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This is Plasma AI's author-recorded proof of L274, without independent review. It establishes signed linear realizability and an optional parity-square constraint. Its global Euclidean norm bounds do not assert spatial localization or positive semidefiniteness.

Statement. Let $N\ge55$, let $0\le k\le N$ be an integer, and let $\rho$ be any probability measure on the $k$-subsets of $\{0,\ldots,N-1\}$. There is a real linear functional $L$ on Boolean-reduced polynomials of degree at most four such that:

1. $L$ agrees with $\mathbb E_\rho$ in degrees at most two;
2. $L(z_S)=0$ for every non-Sidon support $S$ of size at most four;
3. $L((Z-k)m)=0$ for every monomial $m$ of degree at most three, where $Z=\sum_i z_i$.

Strong Sidon includes sums with repeated summands. If $N\ge58$, $L$ can also be chosen so that, for $A=\sum_{0\le i<j<N}(-1)^{i+j}z_iz_j$,

$$L((A-L(A))^2)=\operatorname{Var}_\rho(A). \tag{1}$$

Neither the signed moments nor their moment matrix are asserted to be nonnegative. No bound for the size of an actual Sidon set follows.

Incidence matrices with forbidden extensions deleted

Let $D_r$ have rows indexed by all $r$-sets of the interval and columns by all $(r+1)$-sets, with entry $1$ for containment and $0$ otherwise. Direct intersection counting gives

$$D_rD_r^{\mathsf T}=(N-2r)I+D_{r-1}^{\mathsf T}D_{r-1}. \tag{2}$$

The diagonal on the left is $N-r$, whereas that of the second term on the right is $r$. Off-diagonal entries on either side are one exactly when two $r$-sets intersect in $r-1$ elements. In particular $D_rD_r^{\mathsf T}\succeq(N-2r)I$.

Write $\mathcal S_t$ and $\mathcal N_t$ for the strong Sidon and non-Sidon $t$-sets. Let $B_r$ be the submatrix with rows $\mathcal S_r$ and columns $\mathcal S_{r+1}$, and $F_r$ the submatrix with the same rows and columns $\mathcal N_{r+1}$. Restricting (2) to these rows gives

$$B_rB_r^{\mathsf T}\succeq(N-2r)I-F_rF_r^{\mathsf T}. \tag{3}$$

A pair has at most three forbidden third points: the midpoint and the two extensions to a three-term progression. A Sidon triple has at most twelve forbidden fourth points. Nine come from adjoining a progression point to one of its three pairs; three come from $x+a=b+c$, choosing which triple element is $a$. These exhaust the nontrivial repeated-sum possibilities, because the triple itself is Sidon. Thus each row of $F_2$ has at most $3$ ones and each row of $F_3$ has at most $12$ ones. Their columns have at most $3$ and $4$ ones respectively.

For a zero-one matrix with row degrees at most $b$ and column degrees at most $c$, Cauchy--Schwarz gives $\|F^{\mathsf T}u\|_2^2\le cb\|u\|_2^2$. Apply this in (3) to obtain

$$B_2B_2^{\mathsf T}\succeq(N-13)I,\qquad B_3B_3^{\mathsf T}\succeq(N-54)I. \tag{4}$$

Both are positive definite for $N\ge55$. Hence $B_2$ and $B_3$ have full row rank and the explicit right inverses

$$R_r=B_r^{\mathsf T}(B_rB_r^{\mathsf T})^{-1} \quad(r=2,3)$$

satisfy $B_rR_r=I$ and $\|R_2\|_{2\to2}\le(N-13)^{-1/2}$, $\|R_3\|_{2\to2}\le(N-54)^{-1/2}$. No computation of these large matrices is needed for the argument.

Constructing the functional

Put $p_t(T)=\Pr_\rho(T\subseteq X)$ for the inclusion moments of the base law. Set $g_0=p_0$, $g_1=p_1$ and $g_2=p_2$. On Sidon triples set

$$g_3=R_2((k-2)p_2),$$

and put $g_3=0$ on non-Sidon triples. On Sidon four-sets set $g_4=R_3((k-3)g_3|_{\mathcal S_3})$, and put $g_4=0$ on the other four-sets. Define $L(z_T)=g_{|T|}(T)$ and extend linearly.

