Why the Golden Ratio Selects the Prime Three

For denominators n = 3m with m = 2a5c and nonnegative integers a and c, the proper fractions split into three residue classes. Counting terminating fractions together with those whose repetend matches 1/n gives alignment (2m - 1)/(3m - 1). This exceeds 1/φ, where φ = (1 + √5)/2, exactly when m ≥ 4 in the supported family. The real crossing occurs at m = φ2. In a base b ≡ 1 (mod p) with p prime, the same count for b-supported m gives (2m - 1)/(pm - 1). Among odd primes only p = 3 can cross the golden threshold; prime two has alignment one. The reciprocal-square condition on the real critical resolution yields the cubic τ3 - τ2 - pτ + 2 = 0. For prime parameters it factors over Q exactly at 2, 3, and 5. Its thresholds at 3 and 5 are 1/φ and 1/φ2; at primes p ≥ 7 they have degree three and lie in no quadratic field. Among prime-associated thresholds satisfying this condition, 1/φ is the unique value in (2/5, 2/3).