Counter-example of your conjecture, with 4 cubes of primes: 41^3 - 2^3 = 89^3 - 86^3 = 40^3 + 17^3 41, 2, 89 and 17 are primes. 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 David Wilson Envoyé : mercredi 10 janvier 2007 09:41 À : math-fun Objet : Re: [math-fun] Taxicab and prime numbers My conjecture is that in a 3-way sum, at most two of the cubes can be cubes of primes. ----- Original Message ----- From: "Christian Boyer" <cboyer@club-internet.fr> To: "'math-fun'" <math-fun@mailman.xmission.com> Sent: Wednesday, January 10, 2007 3:08 AM Subject: RE: [math-fun] Taxicab and prime numbers Worst! It seems impossible to find solutions of the more general equations: (5) p^3 + q^3 = r^3 + s^3 = m^3 + n^3 or (6) p^3 +/- q^3 = r^3 +/- s^3 = m^3 +/- n^3 with p, q, r, s primes m, n integers (primes or non-primes) Why? Christian. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun