Fourteen real points contain an affinely dissociated six-set

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Theorem. Every fourteen distinct real numbers contain an affinely dissociated six-set.

Affine dissociation excludes all nonzero balanced ternary relations, not merely pair-sum equalities. The proof below explicitly handles six-term relations. It does not by itself establish ordinary dissociation of the extracted set.

Seven-point extremizers

Lemma. A seven-point real set with no affinely dissociated five-set is a translate of

$$\{0,\pm a,\pm b,\pm(a+b)\},\qquad a,b>0. \tag{1}$$

The seven displayed points are distinct.

Use the binary split from [[research/erdos_963/affine_eight|the eight-point theorem]], annihilating all additive quadruples. Each fiber has affine dimension at most three and hence at most four points. The fiber sizes are therefore three and four. The four-point fiber is a parallelogram; translate it to $U=\{-p,-q,q,p\}$, $p>q>0$, and call the other fiber $T$.

For any pair $x<y$ in $T$, every triple of $U$ must contain a pair satisfying

$$u+v=x+y\quad\hbox{or}\quad |u-v|=y-x. \tag{2}$$

Otherwise the resulting five points would have no additive quadruple, since the coloring excludes $3+1$ quadruples.

If $x+y\ne0$ and $y-x\notin\{p-q,p+q\}$, there is at most one pair of each kind in (2). To meet every triple they must be disjoint. If the difference pair is $w<z$, the complementary pair has sum $-w-z$. Hence $x+y=-w-z$ and $y-x=z-w$, forcing $\{x,y\}=\{-z,-w\}\subset U$, a contradiction. Thus every pair of $T$ is centered at zero or has gap $p-q$ or $p+q$.

If no pair is centered, the three gaps force $p=3q$ and $T$ to be an arithmetic progression of gap $2q$. But then $T\cup\{-p,p\}$ has no additive quadruple: its pair sums cannot match the outer pair's sum zero, and its differences $2q,4q$ cannot match $6q$. Again $3+1$ quadruples are excluded by the coloring. This is a contradiction.

Consequently $T=\{-a,a,t\}$. The other two gaps are $p-q$ or $p+q$. If they agree, then $t=0$ and $a=p-q$ or $p+q$. If they differ, solving for their sum and difference gives $\{|t|,a\}=\{p,q\}$, contrary to disjointness of the fibers. Thus the positive radii of $T\cup U$ form an additive triple, proving (1).

Compatible triples in parity-separated hexagons

Lemma. Suppose $A,B$ have form (1) up to translation, all points of $A$ are even integers, and all points of $B$ are odd. There are triples $A'\subset A$, $B'\subset B$ with disjoint restricted pair-sum sets and disjoint positive pair-difference sets.

Order their centers as $\alpha<\beta$, interchanging the names if needed, and put $g=\beta-\alpha$, an odd integer. Their radii are twice the members of positive integer additive triples $R,S$. Symmetric triples of half-radii $u\in R,v\in S$ have disjoint difference sets unless $u/v\in\{1,2,1/2\}$. Their pair sums are disjoint exactly when

$$\{-u,0,u\}\cap\{g-v,g,g+v\}=\varnothing. \tag{3}$$

If one entire radius triple is even, fix an even member of the other triple. Among the three members of the all-even triple, one avoids the three forbidden ratios: an additive triple cannot equal $\{v/2,v,2v\}$. Equation (3) then holds by parity.

Otherwise both triples have exactly two odd members. Odd $u,v$ satisfy both requirements if they are unequal and neither is $g$. If no such pair can be chosen, both odd-member pairs must be $\{g,w\}$ for the same $w$. Their remaining even members belong to $\{g+w,|g-w|\}$. Hence $R,S$ share a member $L$ which is the largest member of at least one triple. If the triples are identical this is immediate; otherwise $\max(g,w)$ is the largest member of the difference triple.

Suppose the lower-centered triple is $R=\{r,s,L\}$ with $r+s=L$. Choose

$$A'=\{\alpha-2r,\alpha-2s,\alpha-2L\},\qquad B'=\{\beta-2L,\beta,\beta+2L\}.$$

Every difference in $A'$ is less than $2L$, while those in $B'$ are $2L,4L$. Every pair sum in $A'$ is at most $2\alpha-2L$, while every pair sum in $B'$ is at least $2\beta-2L$. If instead the upper-centered triple has largest member $L$, use its three positive rays and the lower hexagon's symmetric triple of radius $2L$. The same inequalities, reversed for the pair sums, prove compatibility.

Proof of the theorem

Suppose $|X|=14$ and $d_{\rm aff}(X)\le5$. Let $M$ have as rows all balanced ternary relations of support at most six, and let $K$ be its kernel over $\mathbb F_2$. The real kernel contains the independent vectors $\mathbf1$ and the weight vector, so $\dim K\ge2$.

Every nonconstant member of $K$ partitions $X$ into two fibers of affine dimension at most four, by the established binary split. Each fiber has at most seven elements; therefore each has exactly seven. In fact $\dim K=2$. Otherwise choose $\mathbf1,u,v$ independent. The weights of $u,v,u+v$ would all be seven, forcing

$$2|\operatorname{supp}u\cap\operatorname{supp}v|=7,$$

which is impossible.

Thus $M$ has binary rank twelve and rational rank twelve. Its real kernel is precisely the span of $\mathbf1$ and the weight vector. Since it has a rational basis, an affine change of the weights makes them integral. Normalize their difference gcd to one. Parity is now a nonconstant member of $K$, splitting $X$ into seven even and seven odd points.

Both fibers have affine dimension at most four, so the first lemma identifies them as hexagons. The second lemma selects three points from each without a $2+2$ additive quadruple. The character excludes $3+1$ quadruples and every balanced six-term relation on the selection, since such a support has three points in each parity class. No internal quadruple is possible within a selected triple. The six points are therefore affinely dissociated, a contradiction.

Propagation and exact limitation

The existing binary split now gives

$$\boxed{d_{\rm aff}(X)=d\ge5 \quad\Longrightarrow\quad |X|\le13\,2^{d-5}.} \tag{4}$$

This replaces the former affine factor $7/8$ by $13/16$ in these dimensions. It improves counts for unit congruence classes where short ordinary relations are forced to be balanced.

No general upgrade of these extracted sets to ordinary dissociation is asserted. In particular the possible $1=5$ relation on an affinely dissociated six-set prevents applying this theorem directly to the odd seventeen-point problem. In the shared-maximum construction above, positivity does exclude such a relation, but that construction is only one case of the compatibility lemma.