Can the base B number 2008 be a perfect fourth power, where B is a positive integer ≥ 9?
If so, give an example. If not, prove that the base B number 2008 can never be a perfect fourth power.
There is no such number.
2008 base B = 8 + 2*B^3 = k^4 for some k
8+2*B^3 = 2(B^3 + 4).
Now, since this side is even, then k^4 is also even, which means k itself is even and k^4 must be divisible by 16 (each k contributes at least a factor of 2)
So we'd have 2(B^3+4) = 16x for some x or
B^3 + 4 = 8x for some x.
If B is odd, the l.h.s. is odd and can't ever be a multiple of 8. But if B is even then B^3 is itself a multiple of 8 so B^3+4 is still not divisible by 8. Thus, whether B is odd or even, there is no solution.

Posted by Paul
on 20101009 17:54:53 