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Wilson定理基礎 · 脈衝 R04

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本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留;AI 覆核、有限計算與 Lean 形式證明分開標示。

batch-20261015-advance-20261118-r04-wilson-theorem-foundations · 2026-10-15 · 100 題

欄序為來源綁定的呈現補充;原始 JSONL 與數學答案不變。

欄序與來源綁定資料

1. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":2}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(2, 是, [[1, 1]], 1, 1, 是, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(2, 是, [[1, 1]], 1, 1, 是, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

2. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":3}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(3, 是, [[1, 1], [2, 2]], 2, 2, 是, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(3, 是, [[1, 1], [2, 2]], 2, 2, 是, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

3. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":4}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(4, 否, [[1, 1], [2, 2], [3, 2]], 2, 3, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(4, 否, [[1, 1], [2, 2], [3, 2]], 2, 3, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

4. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":5}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(5, 是, [[1, 1], [2, 2], [3, 1], [4, 4]], 4, 4, 是, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(5, 是, [[1, 1], [2, 2], [3, 1], [4, 4]], 4, 4, 是, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

5. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":6}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(6, 否, [[1, 1], [2, 2], [3, 0], [4, 0], [5, 0]], 0, 5, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(6, 否, [[1, 1], [2, 2], [3, 0], [4, 0], [5, 0]], 0, 5, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

6. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":7}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(7, 是, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 1], [6, 6]], 6, 6, 是, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(7, 是, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 1], [6, 6]], 6, 6, 是, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

7. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":8}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(8, 否, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0]], 0, 7, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(8, 否, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0]], 0, 7, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

8. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":9}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(9, 否, [[1, 1], [2, 2], [3, 6], [4, 6], [5, 3], [6, 0], [7, 0], [8, 0]], 0, 8, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(9, 否, [[1, 1], [2, 2], [3, 6], [4, 6], [5, 3], [6, 0], [7, 0], [8, 0]], 0, 8, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

9. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":10}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(10, 否, [[1, 1], [2, 2], [3, 6], [4, 4], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0]], 0, 9, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(10, 否, [[1, 1], [2, 2], [3, 6], [4, 4], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0]], 0, 9, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

10. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":11}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(11, 是, [[1, 1], [2, 2], [3, 6], [4, 2], [5, 10], [6, 5], [7, 2], [8, 5], [9, 1], [10, 10]], 10, 10, 是, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(11, 是, [[1, 1], [2, 2], [3, 6], [4, 2], [5, 10], [6, 5], [7, 2], [8, 5], [9, 1], [10, 10]], 10, 10, 是, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

11. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":12}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(12, 否, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0]], 0, 11, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(12, 否, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0]], 0, 11, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

12. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":13}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(13, 是, [[1, 1], [2, 2], [3, 6], [4, 11], [5, 3], [6, 5], [7, 9], [8, 7], [9, 11], [10, 6], [11, 1], [12, 12]], 12, 12, 是, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(13, 是, [[1, 1], [2, 2], [3, 6], [4, 11], [5, 3], [6, 5], [7, 9], [8, 7], [9, 11], [10, 6], [11, 1], [12, 12]], 12, 12, 是, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

13. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":14}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(14, 否, [[1, 1], [2, 2], [3, 6], [4, 10], [5, 8], [6, 6], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0]], 0, 13, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(14, 否, [[1, 1], [2, 2], [3, 6], [4, 10], [5, 8], [6, 6], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0]], 0, 13, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

14. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":15}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(15, 否, [[1, 1], [2, 2], [3, 6], [4, 9], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0]], 0, 14, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(15, 否, [[1, 1], [2, 2], [3, 6], [4, 9], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0]], 0, 14, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

15. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":16}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(16, 否, [[1, 1], [2, 2], [3, 6], [4, 8], [5, 8], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0]], 0, 15, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(16, 否, [[1, 1], [2, 2], [3, 6], [4, 8], [5, 8], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0]], 0, 15, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

16. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":17}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(17, 是, [[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, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(17, 是, [[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, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

17. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":18}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(18, 否, [[1, 1], [2, 2], [3, 6], [4, 6], [5, 12], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0]], 0, 17, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(18, 否, [[1, 1], [2, 2], [3, 6], [4, 6], [5, 12], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0]], 0, 17, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

18. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":19}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(19, 是, [[1, 1], [2, 2], [3, 6], [4, 5], [5, 6], [6, 17], [7, 5], [8, 2], [9, 18], [10, 9], [11, 4], [12, 10], [13, 16], [14, 15], [15, 16], [16, 9], [17, 1], [18, 18]], 18, 18, 是, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(19, 是, [[1, 1], [2, 2], [3, 6], [4, 5], [5, 6], [6, 17], [7, 5], [8, 2], [9, 18], [10, 9], [11, 4], [12, 10], [13, 16], [14, 15], [15, 16], [16, 9], [17, 1], [18, 18]], 18, 18, 是, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

19. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":20}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(20, 否, [[1, 1], [2, 2], [3, 6], [4, 4], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0]], 0, 19, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(20, 否, [[1, 1], [2, 2], [3, 6], [4, 4], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0]], 0, 19, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

20. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":21}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(21, 否, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 15], [6, 6], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0]], 0, 20, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(21, 否, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 15], [6, 6], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0]], 0, 20, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

21. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":22}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(22, 否, [[1, 1], [2, 2], [3, 6], [4, 2], [5, 10], [6, 16], [7, 2], [8, 16], [9, 12], [10, 10], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0]], 0, 21, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(22, 否, [[1, 1], [2, 2], [3, 6], [4, 2], [5, 10], [6, 16], [7, 2], [8, 16], [9, 12], [10, 10], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0]], 0, 21, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

22. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":23}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(23, 是, [[1, 1], [2, 2], [3, 6], [4, 1], [5, 5], [6, 7], [7, 3], [8, 1], [9, 9], [10, 21], [11, 1], [12, 12], [13, 18], [14, 22], [15, 8], [16, 13], [17, 14], [18, 22], [19, 4], [20, 11], [21, 1], [22, 22]], 22, 22, 是, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(23, 是, [[1, 1], [2, 2], [3, 6], [4, 1], [5, 5], [6, 7], [7, 3], [8, 1], [9, 9], [10, 21], [11, 1], [12, 12], [13, 18], [14, 22], [15, 8], [16, 13], [17, 14], [18, 22], [19, 4], [20, 11], [21, 1], [22, 22]], 22, 22, 是, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

23. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":24}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(24, 否, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0]], 0, 23, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(24, 否, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0]], 0, 23, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

24. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":25}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(25, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 20], [6, 20], [7, 15], [8, 20], [9, 5], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0]], 0, 24, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(25, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 20], [6, 20], [7, 15], [8, 20], [9, 5], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0]], 0, 24, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

25. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-domain-factorial;資料={"n":26}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope] 欄序回答。

instance_equivalence可在合數時為是,因兩邊同為否;不表示該n為質數。

查看答案與解釋

(26, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 16], [6, 18], [7, 22], [8, 20], [9, 24], [10, 6], [11, 14], [12, 12], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0]], 0, 25, 否, 是, Wilson_equivalence_instance)

Wave6第十二層核對=(26, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 16], [6, 18], [7, 22], [8, 20], [9, 24], [10, 6], [11, 14], [12, 12], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0]], 0, 25, 否, 是, Wilson_equivalence_instance);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-domain-factorial · candidate

26. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":2}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(2, [[1, 1]], [1], [], single_self_inverse_residue_1_equals_minus_1, 1, 是)

Wave6第十二層核對=(2, [[1, 1]], [1], [], single_self_inverse_residue_1_equals_minus_1, 1, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

27. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":3}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(3, [[1, 1], [2, 2]], [1, 2], [], self_inverse_residues_1_and_p_minus_1, 2, 是)

Wave6第十二層核對=(3, [[1, 1], [2, 2]], [1, 2], [], self_inverse_residues_1_and_p_minus_1, 2, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

28. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":5}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(5, [[1, 1], [2, 3], [3, 2], [4, 4]], [1, 4], [[2, 3, 1]], self_inverse_residues_1_and_p_minus_1, 4, 是)

Wave6第十二層核對=(5, [[1, 1], [2, 3], [3, 2], [4, 4]], [1, 4], [[2, 3, 1]], self_inverse_residues_1_and_p_minus_1, 4, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

29. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":7}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(7, [[1, 1], [2, 4], [3, 5], [4, 2], [5, 3], [6, 6]], [1, 6], [[2, 4, 1], [3, 5, 1]], self_inverse_residues_1_and_p_minus_1, 6, 是)

Wave6第十二層核對=(7, [[1, 1], [2, 4], [3, 5], [4, 2], [5, 3], [6, 6]], [1, 6], [[2, 4, 1], [3, 5, 1]], self_inverse_residues_1_and_p_minus_1, 6, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

30. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":11}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(11, [[1, 1], [2, 6], [3, 4], [4, 3], [5, 9], [6, 2], [7, 8], [8, 7], [9, 5], [10, 10]], [1, 10], [[2, 6, 1], [3, 4, 1], [5, 9, 1], [7, 8, 1]], self_inverse_residues_1_and_p_minus_1, 10, 是)

Wave6第十二層核對=(11, [[1, 1], [2, 6], [3, 4], [4, 3], [5, 9], [6, 2], [7, 8], [8, 7], [9, 5], [10, 10]], [1, 10], [[2, 6, 1], [3, 4, 1], [5, 9, 1], [7, 8, 1]], self_inverse_residues_1_and_p_minus_1, 10, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

31. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":13}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(13, [[1, 1], [2, 7], [3, 9], [4, 10], [5, 8], [6, 11], [7, 2], [8, 5], [9, 3], [10, 4], [11, 6], [12, 12]], [1, 12], [[2, 7, 1], [3, 9, 1], [4, 10, 1], [5, 8, 1], [6, 11, 1]], self_inverse_residues_1_and_p_minus_1, 12, 是)

Wave6第十二層核對=(13, [[1, 1], [2, 7], [3, 9], [4, 10], [5, 8], [6, 11], [7, 2], [8, 5], [9, 3], [10, 4], [11, 6], [12, 12]], [1, 12], [[2, 7, 1], [3, 9, 1], [4, 10, 1], [5, 8, 1], [6, 11, 1]], self_inverse_residues_1_and_p_minus_1, 12, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

32. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":17}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(17, [[1, 1], [2, 9], [3, 6], [4, 13], [5, 7], [6, 3], [7, 5], [8, 15], [9, 2], [10, 12], [11, 14], [12, 10], [13, 4], [14, 11], [15, 8], [16, 16]], [1, 16], [[2, 9, 1], [3, 6, 1], [4, 13, 1], [5, 7, 1], [8, 15, 1], [10, 12, 1], [11, 14, 1]], self_inverse_residues_1_and_p_minus_1, 16, 是)

Wave6第十二層核對=(17, [[1, 1], [2, 9], [3, 6], [4, 13], [5, 7], [6, 3], [7, 5], [8, 15], [9, 2], [10, 12], [11, 14], [12, 10], [13, 4], [14, 11], [15, 8], [16, 16]], [1, 16], [[2, 9, 1], [3, 6, 1], [4, 13, 1], [5, 7, 1], [8, 15, 1], [10, 12, 1], [11, 14, 1]], self_inverse_residues_1_and_p_minus_1, 16, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

33. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":19}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(19, [[1, 1], [2, 10], [3, 13], [4, 5], [5, 4], [6, 16], [7, 11], [8, 12], [9, 17], [10, 2], [11, 7], [12, 8], [13, 3], [14, 15], [15, 14], [16, 6], [17, 9], [18, 18]], [1, 18], [[2, 10, 1], [3, 13, 1], [4, 5, 1], [6, 16, 1], [7, 11, 1], [8, 12, 1], [9, 17, 1], [14, 15, 1]], self_inverse_residues_1_and_p_minus_1, 18, 是)

Wave6第十二層核對=(19, [[1, 1], [2, 10], [3, 13], [4, 5], [5, 4], [6, 16], [7, 11], [8, 12], [9, 17], [10, 2], [11, 7], [12, 8], [13, 3], [14, 15], [15, 14], [16, 6], [17, 9], [18, 18]], [1, 18], [[2, 10, 1], [3, 13, 1], [4, 5, 1], [6, 16, 1], [7, 11, 1], [8, 12, 1], [9, 17, 1], [14, 15, 1]], self_inverse_residues_1_and_p_minus_1, 18, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

34. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":23}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(23, [[1, 1], [2, 12], [3, 8], [4, 6], [5, 14], [6, 4], [7, 10], [8, 3], [9, 18], [10, 7], [11, 21], [12, 2], [13, 16], [14, 5], [15, 20], [16, 13], [17, 19], [18, 9], [19, 17], [20, 15], [21, 11], [22, 22]], [1, 22], [[2, 12, 1], [3, 8, 1], [4, 6, 1], [5, 14, 1], [7, 10, 1], [9, 18, 1], [11, 21, 1], [13, 16, 1], [15, 20, 1], [17, 19, 1]], self_inverse_residues_1_and_p_minus_1, 22, 是)

Wave6第十二層核對=(23, [[1, 1], [2, 12], [3, 8], [4, 6], [5, 14], [6, 4], [7, 10], [8, 3], [9, 18], [10, 7], [11, 21], [12, 2], [13, 16], [14, 5], [15, 20], [16, 13], [17, 19], [18, 9], [19, 17], [20, 15], [21, 11], [22, 22]], [1, 22], [[2, 12, 1], [3, 8, 1], [4, 6, 1], [5, 14, 1], [7, 10, 1], [9, 18, 1], [11, 21, 1], [13, 16, 1], [15, 20, 1], [17, 19, 1]], self_inverse_residues_1_and_p_minus_1, 22, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

35. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":29}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(29, [[1, 1], [2, 15], [3, 10], [4, 22], [5, 6], [6, 5], [7, 25], [8, 11], [9, 13], [10, 3], [11, 8], [12, 17], [13, 9], [14, 27], [15, 2], [16, 20], [17, 12], [18, 21], [19, 26], [20, 16], [21, 18], [22, 4], [23, 24], [24, 23], [25, 7], [26, 19], [27, 14], [28, 28]], [1, 28], [[2, 15, 1], [3, 10, 1], [4, 22, 1], [5, 6, 1], [7, 25, 1], [8, 11, 1], [9, 13, 1], [12, 17, 1], [14, 27, 1], [16, 20, 1], [18, 21, 1], [19, 26, 1], [23, 24, 1]], self_inverse_residues_1_and_p_minus_1, 28, 是)

Wave6第十二層核對=(29, [[1, 1], [2, 15], [3, 10], [4, 22], [5, 6], [6, 5], [7, 25], [8, 11], [9, 13], [10, 3], [11, 8], [12, 17], [13, 9], [14, 27], [15, 2], [16, 20], [17, 12], [18, 21], [19, 26], [20, 16], [21, 18], [22, 4], [23, 24], [24, 23], [25, 7], [26, 19], [27, 14], [28, 28]], [1, 28], [[2, 15, 1], [3, 10, 1], [4, 22, 1], [5, 6, 1], [7, 25, 1], [8, 11, 1], [9, 13, 1], [12, 17, 1], [14, 27, 1], [16, 20, 1], [18, 21, 1], [19, 26, 1], [23, 24, 1]], self_inverse_residues_1_and_p_minus_1, 28, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

36. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":31}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(31, [[1, 1], [2, 16], [3, 21], [4, 8], [5, 25], [6, 26], [7, 9], [8, 4], [9, 7], [10, 28], [11, 17], [12, 13], [13, 12], [14, 20], [15, 29], [16, 2], [17, 11], [18, 19], [19, 18], [20, 14], [21, 3], [22, 24], [23, 27], [24, 22], [25, 5], [26, 6], [27, 23], [28, 10], [29, 15], [30, 30]], [1, 30], [[2, 16, 1], [3, 21, 1], [4, 8, 1], [5, 25, 1], [6, 26, 1], [7, 9, 1], [10, 28, 1], [11, 17, 1], [12, 13, 1], [14, 20, 1], [15, 29, 1], [18, 19, 1], [22, 24, 1], [23, 27, 1]], self_inverse_residues_1_and_p_minus_1, 30, 是)

Wave6第十二層核對=(31, [[1, 1], [2, 16], [3, 21], [4, 8], [5, 25], [6, 26], [7, 9], [8, 4], [9, 7], [10, 28], [11, 17], [12, 13], [13, 12], [14, 20], [15, 29], [16, 2], [17, 11], [18, 19], [19, 18], [20, 14], [21, 3], [22, 24], [23, 27], [24, 22], [25, 5], [26, 6], [27, 23], [28, 10], [29, 15], [30, 30]], [1, 30], [[2, 16, 1], [3, 21, 1], [4, 8, 1], [5, 25, 1], [6, 26, 1], [7, 9, 1], [10, 28, 1], [11, 17, 1], [12, 13, 1], [14, 20, 1], [15, 29, 1], [18, 19, 1], [22, 24, 1], [23, 27, 1]], self_inverse_residues_1_and_p_minus_1, 30, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

37. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":37}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(37, [[1, 1], [2, 19], [3, 25], [4, 28], [5, 15], [6, 31], [7, 16], [8, 14], [9, 33], [10, 26], [11, 27], [12, 34], [13, 20], [14, 8], [15, 5], [16, 7], [17, 24], [18, 35], [19, 2], [20, 13], [21, 30], [22, 32], [23, 29], [24, 17], [25, 3], [26, 10], [27, 11], [28, 4], [29, 23], [30, 21], [31, 6], [32, 22], [33, 9], [34, 12], [35, 18], [36, 36]], [1, 36], [[2, 19, 1], [3, 25, 1], [4, 28, 1], [5, 15, 1], [6, 31, 1], [7, 16, 1], [8, 14, 1], [9, 33, 1], [10, 26, 1], [11, 27, 1], [12, 34, 1], [13, 20, 1], [17, 24, 1], [18, 35, 1], [21, 30, 1], [22, 32, 1], [23, 29, 1]], self_inverse_residues_1_and_p_minus_1, 36, 是)

Wave6第十二層核對=(37, [[1, 1], [2, 19], [3, 25], [4, 28], [5, 15], [6, 31], [7, 16], [8, 14], [9, 33], [10, 26], [11, 27], [12, 34], [13, 20], [14, 8], [15, 5], [16, 7], [17, 24], [18, 35], [19, 2], [20, 13], [21, 30], [22, 32], [23, 29], [24, 17], [25, 3], [26, 10], [27, 11], [28, 4], [29, 23], [30, 21], [31, 6], [32, 22], [33, 9], [34, 12], [35, 18], [36, 36]], [1, 36], [[2, 19, 1], [3, 25, 1], [4, 28, 1], [5, 15, 1], [6, 31, 1], [7, 16, 1], [8, 14, 1], [9, 33, 1], [10, 26, 1], [11, 27, 1], [12, 34, 1], [13, 20, 1], [17, 24, 1], [18, 35, 1], [21, 30, 1], [22, 32, 1], [23, 29, 1]], self_inverse_residues_1_and_p_minus_1, 36, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

38. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":41}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(41, [[1, 1], [2, 21], [3, 14], [4, 31], [5, 33], [6, 7], [7, 6], [8, 36], [9, 32], [10, 37], [11, 15], [12, 24], [13, 19], [14, 3], [15, 11], [16, 18], [17, 29], [18, 16], [19, 13], [20, 39], [21, 2], [22, 28], [23, 25], [24, 12], [25, 23], [26, 30], [27, 38], [28, 22], [29, 17], [30, 26], [31, 4], [32, 9], [33, 5], [34, 35], [35, 34], [36, 8], [37, 10], [38, 27], [39, 20], [40, 40]], [1, 40], [[2, 21, 1], [3, 14, 1], [4, 31, 1], [5, 33, 1], [6, 7, 1], [8, 36, 1], [9, 32, 1], [10, 37, 1], [11, 15, 1], [12, 24, 1], [13, 19, 1], [16, 18, 1], [17, 29, 1], [20, 39, 1], [22, 28, 1], [23, 25, 1], [26, 30, 1], [27, 38, 1], [34, 35, 1]], self_inverse_residues_1_and_p_minus_1, 40, 是)

