Yes, I permit any two (or even all three) of x,y,z to be equal when I consider solutions to x+y=z. (So in a way including 0 in {0,1,2,...,2n} was silly of me.) Jim On Thursday, October 12, 2017, Allan Wechsler <acwacw@gmail.com> wrote:
To clarify, you permit x=y or y=z, correct? Otherwise, for n=1, {0,1} and {1,2} are counterexamples.
On Thu, Oct 12, 2017 at 7:26 PM, James Propp <jamespropp@gmail.com <javascript:;>> wrote:
Prove that if S is a subset of {0,1,2,...,2n} containing more than n elements, then there must exist x,y,z in S such that x+y=z.
(This smells like an Olympiad problem to me. In fact, I may well have solved this very problem forty years ago, but I don't see how to solve it now. Some sort of pigeonhole argument, sure, but how to set it up?)
The bound is tight; both 1,3,5,...,2n-1 and n+1,...,2n are sum-free.
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