Ooops, sorry. Please read: "list of 10,597,218 numbers < 9.5 * 10^20 that are differences of cubes in THREE or more ways" -----Message d'origine----- De : Christian Boyer [mailto:cboyer@club-internet.fr] Envoyé : dimanche 15 juin 2008 15:13 À : 'math-fun' Objet : RE: [math-fun] System of 2 diophantine equations Integers < 300,000^3 was a computation done by Frank Rubin in 2006. Latest news on this problem: using tables built during his computation on Cabtaxi(10) http://cboyer.club.fr/Taxicab.htm, Uwe Hollerbach worked on his list of 10,597,218 numbers < 9.5 * 10^20 that are differences of cubes in two or more ways. If his computation is correct, the system a^3 - d^3 = b^3 - e^3 = c^3 - f^3 = S a^3 - g^3 = b^3 - h^3 = c^3 - i^3 = S' has no solution with S and S' < 9.5 * 10^20. Deceptive! But not a proof that the problem is impossible. Christian. -----Message d'origine----- De : math-fun-bounces@mailman.xmission.com [mailto:math-fun-bounces@mailman.xmission.com] De la part de Joshua Zucker Envoyé : dimanche 15 juin 2008 06:44 À : math-fun Objet : Re: [math-fun] System of 2 diophantine equations What methods have you tried? My idea: make a list of numbers that are a difference of cubes in 2 or more ways, (is there a quick test for this, or do we have to slowly build up a big hash table?) then look for a,b,c such that a^3 - b^3 is a difference of cubes in 2 ways, same with a^3 - c^3, b^3 - c^3. --Joshua Zucker On Sat, Jun 14, 2008 at 2:17 AM, Christian Boyer <cboyer@club-internet.fr> wrote:
I am looking for at least one integer solution of this system: a^3 - d^3 = b^3 - e^3 = c^3 - f^3 a^3 - g^3 = b^3 - h^3 = c^3 - i^3
Quite easy to find near solutions, for example: 165^3 - 72^3 = 178^3 - 115^3 = 162^3 - 51^3 165^3 - 618^3 = 178^3 - 619^3 = 162^3 - 235788435 Unfortunately 235788435 is not a cube... It seems that there is no solution with integers < 300000^3.
Any idea?
Christian.
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