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The deductions and obstructions here are author-recorded by Plasma AI. Source-dependent conclusions retain the named premises and reading limits. No independent review, novelty claim or complete solution of the linked problems follows from this account.

The relevant questions are E0503, E0670, E0668, E1085, E0660.

Dimension, extremal constructions and realization rank

E0503: the elementary family and the rank condition

For $d\ge2$, take the vectors $e_i+e_j$, $i<j$, in $\mathbb R^{d+1}$. They lie in the affine hyperplane $\sum x_i=2$, of dimension $d$. Distinct vectors have squared distance two or four, according as their two-element supports intersect or are disjoint. Their centroid is $c=(2/(d+1),\ldots,2/(d+1))$; each original vector has squared distance $2-4/(d+1)>0$ to it. Adding $c$ preserves the isosceles-triple property. This proves $f(d)\ge\binom{d+1}{2}+1$. Blokhuis's held upper bound $\binom{d+2}{2}$ leaves exactly $d$ points of slack.

To turn a proposed two-distance edge coloring into Euclidean points, let $S$ be its proposed squared-distance matrix and $H=I-\mathbf1\mathbf1^T/N$. The necessary and sufficient realization condition is $-HSH/2\succeq0$ of rank at most $d$, with positive off-diagonal distances and zero diagonal. Necessity follows from centering coordinates; for sufficiency factor the PSD matrix into a Gram matrix and recover $S_{ij}=B_{ii}+B_{jj}-2B_{ij}$. Thus the residual step is Euclidean rank and positivity, not simply a triangle coloring. The Blokhuis result page was read at statement/account depth, not the tract proof. The pending general reduction was not re-audited; this account does not depend on accepting it.

E0670: a dimension obstruction and an exact line bound

This problem separates all $\binom n2$ pair lengths by at least one, in a fixed ambient dimension. For $n\ge3$, take any pair $a,b$ and a third point $c$. The two distinct lengths $|a-c|,|b-c|$ differ by at least one; the reverse triangle inequality gives $|a-b|\ge1$. Sorting all pair lengths now gives the elementary diameter bound $\binom n2$, with coefficient one half.

In dimension one write $x_1<\cdots<x_n$. For any integer $1\le k<n$, take all lengths $x_{i+j}-x_i$ with $1\le j\le k$. There are $T=kn-k(k+1)/2$ of them. They are distinct, mutually one-separated and at least one, so their sum is at least $T(T+1)/2$. Each adjacent gap $x_{i+1}-x_i$ enters at most $j$ terms with index gap $j$, hence at most $k(k+1)/2$ terms altogether. If $\Delta=x_n-x_1$, then

$$\Delta\ge\frac{T(T+1)}{k(k+1)}.$$

Taking $k=\lfloor\sqrt n\rfloor$ proves $\Delta\ge n^2-O(n^{3/2})$. This is a complete line argument, not a proof in an arbitrary fixed dimension.

Ho's inspected construction uses all nontrivial Fourier modes on an odd cyclic group of order $m$, with positive weights $W_r$:

$$v_t=(\sqrt{W_r}\cos(2\pi rt/m), \sqrt{W_r}\sin(2\pi rt/m))_{r=1}^{(m-1)/2}.$$

Any affine relation among a selected subset extends to real coefficients $\alpha_t$ on all $m$ indices, zero elsewhere, with $\sum_t\alpha_t=0$. The relation among the vectors annihilates all nonzero Fourier coefficients because every $W_r$ is positive; reality also supplies the conjugate modes. The zero coefficient vanishes by the affine condition. Fourier inversion gives $\alpha_t=0$ for all $t$. Thus every selected $n$-point subset is affinely independent and has intrinsic dimension $n-1$.

The construction can therefore be compressed from its displayed $n^2-n$ ambient dimension to $n-1$, but not below that. The pair-distance matrix determines the centered Gram rank, so a different isometric embedding cannot repair the dimension. This is a precise obstruction to using this construction against the fixed-$d$ question. The entire short primary text was read, and positivity of the weights was checked on PDF page 3. Its arXiv record was consulted for identity; no independent review or Lean check was performed.

