From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: Re: diophantine eqn. a*x^2+b*y^3=z^6 Date: 25 Oct 1999 06:53:18 GMT Newsgroups: sci.math Keywords: The elliptic curves y^2=x^3+D In article <7utqvq$iqv$1@nnrp1.deja.com>, David Bernier wrote: >I am interested in the Diophantine equation: > a*x^2 + b*y^3 = z^6, where (say) a>0, b>0 are integers >and (x,y,z) \in Z^3 - {(0,0,0)}. I know that this eqn. >is non-homogeneous. True, but it's close enough to homogeneous to help: set x'=x/z^3 and y'=y/z^2 and you get a rational solution to the equation a x'^2 + b y'^3 = 1 and conversely from any such rational solution you may recover x,y,z. >I've heard of Faltings' theorem, but I can't say if it >says anything about such an equation. No, not if you mean The Theorem Formerly Known as Mordell's Conjecture; that asserts the finitude of the set of rational solutions to equations which describe curves of genus greater than 1. Your curves are all of genus 1 (assuming a, b nonzero), and indeed are reduced to Weierstrass Normal Form with the substitutions x' = Y/(a^2*b), y'=X/(-a*b), yielding Y^2 = X^3 + (a^3*b^2) >I plan to look for solutions >in the rings Z_{p^k} for p prime and k=>0, and, if all >goes well, choose an a,b that pass the test for these >rings and look for solutions in Z^3. OK. What does it mean to look in Z_{p^k} ? There are infinitely many such rings... And what does "all goes well" mean? And how long should you look for solutions in Z^3 before giving up (if you find none)? And if you should find some solutions, what will that say about the existence of other solutions to the given equation? Your plan of attack is incomplete. It's unclear whether or not you intended a,b to be integral or rational (or something else?) but in fact you are describing the same family of elliptic curves in each case: these are the curves Y^2 = X^3 + D where D can be any integer (with the understanding that Y^2 = X^3 + D*k^6 describes the same curve for any integer k). This is a well-studied family of elliptic curves. You will find several refences to them at index/14H52.html There are some fun ones which serve as a warning: it's not always easy to know how far to search to find solutions. For example, with D=1090 we find the solution X=28 187 351, Y=149 651 610 621. (!) I don't have a table of values handy but with today's software it is easy to generate lots of data. Here are some small values of D showing in each case the rank of the elliptic curve. (The rank is positive precisely when there are infinitely many solutions. There are also "torsion" solutions (-n,0) when D=n^3 and when D=n^2 the solutions (0,n) and (0,-n). The only other torsion is when D=n^6; we then add points (2n^2, +-3n^3).) Also shown are the necessary number of independent solutions. From these generators one may compute all the other solutions; for example, when D=2, for all the solutions (x,y), either (x,y) or (x,-y) is one of: [-1, 1], [17/4, -71/8], [127/441, 13175/9261], [66113/80656, -36583777/22906304], [108305279/48846121, 1226178094681/341385539669], [-174016613231/306196222500, -228355009922164103/169433679720375000], ... so as the original poster suggested, these points are getting "sparse" in Z^3. dave Torsion-free part of Y^2 = X^3 + D: D Rk Generators 1 0 2 1 [-1, 1] 3 1 [1, 2] 4 0 5 1 [-1, 2] 6 0 7 0 8 1 [1, 3] 9 1 [3, 6] 10 1 [-1, 3] 11 1 [-7/4, 19/8] 12 1 [-2, 2] 13 0 14 0 15 2 [1, 4] [1/4, 31/8] 16 0 17 2 [-2, 3] [4, 9] 18 1 [7, 19] 19 1 [5, 12] 20 0 -1 0 -2 1 [3, 5] -3 0 -4 1 [2, 2] -5 0 -6 0 -7 1 [2, 1] -8 0 -9 0 -10 0 -11 2 [3, 4] [9/4, 