Jean-Charles Meyrignac thinks that a^n + b^n = c^n + d^n, with n>4, is impossible. On a^5 + b^5 = c^5 + d^5 + e^5, he says that the ONLY known solution is 14132^5 + 220^5 = 14068^5 + 6237^5 + 5027^5 = 563661204304422162432 So, for sure, it will be incredibly difficult to find solutions with e=0. On a^6 + b^6 = c^6 + d^6 + e^6 + f^6, there is no known solution after... 184 years of computation time! ( http://euler.free.fr/top.htm ) So, for sure, it will be incredibly difficult to find solutions with e=f=0. Christian. -----Message d'origine----- De : math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com [mailto:math-fun-bounces+cboyer=club-internet.fr@mailman.xmission.com] De la part de Christian Boyer Envoyé : vendredi 16 juin 2006 20:06 À : 'David Wilson'; 'math-fun' Objet : RE: [math-fun] x^n + y^n = z^n + w^n David wrote:

I don't recall ever hearing of a solution to a_1^k +- a_2^k +- a_3^k +- ... +- a_j^k = 0 in distinct positive integers a_i with 1 <= j < k.

Yes, you are right. My question is specifically on (j,k)=(4,>4): what are the already checked ranges of integers? Only Blair Kelly's results? x^5 + y^5 < 1.02 x 10^26 means a not very large range: 0 < x < y < 160,000. I will ask Jean-Charles Meyrignac, http://euler.free.fr/, he has perhaps some answers. Christian. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun

Gary McGuire wrote

The first counterexample to a conjecture of Euler was 84^5+110^5=144^5+(-27)^5+(-133)^5. Do you want solutions in the integers or in the positive integers?

Yes, positive distinct integers. About a^n = b^n + c^n + d^n + e^n, there is your solution: 144^5 = 133^5 + 110^5 + 84^5 + 27^5 but also this other one found by Jim Frye in 2004: 85359^5 = 85282^5 + 28969^5 + 3183^5 + 55^5 Christian.

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.

I'm wondering if any progress has been made on the non-attacking queens game described in the May 2006 issue of The College Mathematics Journal. "Two players successively place queens on the board so that no two attack each other. The winner is the player who places a queen so that all remaining squares are attacked, ..." For square boards, n x n, with n odd, there is a simple strategy for player 1 to win. The authors report that player 1 also wins with n = 2, 4, 6 and 8, but that a winning strategy exists for player 2 for n = 10! That was as far as they took the analysis. Paul