From: graph@worldnet.fr (Jean-Charles Meyrignac) Subject: New results in equal sums of like powers Date: 25 Jul 00 17:03:12 GMT Newsgroups: sci.math.numberthy Summary: [missing] A distributed computing project about equal sums of like powers (http://euler.free.fr) recently led to some new discoveries using various methods. We use the notation (k,m,n) to express that a sum of m powers of k is equal to a sum of n powers of k. (6,2,5) 1117^6+770^6=1092^6+861^6+602^6+212^6+84^6 discovered by Edward Brisse on 04/20/1999 and independently by Giovanni Resta -using his own method- just a few weeks before. We collected a list of 40 other solutions to the equation a^6+b^6=c^6+d^6+e^6+f^6+g^6. (7,1,7) 568^7=525^7+439^7+430^7+413^7+266^7+258^7+127^7 (Mark Dodrill, 03/20/1999) (7,2,6) 125^7+24^7=121^7+94^7+83^7+61^7+57^7+27^7 (Jean-Charles Meyrignac, 1997) (8,1,9) 1167^8=1094^8+1040^8+560^8+558^8+366^8+348^8+284^8+271^8+190^8 (Nuutti Kuosa, 05/20/2000) Nuutti Kuosa also found (8,1,10) 235^8=226^8+184^8+171^8+152^8+142^8+66^8+58^8+34^8+16^8+6^8 in 1999 using his own program. (9,2,10) 1379^9+699^9=1219^9+1169^9+1169^9+1159^9+899^9+529^9+289^9+269^9+149^9+99^9 (Luigi Morelli, 03/12/1999) (10,1,14) 120^10=115^10+104^10+91^10+80^10+80^10+74^10+63^10+55^10+40^10+35^10+33^10+1 3^10+10^10+5^10 (Torbj�rn Alm, 10/14/1999) (10,6,6) 151^10+140^10+127^10+86^10+61^10+22^10=148^10+146^10+121^10+94^10+47^10+35^1 0 (Nuutti Kuosa, 09/03/1999) which lead to the discovery of the (k=1,2,3,4,5,6,7,8,9,10,11) case of Tarry's problem by Chen Shuwen. We also collected results upto 32th power. Some open questions remain: - In 1966, Lander, Parkin and Selfridge found: 144^5-133^5-110^5-84^5-27^5=0 In 1997, Bob Scher and Ed Seidl found: 14132^5+220^5-14068^5-6237^5-5027^5=0 Does another sum of five fifth powers equal to 0 ? A search below 100,000 didn't yield to another solution. - In 1966, Lander, Parkin and Selfridge found (6,1,7) 1141^6=1077^6+894^6+702^6+474^6+402^6+234^6+74^6 Can a sixth power be equal to six sixth powers ? Similarly, does a sum of two sixth powers can be equal to the sum of four sixth powers ? - On the 7th power, we have: (7,1,7) 568^7=525^7+439^7+430^7+413^7+266^7+258^7+127^7 (Mark Dodrill, 03/20/1999) (7,2,6) 125^7+24^7=121^7+94^7+83^7+61^7+57^7+27^7 (Jean-Charles Meyrignac) (7,3,5) 96^7+41^7+17^7=86^7+77^7+77^7+68^7+56^7 (Randy Ekl, Rizos Sakellariou) (7,4,4) 149^7+123^7+14^7+10^7=146^7+129^7+90^7+15^7 (Randy Ekl) meaning that a sum of eight powers of 7 can be equal to 0. Can a sum of seven seventh powers be equal to 0 ? - More generally, on any given power k, is it possible to find k terms suming to 0 by addition/substraction ?