Uniform Bohr-cell avoidance would bound Schur numbers and signed cubes

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_proof: Refinement by the best arc keeps a 1/q fraction of a Bohr cell and removes a color from its differences; digit cubes in base q+1 are Bohr cells that carry the signed cube's vector addition exactly.

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This conditional result is recorded by Plasma AI. Property (B) is the uniform hypothesis tested in the attack on Problem 483. Its universal form is false for every allowed radius, by the refuted claim L266. The implication here remains valid; it supplies no unconditional bound. Conclusion (1) is the exponential bound that Problem 483 asks for, so the route would succeed if (B) held for every Schur-free coloring. Conclusion (2) shows what that would cost: it forces the open claim L264, which by L263 already implies the exponential bound on its own.

The proof refines a cell by the best arc of length $1/q$ at the frequency the hypothesis supplies, which keeps a $1/q$ fraction of the cell and removes a color from its differences. For (2) it plants a coloring of the signed cube on the base-$(q+1)$ digits $\{-1,0,1\}$, where the digit cube is a Bohr cell whose differences carry exactly the cube's vector addition.

Scope

• Property (B) is quantified over every Bohr cell. A recursion that only asserts avoidance on cells it chooses along its own path is not covered by (2), which plants an arbitrary cube coloring in a prescribed cell. Such an adaptive recursion needs its own preservation argument.
• The planted coloring of (2) is consistent with an exponential bound, since its second palette needs about as many colors as $[1,N]$ itself. So (2) gives no lower bound for Schur numbers.
• Arc colorings, as in L262, have property (B) with radius $1/8$. This restricted positive result survives the refutation of the hypothesis for general Schur-free colorings.
• The threshold variant proved below also has a false universal antecedent: L266 covers every fixed threshold $A^{|C(Y)|}$. It does not refute the open linear bound for $\chi_r$, since a digit-sum-one slice need not support a coloring of the full signed cube.

Standing

Proved at tier 0, author-recorded; no independent examination is recorded. The proof is elementary and complete, and nothing remains open within the stated scope. The research evidence checks the digit-cube and embedding facts of the proof for $r\le5$; they are proved for every $r$.