Wave6第十二層核對=(41, [[1, 1], [2, 21], [3, 14], [4, 31], [5, 33], [6, 7], [7, 6], [8, 36], [9, 32], [10, 37], [11, 15], [12, 24], [13, 19], [14, 3], [15, 11], [16, 18], [17, 29], [18, 16], [19, 13], [20, 39], [21, 2], [22, 28], [23, 25], [24, 12], [25, 23], [26, 30], [27, 38], [28, 22], [29, 17], [30, 26], [31, 4], [32, 9], [33, 5], [34, 35], [35, 34], [36, 8], [37, 10], [38, 27], [39, 20], [40, 40]], [1, 40], [[2, 21, 1], [3, 14, 1], [4, 31, 1], [5, 33, 1], [6, 7, 1], [8, 36, 1], [9, 32, 1], [10, 37, 1], [11, 15, 1], [12, 24, 1], [13, 19, 1], [16, 18, 1], [17, 29, 1], [20, 39, 1], [22, 28, 1], [23, 25, 1], [26, 30, 1], [27, 38, 1], [34, 35, 1]], self_inverse_residues_1_and_p_minus_1, 40, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

39. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":43}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(43, [[1, 1], [2, 22], [3, 29], [4, 11], [5, 26], [6, 36], [7, 37], [8, 27], [9, 24], [10, 13], [11, 4], [12, 18], [13, 10], [14, 40], [15, 23], [16, 35], [17, 38], [18, 12], [19, 34], [20, 28], [21, 41], [22, 2], [23, 15], [24, 9], [25, 31], [26, 5], [27, 8], [28, 20], [29, 3], [30, 33], [31, 25], [32, 39], [33, 30], [34, 19], [35, 16], [36, 6], [37, 7], [38, 17], [39, 32], [40, 14], [41, 21], [42, 42]], [1, 42], [[2, 22, 1], [3, 29, 1], [4, 11, 1], [5, 26, 1], [6, 36, 1], [7, 37, 1], [8, 27, 1], [9, 24, 1], [10, 13, 1], [12, 18, 1], [14, 40, 1], [15, 23, 1], [16, 35, 1], [17, 38, 1], [19, 34, 1], [20, 28, 1], [21, 41, 1], [25, 31, 1], [30, 33, 1], [32, 39, 1]], self_inverse_residues_1_and_p_minus_1, 42, 是)

Wave6第十二層核對=(43, [[1, 1], [2, 22], [3, 29], [4, 11], [5, 26], [6, 36], [7, 37], [8, 27], [9, 24], [10, 13], [11, 4], [12, 18], [13, 10], [14, 40], [15, 23], [16, 35], [17, 38], [18, 12], [19, 34], [20, 28], [21, 41], [22, 2], [23, 15], [24, 9], [25, 31], [26, 5], [27, 8], [28, 20], [29, 3], [30, 33], [31, 25], [32, 39], [33, 30], [34, 19], [35, 16], [36, 6], [37, 7], [38, 17], [39, 32], [40, 14], [41, 21], [42, 42]], [1, 42], [[2, 22, 1], [3, 29, 1], [4, 11, 1], [5, 26, 1], [6, 36, 1], [7, 37, 1], [8, 27, 1], [9, 24, 1], [10, 13, 1], [12, 18, 1], [14, 40, 1], [15, 23, 1], [16, 35, 1], [17, 38, 1], [19, 34, 1], [20, 28, 1], [21, 41, 1], [25, 31, 1], [30, 33, 1], [32, 39, 1]], self_inverse_residues_1_and_p_minus_1, 42, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

40. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":47}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(47, [[1, 1], [2, 24], [3, 16], [4, 12], [5, 19], [6, 8], [7, 27], [8, 6], [9, 21], [10, 33], [11, 30], [12, 4], [13, 29], [14, 37], [15, 22], [16, 3], [17, 36], [18, 34], [19, 5], [20, 40], [21, 9], [22, 15], [23, 45], [24, 2], [25, 32], [26, 38], [27, 7], [28, 42], [29, 13], [30, 11], [31, 44], [32, 25], [33, 10], [34, 18], [35, 43], [36, 17], [37, 14], [38, 26], [39, 41], [40, 20], [41, 39], [42, 28], [43, 35], [44, 31], [45, 23], [46, 46]], [1, 46], [[2, 24, 1], [3, 16, 1], [4, 12, 1], [5, 19, 1], [6, 8, 1], [7, 27, 1], [9, 21, 1], [10, 33, 1], [11, 30, 1], [13, 29, 1], [14, 37, 1], [15, 22, 1], [17, 36, 1], [18, 34, 1], [20, 40, 1], [23, 45, 1], [25, 32, 1], [26, 38, 1], [28, 42, 1], [31, 44, 1], [35, 43, 1], [39, 41, 1]], self_inverse_residues_1_and_p_minus_1, 46, 是)

