From: jr@redmink.demon.co.uk (John R Ramsden) Subject: Re: x^2+1=2y^4 Date: Thu, 08 Feb 2001 22:20:28 GMT Newsgroups: sci.math Summary: Integral solutions to some 2-variable equations of degree 3 and 4 buddhist wrote: > > Prove : > x^2+1=2y^4 only has two positive integer solutions > x=y=1 , (x=239 and y=13) . > Thx. I assume "Thx" is short for "thanks", but if it's "thinks..." you'll probably have to apply your buddhist meditation techniques for quite some time to prove this! The following isn't actually a proof, but goes a small part of the way towards analysing this equation. Firstly, there must be integers a, b, c, d with (a, d) = (b, c) = 1 such that: +/-x = a.b - c.d y^2 = |a.b + c.d| = |a.c - b.d| 1 = |a.c + b.d| By performing the following in sequence as necessary, we can ensure that every expression in || is non-negative: 1 If a.b + c.d < 0 then reverse signs of b, d. 2 If a.c + b.d < 0 then reverse signs of c, d (which leaves the sign of a.b + c.d unchanged). 3 If a.c - b.d < 0, then replace a, b, c, d by either b, a, d, c or d, c, b, a respectively (either of which leaves a.b + c.d and a.c + b.d both unchanged). Exactly one of a, b, c, d is even, and, if necessary, applying *both* the permutations in 3 we can assume that a, b, c is odd and d is even. The two expressions for y^2 then give: b, c = a - d, a + d so that: y^2 = a^2 + d^2 from which we conclude that there must exist integers m, n with (m, n) = 1 and m + n odd such that: y, a, d = m^2 + n^2, m^2 - n^2, 2.m.n But also, a.c + b.d = 1 gives: (a + d)^2 - 2.d^2 = 1 and replacing in this the preceding expressions for a, d gives: (m^2 + 2.m.n - n^2)^2 - 2.(2.m.n)^2 = 1 The latter is what is called a norm equation, and being of degree 4 doesn't lend itself to elementary techniques. But its two solutions that give the two integer solutions to your original equation are: m, n = 1, 0 and 3, 2. and presumably one can use more advanced methods to prove that there are no other integer solutions. In fact I'm sure this must have been done by now for your equation. The book to check is "Diophantine Analysis" by L J Mordell. ============================================================================== From: Gerry Myerson Subject: Re: x^2+1=2y^4 Date: Fri, 09 Feb 2001 10:47:42 +1000 Newsgroups: sci.math In article <3a831b5c.45230576@news.demon.co.uk>, jr@redmink.demon.co.uk (John R Ramsden) wrote: => buddhist wrote: => > => > Prove : => > x^2+1=2y^4 only has two positive integer solutions => > x=y=1 , (x=239 and y=13) . => The book to check is "Diophantine Analysis" by L J Mordell. Page 271. GM ============================================================================== From: "buddhist" Subject: Re: x^3-y^2=+-1 Date: Wed, 14 Feb 2001 10:22:02 +0800 Newsgroups: sci.math ray steiner wrote : > (A copy of this message has also been posted to the following newsgroups: > sci.math) > > In article <96b1n6$npt@netnews.hinet.net>, "buddhist" > wrote: > > > Gerry Myerson : > > > In article <9691r0$d4g@netnews.hinet.net>, "buddhist" > > > wrote: > > > > > > > How about the equations > > > > x^3-y^2=+-1 where x,y are positive integers ? > > > > > > > > In the case x^3-y^2=1 , I use some methods and > > > > QB program test all x<10^3 , and there's no solutions ! > > > > > > There is an enormous literature on equations of the form > > > y^2 = x^3 + k. Give QB a rest & do a search for Mordell's > > > equation. You will find that for some values of k there > > > are elementary solutions in introductory number theory > > > text books, for others there are solutions using more > > > advanced techniques from algebraic number theory or even > > > more advanced techniques involving Baker's method and > > > "linear forms in logarithms." > > > > > > The solutions are known for all small values of k, where > > > I'm not sure what small means. Search the web. Read some > > > books. > > > > "Diophantine Equations",1970, Mordell , page 246: > > 0< k =< 100 have been solved ! > > What's the informations of k about "today"? > > What kinds of k are unsolved today ? > > Would somebody tell me if he knows ? > Dear Buddhist: > I am replying to some of your recent inquiries. I wonder if you would > also please post > this to Sci.math. My newsfeed is out of whack and right now I have no > posting privileges. > 1). The eqn y^2= x^3+k. The complete solution is now known for |k| < 10000 and > for most k such that |k| < 100000. > The authors use the theories of elliptic curves and linear forms in > elliptic logarithms. > Ref: Gebel, Petho and Zimmer: Compositio Mathematica 110(1998), 335-367. > > 2). The eqn x^2+1= 2y^4. The only positive integer solutions are (x,y)= > (1,1) and (239,13). > There are 3 proofs of this in the literature. > The earliest is due to Ljunggren(1942). His proof is exceedingly complicated and > uses Skolem's p-adic method applied to units in an octic field. > Ref: Avhandliger Norske Videnskaps Akademia Oslo(1942), 1-27. > > Tzanakis and I gave a simpler proof of this about 10 years ago, using the > theory of linear > forms in algebraic numbers(a la Baker). Fortunately, we reduced the > problem to estimating > a linear form in only 2 logarithms, so we were able to use continued > fractions to finish > the proof. > Ref: Journal of Number Theory 37(1991), 123-132. > > Finally, there is a proof by Chen which uses the theory of Diophantine > approximation > and hypergeometric functions. > Ref: JNT 48(1994), 62-74 > > > 3). The equation 3A^2-2B^4= 1. The only solution of this eqn in positive > integers is (A, B) = (1,1). > The result follows from theorem 3.6 of Walsh's paper, which states in part > that under certain conditions the equation ax^2- by^4=1 has at most one solution > in positive integers. > Ref: Algebraic Number theory and Diophantine analysis (Graz, 1998), > 531--554, de Gruyter, Berlin, 2000. > > 4). The equation A^4-3B^4= -2. The only solution of this eqn in positive > integers is (A, B) = (1,1). > This follows from a theorem of Ljunggren(1938) which states that the > equation Ax^4-By^4= plus > or minus C(C=1,2,4,8) has at most one solution in positive integers and > this solution > may be effectively computed from the fundamental unit of the quadratic > ring Z(sqrt(AB)). > Ref: Archiv for Mathematik og Naturvidenskab 41(1938), No. 10, 1-18. > > 5). Your other 3 quartics: I have only partial results. Will send another > e-mail when > time permits. > > Question: How are these quartics related to 1^a+... + n^a= q^a? > > Hope this helps a bit. > Regards, > Ray Steiner > > -- > steiner@math.bgsu.edu Your results 3) and 4) give me that : 1^5+2^5+...+n^5=x^4 where n,x are positive integers if and only if n=x=1 . The results of 5) may give me about 1^4+...+n^4=x^2 or 1^6+...+n^6=x^2 or 1^7+...+n^7=x^2 -- Shyu Jeng-Chyuan N.C.U. Taiwan, R.O.C.