From: "Iain Davidson" <Sttscitrans@tesco.net>
Subject: Re: Sum of Three Cubes is a Square (the 4 families)
Date: Thu, 19 Jul 2001 00:58:31 +0100
Newsgroups: sci.math
Summary: Parameterized solutions in integers 

Tito Piezas III <tpiezas@uap.edu.ph> wrote in message
news:ucfdqgvx4t4h@forum.mathforum.com...
> Hello all,
>
> The identities,
>
>          40^3 +  2^3 +  1^3 = 253^2
>          17^3 +  4^3 +  4^3 = 71^2
>          16^3 +  6^3 +  9^3 = 71^2
>          25^3 + 12^3 + 36^3 = 253^2
> and
>       22520^3 + 10^3 +  1^3 = 3379501^2
>        5665^3 + 20^3 +  4^3 = 426383^2
>        2560^3 + 30^3 +  9^3 = 129527^2
>         745^3 + 60^3 + 36^3 = 20341^2
>
> have the parametrization,
>
> (36a^4 + 4ab^3)^3 + (2ab^3)^3 + (b^4)^3 = (216a^6 + 36a^3b^3 + b^6)^2
>
> (9a^4 + 8ab^3)^3 + (4ab^3)^3 + (4b^4)^3 = (27a^6 + 36a^3b^3 + 8b^6)^2
>
> (4a^4 + 12ab^3)^3 + (6ab^3)^3 + (9b^4)^3 = (8a^6 + 36a^3b^3 + 27b^6)^2
>
> (a^4 + 24ab^3)^3 + (12ab^3)^3 + (36b^4)^3 = (a^6 + 36a^3b^3 +216b^6)^2

These seem to be basically equations of the form

x^3 + (py)^3 + (qy)^3 = s^2

x^3 +(p^3 +q^3)y^3 = s^2

So basically solutions to x^3 +Ay^3 = s^2,
where A is a sum of two cubes are required

The cubic form
x^3 +Ay^3 +A^2*z^3 -3Axyz can be factorised
as
(x +yr + zr^2)(x +ywr + zwwr^2)(x +ywwr + zwr^2)
 r cubrt(a), w primitive cube root unity

The form can be made a square by assuming
(x +ywr + zwwr^2) = (t +uwr + vwwr^2)^2

Setting  z =0,
x^3 +Ay^3 = s^2
A =p^3 +q^3,

Then,
x = 4t(t^3 +(p^3+q^3)u^3)
y = -u(8t^3 -(p^3+q^3)u^3)

t =1,u=1, p=2,q=1

40^3 +2^3 +1^3 =253^2

 t =1,u=2, p=2,q=1

292^3 + 256^3 + 128^3 =6616^3

The method for x^3 +Ay^3 =s^2, does not
seem to give all solutions though, e.g
x^3 +2y^3 = s^2, x =17, y=4, s =71
==============================================================================

From: ayatollah potassium <sanguis@bibimus.edu>
Subject: Re: Sum of Three Cubes is a Square
Date: Thu, 26 Jul 2001 19:28:54 -0400
Newsgroups: sci.math

All rational solutions of   x^3 + y^3 + z^3 = w^2  are parametrized by:

x = a (a^3+b^3+c^3) t^2
y = b (a^3+b^3+c^3) t^2
z = c (a^3 +b^3+c^3) t^2
w = (a^3+b^3+c^3)^2 t^3.

James Buddenhagen wrote:

> I'd be interested in parameterizations where d is not a cube.
>
> --Jim Buddenhagen