Counting regularity

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Write $A(x)=|A\cap[1,x]|$, $B(x)=|B\cap[1,x]|$, and suppose

$$a_n/b_n\longrightarrow1,\qquad 1\le r_{A,B}(n)\le M \quad(n\text{ sufficiently large}).$$

Increasing $M$ if necessary makes the upper bound hold for every $n$. All claims below are proved elementary consequences, not a resolution of the problem.

Section 1 of the independent investigation recovers these estimates. Its explicit rank constants, uniform local form, and coverage-only gap restriction have been checked and are included below.

Sections 2--3 of the later independent input also check out. Its common limiting measure and $dt\le\mu*\mu\le M\,dt$ are already included in the energy formulation; the continuous square-root model does not settle the discrete question. Its Narkiewicz-based product bound $\limsup A(x)B(x)/x>1$ is weaker than the established $4/\pi$ bound in collision density, proved without a representation upper bound. That note also excludes eventual uniqueness under balance, without classifying eventually unique pairs.

Comparison with the third independent input

Sections 2.1--2.4 of the third independent input correctly recover the square-root bounds, equal counting functions to first order, local continuity, and the common convolution limit. Its conclusion under eventual uniqueness,

$$A(x)\sim B(x)\sim\frac2{\sqrt\pi}\sqrt x,$$

is a valid necessary condition, but does not leave eventual uniqueness as an open subcase of the current research. The established collision-density argument already excludes it; nonconstant representations forces disagreement with every nonzero periodic pattern on a set of positive lower density. No classification of eventually unique complements is needed for either conclusion.

Its continuous square-root model correctly shows that $dx\le\mu*\mu\le M\,dx$ alone is insufficient. This is consistent with, but weaker than, the checked bounded models for the additional second-moment constraints in energy deficit. Likewise, its bounded-displacement thickening argument in Section 2.5 is valid; the finite-shift union reduction below already gives a more general statement and a smaller bound when $D\ge1$. These comparisons do not change the unresolved status of the unrestricted problem.

Square-root size

Counting covered integers and all pairs below $x$ gives

$$x-O(1)\le A(x)B(x)\le2Mx.$$

Balance gives $A(x)\le B(2x)+O(1)$. Thus

$$A(x)(A(x)-O(1))\le A(2x)B(2x)\le4Mx,$$

and symmetrically $B(x)=O(\sqrt x)$. The lower product bound now gives

$$A(x),B(x)=\Theta(\sqrt x),\qquad a_n,b_n=\Theta(n^2).$$

More precisely, the first $n$ elements of each set supply $n^2$ pairs with sums at most $a_n+b_n$, whereas every represented integer below $\min(a_n,b_n)$ uses only the first $n-1$ elements of each set. Thus

$$n^2\le M(a_n+b_n),\qquad \min(a_n,b_n)\le(n-1)^2+O(1).$$

Termwise balance and inversion of these bounds give

$$\frac{1-o(1)}{2M}n^2\le a_n,b_n\le(1+o(1))n^2, \qquad (1-o(1))\sqrt x\le A(x),B(x) \le(\sqrt{2M}+o(1))\sqrt x.$$

Same counting function to first order

Fix $0<\varepsilon<1/2$, and take $x$ sufficiently large that all relevant ranks satisfy $(1-\varepsilon)b_i\le a_i\le(1+\varepsilon)b_i$. If $K=A(x)-B(x)>0$, each rank $B(x)<i\le A(x)$ has

$$(1-\varepsilon)x<a_i\le x<b_i\le x/(1-\varepsilon).$$

The $K^2$ pairs drawn from these two rank blocks have sums in an interval of length $(\varepsilon+\varepsilon/(1-\varepsilon))x$. Consequently

$$K^2\le M\bigl((\varepsilon+\varepsilon/(1-\varepsilon))x+2\bigr).$$

Interchanging $A,B$ handles negative $K$. Taking the limsup as $x\to\infty$ and then $\varepsilon\downarrow0$ proves

$$A(x)-B(x)=o(\sqrt x),\qquad A(x)/B(x)\longrightarrow1.$$

Without the representation bound, replacing termwise balance by equal counting functions is not justified in this way.

