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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