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NMO Selection Tests for BMO and IMO

Combinatorics Difficulty 7.6 National Olympiad, round 2 Prove it Romania

Given a positive integer nn, show that for no set of integers modulo nn, whose size exceeds 1+n+41 + \sqrt{n+4}, is it possible that the pairwise sums of unordered pairs be all distinct.

Solution

Let S⊆Z/nZS \subseteq \mathbb{Z}/n\mathbb{Z} be a set whose pairwise sums of unordered pairs are distinct, and consider the pairwise differences of ordered pairs.
The crucial observation is that if a difference d≠0d \neq 0 occurs twice in SS, then these two occurrences must be adjacent in a 3-term arithmetic progression in SS (with difference dd). Indeed, if a,a+d,a′,a′+da, a+d, a', a'+d are all in SS, where a≠a′a \neq a', then a+(a′+d)=a′+(a+d)a+(a'+d) = a'+(a+d) so, by assumption on SS, a′=a±da' = a \pm d, i.e. the ordered pairs (a,a+d)(a, a+d) and (a′,a′+d)(a', a'+d) are adjacent in a 3-term arithmetic progression.
We also observe that, excepting the arithmetic progression of common difference n/2n/2, in case nn is even, no two 3-term arithmetic progressions can have the same central term, since if a,a±d,a±d′a, a \pm d, a \pm d' are all in SS, where d,d′≠0,n/2d, d' \neq 0, n/2 are distinct, then (a+d′)+(a−d′)=(a+d)+(a−d)(a+d') + (a-d') = (a+d) + (a-d) would violate the assumption on SS.
It follows that the number of 3-term arithmetic progressions in SS is at most ∣S∣+2|S|+2. Now any non-zero difference not occurring in a 3-term arithmetic progression can appear at most once in SS, and those which do appear in 3-term arithmetic progressions can appear at most twice, except ±n/3\pm n/3 in case nn is divisible by 3, which can appear three times.
Consequently, the total number of non-zero differences appearing in SS is at least ∣S∣(∣S∣−1)−(∣S∣+2)−2=∣S∣2−2∣S∣−4|S|(|S|-1) - (|S|+2) - 2 = |S|^2 - 2|S| - 4. If ∣S∣>1+n+4|S| > 1 + \sqrt{n+4}, this quantity is greater than n−1n-1 — a contradiction.

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