Local continuity

For fixed $0\le s<t$, let $K=A(tX)-A(sX)$. The corresponding $B$-ranks lie between $sX/(1+\varepsilon)$ and $tX/(1-\varepsilon)$, apart from a fixed finite initial block. Their $K^2$ sums lie in an interval of length

$$\left(t-s+\frac{t}{1-\varepsilon}-\frac{s}{1+\varepsilon}\right)X.$$

Hence

$$\limsup_{X\to\infty}\frac{A(tX)-A(sX)}{\sqrt X} \le\sqrt{2M(t-s)}.$$

The same holds for $B$. In particular, subsequential rescaled counting measures have a common, nonatomic limit with a (1/2)-Hölder distribution function. This regularity alone does not contradict bounded multiplicity: the measure $c x^{-1/2}\,dx$ has a constant self-convolution density.

There is also the uniform formulation

$$A(x+h)-A(x)\le\sqrt{2Mh+o(x)}\qquad(0\le h\le x),$$

and the analogous bound for $B$, where the error is independent of $h$. Indeed, the corresponding ranks have $a_i\in(x,x+h]$ and $b_i=a_i+o(x)$, uniformly over $a_i\in(x,2x]$. All cross pairs from these rank blocks therefore lie in an interval of length $2h+o(x)$. The representation bound proves the assertion, with the endpoint error absorbed into $o(x)$.

A gap restriction using only balance and coverage

Drop the representation upper bound. For every sufficiently large $n$, coverage of $a_n+b_n+1$ forces

$$\min(a_{n+1},b_{n+1})\le a_n+b_n+1.$$

Otherwise a representation could use neither a later $A$-element nor a later $B$-element, and its sum would be at most $a_n+b_n$. Using balance at ranks $n$ and $n+1$ yields

$$\limsup_{n\to\infty}\frac{a_{n+1}}{a_n}\le2, \qquad \limsup_{n\to\infty}\frac{b_{n+1}}{b_n}\le2.$$

This necessary condition does not control multiplicities.

Bounded displacement reduces to the diagonal problem

Section 7.4 of the later independent input gives a correct reduction which works at every fixed capacity, unlike the stronger exclusions at capacities two or three in discrepancy fibers. Here is a slightly more general version, not requiring rank balance.

Suppose $A+B$ is cofinite, $r_{A,B}\le M$ eventually, and finite sets $E,F\subset\mathbb Z$ satisfy

$$A\subseteq\bigcup_{t\in E}(B+t),\qquad B\subseteq\bigcup_{s\in F}(A+s).$$

Choose one witnessing shift for each element. Mapping $(a,a')$ to $(a,a'-t(a'))$, separately for each fixed shift, gives

$$r_{A,A}(n)\le\sum_{t\in E}r_{A,B}(n-t).$$

The map is injective on each shift fiber. Similarly, mapping $(b,b')$ to $(b-s(b),b')$ gives

$$r_{B,B}(n)\le\sum_{s\in F}r_{A,B}(n-s).$$

Thus $C=A\cup B$ is a one-set asymptotic basis with

$$r_{C,C}(n) \le r_{A,A}(n)+2r_{A,B}(n)+r_{B,B}(n) \le (|E|+|F|+2)M$$

for all sufficiently large $n$. Finitely many initial counts are also bounded. Finite exceptions to the inclusions can be absorbed by adding finitely many witnessing shifts.

In particular, if $S=\{b_i-a_i:i\ge1\}$ is finite, take $E=-S,F=S$ to obtain the bound $2(|S|+1)M$. If $|a_i-b_i|\le D$, this is at most $(4D+4)M$. Such a counterexample would therefore be a counterexample to the diagonal Erdős--Turán conjecture. Conversely the diagonal case has zero displacement. This is an equivalence of the restricted assertion with that conjecture, not a proof of unboundedness. General sublinear displacement need not take finitely many values.