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

(In reply to Hint by K Sengupta)

2*N³+7=y³
Taking mod 7 on both sides
Case (i): N,y are relatively prime to 7.
By Fermat's theorem, x^6 mod 7 = 1 => x³ mod 7 =+1 or -1
LHS=2*(+/-)1=+2,-2
RHS=+1,-1.
There are no common values for LHS and RHS. So, no solutions
for this case.

Case(ii): If one of N,y is a multiple of 7, other has to be
multiple of 7. 2*N³-y³ will be multiple of 7³. So, their 
difference can never be 7. So, no solutions for this case.

I thought mod p(=prime) should be taken for p values
such that p-1 is a multiple of 3.
I tried it for p=13 and got an exception. I gave it up
there but I missed out 7. Very bad on my part :(

Posted by Praneeth on 2007-09-11 01:02:36