Date: Fri, 2 May 97 07:25:39 CDT
From: rusin (Dave Rusin)
To: kuwata@matin.math.unicaen.fr
Subject: Re: Seeking help with an elliptic curve


Dear Prof Kuwata,

Your timing is unbelievable! Just this past weekend I was able to return
to this surface and found the same point in  E(Q(T)) (it appears to be
a generator). I am very anxious to hear of your methods. What I was able
to do was to compute many points on the specializations of the curve and
then piece them together -- not an interesting strategy except that the
methods of finding the points on the curves over  Q  are apparently not
well known. I've spent the Spring working on this, principally with John
Cremona, who is now rewriting his program  MWRANK. Connell has already
rewritten  APECS  to incorporate this. Below I attach the mail I just
sent them this week.

By the way, the average rank of the specializations (I did calculations
on about a hundred of them) is just over  2.0. This still seems a bit
high since  E(Q(T))  is evidently of rank 1.

I look forward to hearing more details of your method.

dave

----------------------------------------------------------------------
Date: Tue, 29 Apr 97 20:12:51 CDT
From: rusin (Dave Rusin)
To: cremona@math.u-bordeaux.fr
Subject: An application

I thought you might be interested in an application of our efforts this
Spring. When I first wrote you last Fall it was because I was trying to
find solutions to the equation
	y^2= x ( x^2 - x + (1-T^10)/5 )
which is really a special case of the curve
	U y^2= x ( x^2 - x + (1-U^5)/5 ).
I had points for a few  T  and evidently there were points for every  T, but
I couldn't find a parameterization. By Christmas I was stumped, as MWRANK
and APECS were painfully slow and hampered by large  Sha's.

As you know, the search for methods to avoid these problems has led to
much improved algorithms which this week I turned loose on the above curve.

Thanks to the New Improved methods of estimating ranks it was only a day's
task for the machine to find a couple dozen values of  U  for which the rank 
was clearly  1. With New Improved quartic reduction and   ratpoint   it was 
another afternoon's work to actually compute these points which  APECS  easily 
transformed to native coordinates (above). (A uniform choice of generator
may be made). As luck would have it, the generators found for individual
U's  appeared close to each other on the real number line (that is, I was
not looking at  2P0  on one curve and  3P0  on another, say).

Factoring the  x-coordinates showed clearly that each differed from a
square by  (1-U)*U . Dividing and extracting square roots gave some fairly
small numbers x'(U) which, however, did appear to grow in height with  U. 
Having plenty of points to work with I computed heights of  U  and  x'(U); a 
scatter plot clearly showed a clumping along the line with slope  5, so I 
expected  x'(U)  to be a ratio of polynomials of degree at most 5. With
just a dozen of the points it was easy to determine the unknown coefficients
and find the x-coordinates of the points on the curve to be

U*(1-U^5)/(1-U)*(1+6*U+6*U^2+11*U^3+U^4)^2/(-5*U-10*U^2-10*U^3-5*U^4+5*U^5)^2

The result is easily checked to hold for all  U.


This enables me to fill in some holes in the tables I was developing.
(I had tried to compute ranks and generators for about a hundred of
the initial curves with  T  ranging in complexity up to a worst of  53/5).
I now know the ranks precisely in all but about half a dozen of the worst
curves, and have all the generators I need for the great majority of
those  T's.

I don't know how I would have been able to find this point in
E(Q(U))  without all the algorithms we've developed lately!

dave