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 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