E0668: finite families of optimal unit-distance sets

Let $T$ be a unit equilateral triangle. For a unit segment with direction $v$, the Minkowski sum $T+\{0,v\}$ generically has six distinct points, six unit edges inside the two copies of $T$, and three matching unit edges. The inspected value $u(6)=9$ makes it optimal. Relative rotation changes some cross distances continuously, so the family has continuum many congruence classes: a fixed finite distance multiset cannot contain a continuously varying edge value.

Likewise let $W_7$ be the centered regular unit hexagon, with 12 unit edges. Generically $T+R_\theta W_7$ has 21 distinct points, $3\cdot12$ unit edges within copies of the wheel and $3\cdot7$ matching triangle edges. The inspected value $u(21)=57$ makes this family optimal. A relative rotation again supplies continuum many congruence classes. These constructions and the flexibility warning were inspected in Alexeev--Mixon--Parshall's Appendix A, with the extremal counts in Theorem 1/Table 1. The source's exhaustive computation was not rerun. More generally, an equality $u(3m)=3u(m)+3m$ would make this product strategy useful, subject to distinctness of the points. No eventual family of such equalities was obtained.

At order four, five unit edges are attainable by two equilateral triangles on opposite sides of a shared unit edge. Six are impossible: the fourth point cannot be unit distance from all vertices of a unit triangle. Any five-edge graph is $K_4$ minus one edge, so its two remaining vertices must occupy the two intersections of the unit circles about the common edge's endpoints. They must be distinct, hence on opposite sides. The resulting diamond is unique up to congruence. This disproves the all-$n>3$ subassertion already discussed on the problem page, while the eventual-divergence question remains. No standing promotion is made.

E1085: why the high-dimensional construction does not descend

Swanepoel's inspected Lenz theorem has $d\ge4$ as a hypothesis. Its populated orthogonal circles need two orthogonal two-dimensional linear subspaces. They cannot both occur in $\mathbb R^3$. Replacing one circle by the orthogonal axis does not give a large second part: if the circle has radius $r$, every axis point at unit distance from all its points satisfies $z^2=1-r^2$, allowing at most two points. Thus this complete-bipartite mechanism does not give quadratic unit-distance growth in dimensions two or three. A different construction is needed; this does not exclude all other geometries. Minimum-one packing contact bounds are linear because of the extra separation hypothesis, which E1085 does not have.

E0660: a spatial source gap

E0660 is an explicit source gap. The held planar Altman theorem does not state the three-dimensional convex-polyhedron result. Erdős's 1975 Section 1, page 101, reports a bound $cn$ in three dimensions but gives neither the constant needed here nor a located proof. The target coefficient is $1/2-o(1)$; an unspecified positive $c$ does not imply it. Nor does projection provide a transfer: projecting can split equal spatial lengths and can turn vertices into interior planar points. The scoped primary-source search found the historical attribution, not the original three-dimensional argument. The needed primary spatial argument remains unidentified here.

Source interfaces

These source locators identify the uses in this account. Statement and account reading do not certify a source proof; proof reading here is author inspection and confers no independent standing.

• [[../library/distance_problems/alexeev_2024_erdos_unit_distance_problem_small_point/_index|alexeev 2024 erdos unit distance problem small point]]: Appendix A geometric discussion of n=6 and n=21 Minkowski sums; not the computational extremal-count proof (proof reading). Section 1 Theorem 1 and Table 1 (statement reading).
• [[../library/distance_problems/altman_1963_problem_p_erdos/_index|altman 1963 problem p erdos]]: Theorem on printed p.149; planar domain and proof scope (account reading).
• [[../library/distance_problems/blokhuis_1984_few_distance_sets/_index|blokhuis 1984 few distance sets]]: Statement and equality condition of Theorem 7.2.5; source proof pointer (account reading).
• [[../library/distance_problems/erdos_1975_problems_elementary_combinatorial_geometry/_index|erdos 1975 problems elementary combinatorial geometry]]: Section 1, higher-dimensional attribution on printed p.101 (account reading).
• [[../library/distance_problems/ho_2026_erdos_s_diameter_conjecture_separated_distances/_index|ho 2026 erdos s diameter conjecture separated distances]]: Sections 2-4, Proposition 4, equation (4), Propositions 5 and 8; Remark 9 (proof reading).
• [[../library/distance_problems/swanepoel_2009_unit_distances_diameters_euclidean_spaces/_index|swanepoel 2009 unit distances diameters euclidean spaces]]: Section 1.1; Section 2 Lenz definitions and Theorem 1 (statement reading).