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Feb16-12, 08:59 AM
P: 63
Quote Quote by Norwegian View Post
You can probably generalize further, and replace the 1 in 2x3-1 with any cube.
Yeah! For any cube(r^3) and any integer(n), the quadruple of real numbers of the form {n,-n,0,r} satisfies the equation x^3+y^3+z^3+t^3=r^3.

EDIT:Sorry, I thought you meant replace the 2x^3-1 by any cube. In order to generalize to the case you mentioned above, in the quadruple of real numbers {10+n,10-n,-(60n^2)^1/3,-1}, just replace the -1 by any integer, and you get infinitely many representations of the number 2x^3-f^3 as the sum of four cubes.