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A bound on consecutive uniquely represented differences


The proposed difference-basis route in the integer transfer analysis asks for a strong Sidon set BB of mm integers with B−B⊇[−L,L]B-B\supseteq[-L,L] and L≍m2L\asymp m^2. The following author-recorded argument of Plasma AI rules out that route without restricting the diameter of BB. It does not bound the cost of coordinated completion and does not resolve Problem 156.

Proposition. Let B⊂ZB\subset\mathbb Z have m≥2m\ge2 elements. Suppose that every integer 1≤d≤L1\le d\le L has exactly one representation d=b′−bd=b'-b with b,b′∈Bb,b'\in B. Then

L≤(3+22)m5/3−2m+2.L\le (3+2\sqrt2)m^{5/3}-2m+2.

In particular the conclusion holds for a strong Sidon set whose difference set contains [1,L][1,L]. Uniqueness of differences greater than LL is not needed.

The complete proof is [[theory/additive_bases/L239_consecutive_unique_differences/_proof|the sliding-count and smoothing argument of L239]]. Its exact conclusion is recorded at tier 0 on the [[theory/additive_bases/L239_consecutive_unique_differences/_index|claim card]]. The extension below uses the same counts and keeps its own research standing.

Stability when differences are missing

Let BB be a strong Sidon set of m≥2m\ge2 integers, fix an integer u≥1u\ge1, and let HH count its missing positive differences among 1,…,2u−11,\ldots,2u-1. Use F(j)F(j), fa(n)f_a(n), TaT_a and EaE_a as defined in the proof of L239. A missing difference removes at most 2u2u from ∑jF(j)2\sum_j F(j)^2. Its change to ∑j(F(j+u)−F(j))2\sum_j(F(j+u)-F(j))^2 is −4u+6d-4u+6d for d<ud<u, and 2(2u−d)2(2u-d) for u≤d<2uu\le d<2u; either change is at most 2u2u. Hence

1u∑aTa≥m+u−1−2H,1u∑aEa≤2(m−1+H).\frac1u\sum_a T_a\ge m+u-1-2H,\qquad \frac1u\sum_a E_a\le2(m-1+H).

Hence, whenever the left radicand is nonnegative,

m+u−1−2H≤mk+(k−1)m−1+H2(k≥1).(1)\sqrt{m+u-1-2H} \le\frac{m}{\sqrt k}+(k-1)\sqrt{\frac{m-1+H}{2}} \quad(k\ge1). \tag{1}

In particular, H=O(m)H=O(m) implies u=O(m5/3)u=O(m^{5/3}). More generally, if u≍m2u\asymp m^2, then HH cannot be o(m2)o(m^2). To see this, suppose it were, and choose integers k→∞k\to\infty with km+H=o(m)k\sqrt{m+H}=o(m), which is possible since m+H=o(m2)m+H=o(m^2). The right side of (1) is o(m)o(m) while the left side has order mm, a contradiction.

These statements concern differences of one set. Several translated difference sets may cover each other's gaps, and blockers using two or three new points need not be translates of the original difference set. Those mechanisms remain available in the editable proof sketch.

Standing. This is a complete author-recorded argument. No independent review or novelty determination is recorded. The exact unique-difference bound is L239; the stability extension is the double count and smoothing inequality above, not an inference from the small rulers in the integer transfer analysis.