Rich-distance degree profiles and concentric circles

---

This records the checked part of the partial-permutation and equitable-partition approach. It does not resolve either unrestricted assertion of Problem 132.

For $p\in A$ and an occurring distance $r$, put

$$d_r(p)=|\{q\in A:|p-q|=r\}|.$$

The rich-degree profile of $p$ is the vector of these degrees indexed only by distances having more than $n=|A|$ unordered pairs. No assumption of a group action or of an equitable partition is made below.

Profile lemma

Suppose $R(A)\le K$ and $S\subseteq A$ has size $\delta n>0$, with all its points having the same rich-degree profile. Put

$$L=\lfloor4K/\delta\rfloor,\qquad B=\binom{L+K}{K}.$$

There is a subset $T\subseteq S$ of size at least $\delta n/(2B)$ on which the distance to the centroid of $A$ is constant. If $|T|\ge2$, it lies on a positive-radius circle centered at that centroid. In all cases some positive-radius circle contains at least $\delta n/(2B)$ points of $S$, allowing an arbitrary circle through a singleton when necessary.

Proof. The total degree in the rare classes satisfies

$$\sum_{p\in A}\sum_{r\text{ rare}}d_r(p) =2\sum_{r\text{ rare}}m_A(r)\le2Kn.$$

At least half the points of $S$ therefore have total rare degree at most $4K/\delta$, and hence at most $L$. Their rare-degree vectors, padded with zeros to $K$ coordinates if necessary, have at most $\binom{L+K}{K}$ possibilities. On a subset $T$ of the asserted size, all rare degrees agree. The rich degrees already agree by hypothesis, so

$$\sum_{q\in A}|p-q|^2=\sum_r r^2d_r(p)$$

is constant for $p\in T$. If $c=n^{-1}\sum_{q\in A}q$, then

$$\sum_{q\in A}|p-q|^2 =n|p-c|^2+\sum_{q\in A}|q-c|^2.$$

Thus $|p-c|$ is constant on $T$. Distinct points rule out radius zero when $|T|\ge2$. This proves the lemma.

Consequences and limitation

Together with the checked positive-circle theorem, the lemma implies that in any hypothetical bounded-rarity sequence $|A|\to\infty$, every rich-degree-profile fiber has size $o(n)$, uniformly over the fibers. Otherwise a subsequence would have fixed $K,\delta>0$ and hence a positive circular fraction, forcing linearly many rare distances.

In particular, boundedly many equitable cells for the rich distance graphs would suffice to prove rarity divergence: within each cell, the total degree in each such graph is constant. If full distance profiles agree on a cell, the centroid identity puts that entire cell on one concentric circle without the rare-degree pigeonhole step.

The missing implication is substantial. The inequalities $m_A(r)>n$ control the column sums $\sum_p d_r(p)>2n$, not equality of the row profiles. No positive-sized profile fiber, bounded equitable partition, or extension of partial isometries was deduced from bounded rarity. Standard distance-energy counts do not supply such an extension; their separate limitation remains in motion_incidence.

The profile lemma and its integer counting factors were checked by the author and are author-recorded. No further extension criterion has been developed, and the missing implication is not treated as a theorem.