Core completion

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The paired core controls every repeated difference. This gives a reflection-completion criterion, a symmetrization retaining all but the central-third outsiders, and additional residue constraints when an outsider's triple lies in the core.

Reflection-completion criterion

Conditional reduction when completion is admissible

Suppose the proposed sharp bound has been proved for all integer sets that are symmetric about an integer or half-integer. Then it also holds for every admissible set whose reflection closure is admissible. The qualification about half-integer centers is part of this hypothesis.

Indeed, center the exceptional sum at zero and write the original interval as $[-L,R]$, where $R\ge L\ge0$ after reflection and $L+R=N-1$. The centered coordinates can be half-integers. Put $C=A\cap(-A)$ and $T=A\cup(-A)$. Then $C$ and $T$ are symmetric admissible sets of diameters at most $2L$ and $2R$, respectively, and

$$2|A|=|C|+|T|.$$

The hypothesized symmetric bound therefore gives, with $K=2/\sqrt3$,

$$|A|\le\frac K2\bigl(\sqrt{2L}+\sqrt{2R}\bigr)+o(\sqrt N) \le K\sqrt N+o(\sqrt N).$$

The last step is Cauchy--Schwarz. The error remains uniform when one of the two diameters is small: for every fixed $\varepsilon>0$, the symmetric hypothesis gives the bound $(K+\varepsilon)\sqrt h+O_\varepsilon(1)$ at every diameter $h$. Ordinary Sidon sets, for which no exceptional center is needed, already satisfy the stronger classical bound.

Thus off-center geometry causes no additional loss in the unobstructed completion case. Neither the hypothesized symmetric bound nor admissibility of reflection closure is asserted for general sets.

Reflection closure and four-term relations

Translate the exceptional sum to zero, allowing half-integer coordinates, and suppose the central element has been omitted. Write

$$C=A\cap(-A),\qquad D=A\setminus C.$$

Thus $C=-C$, $D\cap(-D)=\varnothing$, and $D$ is ordinary Sidon. Then the reflection closure $C\cup D\cup(-D)$ is admissible with exceptional sum zero if and only if

$$C\cap3D=\varnothing,\qquad D\cap3D=\varnothing, \qquad 0\notin4D. \tag{1}$$

Here $jD$ means the $j$-fold sumset, allowing repetitions, not a dilation. The omission of the central element is part of this statement; deleting it costs at most one point in asymptotic applications.

For sufficiency, a forbidden equality in the closure is a nontrivial signed four-term relation. Classify it by the number of occurrences from $D$, counting multiplicity and absorbing signs of core terms into $C=-C$. With at most two outsider occurrences, it can be written as a forbidden equality of two pairs already in $A$, unless its coefficients cancel. With three outsider occurrences and mixed signs, the same is true; equal signs instead give $C\cap3D\ne\varnothing$. With four outsider occurrences, the two-against-two case is an ordinary Sidon equality in $D$. The remaining cases are precisely the last two obstructions in (1). The pair equalities involving outsiders cannot have common sum zero, since outsiders have neither an opposite outsider nor an opposite core element. These observations also prove necessity: each obstruction in (1) gives a nontrivial signed four-term relation.

There is some sparsity in the first obstruction. Each unordered pair of outsiders belongs to at most one unordered triple whose sum is in $C$: two distinct completions would repeat a difference $c-d$ involving an outsider, contrary to the difference-reflection lemma. Distinct triples with the same core target have disjoint supports, by pair-sum uniqueness. In particular, the distinct-vertex triples form a partial Steiner triple system. Such a condition alone does not supply the large independent set that a symmetrization argument would require.

New closure collisions lie in $\pm(D+D)$

With the same hypotheses, the four sets

$$(C+C)\setminus\{0\},\qquad C+D,\qquad C-D,\qquad (D-D)\setminus\{0\}$$

are pairwise disjoint. Indeed, since $C=-C$, they represent differences of types core--core, outsider--core, core--outsider, and outsider--outsider, respectively. The difference-reflection lemma allows a repeated nonzero difference only between two core--core pairs. It also shows that the representations in each of the last three types are unique. In the first type, the original sum condition gives a unique unordered representation.

The only remaining pair-sum types in the closure $T=C\cup D\cup(-D)$ are $D+D$ and $-D-D$, each with unique unordered representations. The first is disjoint from $C+C$ and $C+D$, by admissibility of $A$; the second is disjoint from $C+C$ and $C-D$, by admissibility of $-A$. Consequently every new nonzero collision lies in $\pm(D+D)$. There are at most $d(d+1)$ such values, where $d=|D|$, and each has at most three unordered representations. Three representations can occur only through the types $D+D$, $-D-D$, and $D-D$.

