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.
2008 in any base is simply 2x^3+8 where x is the base.
The winning candidate will be one where x^3+8 = z^4; and for this purpose the notation of the base is irrelevant - we can always 'translate' the candidate into an appropriate base later.
So, working with base 10 for convenience, 2(x^3+4) is always even, while z^4 may take the value 0,1,5,6, base 10. Thus there can only be a match if z is of the form (2n).Substituting (2n) we have 2(x^3+4) = (2n)^4; 2x^3 +8 = 16n^4 , or x^3+4 = 2n^4 - where x must be even, for the equality to hold good - so,(2y)^3+4 = 2n^4, or 4(2y^3 +1) = 2n^4; but now, 2 must be a factor of n, so n is itself even....and so ad infinitum without solution.
However, 2:0:10:4 (=10000(base10)) is a reasonable representation of the year 2014 in base 17, while 2000 is the base 32 equivalent of 65536(base10)
Edited on October 10, 2010, 6:06 am
Posted by broll
on 2010-10-10 03:42:26