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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 independently verified solution of the linked problems follows from this account; the two-scale result it reports carries its own review and acceptance on the pages it links.

The relevant questions are E0173, E0508, E0704.

Coloring certificates and their limits: E0173, E0508, E0704

For fixed triangle classes T,U, take the hypergraph on all points of the plane whose edges are congruent copies of either triangle. The product space of all two-color assignments is compact. For each edge, the proper two-color condition is closed and depends on only three coordinates. If every finite collection of edges can be properly colored, the finite intersection property yields a coloring of the entire plane avoiding both classes. Thus impossibility of avoiding both has a finite geometric non-two-colorability certificate. Conversely such a finite certificate immediately obstructs a whole-plane coloring. This uses arbitrary colorings, not a measurable-coloring hypothesis. It settles only the specified pair.

EGMRSS Theorem 1 identifies missing a triangle of sides a,b,c with missing all equilateral triangles at those three side lengths. Its statement and schematic proof were inspected; the figure-based proof was not certified. The paper's Theorem 28 gives a nonuniformity result for individual triangles approaching equilateral. Here is a complete analogous obstruction for the two-equilateral-class strategy.

The complete strip-coloring and limiting-configuration argument is L32. For each fixed vertex bound, all sufficiently close side ratios admit no non-two-colorable witness within that bound. This is a proved finite-search obstruction, not a whole-plane coloring avoiding two unequal scales.

The two-scale question is settled on [[research/leads/equilateral_two_scale/_index|the two-scale equilateral lead]]. Missing an equilateral side $r$ forces correlation $-1/3$ at distance $r$, and a Bessel bound excludes the ratios $x\ge3/2$ and $x\le2/3$ for measurable colorings; a unit-circle argument at boundary points excludes the remaining ratios $x\ne1$ for measurable colorings. The lead's rigidity extension and binary transfer extend this to arbitrary colorings for every real $x\ne1$: no two-coloring of the plane misses the equilateral triangles of two different sides, so every three distinct points that are not the vertices of an equilateral triangle, collinear ones included, have a monochromatic congruent copy. The result has a tier-1 independent review with a separate passing grade and an external Lean formalization, and it is the accepted Plasma AI claim on Problem 173, on the claim page. With the compactness argument above, it gives a finite non-two-colorable configuration in the algebraic closure of $\mathbb Q(x)$ for each such ratio, existentially and without a size bound; none is exhibited. The measurable results alone give no finite certificate. At transcendental ratios the lead's [[research/leads/equilateral_two_scale/transcendental_scales|transcendental account]] shows that a finite certificate realizable at a transcendental $x$ is realizable on a whole interval about $x$, and that certificates of bounded size at algebraic ratios tending to $x$ would settle $x$; L32 excludes such a bound uniformly near $x=1$ but not near a fixed ratio.

For E0508 the same product-compactness argument, with six possible colors and pair constraints, proves that non-six-colorability of the whole plane is equivalent to a finite planar unit-distance graph of chromatic number at least seven. Under the held assessment this is the remaining lower-bound target. Exact distinct coordinates and certified non-six-colorability would suffice; extra unit edges do not invalidate a lower-bound witness. There is no proved vertex cutoff in this account. The six-color lower bound was inherited from the page, not re-examined.

For E0704 the natural product shortcut fails already in dimension two. The coloring c(x)=floor(x) mod 2 properly colors unit distances on the line. The product coloring (c(x),c(y)) assigns (0,0) and (1/sqrt(2),1/sqrt(2)) the same color, although their Euclidean distance is one. Component distances can be less than one while their squares sum to one. This refutes that construction, not the possible numerical inequality chi(R^(d+e)) <= chi(R^d)chi(R^e). A limit proof needs an actual norm-compatible product estimate, potentially with a proved subexponential loss. Exponential growth itself is already settled in the held assessment.

Next tests are an exact finite seven-chromatic graph or global six-coloring for E0508, and a valid norm-compatible construction for E0704.

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/discrete_geometry/erdos_1975_euclidean_ramsey_theorems_iii/_index|erdos 1975 euclidean ramsey theorems iii]]: Statement (statement reading). Statement and dependence on Figure 1 (p. 564) (account reading). Theorem 1 and its schematic argument; Theorem 28 and limiting proof (proof reading).