Wave6第十二層核對=(47, [[1, 1], [2, 24], [3, 16], [4, 12], [5, 19], [6, 8], [7, 27], [8, 6], [9, 21], [10, 33], [11, 30], [12, 4], [13, 29], [14, 37], [15, 22], [16, 3], [17, 36], [18, 34], [19, 5], [20, 40], [21, 9], [22, 15], [23, 45], [24, 2], [25, 32], [26, 38], [27, 7], [28, 42], [29, 13], [30, 11], [31, 44], [32, 25], [33, 10], [34, 18], [35, 43], [36, 17], [37, 14], [38, 26], [39, 41], [40, 20], [41, 39], [42, 28], [43, 35], [44, 31], [45, 23], [46, 46]], [1, 46], [[2, 24, 1], [3, 16, 1], [4, 12, 1], [5, 19, 1], [6, 8, 1], [7, 27, 1], [9, 21, 1], [10, 33, 1], [11, 30, 1], [13, 29, 1], [14, 37, 1], [15, 22, 1], [17, 36, 1], [18, 34, 1], [20, 40, 1], [23, 45, 1], [25, 32, 1], [26, 38, 1], [28, 42, 1], [31, 44, 1], [35, 43, 1], [39, 41, 1]], self_inverse_residues_1_and_p_minus_1, 46, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

41. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":53}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(53, [[1, 1], [2, 27], [3, 18], [4, 40], [5, 32], [6, 9], [7, 38], [8, 20], [9, 6], [10, 16], [11, 29], [12, 31], [13, 49], [14, 19], [15, 46], [16, 10], [17, 25], [18, 3], [19, 14], [20, 8], [21, 48], [22, 41], [23, 30], [24, 42], [25, 17], [26, 51], [27, 2], [28, 36], [29, 11], [30, 23], [31, 12], [32, 5], [33, 45], [34, 39], [35, 50], [36, 28], [37, 43], [38, 7], [39, 34], [40, 4], [41, 22], [42, 24], [43, 37], [44, 47], [45, 33], [46, 15], [47, 44], [48, 21], [49, 13], [50, 35], [51, 26], [52, 52]], [1, 52], [[2, 27, 1], [3, 18, 1], [4, 40, 1], [5, 32, 1], [6, 9, 1], [7, 38, 1], [8, 20, 1], [10, 16, 1], [11, 29, 1], [12, 31, 1], [13, 49, 1], [14, 19, 1], [15, 46, 1], [17, 25, 1], [21, 48, 1], [22, 41, 1], [23, 30, 1], [24, 42, 1], [26, 51, 1], [28, 36, 1], [33, 45, 1], [34, 39, 1], [35, 50, 1], [37, 43, 1], [44, 47, 1]], self_inverse_residues_1_and_p_minus_1, 52, 是)

Wave6第十二層核對=(53, [[1, 1], [2, 27], [3, 18], [4, 40], [5, 32], [6, 9], [7, 38], [8, 20], [9, 6], [10, 16], [11, 29], [12, 31], [13, 49], [14, 19], [15, 46], [16, 10], [17, 25], [18, 3], [19, 14], [20, 8], [21, 48], [22, 41], [23, 30], [24, 42], [25, 17], [26, 51], [27, 2], [28, 36], [29, 11], [30, 23], [31, 12], [32, 5], [33, 45], [34, 39], [35, 50], [36, 28], [37, 43], [38, 7], [39, 34], [40, 4], [41, 22], [42, 24], [43, 37], [44, 47], [45, 33], [46, 15], [47, 44], [48, 21], [49, 13], [50, 35], [51, 26], [52, 52]], [1, 52], [[2, 27, 1], [3, 18, 1], [4, 40, 1], [5, 32, 1], [6, 9, 1], [7, 38, 1], [8, 20, 1], [10, 16, 1], [11, 29, 1], [12, 31, 1], [13, 49, 1], [14, 19, 1], [15, 46, 1], [17, 25, 1], [21, 48, 1], [22, 41, 1], [23, 30, 1], [24, 42, 1], [26, 51, 1], [28, 36, 1], [33, 45, 1], [34, 39, 1], [35, 50, 1], [37, 43, 1], [44, 47, 1]], self_inverse_residues_1_and_p_minus_1, 52, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

42. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":59}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(59, [[1, 1], [2, 30], [3, 20], [4, 15], [5, 12], [6, 10], [7, 17], [8, 37], [9, 46], [10, 6], [11, 43], [12, 5], [13, 50], [14, 38], [15, 4], [16, 48], [17, 7], [18, 23], [19, 28], [20, 3], [21, 45], [22, 51], [23, 18], [24, 32], [25, 26], [26, 25], [27, 35], [28, 19], [29, 57], [30, 2], [31, 40], [32, 24], [33, 34], [34, 33], [35, 27], [36, 41], [37, 8], [38, 14], [39, 56], [40, 31], [41, 36], [42, 52], [43, 11], [44, 55], [45, 21], [46, 9], [47, 54], [48, 16], [49, 53], [50, 13], [51, 22], [52, 42], [53, 49], [54, 47], [55, 44], [56, 39], [57, 29], [58, 58]], [1, 58], [[2, 30, 1], [3, 20, 1], [4, 15, 1], [5, 12, 1], [6, 10, 1], [7, 17, 1], [8, 37, 1], [9, 46, 1], [11, 43, 1], [13, 50, 1], [14, 38, 1], [16, 48, 1], [18, 23, 1], [19, 28, 1], [21, 45, 1], [22, 51, 1], [24, 32, 1], [25, 26, 1], [27, 35, 1], [29, 57, 1], [31, 40, 1], [33, 34, 1], [36, 41, 1], [39, 56, 1], [42, 52, 1], [44, 55, 1], [47, 54, 1], [49, 53, 1]], self_inverse_residues_1_and_p_minus_1, 58, 是)