Outer-sector symmetrization loses only central-third outsiders

Let $A\subset[-R,R]$, with $R>0$, have no repeated unordered pair sum except possibly zero, and suppose $0\notin A$. Set

$$C=A\cap(-A),\qquad D=A\setminus C,\qquad d_0=|D\cap[-R/3,R/3]|.$$

Then there is a symmetric admissible set $T\subset[-R,R]$ such that

$$\boxed{\quad |T|\ge |A|-d_0.\quad}$$

The assertion holds for integer or half-integer centered coordinates. Deleting a central element before applying it costs at most one point.

To prove it, choose the larger of the two outsider sectors

$$D\cap(R/3,R],\qquad D\cap[-R,-R/3),$$

and reflect the entire set if necessary so that the selected set $H$ is positive. The subset $C\cup H\subset A$ has paired core exactly $C$. Every sum of three elements of $H$ exceeds $R$, including sums with repeated terms. Therefore

$$C\cap3H=\varnothing,\qquad H\cap3H=\varnothing, \qquad 0\notin4H.$$

The completion criterion proves that $T=C\cup H\cup(-H)$ is admissible. Its pieces are disjoint, and hence

$$|T|=|C|+2|H|\ge |C|+|D|-d_0=|A|-d_0.$$

Reflecting back, when necessary, gives the claimed set in the original centered interval.

More precisely, this construction is lossless whenever either outer sector contains at least $|D|/2$ outsiders. In particular, all sets with $d_0=0$ reduce losslessly to the symmetric case, whether or not their full reflection closures are admissible. On either fixed centered lattice, if $M_{\mathrm{sym}}(R)$ denotes the maximum size of a symmetric admissible subset of $[-R,R]$, the finite inequality is

$$|A|\le M_{\mathrm{sym}}(R)+d_0.$$

Thus a sharp symmetric bound would also cover centered families with $d_0=o(\sqrt R)$. Any excess of size $\varepsilon\sqrt R$ over the symmetric maximum forces at least that many central-third outsiders.

The interval in this statement is centered at the reflection center determined by the exceptional sum. If an original interval becomes $[-L,R]$ with $L<R$, its symmetric hull has diameter $2R$, rather than $L+R$. Applying this construction to the hull therefore does not by itself preserve the original interval length. The earlier conditional completion argument avoids that loss when the full closure is admissible.

One-sided outsiders and marked Sidon sets

Let $B\subset E\subset(0,\infty)$ be finite sets, and put $A=E\cup(-B)$. Then

$$\boxed{\quad A\text{ has no repeated unordered pair sum except possibly zero} \quad\Longleftrightarrow\quad E\text{ is Sidon and }E\cap(E+E+B)=\varnothing. \quad}$$

All sums include repeated terms. This describes precisely the sets with positive outsiders after centering and deleting zero: their positive part is $E$, their negative part is $-B$, and their core is $B\cup(-B)$.

For necessity, all sums from $E+E$ are positive, so admissibility forces $E$ to be Sidon. A relation $e_3=e_1+e_2+b$, with $e_i\in E$ and $b\in B$, would give the forbidden collision

$$e_1+e_2=e_3+(-b)>0.$$

The two unordered pairs are different because only one contains a negative element.

Conversely, suppose the two conditions hold. Positive same-sign sums are unique by Sidonicity of $E$, and negative same-sign sums are unique because $B\subset E$. If two mixed sums coincide, then

$$e-b=e'-b'\quad\Longrightarrow\quad e+b'=e'+b.$$

Sidonicity says either the mixed pairs are identical or $e=b$ and $e'=b'$, in which case their common sum is zero. A positive same-sign sum colliding with a mixed sum would give $e_1+e_2=e_3-b$, contrary to $E\cap(E+E+B)=\varnothing$. A negative same-sign sum colliding with a mixed sum would give

$$-b_1-b_2=e-b_3 \quad\Longrightarrow\quad b_3=e+b_1+b_2,$$

which contradicts the same condition. Positive and negative same-sign sums cannot coincide. This exhausts all pair types and proves sufficiency.

Equivalently, mark the elements of $B$ inside an ordinary Sidon set $E$. No marked element may occur in any of the three input positions of a relation $x+y+z=w$ in $E$. The cardinality of $A$ is the weighted count

$$|A|=|E|+|B|=\sum_{e\in E}\bigl(1+1_B(e)\bigr).$$

For nonempty $B$, its span is $\max E+\max B$. The missing weighted extremal statement for this subclass is therefore

$$(|E|+|B|)^2 \le \left(\frac43+o(1)\right)(\max E+\max B)$$

under the displayed marked-Sidon conditions, for positive integer or half-integer centered coordinates as the span tends to infinity. The symmetric problem is only the case $B=E$, when the condition becomes $E\cap3E=\varnothing$. Its conclusion does not directly control arbitrary markings; in particular, the unmarked elements may participate in three-input relations. If $B=\varnothing$, the ordinary Sidon bound already applies.

