From: steiner@math.bgsu.edu (ray steiner) Subject: Re: number theorie - only one solution ? Date: Sun, 08 Aug 1999 15:54:14 -0500 Newsgroups: sci.math Keywords: y^q=1+x+x^2+...+x^r (Ljunggren) Hi, all! I would like to rescind a previous statement I made on this topic. Let's consider the solution of the more general equation (x^n-1)/(x-1)= y^q, (I) where (x,n,y,q) are positive integers and we may suppose WLOG that q is prime. It was proved by Ljunggren in 1943 in an obscure Norwegian journal(Norsk. Mat. Tidsskrift, 25(1943), 17-20) that if q=2 the only solutions are (x,n,y.q)= (3,5,11,2) and(7,4,20,2). As soon as I can secure a copy of Ljunggren's paper I will post an outline of his proof(or post a new one!). If q > 2 the only solution known is (18,3,7,3). Unsolved problem: Are there any others? A very recent paper(really a note) on this topic is by Bugeaud and Mignotte: Sur l'equation diophantienne (x^n-1)/(x-1)= y^q, II Comptes Rendus Acad Sci. Paris Mathematiques 328(1999), 741-744. Among other things, they outline a proof of the result that eqn (1) has no solutions for x=10. In particular, no repunit can be a power! More on this topic later! Regards, Ray Steiner -- steiner@math.bgsu.edu ============================================================================== From: Roger Alperin Subject: Re: number theorie - only one solution ? Date: Sat, 14 Aug 1999 11:19:51 -0800 Newsgroups: sci.math I saw an article recently on this equation in J of London Math Soc. and the authors classified all solutions where one of the variables was square-free. I think Baker may have been one of the authors. Check MathSciNet for details. These exponential diophantine equations are notoriously difficult. ============================================================================== From: steiner@math.bgsu.edu (ray steiner) Subject: Re: "Number Theorie: only one solution?" Some comments and questions Date: Mon, 23 Aug 1999 16:22:28 -0500 Newsgroups: sci.math Hi, all! An expanded version of the original question is the following: Find all positive integer solutions with x > 1 of the equation: (x^n -1)/(x-1) = y^2. (*) I've grown tired of waiting for Ljunggren's paper(via super slow mail) and I want to work out a proof of my own. So, here are some comments and questions: CASE I. n = 2m+1. I found an article by Kanold in J. fur die Reine und angew. Math(245(1972), 165-171), in which he proves the following result: If (*) has an integer solution x, y with x>= 2^(2m-1)/sqrt(3m+3) then m=2, x=3 and y= 11. Question: How can we do this problem with x < 2^(2m-1)/sqrt(3m+3) ? CASE 2. n= 2m. Now (*) can be written (x*m)^2- (x-1)y^2=1, or, letting d= x-1. ( (d+1)^m)^2-dy^2=1. Is there any way to prove, using continued fractions perhaps, that this can happen only for d=6 and m=2? Probably the hardest case occurs when d is squarefree. Then (d+1)^m/y must be a convergent in the continued fraction expansion of sqrt(d). Why is this impossible if d>6? Regards, Ray Steiner -- steiner@math.bgsu.edu