Subsets of F_p with two prescribed moments, Part II: every exponent — the classification, the Cauchy-Liouville-Mirimanoff polynomials that decide the column n=3, and where this moment subset sum stops being PORC

Fix k ≥ 3 and an odd prime p, and let Ck(n,p) count the n-subsets T of Fp with ∑ t = ∑ tk = 0. Part I evaluates k = 3; this note settles every k. The criterion for the deciding stratum to be singular depends on k only through gcd(k−1,6), by Mann's theorem on relations between roots of unity; but the correct condition in general is that its normalisation have genus zero, and with it the classification collapses: Ck(n,p) is Artin–Tate precisely for n in {3,4,6} when k = 3, for n in {3,4} when k = 5, and for n = 3 alone for every other k. The column n = 3 is described completely. Writing u = e23/e32, the count is governed by the roots of an explicit polynomial Qk(u) of Cauchy–Liouville–Mirimanoff type and the cubic Y3+uY−u attached to each root; it is expressible in quadratic characters if and only if Qk is constant, which happens exactly for k in {3,4,5,7}. That list is a dimension: deg Qk + 1 = dim M2k(SL2(Z)), so the count is elementary exactly at the weights 6, 8, 10, 14. Where it is not, some fibre has Galois group S3, and a permutation-character identity shows the nonabelian part cannot cancel, so the count is not PORC; for 6 ≤ k ≤ 46 this rests on forty printed irreducibility certificates, and beyond on a proved bound. The two ends of the classification fail to be polynomial on residue classes for different reasons: at n = 3, k = 6 the obstruction is a nonabelian Artin-type class function and the count is polynomial on Frobenius sets; for n ≥ 4 it is a surviving weight-one Frobenius trace, and there the count is not even POFS. The note ships with the scripts behind every measured statement, the data behind every figure, and the table of certificates.