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 <alperin@jps.net>
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