A weighted residue-fiber inequality

Return to integer coordinates $A\subset[1,N]$, with exceptional sum $s$, core $C=A\cap(s-A)$, and outsiders $D=A\setminus C$. Put

$$k=|A|,\quad c=|C|,\quad d=|D|,\qquad w_a=\begin{cases}1/\sqrt2,&a\in C,\\1,&a\in D,\end{cases}$$ $$W=\sum_aw_a=k-(1-1/\sqrt2)c, \qquad Q=\sum_aw_a^2=k-c/2.$$

For every positive integer $q$, let

$$H=\lceil N/q\rceil,\qquad W_r=\sum_{a\equiv r\pmod q}w_a.$$

For every integer $h\ge1$,

$$\boxed{\quad \sum_{r\bmod q}W_r^2 \le (H+h-1)\left(1+\frac{Q-1}{h}\right). \quad} \tag{2}$$

To prove this, divide the coordinates in each residue fiber by $q$ after a translation, obtaining an interval of at most $H$ integer positions. Let $X_{r,h}(t)$ be its weighted length-$h$ window count. The repeated-difference structure gives

$$\sum_{a-b=j}w_aw_b\le1\qquad(j>0):$$

a repeated difference has two core-core representations of weight $1/2$ each, and an unrepeated difference has weight at most one. Consequently

$$\begin{aligned} \sum_{r,t}X_{r,h}(t)^2 &=hQ+2\sum_{j=1}^{h-1}(h-j) \sum_{a-b=qj}w_aw_b\\ &\le hQ+h(h-1). \end{aligned}$$

Each fiber has at most $H+h-1$ nonzero window positions and total window mass $hW_r$. Applying Cauchy--Schwarz separately in the fibers gives (2).

When $Q\ge1$, choose

$$h=\max\{1,\lceil\sqrt{(H-1)(Q-1)}\rceil\}.$$

Writing the right side of (2) as $H+Q-2+h+(H-1)(Q-1)/h$ shows that

$$\sum_rW_r^2 \le\bigl(\sqrt{H-1}+\sqrt{Q-1}\bigr)^2+1. \tag{3}$$

In particular, for fixed $q$, the established $k=O(\sqrt N)$ bound implies

$$q\sum_rW_r^2\le N+O_q(N^{3/4}). \tag{4}$$

The finite form (2) remains valid for growing moduli; no uniform fixed-modulus asymptotic is being asserted for them.

Outsiders with a unary obstruction

Call $a\in D$ unary-obstructed when $3a-s\in C$, and let $U$ be their set, $u=|U|$. In centered coordinates this is the relation $3a\in C$, which prevents adjoining the individual reflected point $-a$.

Each $a\in U$ forces the core pair

$$3a-s,\qquad 2s-3a.$$

These $2u$ points are all distinct. A within-pair coincidence would put $a=s/2$ in the core; a coincidence between different pairs would give two outsiders summing to $s$. All lie in the single residue class

$$r_0=-s\pmod3.$$

Thus $W_{r_0}\ge\sqrt2u$. Minimizing three squared masses with given total $W$ and this lower bound, then using (4), proves

$$\boxed{\quad W^2+\frac92\left(\sqrt2u-\frac W3\right)_+^2 \le N+O(N^{3/4}). \quad} \tag{5}$$

The finite right side is $3(H+h-1)(1+(Q-1)/h)$, where $H=\lceil N/3\rceil$. This works without doubling the interval when $s$ is odd.

The residue formulation is as follows. Label the other two residues $r_1,r_2$, and write

$$c_0=|C\cap(r_0\bmod3)|,\qquad d_i=|D\cap(r_i\bmod3)|.$$

Reflection interchanges the other two core fibers. The elementary identity for the sum of three squares therefore gives

$$W^2+\frac92\left(\frac{c_0}{\sqrt2}+d_0-\frac W3\right)^2 +\frac32(d_1-d_2)^2 \le N+O(N^{3/4}). \tag{6}$$

If $u_0=|U\cap(r_0\bmod3)|$ is known, $\sqrt2u$ in (5) can be replaced by $\sqrt2u+u_0$.

