From: "Achava Nakhash, the Loving Snake" Subject: Re: x^2-1=y^m Date: Wed, 10 Mar 1999 15:56:22 -0800 Newsgroups: sci.math Gerry Myerson wrote: > In article , blang@club-internet.fr > (Bruno Langlois) wrote: > > > Who can solve the diophantine equation : > > x^2-1=y^m (x,y,m > 1) > > Chao Ko, On the diophantine equation x^2 = y^n + 1, xy \ne 0, Scientia > Sinica (Notes) 14 (1965) 457--460, MR 32 #1164. > > See also Mordell, Diopantine Equations, pp. 302--304. > > Gerry Myerson (gerry@mpce.mq.edu.au) See also Paolo Ribenboim's book, Catalan's Conjecture. There is a lot of great stuff in this book, but there are also a huge number of typos's and statements that aren't very clear. I think that the proof of Chao Ko's result is pretty clean, though. Regards, Achava ============================================================================== From: Kurt Foster Subject: Re: x^2-1=y^m Date: 11 Mar 1999 17:00:41 GMT Newsgroups: sci.math In , Bruno Langlois said: . Who can solve the diophantine equation : . x^2-1=y^m (x,y,m > 1) [References to literature already posted by Gerry Myerson] Since x-1 and x+1 cannot have a common divisor greater than 2, and two consecutive odd positive integers can't both be m-th powers, the only possiblities are when x is odd. Either x+1 is twice an m-th power and x-1 half an (even) m-th power, or vice-versa. One obtains the equation W^m - 2^(m-2) * Z^m = +/- 1. There are no solutions other than xy = 0 when m = 2. The equation implies |W/Z - 2^(1 - 2/m)| < 1/[C(m) * Z^m] where C(m) > 1. This indicates an unusually good approximation to the algebraic number 2^(1 - 2/m) by the rational W/Z when m > 2, which makes me suspect there aren't many solutions. The only one I see offhand is the obvious one m = 3, W = Z = 1, which corresponds to x^2 = 9, y^3 = 8.