From: rusin@vesuvius.math.niu.edu (Dave Rusin) Subject: SUms of powers (was: Re: fermat) Date: 27 Mar 1999 02:35:25 GMT Newsgroups: sci.math Keywords: No nontrivial solutions known for x^5 + y^5 + z^5 + t^5 = 0 John R Ramsden wrote: [Wiles proved FLT] >work he had done at the start of his researches (involving Arakelov Theory, >whatever that is), Does anyone know _who_ Arakelov is? A couple of papers around the time of the Vancouver ICM, then -- poof! -- nada. >P.S. Now Fermat's Last Theorem has been settled, despite Luther's fervent >declarations to the contrary, and non-zero integer solutions have been found >for x^4 + y^4 + z^4 = t^4, isn't is about time someone turned their guns on >x^5 + y^5 + z^5 = t^5 ? > >I raised this a while ago here, and only got the sound of chirping crickets! >Anyone game? Try expressing it as x^5 + y^5 = t^5 - z^5 and factor each >side in a cyclotomic field. That's a start, the rest is up to you :-) >Seriously, >I think Noam Elkies, who found the first solution to the fourth power >problem, >also obtained various criteria for shortening a numeric search for fifth >powers. >So if anyone is interested in conducting a search, he might be persuaded to >pass this info on. Yes, Elkies found the first matching set of four integral fourth powers. People have looked for four matching fifth powers, but if they do exist, they're likely to be quite rare; on the other hand, we don't have sufficient tools to show they can't exist. If you're optimistic and think you can find them, try something like setting S=x+y, P=x*y and look for integer solutions to the corresponding equation in S, P, z, t. There _are_ solutions to this one, as I recall, and you can even parameterize lots of them. You need S^2-4P to be a perfect square, though, to be able to recover x and y. So you can keep your computer happy, looping around the possible (S,P,z,t)'s and hoping for S^2-4P to be square. That wouldn't be terribly unlike Elkies's method for fourth powers. If I had to guess, I'd probably suppose there are no matching sets of four fifth powers. (Sets of _five_ fifth powers summing to zero, on the other hand, have been known for a while.) On the other hand, you might have more luck with sixth powers, since you can use all sorts of patterns with sums of squares or cubes. Moreover, we're not even close to what may be the borderline cases with sixth powers: I don't think anyone has found six sixth powers which sum to another sixth power, let alone five or four or three of them. Happy hunting! dave ============================================================================== From: gerry@mpce.mq.edu.au (Gerry Myerson) Subject: Re: a^n + b^n = c^n + d^n (Re: Sum of integer powers) Date: Thu, 22 Apr 1999 15:22:30 +1100 Newsgroups: rec.puzzles,sci.math In article <7fm00s$i28$1@news.fas.harvard.edu>, Noam D. Elkies wrote: => It is a long-standing conjecture that there are no solutions [to the => problem of finding N expressible as the sum of two positive n-th => powers in two different ways] for any n bigger than 4. This is, I => think, one of the Unsolved Problems in Number Theory in R.K.Guy's book => of this name. In D1, Guy notes that no solution is known for n = 5 and that the search has been carried up to N = 1.02 x 10^26. In F30, Guy writes, x^5 is a likely answer to the following unsolved problem of Erdos. Find a polynomial P(x) such that all the sums P(a) + P(b) (0 \le a < b) are distinct. Gerry Myerson (gerry@mpce.mq.edu.au) ============================================================================== From: Public User Subject: Re: Sum of integer powers Date: Fri, 23 Apr 1999 16:37:52 -0500 Newsgroups: rec.puzzles,sci.math Public User wrote: > > Minimum x such that x=a^n+b^n=c^n+d^n > where a,b,c,d are positive integers > and (a,b) not = (c,d) > n > ----------------------------------------------- > > 1 4 = 1^1+3^1 = 2^1+2^1 > > 2 50 = 1^2+7^2 = 5^2+5^2 > > 3 1729 = 1^3+12^3 = 9^3+10^3 > > 4 635,318,657 = 133^4+134^4 = 59^4+158^4 With the assistance of a local number theory university professor, we were able to find the following math article (http://www.ams.org/jourcgi/jour-pbprocess?fn=110&arg1=S0025-5718-98-00979-X). Among other things, it states, as far as I understand it, that there is no solution for x = a^5+b^5 = c^5+d^5 for positive integer x,a,b,c,d for x<=4.01 X 10^30 with (a,b) not = (c,d). Drats! Yours truly, Will