19 Apr
2012
19 Apr
'12
2:43 p.m.
On 4/19/2012 4:12 PM, Allan Wechsler wrote:
Certainly we can prove this theorem for base 2, can't we? How many cubes are there of the form 2^n-1? Something is telling me, "one," but I can't prove it.
If 2^n - 1 = m^3, then 2^n = m^3 + 1 = (m + 1)(m^2 - m + 1). The second factor is odd, so it's only a power of 2 if m^2 - m = 0. Thus 0 and 1 are the only cubes of the form 2^n - 1. (Of course Alan wanted n > 0, so only one solution is relevant to the original question.) By the way, the base-10 problem is easy if you replace n^3 with n^2. -- Fred W. Helenius fredh@ix.netcom.com