From: ray steiner Subject: Re: k^m +1 =n^2 Date: Wed, 05 Jul 2000 17:12:29 GMT Newsgroups: sci.math Summary: Catalan's conjecture In article <3qgmz08jyug0@forum.mathforum.com>, qquet@hotbot.com (Leroy Quet) wrote: > Let k, m, and n be positive integers, m >= 2. > What can be said about solutions to > k^m +1 =n^2, > except that k is even and n is odd? > > The only solution to > 2^m +1 =n^2 > is m = n = 3. > Thanks, > Leroy Quet > A very interesting thread indeed, because it gives me a chance to share the latest results on Catalan's conjecture. Catalan's conjecture is that the only consecutive integer powers are 2^3=8 and 3^2=9. It was proposed by Catalan to Crelle in 1844 and is still open today. First, let's look at the posted question: If k^m +1 = n^2, we can assume WLOG that m is a prime q. In 1964 Chao Ko proved that n^2-1= k^q has no solution in positive integers if q >=5. His proof is given on pp 302-304 of Mordell's book DIOPHANTINE EQUATIONS. Unfortunately, there is a big gap in Mordell's presentation. On p. 302 he states "It easily follows from (66) that"..., but the connection is far from easy. It took 3 pages of a paper by Nagell in Arkiv foer Mathematik to fill this gap. Unfortunately, I don't have the reference handy. There is a simplified version of Chao Ko's proof given by Chein(1976). It can be found on pp. 92-94 of Ribenboim's excellent book CATALAN'S CONJECTURE(Academic Press, New York, 1994). This book also contains most of the results on Catalan's conjecture up to about 1992 or so. I will now give a summary of some of the recent results. Along the way, I'll pose some questions, which, if they could be answered, would go a long way toward helping to resolve Catalan's conjecture. I'll give references to the literature if they are available. Otherwise, all results are from preprints. Suppose we wish to solve x^m-y^n = 1. Again, we may assume m=p, n=q, where p, q are primes and p < q. Here are some of the known results about p and q. 1). 10^7 < p < 3.31*10^12 2). 10^7 < q < 4.13*10^17 The lower bounds were established by Mignotte late last year. The upper bounds were established by O'Neil in 1995 in an unpublished manuscript. By a bit more careful estimate I can reduce these slightly: p< 3.22*10^12 q< 4.122*10^17 Further, if p is not 1(mod 8) it seems that the upper bounds can be reduced substantially. Work in progress! 2). q <= 2.77p*log(p)*(log q- log log p + 2.34)^2. (*). This was established by Mignotte in 1995. See Experimental Math(4) (1995), 259-268. 3). p and q form a DOUBLE WIEFERICH PAIR. That is p^{q-1}= 1(mod q^2) and q^{p-1}= 1(mod p^2). This was established by Mihailescu late last year. The proof uses Stickelberger elements in the cyclotomic field Q(zeta_p). Note that there are only 6 double Wieferich pairs known: (2, 1093), (3, 1006003), (5,1645333507), (83, 4871), (911,318917) and (2903, 18787). There are no other such pairs with p < 10^7 and q satisfying (*). Incidentally, it is easy to show that Mihailescu's result implies that both the congruences q = \pm 1(mod p) are impossible. Questions: a). Are there any other double Wieferich pairs with p, q satisfyingthe conditions above? I would like to do a search, but do not have the computing power to do it alone. b). if p and q are odd primes and form a double Wieferich pair must one of p, q be congruent to 3(mod 4)? This certainly holds for all the above pairs. What can be said in general? 3). q must divide the relative class number of the field Q(zeta_p), H_p^-. In addition, if q < 1.88*p, then p also divides H_q^-. This is a very recent result of Bugeaud and Hanrot. Questions: If p, q form a Catalan pair, can p-1 and q-1 have an odd prime factor in common? What can be said about p-1 and q+1? If these questions could be answered in the negative, we would have the following remarkable conjecture: Suppose p, q form a Catalan pair. Then either q divides the relative class number of the maximal 2 subfield of Q(zeta_p), h_p^-, or, if q doesn't divide h_p^-, then q^3 divides H_p^-. Note that primes q not dividing h_p^- but dividing H_p^- to the third or higher power are extremely rare! There is only one such case for p < 500: 2939^3 divides H_227^-. Does anyone know of any such instances for p > 500? Well, that does it for now. Any info anyone can provide on any of the questions above would be most welcome! Regards, Ray Steiner -- steiner@math.bgsu.edu Sent via Deja.com http://www.deja.com/ Before you buy. ============================================================================== From: ray steiner Subject: Re: k^m +1 =n^2 Date: Fri, 07 Jul 2000 16:25:19 GMT Newsgroups: sci.math Re: The Super-Catalan problem a^m-b^n= d. The problems of determining the d for which this equation is solvable and determining whether the solutions are bounded are examples of Pillai's conjecture. Not much is known about Pillai's conjecture if d > 1. In fact the only other result of any consequence is that one can effectively bound all the solutions if d consists of powers of a fixed set (finite, of course!) of primes, e. g. , d= 2^a*3^b. The proof is in Sprindzhuk's book CLASSICAL DIOPHANTINE EQUATIONS, Springer Lecture Notes in Mathematics, vol 1559. In fact, the case d=6 is still open. So, let me pose the challenge: Are there any two integer powers whose difference is 6? Regards, Ray Steiner Sent via Deja.com http://www.deja.com/ Before you buy. ============================================================================== From: Jan-Christoph Puchta Subject: Re: k^m +1 =n^2 Date: Fri, 07 Jul 2000 12:23:56 +0200 Newsgroups: sci.math Bill Taylor wrote: > Several folk have observed that Catalan's conjecture is that there is only > that one solution to |a^m - b^n| = 1, for integer a,b,m,n > 1. > > There seems to be an obvious extension of this conjecture, that I've never > seen referred to, but I bet it's true:- > > Super-Catalan: ...similarly, for any integer d > 0, there are only > ============= finitely many solutions to |a^m - b^n| = d. > > Anyone know anything about that one? It is mentioned in Riebenboim's book on Catalan's conjecture, which is probably the best resource on this topic. This equation is more difficult then Catalan's conjecture, for the following reasons: The algebraic approach start with factoring x^n-d, then one uses the fact that x - e d^{1/n}, where e is any n-th root of unity, is the m-th power of some ideal. However, to factor x^n-d one has to use number fields with nonabelian Galois group, and things become nasty very fast. The analytic approach start fom Baker's bound for linear forms in logarithms. Naively one would write n log x - m log y < c x^{-n}, then use Baker to give a lower bound for the left hand side. However, this bound is to weak to give a contradiction. Tijdeman obtained his upper bound by applying Baker's bound to some manipulated equality, and these manipulations fail in the general case d>1. However, the general "abc implies everthing"-theorem can be applied: If the abc-conjecture is true, then for every d there are only finitely mane x, y, m, n with m, n>3 and |x^m-y^n|=d JCP