From: rusin@vesuvius.math.niu.edu (Dave Rusin)
Newsgroups: sci.math
Subject: Re: Rational points on a^3+b^3=k
Date: 18 Nov 1997 20:01:17 GMT
The subject line reads,
Rational points on a^3+b^3=k
The case k=6 has arisen in this newsgroup before.
Thomas Womack wrote:
>I've found non-trivial solutions for this by brute-force search for
>k=6,7,12,13,15,19,20,26,28,30,33,35,37,43.
...
>For some values of k, there is more than one solution:
...
>There are three obvious questions :
Well, you asked four, but that's OK. The last is easiest:
>Where could I find out more about this problem - and, ideally, find answers
>to the three questions? Which region of maths does it fall in?
You want to know about elliptic curves (over the rationals). That applies
usually to single polynomial equations in two variables, with total
degree 3 (or sometimes 4). For more information about elliptic curves, try
http://www.math.niu.edu/~rusin/known-math/index/14H52.html
>Which values of k are possible? (1 clearly isn't; the proof that 3 isn't is
>in Hardy + Wright)
>
>Are there infinitely many solutions for each k?
>
>How do you construct them?
Use the substitution
a=(k/6+Y)/X
b=(k/6-Y)/X
to rewrite the equation in Weierstrass form:
Y^2 = X^3 + D
where D = - k^2/108 . You want to know if there are any rational points
on this curve, and how to construct them. Well, this curve can be
given the structure of a finitely generated abelian group, so there is
a construction (the "chord and tangent process") which will create all
the solutions from a finite number of them. The torsion part of this
group is pretty tame. If you want infinitely many k, you need to know
whether the rank of this group is positive. Well, software exists for
this kind of thing:
k rank generators:(X,Y)->(a,b)
1 0
2 0
3 0
4 0
5 0
6 1 (7/9, 10/27)->(37/21, 17/21)
7 1 (7/3, 7/2)->(2,-1)
8 0
9 1 (1,1/2)->(2,1) [You didn't mention this one]
10 0
11 0
12 1 (13/9,35/27) -> (89/39,19/39)
13 1 (13/9,65/54) -> (7/3,2/3)
14 0
15 1 (49/36, 143/216)->(683/294,397/294)
16 0
17 1 (7/3,19/6)->(18/7,1/7) [Evidently you didn't know about this.]
18 0
19 2 (19/3,95/6)->(3,-2), (19/12,19/24)->(5/2, 3/2)
20 1 (7/3,3)-> (19/7,1/7)
Other rank-1 curves in the range k=1, 2, ..., 50 are those for k=
22 (try a=25469/9954, b=17299/9954), 26, 28, 31 (try a=137/42, b=-65/42), 33,
34 (try a=631/182, b=-359/182), 35, 42 (try a=449/129, b=-71/129), 43,
48 (try a=74/21, b=34/21), 49 (try a=11/3, b=-2/3) and k=50 (try a=23417/6111,
b=-11267/6111).
For k=30 and k=37 the rank is actually 2.
There are also solutions in the torsion subsgroups in some cases, but
I think this happens iff k = m^3 for some m (in which case the torsion
subgroup has order 3 and we have the trivial solutions {a,b}={m,0}) or
k=2m^3 (in which case the torsion subgroup has order 2 and we have the
trivial solutions a=b=m). It is only if the torsion group is nontrivial
and the rank is zero that there can be only finitely many solutions to
the original equation, and that appears to happen iff k or k/2 is a cube.
The question, "Can you predict the rank of the elliptic curves in the
following 1-parameter family..." usually has answer "no". I'm sure
there has been work on this particular family, since it's so well
known, but I don't think there's a conclusive answer. In some sense
the "average" rank is supposed to be 1/2, I believe.
dave
==============================================================================
Newsgroups: sci.math
From: iain@stt.win-uk.net (Iain Davidson)
Date: Wed, 19 Nov 1997 16:56:53 GMT
Subject: Re: Rational points on a^3+b^3=k
In article <64sc62$k3j$1@news.ox.ac.uk>, "Thomas Womack" (mert0236@sable.ox.ac.uk) writes:
>I've found non-trivial solutions for this by brute-force search for
>k=6,7,12,13,15,19,20,26,28,30,33,35,37,43.
>
>For example, (397/294)^3 + (683/294)^3 = 15
>
>Which values of k are possible? (1 clearly isn't; the proof that 3 isn't is
>in Hardy + Wright)
From the identity
(x^3 + y^3 - 3xy)^3 + (3xy(x-y))^3 =
(x^3 +3x^2y -6xy^2 +y^3)*(x^2 -xy +y^2)^3
a^3 + b^3 = k, k of the form (x^3 + 3x^2y -6xy^2 + y^3) will always
have a solution and by chord or tangent method an infinitude of
solutions.
If you expand the roots of u^3 +3u^2 -6u + 1 by continued
fractions, you will get x,y that give some smaller values of k.
Iain Davidson Tel : +44 1228 49944
4 Carliol Close Fax : +44 1228 810183
Carlisle Email : iain@stt.win-uk.net
England
CA1 2QP