A polynomial grid bound for the dimension-six congruence endpoint

---

Theorem. If $O$ is a positive integer set in one odd class modulo four and $d(O)=6$, then $|O|\le31$.

This resolves the concrete endpoint $q=4,d=6$ left in the earlier congruence argument. It does not establish the unrestricted odd-set bound $2^{d-1}$.

Reduction to a Cartesian product

The fourteen-point theorem and binary splitting give affine bounds $7,13,26$ in dimensions $4,5,6$, respectively. Split $O=A\sqcup B$ modulo eight. If both ordinary dimensions are at most five, both affine dimensions are at most five, by congruence and positivity, and $|O|\le26$. Otherwise orient the split so that $d(A)=6$. The same congruence argument gives $d_{\rm aff}(A)=6$, so $|A|\le26$.

For any dissociated six-set $D\subset A$, of total $S_D$, every $b\in B$ has a ternary representation on $D$ with coefficient sum $5$ or $-3$. Positivity leaves only

$$b=S_D-x_i\quad\hbox{or}\quad b=2x_j+x_i-S_D\quad(i\ne j). \tag{1}$$

With $x_1<\cdots<x_6$, the second expression is positive only if $j=6$, or if $i=6,j=5$. Thus $|B|\le12$. If $|B|\le4$, already $|O|\le30$.

Assume henceforth $|B|\ge5$. Apply the binary affine split to $A=U\sqcup V$, using all balanced ternary relations of support at most seven. Both fibers have affine dimension at most five. Every affinely dissociated five-set $E\subset U$ extends by every $a\in V$ to an affinely dissociated six-set: a new relation would use exactly one point of $V$, contrary to the character.

If both fibers have affine dimension at most four, then $|O|\le14+12=26$. Otherwise orient the split so $U$ contains an affinely dissociated five-set

$$E=\{e_1<\cdots<e_5=M\},\qquad N=e_4,\qquad S=\sum E.$$

We have $|U|\le13$. Each $E\cup\{a\}$, $a\in V$, is ordinarily dissociated as well, because it lies in one unit class modulo eight.

Fifteen lines cover the product

Substitute $D=E\cup\{a\}$ into (1). Every $b\in B$ belongs to one of

$$K=\{S,2M-S\},\qquad a+C,\qquad -a+\mathcal D, \tag{2}$$

where

$$C=\{\pm(S-e):e\in E\},$$

and $\mathcal D$ is the five-element multiset

$$\{2M+e-S:e\in E\setminus\{M\}\} \cup\{2N+M-S\}.$$

Indeed, the constant expressions $2e-S>0$ require $e=M$. In a negative-slope expression $2f+e-S-a>0$, $e\ne f$, positivity requires $f=M$, or $f=N,e=M$: otherwise $f-\sum(E\setminus\{e,f\})$ is negative because an omitted element exceeds $f$. These cases exhaust (1).

Therefore

$$P(x,y)=\prod_{c\in C}(y-x-c) \prod_{t\in\mathcal D}(y+x-t) \tag{3}$$

vanishes on $V\times(B\setminus K)$. Retaining repetitions in $\mathcal D$ ensures degree fifteen. Its leading homogeneous part is $(y-x)^{10}(y+x)^5$. The coefficients of $x^{15-j}y^j$, $0\le j\le15$, are

$$1,-5,5,15,-35,-1,65,-45,-45,65,-1,-35,15,5,-5,1;$$

in particular none is zero.

For finite real sets $X,Y$ of sizes $m,n$, a polynomial of total degree at most $m+n-2$ vanishing on $X\times Y$ has zero coefficient of $x^{m-1}y^{n-1}$. To see this directly, apply the univariate interpolation weights

$$\frac{1}{\prod_{x'\in X\setminus\{x\}}(x-x')}$$

and the analogous weights on $Y$, and sum their products times the polynomial over the grid. Each monomial vanishes under this functional if either exponent is too small; under the total-degree constraint only the indicated monomial can survive, with coefficient one. The functional is zero on a polynomial vanishing on the grid.

Both $V$ and $B\setminus K$ are nonempty. If their size sum were at least seventeen, take nonempty subgrids whose side sizes sum to seventeen and apply the preceding fact to (3). This contradicts its nonzero leading coefficients. Consequently

$$|V|+|B\setminus K|\le16.$$

Finally,

$$|O|\le|U|+|V|+|B\setminus K|+|B\cap K| \le13+16+2=31.$$

The argument was checked by the author, including its coefficient list and the need to treat $\mathcal D$ as a multiset, and is author-recorded. It is not an extrapolation from searches of integer intervals.

What remains

Splitting an arbitrary positive odd set of dimension at most six modulo four now bounds it by $62$. The desired bound is $32$. The analogous dimension-five odd bound remains $28$, versus the desired $16$. The grid mechanism has not been generalized to control the many more representation types in those unrestricted odd configurations.

One attempted shortcut was not established: the points individually dependent on a fixed affinely independent set were asserted to form a lower-dimensional set. That does not follow from pointwise dependence. The proof above avoids this claim: $U$ is a genuine binary-character fiber whose dimension bound is proved, not an assumed matroid closure.