An exact obstruction to a mixed dyadic-triadic selector

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The dyadic investigation left open whether every occurring norm-valuation level contains some rare length. A proposed refinement was to select, within a dyadic level, the maximal triadic valuation and then the largest Euclidean length. The following exact example disproves that selector, not the per-level existence question or either assertion of Problem 132.

Put $\omega=e^{i\pi/3}$, so $N(a+b\omega)=a^2+ab+b^2$, and define

$$z=\frac{-95-24\omega}{109},\quad w=\frac{13-48\omega}{43},\quad a=1-w=\frac{30+48\omega}{43}.$$

Both $N(z)$ and $N(w)$ equal one. Take

$$(p_0,p_1,p_2,p_3,p_4,p_5) =(0,1,\omega,a,a+z,a+z\omega).$$

In coordinates, $p_4=(-815+4200\omega)/4687$ and $p_5=(4302+115\omega)/4687$, where $4687=43\cdot109$. The seven unit pairs are

$$01,\ 02,\ 12,\quad34,\ 35,\ 45,\quad13.$$

Every remaining squared distance occurs once:

Thus all six values in the norm shell $v_2=0$ have $v_3=0$. Its largest value is one, with multiplicity seven, greater than the number of vertices six. The other five shell values are rare. The entire set has $R=8$, so is emphatically not an Erdős counterexample.

The example also defeats a geometric shortcut. Color the points by

$$\{p_0,p_3\},\qquad\{p_1,p_4\},\qquad\{p_2,p_5\}.$$

Their residues modulo two are $0,1,\omega$, respectively. Modulo $1+\omega$ they are $0,1,-1$, respectively. Thus the three residue parts agree for both primes. All cross-color distances are at most one. The unit graph consists of two disjoint equilateral triangles of the same color orientation, joined by one bridge. It has no four-cycle but has seven edges on six vertices. Consequently excluding four-cycles and oppositely oriented triangles cannot supply the missing edge bound.

The exact rational certificate checks all fifteen pairs, multiplicities, and valuations. The author recomputed its spectrum.