第 27704 題：Wilson定理

原題與已知條件

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary；資料={"mode":"fermat_confusion","n":7}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary；資料={"mode":"fermat_confusion","n":7}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

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。

• mko-wilson-theorem

逐步推導

source roles = [wilson-boundary]

derived = [[mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary]]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product1 := mul(1, 1)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product1 = 1

source roles = [Wilson boundary ORIGINAL prime, 7, 1, 1, 1]

derived = [0, 1, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product2 := mul(1, 2)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product2 = 2

source roles = [Wilson boundary ORIGINAL prime, 7, 2, 1, 2]

derived = [0, 2, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product3 := mul(2, 3)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product3 = 6

source roles = [Wilson boundary ORIGINAL prime, 7, 3, 2, 6]

derived = [0, 6, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product4 := mul(6, 4)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product4 = 24

source roles = [Wilson boundary ORIGINAL prime, 7, 4, 6, 24]

derived = [3, 3, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product5 := mul(3, 5)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product5 = 15

source roles = [Wilson boundary ORIGINAL prime, 7, 5, 3, 15]

derived = [2, 1, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product6 := mul(1, 6)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product6 = 6

source roles = [Wilson boundary ORIGINAL prime, 7, 6, 1, 6]

derived = [0, 6, 是, 是]

source roles = [7, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 1], [6, 6]], 6]

derived = [6, 是, 否]

逐步反向核對

• S014 · premise_and_result_audit — 回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。
• S013 · premise_and_result_audit — 回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。
• S012 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product6，不以小數近似替代。
• S011 · premise_and_result_audit — 回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。
• S010 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product5，不以小數近似替代。
• S009 · premise_and_result_audit — 回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。
• S008 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product4，不以小數近似替代。
• S007 · premise_and_result_audit — 回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。
• S006 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product3，不以小數近似替代。
• S005 · premise_and_result_audit — 回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。
• S004 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product2，不以小數近似替代。
• S003 · premise_and_result_audit — 回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。
• S002 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product1，不以小數近似替代。
• S001 · premise_and_result_audit — 回查原始資料、操作數、完整範圍與本步前提；有限見證、條件式定理引用與形式證明分開。

結論

(fermat_confusion, 7, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 1], [6, 6]], 6, 6, 是, Wilson:(n-1)!==-1_mod_7, Fermat:a^(7-1)==1_mod_7_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed)

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。

本題覆核與來源綁定

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。