The pure unary template cannot improve the lower constant

Suppose that the entire core consists of the forced pairs:

$$C=\{3a-s,\ 2s-3a:a\in D\}.$$

Then $c=2d$, $k=3d$. After centering, this is precisely the template $D\cup3D\cup(-3D)$ with three disjoint pieces, where $3D$ in this sentence denotes dilation.

The residue weights are

$$W_{r_0}=\sqrt2d+d_0,\qquad W_{r_1}=d_1,\qquad W_{r_2}=d_2.$$

Since $d_0+d_1+d_2=d$,

$$\begin{aligned} \sum_rW_r^2 &\ge(\sqrt2d+d_0)^2+\frac{(d-d_0)^2}{2}\\ &=\frac52d^2+(2\sqrt2-1)dd_0+\frac32d_0^2 \ge\frac52d^2. \end{aligned}$$

Using (3) with $Q=2d$ proves the finite inequality

$$\frac52d^2 \le\bigl(\sqrt{\lceil N/3\rceil-1}+\sqrt{2d-1}\bigr)^2+1,$$

and hence

$$\boxed{\quad k^2\le\frac65N+O(N^{3/4}),\qquad k\le\sqrt{\frac65N}+O(N^{1/4}). \quad} \tag{7}$$

This excludes the whole integer template, independently of any algebraic construction used for its input.

A small extension is also immediate. If every outsider is unary-obstructed and

$$2d\le c\le\frac{31+\sqrt2}{14}\,d,$$

then $k^2\le(4/3)N+O(N^{3/4})$. In this range the active left side of (5) is

$$6d^2+\frac34\bigl(c+(\sqrt2-2)d\bigr)^2,$$

which is at least $3(c+d)^2/4$ exactly when the displayed upper bound on $c/d$ holds. This does not cover a core much larger than the forced $2d$ points, or non-unary completion obstructions.

Why the residue information matters

The corresponding unweighted continuum difference-capacity test at every shift admits a supercritical profile. Put

$$E=\pm\bigl((1/3,1]\cup(1/27,1/9]\cup(1/243,1/81]\bigr), \qquad f_D=\tfrac12 1_E,\quad f_C=\tfrac13 1_{3E}, \quad f_A=f_C+f_D.$$

The supports of $f_C,f_D$ are disjoint, and $f_C$ is exactly the profile obtained from the two reflected dilations of $f_D$. Writing $\widetilde f(t)=f(-t)$, interval-overlap calculation gives

$$\max_t\left[(f_A*\widetilde f_A)(t) -\tfrac12(f_C*\widetilde f_C)(t)\right] =\frac{458}{729},\qquad \int f_A=\frac{182}{81}.$$

The maximum is attained at $t=2/9$. After rescaling the heights to make that maximum one, the squared mass divided by the support span is

$$\frac{(182/81)^2}{6(458/729)} =\frac{8281}{6183}>\frac43.$$

The retained verifier checks all 44 nonnegative differences of interval endpoints. They include every knot of this continuous piecewise-affine correlation, so this is a finite certificate for the stated calculation.

This profile is not an integer construction and is not claimed to satisfy every possible vector-weighted or higher-order relaxation. The residue theorem (7) proves that it cannot come from an admissible integer set of the pure unary form. It illustrates a concrete arithmetic restriction missed by this all-shift macroscopic test.

A separate modular check for a Ruzsa input

There is also a short obstruction for the analogous dilation template in a particular cyclic group. For a prime $p\ge5$, fix a generator $g$ of $\mathbb F_p^*$, put

$$v(x)=(\log_gx,x)\in\mathbb Z_{p-1}\times\mathbb F_p, \qquad D_T=\{v(x):x\in T\},$$

and suppose $D_T\cup3D_T\cup(-3D_T)$ is admissible with exceptional sum zero, where the multiples here are group dilations. For every $w\ne0$,

$$v(w)+3v(-2w)=v(-8w)+3v(w). \tag{8}$$

The logarithmic coordinates agree because $w(-2w)^3=(-8w)w^3$; the other coordinate is $-5w$. The equality is nontrivial: either identification of its unordered pairs would force $p=3$ or $p=2$. Its common sum is nonzero when $p\ne5$; for $p=5$, its first coordinate is $\log_g2\ne0$ modulo four, so it remains nonzero.

Thus $T$ must miss a vertex of each three-element set $\{w,-2w,-8w\}$. Counting these as a multihypergraph, there are $p-1$ edges and each vertex occurs exactly three times. Consequently

$$|T|\le\frac23(p-1).$$

This modular statement is construction-specific. The interval theorem (7) is stronger for its own integer template and requires no Ruzsa input or homomorphic encoding.