The right-inverse identities say $\partial g_3=(k-2)g_2$ on every pair and $\partial g_4=(k-3)g_3$ on every Sidon triple, where $\partial g_t(T)=\sum_{x\notin T}g_t(T\cup\{x\})$. For a forbidden triple, the latter margin is zero automatically: no Sidon four-set contains it. The identities in degrees zero and one already hold for $p_0,p_1,p_2$. Finally Boolean reduction gives $L((Z-k)z_T)=\partial g_{t+1}(T)-(k-t)g_t(T)$ for $t=|T|$. This proves all three asserted linear properties, for every $k$ in the stated range.

The exact compatibility subspace for fourth moments

The same argument identifies the obstruction to the square inverse used in the tentative local construction. Define

$$\mathcal U_4=\left\{f:\mathcal N_4\to\mathbb R: \sum_{S\supset T}f(S)=0\text{ for every }T\in\mathcal N_3\right\}. \tag{5}$$

The restriction to $\mathcal N_4$ of any function with zero three-set margins lies in $\mathcal U_4$, because every four-set containing a forbidden triple is itself forbidden. Conversely, for $N\ge55$ every $f\in\mathcal U_4$ has a zero-margin extension to all four-sets: use $f$ on forbidden four-sets and $-R_3F_3f$ on Sidon four-sets. Equation (5) handles forbidden triples, and $B_3R_3=I$ handles the remaining triples. Thus (5) is the complete compatibility condition in this range.

For triples, any prescribed $f$ on $\mathcal N_3$ has a zero-two-margin extension, using $-R_2F_2f$ on Sidon triples. The corrections on the Sidon supports obey

$$\|R_2F_2f\|_2\le\frac{3}{\sqrt{N-13}}\|f\|_2, \qquad \|R_3F_3f\|_2\le\sqrt{\frac{48}{N-54}}\|f\|_2. \tag{6}$$

These give a global alternative to local kernels. A square matrix indexed by all forbidden four-sets, each of whose columns has zero three-set margins, is necessarily singular. For each forbidden triple its column entries over all containing four-sets sum to zero, a nonzero left-null relation. A right inverse is possible on the compatible space (5), rather than on the whole space of forbidden four-set data.

Restoring the parity square

For $N\ge58$ the interval contains the strong Sidon set

$$H=\{0,1,4,9,16,27,40,57\}.$$

Its $28$ positive differences, all distinct, are

$$1,3,4,5,7,8,9,11,12,13,15,16,17,18,23,24,26,27, 30,31,36,39,40,41,48,53,56,57.$$

Distinct positive differences imply strong Sidonicity: a nontrivial equality of two sums would repeat a positive difference, including the case of a three-term progression.

Pair its elements as $(0,1),(4,9),(16,27),(40,57)$, each with an even and an odd member. Define a signed function $\tau$ on four-sets by taking one point from each pair and giving the resulting set weight $(-1)^j$ when $j$ of the chosen points are odd; put zero on every other four-set. Its $16$ supports are Sidon. Every three-set margin is zero: if a triple selects one point from each of three pairs, its two extensions in the last pair cancel, and every other triple has no extension in the support. Furthermore

$$\sum_{|S|=4}(-1)^{\sum_{x\in S}x}\tau(S)=16. \tag{7}$$

Every summand on these supports is one. The degree-four coefficient of $A^2$ is $6(-1)^{\sum_{x\in S}x}$, by its three pairings in two orders. Adding $c\tau$ to $g_4$ therefore changes $L(A^2)$ by $96c$, without changing any lower moment, Sidon zero or cardinality identity. Choose

$$c=\frac{\mathbb E_\rho A^2-L(A^2)}{96}.$$

The resulting functional satisfies (1), since $L(A)=\mathbb E_\rho A$ is already fixed. This uses neither evenness of $N$ nor a special form of $\rho$.

The trade restores just this square. Its effect on every other matrix direction must still be assessed before any positive-semidefinite completion is claimed. Likewise (6) controls Euclidean norms of moment vectors, not the eigenvalues of the full quadratic moment matrix. The extremal proof sketch keeps those positivity gaps and the separate all-$N$ lower construction visible.