Does there exist any positive integer base N ≥ 8 such that 2007 is a perfect cube?

(In reply to another solution by bubu)

The only thing to make sure of is if the inequality 7 + 2*N^3 < (N+1)^3 holds for N. 

A litttle algebra gives N^3 - 3N^2 - 3N + 6 < 0.  This inequality is true for N in the range 1.16 to 3.36 and if N is less than -1.53.

The inequality is never true for any N>=8, and is not usable in a proof.

Posted by Brian Smith on 2007-10-19 10:59:09