a^2 + b^3 + c^4 + d^5...

[Don Reble]

jb at brennen.net wrote:
> Any flaws in that argument, other than the minor bit of hand-waving?

No. Let me make the induction precise.

Let P(N) be the premise
    N<292 or every number from 292 to N is representable.

Calculations show that P(1024) is true.

If N > 1024 and P(N-1), then

    Let a be the integer such that 2^a <= N < 2^(a+1). a >= 10.
    Let M = N - 2^(a-1). M >= 2^a - 2^(a-1) = 2^(a-1) >= 512.
    Then 292 <= M < N, so M is representable.

    If M has a representation without an (a-1)'st power,
        then N is representable as M+2^(a-1).
    If M has a representation with an (a-1)'st power, then that 
        power must be 2^(a-1), because
            3^(a-1) = 81*3^(a-5) > 64*2^(a-5) = 2^(a+1) > N > M
        (and higher bases yield even higher powers).
        And M's representation does not have an a'th power, because
            M - 2^(a-1) = N - 2*2^(a-1) < 2^(a+1) - 2^a = 2^a
        So N is representable as 2^a+[M-2^(a-1)].

    Either way, P(N).


Thanks, Jack.

-- 
Don Reble  djr at nk.ca