From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: rational solution to cubic equation?
Date: 1 Nov 1996 21:18:48 GMT

In article <55cpru$qt8@rzsun02.rrz.uni-hamburg.de>,
Hauke Reddmann <fc3a501@GEO.math.uni-hamburg.de> wrote:
>Joel Samson Galang (jgalang@frogger.rs.itd.umich.edu) wrote:
>: How do I go about finding [other than trial-and-error] the rational
>: solutions for:
>: 
>:      x^3 + y^3 = 6
>: 
>AFAIK (if I understood Dave Rusin correctly) all mixed cubics
>can be transformed to a standard elliptic curve z^2=w^3+Aw+B.

"Almost all". An elliptic curve is among other things a group, which means
it has a distinguished element (the identity element). This means:
	(1) It's not an EC if it's empty, and
	(2) It's not _an_ EC but rather an isomorphism class of ECs until
		you pick an element to serve as the identity.
(To a topologist this is the distinction between a space and a pointed
space.) The standard cubics  z^2=w^3+Aw+B  all have a "point at infinity"
which is universally chosen as the identity element of the group.

Your run-of-the-mill rational cubic equation in two variables may not have any
rational solution, which means you won't be able to transform it to that
special form. (Some cubics which appear to have no solutions may have some
at infinity, which happens iff the terms of degree three form a
homogeneous cubic with a linear factor. This applies to the original
equation posted above, since  (x^3+y^3)  has  x+y  as a factor.)

For example, the equation   3 x^3 + 4 y^3 = 5  has no rational solutions
(including the points at infinity) and so cannot be reduced to the
Weierstrass normal form using a birational transformation with rational
coefficients. (This example is due to Selmer. What's interesting about
it is that it _is_ solvable mod  p^n  for any prime power  p^n, that is,
the Shafarevich group  Sha  is nontrivial for this curve.)

>Loads of knowledge on this theme. But it's higher math.

Well, yes, although I have to say we're still missing some critical pieces
of the puzzle. Given a quadratic in two variables, I know how I would go
about finding all solutions. That's really secondary-school mathematics.
Given  a cubic in two variables _and_ a list of generators of the free part
of the Mordell-Weil group, I can also find all solutions with the
same level of mathematical expertise (even if a teenager would have no idea
what those words mean). The "higher math" you refer to gives considerable
information about the number of such generators, but I would say that 
we don't have sufficient techniques yet to take any elliptic curve with
rational coefficients and even know for sure how many generators there are,
much less how to find them. The experts may have a leg up on the high
school students but they cannot yet claim the subject is finished.

Incidentally, the original cubic shown above was discussed at length in
this group when posed as a question (with a 100 pound prize!) by one of
the London newspapers for New Year's Day 1995. I saved a couple of the
relevant posts at that time; they're at
    http://www.math.niu.edu/~rusin/known-math/95/cubic.prize
[URL updated 1999/01 -- djr]

dave