From: Dave Rusin Subject: Re: Methods of solving x^5 - Dy^5 = 1? Date: Mon, 27 Mar 2000 10:30:11 -0600 (CST) To: joecr@microsoft.com In article <53C1BD5F0C42D2118FC700805F9F2EAC0F993BF5@CLT-MSG-02> you write: > I'm aware of the methods of solving the Pell/Brouncker equation >x^2 - Dy^2 = 1 for non-trivial integer {x,y,D} given D, based on the >properties >of continued fractions. > > Anyone know or could point me to references, on solving >such equations of the form x^n - Dy^n = 1 for n > 2? Or at least >the cases where n = 3 & 5? These are very different cases! For n=2 there are infinitely many integer solutions. For n=3 there may be infinitely many rational solutions for some D but only finitely many integer solutions for each D. For n>3 there are only finitely many rational solutions (and thus only finitely many integer solutions too) for each choice of D. When n=3 you have described an elliptic curve. There are procedures which often describe the complete set of rational solutions, although it is known that these procedures may fail to give a definitive answer to the existence of solutions in some cases. I find for example that there are just a couple of rational points for D=1, D=2, and only the point (1,0) when D=3, 4, 5 but a countably infinite set when D=6, starting with [-17/37, -21/37], [2237723/1805723, 960540/1805723], [84691068680987/209143555850753, -112490043311709/209143555850753], ... and their "inverses" e.g. [-37/17, -21/17]. If you read more about elliptic curves you'll find the solution set has the structure of a finitely-generated abelian group; usually in such a situation you'll find that by hunting around with various D you'll eventually find a case with rank larger than 1 and indeed D=19 gives a curve of rank 2, that is, there are "more" solutions in this case, starting with the simple generators [2/3, -1/3] and [-3/5, -2/5]. > E.g. One such equation I'm interested in is x^5 - 625y^5 = 1. As I say there are only finitely many rational solutions, although the proof of this statement (Faltings' theorem) is not constructive. There are constructive methods which apply in some cases; I don't know very much about that, but would be pleased to hear what you learn. You're doing arithmetic algebraic geometry. For some information see e.g. index/14HXX.html (also section 11G) dave ============================================================================== From: Joe Crump Subject: RE: Methods of solving x^5 - Dy^5 = 1? Date: Mon, 27 Mar 2000 08:38:52 -0800 To: Dave Rusin Hi Dave, Thanks for your reply. I got a lot of feedback pointing me to helpful articles & such. The best references are apparently: ---------------------------------------------------------------------- Yu. Bilu and G. Hanrot, ``Solving Thue equations of high degree'', J. Number Theory 60 [1996], 373-392. CMP 97:02 Mike Bennett & Benne de Weger in Math. Comp. 67 (1998), 413-438. ---------------------------------------------------------------------- There are only a finite number of integer solutions for n > 3, and there are routines readily available for calculating these. Several people pointed out that the x^5 - 625y^5 case has only the trivial solution [1,0], calculated via methods described by Thue (the "pioneer of these types of equations") and implemented in packages such as GP/PARI and KASH. - Joe [quote of previous message deleted --djr]