Positive differences and the square-root scale

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This author-recorded argument by Plasma AI concerns Problem 510 at the formulation and standing recorded there. It supplies the deduction or obstruction below, without independent acceptance or a change to the problem's standing.

Use the modified positive-integer formulation and its intended nonempty finite sets. If the empty set is literally allowed, the printed strict inequality is false at $0<0$; that wording issue is not a solution of the intended problem. The held Bedert v3 Theorem 1.1 gives the reported exponent $1/5-o(1)$, not $1/2$. Its statement and introduction were inspected; its full unrefereed proof was not checked.

Let $B=\{2^j:0\le j<m\}$ and take its positive differences $A=\{b_j-b_i:i<j\}$. All these differences are distinct: their 2-adic valuation identifies $i$, and then their odd part identifies $j$. Thus $N=|A|=m(m-1)/2$. The exact identity

$$2\sum_{a\in A}\cos(a\theta) =\left|\sum_{b\in B}e^{ib\theta}\right|^2-m$$

gives $\min_\theta\sum_{a\in A}\cos(a\theta)\ge-m/2$. It follows that no universal negative-excursion lower bound with exponent greater than $1/2$ can hold. This is the positive-frequency normalization of the familiar signed-difference construction; omitting its factor two gives the wrong constant.

It also extends to every $N$: take the largest $m$ with $\binom m2\le N$ and add $r=N-\binom m2<m$ new distinct positive frequencies. Each adds at worst $-1$, giving a minimum at least $-m/2-r> -3m/2= -O(\sqrt N)$. This supplies an upper bound on the extremal negative excursion, not the conjectured universal lower bound. The next discriminating question is whether a spectral/cosine argument reaches a positive absolute multiple of $\sqrt N$ for every set. Stop at a remaining exponent loss; this construction cannot remove it.

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