R0045-23306 · q-20261014-advance-20261111-r06-quadratic-residue-0081
odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"composite_modulus","modulus":15,"a":6}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。
G001 — odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"composite_modulus","modulus":15,"a":6}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。
操作前
source operation=quadratic-domain-boundary;original snapshot and mathematics retained
操作後
explicit ordered source output fields=[mode, modulus, normalized_class, declared_model_qualified, boundary]
條件：無
原題快照保留歷史介面；此處直接列來源欄序，不依賴JavaScript。tag與No保聲稱範圍，固定來源排序不冒充數學證明。
規則：source_ordered_interface_explicit_without_javascript · 依賴：G001
操作前
Quadratic ORIGINAL boundary class Euclidean quotient times modulus := mul(0, 15)
操作後
Quadratic ORIGINAL boundary class Euclidean quotient times modulus = 0
條件：無
依本題原始角色、次序與已核中間量精確代值，不用浮點近似或另一題的資料。
規則：exact_rational_arithmetic · 依賴：G001, S001
操作前
Quadratic ORIGINAL boundary class ORIGINAL signed value minus multiple := sub(6, 0)
操作後
Quadratic ORIGINAL boundary class ORIGINAL signed value minus multiple = 6
條件：無
依本題原始角色、次序與已核中間量精確代值，不用浮點近似或另一題的資料。
規則：exact_rational_arithmetic · 依賴：G001, S002
操作前
Quadratic ORIGINAL boundary class:ORIGINAL signed value=6;positive modulus=15;actual quotient=0
操作後
canonical residue=6;value=quotient*modulus+residue and0<=residue<modulus
條件：無
用Euclidean形式保留負輸入的原值與同餘類，不把語言的負餘數直接當canonical class。
規則：signed_mod_ORIGINAL_class_normalized_not_changed · 依賴：G001, S003
操作前
Quadratic composite boundary modulus:ORIGINAL integer=15;complete possible trial divisors=[2, 3]
操作後
ALL [divisor,remainder] rows=[[2, 1], [3, 0]];prime=否;a composite>1 has a factor at most its square root
條件：無
逐完整平方根範圍核整除；prime2的範圍為空而通過，0/1不通過。不能由少數取樣、平方表或Euler判準代替此資格。
規則：finite_prime_ALL_trial_divisors_through_square_root · 依賴：G001, S004
操作前
ORIGINAL mode=composite_modulus,modulus=15,a=6;normalized class=6
操作後
declared_model_qualified=否;zero,p2 and composite conventions handled separately;NOT a claim this class has no square root under another convention
條件：無
zero雖有root0但排除於本nonzero慣例；p2或composite亦可能有root。這裡No只界定model，不把域外輸入硬套兩個相反root的定理。
規則：quadratic_boundary_No_only_declared_nonzero_odd_prime_model · 依賴：G001, S005
S006 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。S005 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。S004 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。S003 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL boundary class ORIGINAL signed value minus multiple，不以小數近似替代。S002 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL boundary class Euclidean quotient times modulus，不以小數近似替代。S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。(composite_modulus, 15, 6, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)
依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。
覆核日期：2026-10-26
原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。
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