs/10.4169/amer.math.monthly.118.07.584   (It is available free on arXiv and Jstor.)

All the best,
Andy 
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Brad Klee [bradklee at gmail.com]
Sent: 21 August 2018 17:17
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: Pythagorean triples of triangular numbers?

Hi Jon,

Thanks for including a question about geometry! From the
Diophantine equation,

T(x)^2 + T(y)^2 = T(z)^2,

We can apply a birational change of coordinates and write that,

a = 2*H = x^2 + y^2 - (1/2)*(x^4+y^4) ,

with a new linear parameter "a". H(x,y) is a Hamiltonian family of
algebraic plane curves, mostly with genus g>1, i.e. not elliptic.
For curves of higher genus, Falting's theorem says that addition
rules will not generate an infinite set of points.

However these curves a = 2H(x,y) have interesting symmetry,
and are really worth more investigation. The singular points are:

    a                       (x,y)                             type
============================================
    0                       (0,0)                           Circular
   1/2             (0,+/-1),  (+/-1,0)                 Hyperbolic
    1               +/-(1,1),  +/-(1,-1)                 Circular

And if you plot the function, it's easy to see that the four
hyperbolic points fall on the intersection of two ellipses,
and that the circular points fall interior to the ellipses.

Since H(x,y) is an inversion-invariant quartic, we can quickly
derive the period integral, and then to compute its defining
differential equation. In less than 1/10 of a second:

3*(2*a-1)*T(a)+4*(6*a^2-6*a+1)*T'(a)+4*a*(2*a-1)*(a-1)*T''(a)=0,

with integer-scaled solution around a=0 starting:

1 + 12*a + 228*a^2 + 5040*a^3 + 121380*a^4 + . . .
 ( Should this integral be included in OEIS? )

The interesting thing about the D.E. for T(a) is that it has only
three terms, as do the D.E.'s for elliptic curves. As far as I can
tell, this means that despite g>1, the complex-valued Riemann
surfaces a=2*H(x,y) are constructed by joining *identical* toric
sections. If we can figure out this fact, it should lead to some
really interesting ideas for new, exciting, genus-defying maps.
This is one of those questions with implications for physics
as well as number theory.

Cheers,

Brad



On Tue, Aug 21, 2018 at 3:33 AM <jonscho at hiwaay.net> wrote:

[ Content Trimmed ]

>

> -- elliptic curves offer a smart way to search for solutions?
>
> Thanks!
>
> -- Jon
>
>

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