5/8] -12 0 -13 1 [17, 70] -14 0 -15 1 [4, 7] -16 0 -17 0 -18 1 [3, 3] -19 1 [7, 18] -20 1 [6, 14] ============================================================================== From: Dave Rusin Subject: Re: your reply to my post on "a*x^2 + b*y^3 = z^6" Date: Tue, 26 Oct 1999 02:33:22 -0500 (CDT) Newsgroups: [missing] To: bernier@my-deja.com I had written > and conversely from any such rational solution you may recover x,y,z. which is perhaps too strong of a statement: we cannot uniquely recover (x,y,z) from a rational solution (x', y') of a x'^2 + b y'^3 = 1 , since for any nonzero integer t we have another integer solution (t^3 x, t^2 y, t z) which leads to the same rational point (x', y'). Perhaps it would be better if I were more precise: We have two equations (1) a X^2 + b Y^3 = Z^6 (2) a X^2 + b Y^3 = 1 If (x,y,z) is an integer solution to (1) with z<> 0, then f(x,y,z) = (x',y')=(x/z^3, y/z^2) is a rational solution to (2). If (x', y') is a rational solution to (2) with x'=p/q, y'=r/s in lowest terms (including the restriction q>0, s>0) then let z be the smallest positive integer with q | z^3, s | z^2 (e.g. z=q*s is such an integer, and if z1 and z2 are such integers, it is easily checked that gcd(z1,z2) is too; so this minimum exists). We then set x=z^3 * x', y=z^2 * y' ; these are both integers too, and g(x',y')=(x,y,z) is a solution to (1). Clearly f(g(x',y'))=(x',y'). Now, it need not be true that g(f(x,y,z))=(x,y,z). Rather, we compute g(f(x,y,z)) by writing x/z^3=p/q, y/z^2=r/s in lowest terms and computing the least positive zz with q | zz^3, s | zz^2; note zz | z. Then g(f(x,y,z))=((zz/z)^3*x,(zz/z)^2*y,zz)=(x/n^3, y/n^2, z/n) for n = z/zz. In particular, this _will_ be (x,y,z) again if (x,y,z) is a "primitive" solution, that is, one for which it is impossible to find an integer n with n | z, n^2 | y, n^3 | x, and z/n > 0 except n=1. I'm not sure if this quite answers your question here but I hope it clears up the problem anyway: You wrote >If a*(p/q)^2 + b*(r/s)^3 = 1, I get: > >a*(p*q^2*s^3)^2 + b*(r*s*q^2)^3 = (q*s)^6 > >Supposing as usual that (p,q)=1 and (r,s)=1, >might we get q*s to be a multiple of >t^6 with some integer t>1? To illustrate, I compute all the primitive integer solutions of 2x^2+4y^3=z^6. As in my previous post, set x=z^3*Y/2, y=-z^2*X/2 to get Y^2 = X^3 + 2 , whose rational solutions are in the table in my previous post: > [-1, 1], > [17/4, -71/8], etc. The first point thus corresponds to triples (x,y,z) with x=z^3/2, y=z^2/2. The smallest positive integer z for which this is an integral triple is z=2, giving the point (x,y,z) = (4,2,2) which is primitive; all other integral points correponding to this rational point [-1,1] are of the form (4t^3, 2t^2, 2t) for some integer t. (Note that as I mentioned, the elliptic curve includes a rational point [X,-Y] for each [X,Y]; from [X,Y]=[-1,-1] we similarly get the integer solutions (-4t^3, 2t^2, 2t).) The second point corresponds to triples (x,y,z) with x=-z^3*71/16, y=-z^2*17/8. The primitive solution is (-284, -34, 4) so that we have the collection of integer points (-284 t^3, -34 t^2, 4 t) (as well as (284 t^3, -34 t^2, 4 t) as in the previous paragraph). Similarly we compute primitive integer solutions (x,y,z)= (+-52700 t^3, -254 t^2, 42 t) from the rational point > [127/441, 13175/9261]; integer solutions (x,y,z)=(+- 146335108 t^3 , - 132226 t^2 , 568 t) from > [66113/80656, -36583777/22906304]; (x,y,z)=(+-4904712378724 t^3 , - 216610558 t^2 , 13978 t) from > [108305279/48846121, 1226178094681/341385539669], and so on. dave