From: lerma@math.nwu.edu (Miguel A. Lerma) Subject: powers in arithmetic progression Date: 2 Jun 1999 13:08:56 GMT Newsgroups: sci.math It is possible to find many arithmetic progressions made up of three perfect squares, such as 1, 25, 49, but are there any arithmetic progressions made up of four or more squares (with positive difference)? Three (strictly positive) cubes? Higher powers? The question about the cubes is equivalent to look for non trivial solutions for the Diophantine equation x^3 + z^3 = 2 y^3 (x,y,z being different positive integers). My guess is that there are no such solutions, and the proof is probably similar to the one used to prove Fermat's Last Theorem for exponent n=3, I just wanted to know if the problem has already been solved - my books on Number Theory mention other Diophantine equations but not this one. Miguel A. Lerma ============================================================================== From: gerry@mpce.mq.edu.au (Gerry Myerson) Subject: Re: powers in arithmetic progression Date: Thu, 03 Jun 1999 09:34:28 +1100 Newsgroups: sci.math In article <7j3ad8$3t1@news.acns.nwu.edu>, lerma@math.nwu.edu (Miguel A. Lerma) wrote: => ...are there any arithmetic => progressions made up of four or more squares (with positive difference)? => Three (strictly positive) cubes? Higher powers? => => The question about the cubes is equivalent to look for non trivial => solutions for the Diophantine equation x^3 + z^3 = 2 y^3 (x,y,z being => different positive integers). There are no non-trivial solutions to x^3 + y^3 = 2 z^3. A more general result appears in Mordell, Diophantine Equations, page 126. Fermat proved that there can be no four squares in arithmetic progression. This is also in Mordell, pages 21--22. Gerry Myerson (gerry@mpce.mq.edu.au) ============================================================================== From: rusin@shavano.math.niu.edu (Dave Rusin) Subject: Re: powers in arithmetic progression Date: 3 Jun 1999 15:30:36 GMT Newsgroups: sci.math In article , Gerry Myerson wrote: >In article <7j3ad8$3t1@news.acns.nwu.edu>, lerma@math.nwu.edu (Miguel A. >Lerma) wrote: > >=> ...are there any arithmetic >=> progressions made up of four or more squares (with positive difference)? >=> Three (strictly positive) cubes? Higher powers? >=> >=> The question about the cubes is equivalent to look for non trivial >=> solutions for the Diophantine equation x^3 + z^3 = 2 y^3 (x,y,z being >=> different positive integers). > >There are no non-trivial solutions to x^3 + y^3 = 2 z^3. A more general >result appears in Mordell, Diophantine Equations, page 126. There are none with higher exponents, either. See Darmon, Henri; Merel, Loïc Winding quotients and some variants of Fermat's last theorem. J. Reine Angew. Math. 490 (1997), 81--100. MR98h:11076 The proof is of the flavour of Wiles' proof for FLT. Ribet had done substantial earlier work; see 96/fermatlike dave ============================================================================== From: jpr2718@my-deja.com Subject: Re: powers in arithmetic progression Date: Thu, 03 Jun 1999 16:33:28 GMT Newsgroups: sci.math In article <7j3ad8$3t1@news.acns.nwu.edu>, lerma@math.nwu.edu (Miguel A. Lerma) wrote: > It is possible to find many arithmetic progressions made up of three > perfect squares, such as 1, 25, 49, but are there any arithmetic > progressions made up of four or more squares (with positive difference)? > Three (strictly positive) cubes? Higher powers? > > Miguel A. Lerma One cannot have 4 different squares in arithmetic progression. One cannot have a nontrivial arithmetic progression of n-th powers for n >= 3. Nontrivial means no term is zero and the progression is not a constant multiple of 1, 1, 1. The cases n=3 and n=4 are mentioned in Carmichael's ?1908 book on diophantine equations. n=5 to n=31 were done be Denes (?in the early 1950's). n=17 and all higher were done by Ribet (about 1996). John Sent via Deja.com http://www.deja.com/ Share what you know. Learn what you don't.