The Golden Ratio Is Not an Inflation Factor of Any Regular Tessellation, and the Factors Inside Its Own Ring Are Exactly Its Even Powers: a unit-norm obstruction, the empty interval below φ², and the Lucas traces at which φ^(2n) appears

No regular tessellation or polyhedron {p, q} has conformal-quasicrystal inflation factor φ = (1 + √5)/2, and the obstruction is a unit norm rather than a numerical accident. The inflation factor and its reciprocal are the two roots of x² - kx + 1 with integer trace k = (p - 2)(q - 2) - 2, so every factor is a quadratic Pisot unit of norm +1. The golden ratio is a quadratic Pisot unit of norm -1, with minimal polynomial x² - x - 1, and a unit of one norm is not a unit of the other. Equivalently, φ + 1/φ = √5 forces (p - 2)(q - 2) = 2 + √5 = 4.2360679774997896..., and the left side is a positive integer. The exclusion is therefore simultaneous in p and q and needs no search. The same norm argument places φ exactly. Every integer trace k ≥ 3 is realized, by {k + 4, 3}, and k = 2 is the Euclidean case with factor 1. The smallest hyperbolic factor is λ₃ = (3 + √5)/2 = 2.618033988749895..., which is φ², realized by {7, 3} and its dual {3, 7}. Since 1 < φ < φ² and the trace is an integer, the open interval (1, φ²) contains no inflation factor at all. The golden ratio is excluded for being too small as much as for being the wrong kind of unit: there is no tessellation between flat space and {7, 3}. Two consequences complete the statement. The factors lying in Z[φ] are exactly the even powers φ^(2n), at the Lucas traces L_2n = 3, 7, 18, 47, 123, ..., because φ^(2n) + φ^(-2n) = L_2n and the norm of φ^m is (-1)^m. Every odd power of φ is barred for the same reason φ is. More generally the square of the norm -1 unit of trace t is the inflation factor of trace t² + 2, so 1 + √2 and (3 + √13)/2 appear only as the factors at k = 6 and k = 11. A script recomputes all 43 checkable statements; all 43 pass.