Date: Fri, 25 Oct 1996 13:12:17 +0200
From: [Permission pending]
To: rusin@vesuvius.math.niu.edu (Dave Rusin)
Subject: Re: ...and another elliptic curve (LONG attachment)
Newsgroups: sci.math.research

In article <549sc7$5aa@gannett.math.niu.edu> you wrote:
<snipped what I didn't understand :-) >

Thanks for your answer! I will try to get the package
from the Internet, although the theory of elliptic
curves is totally greek to me. (I understand the basic
concepts, but that's all.)
[sig deleted]
P.S. The trick to convert a22x^2y^2+...+a00=0 into a
cubic - what's the general form?
P.P.S. Is there a trick for a biquadratic too?
I have u^4-44u^2-16v^2-16=0, but NO rational point
that I know of.
P.P.P.S. I have formalized the process of turning
v^2=a4u^4+...+a0, with known rational point u0,v0, 
into a standard cubic.
Are you interested in my formulas or is that
well known already?
[sig deleted]
==============================================================================

Date: Fri, 25 Oct 96 15:29:41 CDT
From: rusin (Dave Rusin)
To: [Permission pending]
Subject: Re: ...and another elliptic curve (LONG attachment)

[deletia -- djr]

>P.S. The trick to convert a22x^2y^2+...+a00=0 into a
>cubic - what's the general form?

Focus on the dependence on  y: the equation is
	A(x) y^2 + B(x) y + C(x) =0.
Now multiply by  4A(x)  and complete the square:
	[ 2A(x)y + B(x) ]^2 = B(x)^2 - 4 A(x) C(x)
so you have an equation of the form  u^2 = quartic in  x.

Several classes of equations can be made to reduce to ECs like this.
A very nice reference is Cassel's book on elliptic curves. It explains
how to do the transformations in some of these cases. I took the
procedure verbatim and wrote it up as maple code; it transforms
any cubic in two variables to canonical form. It's available at
http://www.math.niu.edu/~rusin/known-math/96/elliptic.maple
 [That's an updated URL -- djr 1999/01]
There's a companion file ../quartic.maple  to handle the case
y^2=quartic.

>P.P.S. Is there a trick for a biquadratic too?
>I have u^4-44u^2-16v^2-16=0, but NO rational point
>that I know of.

Note that part of the definition of an elliptic curve is that you
have a curve _with a distinguished point_ to be taken as the
identity element of the group (a torus is homogeneous; you have to
specify which element is to be "0"). In particular, you don't have
an elliptic curve unless you have at least one point. All the
algorithms for reduction to canonical form will assume a rational
point (which will ultimately be moved to infinity).

In your example, there already is a point at infinity which may be
used as the identity element. Apecs has a routine ("Quar0") to transform
any quartic to canonical form if it can find a rational point. In this
example it also notices the point at infinity and says that the curve
you propose is equivalent to the EC
	y^2=x^3+x-581x-7751
(and it gives a birational equivalence). This curve has no torsion
points and, if I interpret various programs properly, it has rank  0  as
well, so there are no rational points to be had.

dave

==============================================================================

From: [Permission pending]
Subject: Re: ...and another elliptic curve (LONG attachment)
To: rusin@math.niu.edu (Dave Rusin)
Date: Mon, 28 Oct 1996 00:59:22 +0200 (DFT)

Thanks again.
I already suspected that: a rational point on that curve
probably would have led me to a perfect cuboid.
And at least I have found something what gives my hunch
that none exists some fresh evidence:
I have some biquadratic with an inner discriminant which
is quadratic in one parameter, and the discriminant of
THAT again is 2*the space diagonal. Proves nothing, but
smells VERY funny. (Without any conditions, I also have

      g^2=2.*(f**6-5.*f**4+3.*f**2+1.)

and if you could splat THAT curve (f^2=h, maybe already
this has no solutions or countably many) I'm done.)
[sig deleted]