第 27707 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":17}。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":17}。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, 17, 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, 17, 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, 17, 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, 17, 4, 6, 24]

derived = [1, 7, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product5 := mul(7, 5)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product5 = 35

source roles = [Wilson boundary ORIGINAL prime, 17, 5, 7, 35]

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, 17, 6, 1, 6]

derived = [0, 6, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product7 := mul(6, 7)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product7 = 42

source roles = [Wilson boundary ORIGINAL prime, 17, 7, 6, 42]

derived = [2, 8, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product8 := mul(8, 8)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product8 = 64

source roles = [Wilson boundary ORIGINAL prime, 17, 8, 8, 64]

derived = [3, 13, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product9 := mul(13, 9)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product9 = 117

source roles = [Wilson boundary ORIGINAL prime, 17, 9, 13, 117]

derived = [6, 15, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product10 := mul(15, 10)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product10 = 150

source roles = [Wilson boundary ORIGINAL prime, 17, 10, 15, 150]

derived = [8, 14, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product11 := mul(14, 11)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product11 = 154

source roles = [Wilson boundary ORIGINAL prime, 17, 11, 14, 154]

derived = [9, 1, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product12 := mul(1, 12)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product12 = 12

source roles = [Wilson boundary ORIGINAL prime, 17, 12, 1, 12]

derived = [0, 12, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product13 := mul(12, 13)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product13 = 156

source roles = [Wilson boundary ORIGINAL prime, 17, 13, 12, 156]

derived = [9, 3, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product14 := mul(3, 14)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product14 = 42

source roles = [Wilson boundary ORIGINAL prime, 17, 14, 3, 42]

derived = [2, 8, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product15 := mul(8, 15)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product15 = 120

source roles = [Wilson boundary ORIGINAL prime, 17, 15, 8, 120]

derived = [7, 1, 是, 是]

Wilson boundary ORIGINAL prime ORIGINAL running factorial product16 := mul(1, 16)

Wilson boundary ORIGINAL prime ORIGINAL running factorial product16 = 16

source roles = [Wilson boundary ORIGINAL prime, 17, 16, 1, 16]

derived = [0, 16, 是, 是]

source roles = [17, [[1, 1], [2, 2], [3, 6], [4, 7], [5, 1], [6, 6], [7, 8], [8, 13], [9, 15], [10, 14], [11, 1], [12, 12], [13, 3], [14, 8], [15, 1], [16, 16]], 16]

derived = [16, 是, 否]

逐步反向核對

• S034 · 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。
• S033 · 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。
• S032 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product16，不以小數近似替代。
• S031 · 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。
• S030 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product15，不以小數近似替代。
• S029 · 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。
• S028 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product14，不以小數近似替代。
• S027 · 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。
• S026 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product13，不以小數近似替代。
• S025 · 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。
• S024 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product12，不以小數近似替代。
• S023 · 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。
• S022 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product11，不以小數近似替代。
• S021 · 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。
• S020 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product10，不以小數近似替代。
• S019 · 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。
• S018 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product9，不以小數近似替代。
• S017 · 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。
• S016 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product8，不以小數近似替代。
• S015 · 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。
• S014 · premise_and_result_audit — 以精確有理數重算 Wilson boundary ORIGINAL prime ORIGINAL running factorial product7，不以小數近似替代。
• 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, 17, [[1, 1], [2, 2], [3, 6], [4, 7], [5, 1], [6, 6], [7, 8], [8, 13], [9, 15], [10, 14], [11, 1], [12, 12], [13, 3], [14, 8], [15, 1], [16, 16]], 16, 16, 是, Wilson:(n-1)!==-1_mod_17, Fermat:a^(17-1)==1_mod_17_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。