Richard Guy wrote:
If n is odd, it is more natural to consider all integers, rather than just positive ones. Are the following the only known solutions for n = 5 ? 133^5+110^5+84^5+27^5-144^5 = 0 14068^5+6237^5+5027^5-14132^5-220^5 = 0 85282^5+28969^5+3183^5+55^5-85359^5 = 0 due to Lander & Parkin, Scher & Seidl, Frye.
Jean-Charles Meyrignac's notation is (n,l,r), with l<=r: n=power, l=number of positive integers on the left side of the equation, r=number of positive integers on the right side of the equation. There are 4 known solutions with n=5: (5,1,4) 144^5=133^5+110^5+84^5+27^5 (Lander, Parkin and Selfridge, 1966) (5,2,3) 14132^5+220^5=14068^5+6237^5+5027^5 (Bob Scher and Ed Seidl, 1997) (5,1,7) 1709^5=1567^5+1373^5+719^5+503^5+431^5+367^5+349^5 (Michael Lau, 07/20/2002) (5,1,4) 85359^5=85282^5+28969^5+3183^5+55^5 (Jim Frye, 08/27/2004) List of known results at http://euler.free.fr/database.txt My initial question was on (n,2,2), with n>4. A symmetrical extension of the Fermat's last theorem, "only" adding w^n on the right side. But, yes, we can add the question (n,1,3), with n>4. Both are probably impossible to get... and probably very difficult to prove. Christian.