Cubes (mod 9) are always 0, 1, or 8. The low order ternary digits are 00, 01, or 22. --Rich ---- Quoting Allan Wechsler <acwacw@gmail.com>:
Wait a sec, I'm being stupid. Ternary odd-digit numbers also have only 1's, so they are of the form (3^n-1)/2. If (3^n-1)/2 is a cube, then 3^n = 2m^3 + 1. OK, I'm stuck again. Fred?
On Thu, Apr 19, 2012 at 4:51 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Well, OK then! Let us move on to ternary and see what we can establish!
On Thu, Apr 19, 2012 at 4:43 PM, Fred W. Helenius <fredh@ix.netcom.com>wrote:
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
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