From: vlibra@tin.it (Vincenzo Librandi) Subject: "Diophantus equation x^3 + y^3 = z^a ?" Date: 13 Oct 1999 15:55:55 -0400 Newsgroups: sci.math Seong Ik CHO wrote: >Would someone can help me about current status of the following >Diophantus equation ? >Is there any proof of the non-existence of solution for a Diophantus >equation x^3 + y^3 = z^p or >x^3 + y^p = z^3, when x, y, and z are all positive integers and p is >greater than 3 ?Seong Ik CHO wrote: Ciao, l'equazione di Diofanto di cui sopra ammette soluzioni: X^3+Y^3 = Z^p 18^3+9^3 = 9^4 1458^3+729^3 = 243^4 201684^3+67228^3 = 9604^4 X^3+Y^p = Z^3 7^3 + 7^4 = 14^3 26^3 + 26^4 = 78^3 63^3+ 63^4 = 252^3 Solo solo alcune soluzioni delle due equazioni, moltissime altre esistono anche con p > 4 Ciao, Vincenzo Librandi ============================================================================== From: Seong Ik CHO Subject: Re: Diophantus equation x^3 + y^3 = z^a ? Date: Thu, 14 Oct 1999 11:22:55 +0900 Newsgroups: sci.math Seong Ik CHO wrote: > Dear colleagues, > > Would someone can help me about current status of the following > Diophantus equation ? > Is there any proof of the non-existence of solution for a Diophantus > equation x^3 + y^3 = z^p or > x^3 + y^p = z^3, when x, y, and z are all positive integers and p is > greater than 3 ? > > Sincerely yours, > > Seong Ik CHO Oh ! I missed one conditional statement. Is there any proof of the non-existence of solution for a Diophantus equation x^3 + y^3 = z^p or x^3 + y^p = z^3, when x, y, and x are all positive integers, p is greater than 3, and gcd(x,y,z) = 1 ? ^^^^^^^^^^^^^^ Thanks. Seong Ik CHO [trailer deleted --djr] ============================================================================== From: jr@redmink.demon.co.uk (John R Ramsden) Subject: Re: Diophantus equation x^3 + y^3 = z^a ? Date: Sat, 16 Oct 1999 05:41:31 GMT Newsgroups: sci.math On Thu, 14 Oct 1999 11:22:55 +0900, Seong Ik CHO wrote: [previous article quoted --djr] If p is divisible by 3 then there are obviously no non-trivial integer solutions. Otherwise p must be == q mod 3, where q = 1 or 2, and dividing each term by (p-q)/3 we can consider rational solutions to X^3 + Y^3 = Z^q. For q = 2 we can assume X + Y, X^2 - X.Y + Y^2 = a, a.b^2 resp, so (completing the square in the second) we see that X, Y are roots of T^2 - a.T + a.(a - b^2)/3 = 0. Then letting T = a.c, and assuming a != 0, this quadratic becomes linear in a. That gives a complete rational solution of X^3 + Y^3 = Z^2, which should hopefully help with your integer problem. Cheers --- John R Ramsden # "Off the top of my head, I'd say that # my father was the better swasher, (jr@redmink.demon.co.uk) # but I was the better buckler." # # Douglas Fairbanks Jr ============================================================================== [The general rational solution to X^3 + Y^3 = Z^2 is [X,Y,Z] = [u*(v^3+u^3)/v^2, (v^3+u^3)/v, (v^3+u^3)^2/v^3] . Similarly the case q=1 can be reduced to the problem X^3 + Y^3 = Z^4, whose general rational solution is [X,Y,Z] = [u*(u^3+v^3), v*(u^3+v^3), (u^3+v^3)]. It is not immediately clear that this is helpful since for example it is conjectured that there are _no_ solutions to the equation x^3+y^3=z^n in relatively prime integers x,y,z except for n=2 or less. --djr]