Can the base M number 2009 be a perfect cube, where M is a positive integer ≥ 2?

If the answer is yes, give an example. Otherwise prove that the base M number 2009 can never be a perfect cube.

first lets look at the cubes mod 9.

1^3 mod 9 = 1

2^3 mod 9 = 8

3^3 mod 9 = 0

4^3 mod 9 = 1

and thus all cubes are either 0,1,8 mod 9

now look at (2m^3+9) mod 9

(2*1^3+9) mod 9 = 2

(2*2^3+9) mod 9 = 7

(2*3^3+9) mod 9 = 0

(2*4^3+9) mod 9 = 2

thus 2m^3+9 is either 0,2,7 mod 9,  but only 0 mod 9 allows it to be a cube and n^3 mod 9 is 0 mod 9 when n is a multiple of 3, thus m must be a multiple of 3.

m=3t

2m^3+9=2(3t)^3+9=54t^3+9=9(6t^3+1)

now 9=3^2 and thus in order for 9(6t^3+1) to be a cube then 6t^3+1 must at least 1 multiple of 3 but 6t^3+1=3(2t^3)+1 and thus is never a multiple of 3, thus 9(6t^3+1) can never be a cube and thus 2009 can never be a cube regardless of what base it is.

Edited on March 9, 2009, 8:05 pm

Posted by Daniel on 2009-03-09 18:42:10