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