From: tpiezas@uap.edu.ph (Tito Piezas III) Subject: The Sum of n nth Powers Conjecture Date: 8 Dec 2000 10:18:29 -0500 Newsgroups: sci.math Summary: [missing] Dec. 8, 2000 Euler had a related conjecture. In 1772, he stated: "If a sum of r positive nth powers equal one nth power, then r is greater than or equal to n." The first counter-example was by Lander, Parkin, and Selfridge in 1966 when they showed an example of r=4, but n=5. 27^5 + 84^5 + 110^5 + 135^5 = 144^5 So even Euler could be wrong. Now, my interest is when we let r=n and phrase Euler's conjecture differently. The "sum of n nth powers conjecture" will be: "There exists a positive integer Z such that beyond Z there is no one nth power equal to the sum of n nth powers." For n=2,3,4,5 2^2 + 3^2 = 5^2 3^3 + 4^3 + 5^3 = 6^3 30^4 + 120^4 + 272^4 + 315^4 = 353^4 19^5 + 43^5 + 46^5 + 47^5 + 67^5 = 72^5 In 1967, Lander (et all) stated than no solutions were known for n>5. As late as 1994, R.K. Guy in "Sums of Like Powers: Euler's Conjecture" stated the same thing, none found for n=6,7,8,9, etc. But in Mar 20, 1999, Mark Dodrill found: 127^7 + 258^7 + 266^7 + 413^7 + 430^7 + 439^7 + 525^7 = 568^7 In Nov. 14, 2000, Maurice Blondot found: 91^7 + 148^7 + 158^7 + 255^7 + 258^7 + 309^7 + 625^7 = 626^7 However, there is still no solution for n=6. By the wording of our conjecture, at least up to this day, we've pushed Z to Z=7. The question to be asked: How far can we push Z? Can we do it indefinitely, or we'll we reach a barrier at some point? So, for those of you with programming skills, here's a chance to use those idle computer time. To find n=6, and any Z > 7, if any. And how do we prove or disprove the conjecture? Note: Some of the material for this post is from the remarkable website of Jean-Charles Meyrignac: euler.free.fr/index.htm (if you discover anything, please post here as well as there) and Eric Weinstein's former mathworld.wolfram.com, the termination of which I am whole-heartedly against. Sincerely, Tito Piezas III University of Asia & the Pacific