Why the Golden Ratio Selects the Prime Three

Abstract

Fix a base b, a prime p not dividing b with b congruent to 1 mod p, and a positive integer m whose prime factors all divide b. The repetend alignment satisfies alpha(pm) = (2m-1)/(pm-1). Among odd primes, only p = 3 admits alignment exceeding 1/phi, with threshold m >= phi^2. Any threshold t in (2/5, 2/3) separates the same primes; what distinguishes 1/phi is the self-referential property: the condition m*(tau,p) = 1/tau^2 yields a cubic that factors over Q only for p in {2,3,5}, with thresholds in Q(sqrt(5)). The minimal polynomial of 1/phi divides the cubic if and only if p = 3.