{"id":525,"job_id":1212,"problem_id":1,"lane_id":3,"type":"explore","user_id":36,"model":"gpt-5.6-sol","provider":"openai","report_md":"# Job1212: a one-prime phase-aware zero threshold\n\nThe arithmetic supply and uniform estimate remain open. I derived a finite second-moment threshold that retains the last prime's two-class geometry, and a hand-worked profile where it succeeds while the generic balanced-integer gate fails. These are author-PROVEN finite statements submitted for manual review, not executions or an infinitude theorem. No new route was admitted or proposed through the API; the earlier daily admission refusal in return515 is preserved without retry.\n\n## Object and known capacity argument\n\nLet p>=5 be prime, m the product of primes below p, W=mp and t0=p+1. For a length L define the retained offsets\n\n D={0<=jmax_v(c_v+c[v+2]) is exactly nonemptiness at every phase in this fiber. This is the known residual-capacity argument, a one-prime instance of the project's TheoremP, not a new general method. Nguyen's Theorem4 uses the same union-bound ingredient for different symmetric-center pairs; its Proposition3 needs complete old blocks, which this local D definition does not assume.\n\nPut M=sum_b Z_b and S2=sum_b Z_b^2. Ordinary expansion gives\n\n M=(p-2)N,\n S2=(p-4)N^2+sum_v(c_v+c[v+2])^2.\n\nThis is a two-point cyclic convolution and integer counting identity. It also specializes return517's CRT-fiber moments to one residual prime. No estimate from Aryan's small-prime-free probabilistic lemma is imported.\n\n## The additional geometry-aware threshold\n\nSuppose Z_b=0 at some phase and N>=1. All N retained offsets must then occupy its two forbidden classes. Write their class masses as u and N-u. The kill sums at the p phases are N once, u once, N-u once, and zero at the remaining p-3 phases, with zero masses allowed. The phases are distinct for p>=5. Thus\n\n S2=(p-3)N^2+u^2+(N-u)^2\n >= (p-3)N^2+B(N,2),\n B(N,2)=ceil(N^2/2).\n\nThe last inequality is integer balancing: transferring one unit from a mass at least two larger than the other lowers the square sum. Equality is attained by the two masses floor(N/2),ceil(N/2) on a forbidden class pair. Attainment here means a generic residue-count vector, not an actual prefix-sieved interval.\n\nTherefore the strict reverse inequality\n\n S2 < (p-3)N^2+ceil(N^2/2)\n\ncertifies nonemptiness at every phase of this one-prime fiber. This is a sharp minimum second moment conditional on a zero within this phase geometry. Equality or failure leaves the gate inconclusive. It is not a complete characterization of positivity or a replacement for the simpler exact capacity test when c is known.\n\n## Hand separation and its limits\n\nTake p=7,N=5,c0=4,c1=1, every other c zero. The two pairs of affected phases are disjoint. The Z values, up to permutation, are\n\n (1,1,4,4,5,5,5), M=25, S2=109, min Z=1.\n\nThe generic one-zero integer threshold is B(25,6)=105, so its strict S2=7 and 5<=N<=p-1 has minZ1 and\n\n S2=(p-2)N^2-4N+4,\n B((p-2)N,p-1)=(p-3)N^2+N,\n S2-B=(N-1)(N-4)>0.\n\nThe new threshold passes this family at N5/6, ties at N7 and need not pass for larger N. I do not infer persistence from the two small values.\n\nFor two distinct reflection-related fibers with duplicated moment profiles, the generic two-zero threshold is B(2M,2p-2)=2B(M,p-1); its failure gap doubles. This addresses the paired-zero convention of return517. A self-reflection fiber has a different two-zero gate and is excluded from this separation claim. The p7 toy is not asserted to be an actual pair of arithmetic mirror fibers or to realize D above.\n\n## Input cost and unresolved route\n\nThe large modulus m can label a fixed fiber without enumerating m or W: constructing D directly uses L offsets and divisibility by primes below p, at most O(L*pi(p)) modular tests, plus O(L+p) work for the residue histogram and moments. This is a finite algorithmic operation count with integer bit costs still payable, not a measured runtime or a uniform lower bound. Avoiding period enumeration does not supply the favorable D or S2 estimate.\n\nFor the actual target, require L<=next_prime(p)^2-p-3. Then t0+j>p and t0+j+2