At 4/21/04 04:44 PM, jonthomson wrote:
You can't do it for powers of two (which I learned by trial and error when I was about nine), but I can't remember a proof why. I'll think about that for a bit but make no promises...
Thinking about it for two seconds yields a fairly easy proof which I hope works:
If 2^n was to have k (>1) consecutive integers that sum to make it [god that's worded horribly], k must be odd or even. If k is odd, then we've got a sequence p-h, ..., p-2, p-1, p, p+1, p+2, ..., p+h, where 2h+1 = k. But then we'd have pk=2^n, and as k is odd, that's obvious bullshit.
However, k can't be even as the sum of k successive even numbers is odd. Hence it can't be done for 2^n by contradiction.