Wave6第十二層核對=(59, [[1, 1], [2, 30], [3, 20], [4, 15], [5, 12], [6, 10], [7, 17], [8, 37], [9, 46], [10, 6], [11, 43], [12, 5], [13, 50], [14, 38], [15, 4], [16, 48], [17, 7], [18, 23], [19, 28], [20, 3], [21, 45], [22, 51], [23, 18], [24, 32], [25, 26], [26, 25], [27, 35], [28, 19], [29, 57], [30, 2], [31, 40], [32, 24], [33, 34], [34, 33], [35, 27], [36, 41], [37, 8], [38, 14], [39, 56], [40, 31], [41, 36], [42, 52], [43, 11], [44, 55], [45, 21], [46, 9], [47, 54], [48, 16], [49, 53], [50, 13], [51, 22], [52, 42], [53, 49], [54, 47], [55, 44], [56, 39], [57, 29], [58, 58]], [1, 58], [[2, 30, 1], [3, 20, 1], [4, 15, 1], [5, 12, 1], [6, 10, 1], [7, 17, 1], [8, 37, 1], [9, 46, 1], [11, 43, 1], [13, 50, 1], [14, 38, 1], [16, 48, 1], [18, 23, 1], [19, 28, 1], [21, 45, 1], [22, 51, 1], [24, 32, 1], [25, 26, 1], [27, 35, 1], [29, 57, 1], [31, 40, 1], [33, 34, 1], [36, 41, 1], [39, 56, 1], [42, 52, 1], [44, 55, 1], [47, 54, 1], [49, 53, 1]], self_inverse_residues_1_and_p_minus_1, 58, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

43. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":61}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(61, [[1, 1], [2, 31], [3, 41], [4, 46], [5, 49], [6, 51], [7, 35], [8, 23], [9, 34], [10, 55], [11, 50], [12, 56], [13, 47], [14, 48], [15, 57], [16, 42], [17, 18], [18, 17], [19, 45], [20, 58], [21, 32], [22, 25], [23, 8], [24, 28], [25, 22], [26, 54], [27, 52], [28, 24], [29, 40], [30, 59], [31, 2], [32, 21], [33, 37], [34, 9], [35, 7], [36, 39], [37, 33], [38, 53], [39, 36], [40, 29], [41, 3], [42, 16], [43, 44], [44, 43], [45, 19], [46, 4], [47, 13], [48, 14], [49, 5], [50, 11], [51, 6], [52, 27], [53, 38], [54, 26], [55, 10], [56, 12], [57, 15], [58, 20], [59, 30], [60, 60]], [1, 60], [[2, 31, 1], [3, 41, 1], [4, 46, 1], [5, 49, 1], [6, 51, 1], [7, 35, 1], [8, 23, 1], [9, 34, 1], [10, 55, 1], [11, 50, 1], [12, 56, 1], [13, 47, 1], [14, 48, 1], [15, 57, 1], [16, 42, 1], [17, 18, 1], [19, 45, 1], [20, 58, 1], [21, 32, 1], [22, 25, 1], [24, 28, 1], [26, 54, 1], [27, 52, 1], [29, 40, 1], [30, 59, 1], [33, 37, 1], [36, 39, 1], [38, 53, 1], [43, 44, 1]], self_inverse_residues_1_and_p_minus_1, 60, 是)

Wave6第十二層核對=(61, [[1, 1], [2, 31], [3, 41], [4, 46], [5, 49], [6, 51], [7, 35], [8, 23], [9, 34], [10, 55], [11, 50], [12, 56], [13, 47], [14, 48], [15, 57], [16, 42], [17, 18], [18, 17], [19, 45], [20, 58], [21, 32], [22, 25], [23, 8], [24, 28], [25, 22], [26, 54], [27, 52], [28, 24], [29, 40], [30, 59], [31, 2], [32, 21], [33, 37], [34, 9], [35, 7], [36, 39], [37, 33], [38, 53], [39, 36], [40, 29], [41, 3], [42, 16], [43, 44], [44, 43], [45, 19], [46, 4], [47, 13], [48, 14], [49, 5], [50, 11], [51, 6], [52, 27], [53, 38], [54, 26], [55, 10], [56, 12], [57, 15], [58, 20], [59, 30], [60, 60]], [1, 60], [[2, 31, 1], [3, 41, 1], [4, 46, 1], [5, 49, 1], [6, 51, 1], [7, 35, 1], [8, 23, 1], [9, 34, 1], [10, 55, 1], [11, 50, 1], [12, 56, 1], [13, 47, 1], [14, 48, 1], [15, 57, 1], [16, 42, 1], [17, 18, 1], [19, 45, 1], [20, 58, 1], [21, 32, 1], [22, 25, 1], [24, 28, 1], [26, 54, 1], [27, 52, 1], [29, 40, 1], [30, 59, 1], [33, 37, 1], [36, 39, 1], [38, 53, 1], [43, 44, 1]], self_inverse_residues_1_and_p_minus_1, 60, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

44. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":67}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(67, [[1, 1], [2, 34], [3, 45], [4, 17], [5, 27], [6, 56], [7, 48], [8, 42], [9, 15], [10, 47], [11, 61], [12, 28], [13, 31], [14, 24], [15, 9], [16, 21], [17, 4], [18, 41], [19, 60], [20, 57], [21, 16], [22, 64], [23, 35], [24, 14], [25, 59], [26, 49], [27, 5], [28, 12], [29, 37], [30, 38], [31, 13], [32, 44], [33, 65], [34, 2], [35, 23], [36, 54], [37, 29], [38, 30], [39, 55], [40, 62], [41, 18], [42, 8], [43, 53], [44, 32], [45, 3], [46, 51], [47, 10], [48, 7], [49, 26], [50, 63], [51, 46], [52, 58], [53, 43], [54, 36], [55, 39], [56, 6], [57, 20], [58, 52], [59, 25], [60, 19], [61, 11], [62, 40], [63, 50], [64, 22], [65, 33], [66, 66]], [1, 66], [[2, 34, 1], [3, 45, 1], [4, 17, 1], [5, 27, 1], [6, 56, 1], [7, 48, 1], [8, 42, 1], [9, 15, 1], [10, 47, 1], [11, 61, 1], [12, 28, 1], [13, 31, 1], [14, 24, 1], [16, 21, 1], [18, 41, 1], [19, 60, 1], [20, 57, 1], [22, 64, 1], [23, 35, 1], [25, 59, 1], [26, 49, 1], [29, 37, 1], [30, 38, 1], [32, 44, 1], [33, 65, 1], [36, 54, 1], [39, 55, 1], [40, 62, 1], [43, 53, 1], [46, 51, 1], [50, 63, 1], [52, 58, 1]], self_inverse_residues_1_and_p_minus_1, 66, 是)

Wave6第十二層核對=(67, [[1, 1], [2, 34], [3, 45], [4, 17], [5, 27], [6, 56], [7, 48], [8, 42], [9, 15], [10, 47], [11, 61], [12, 28], [13, 31], [14, 24], [15, 9], [16, 21], [17, 4], [18, 41], [19, 60], [20, 57], [21, 16], [22, 64], [23, 35], [24, 14], [25, 59], [26, 49], [27, 5], [28, 12], [29, 37], [30, 38], [31, 13], [32, 44], [33, 65], [34, 2], [35, 23], [36, 54], [37, 29], [38, 30], [39, 55], [40, 62], [41, 18], [42, 8], [43, 53], [44, 32], [45, 3], [46, 51], [47, 10], [48, 7], [49, 26], [50, 63], [51, 46], [52, 58], [53, 43], [54, 36], [55, 39], [56, 6], [57, 20], [58, 52], [59, 25], [60, 19], [61, 11], [62, 40], [63, 50], [64, 22], [65, 33], [66, 66]], [1, 66], [[2, 34, 1], [3, 45, 1], [4, 17, 1], [5, 27, 1], [6, 56, 1], [7, 48, 1], [8, 42, 1], [9, 15, 1], [10, 47, 1], [11, 61, 1], [12, 28, 1], [13, 31, 1], [14, 24, 1], [16, 21, 1], [18, 41, 1], [19, 60, 1], [20, 57, 1], [22, 64, 1], [23, 35, 1], [25, 59, 1], [26, 49, 1], [29, 37, 1], [30, 38, 1], [32, 44, 1], [33, 65, 1], [36, 54, 1], [39, 55, 1], [40, 62, 1], [43, 53, 1], [46, 51, 1], [50, 63, 1], [52, 58, 1]], self_inverse_residues_1_and_p_minus_1, 66, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

45. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":71}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(71, [[1, 1], [2, 36], [3, 24], [4, 18], [5, 57], [6, 12], [7, 61], [8, 9], [9, 8], [10, 64], [11, 13], [12, 6], [13, 11], [14, 66], [15, 19], [16, 40], [17, 46], [18, 4], [19, 15], [20, 32], [21, 44], [22, 42], [23, 34], [24, 3], [25, 54], [26, 41], [27, 50], [28, 33], [29, 49], [30, 45], [31, 55], [32, 20], [33, 28], [34, 23], [35, 69], [36, 2], [37, 48], [38, 43], [39, 51], [40, 16], [41, 26], [42, 22], [43, 38], [44, 21], [45, 30], [46, 17], [47, 68], [48, 37], [49, 29], [50, 27], [51, 39], [52, 56], [53, 67], [54, 25], [55, 31], [56, 52], [57, 5], [58, 60], [59, 65], [60, 58], [61, 7], [62, 63], [63, 62], [64, 10], [65, 59], [66, 14], [67, 53], [68, 47], [69, 35], [70, 70]], [1, 70], [[2, 36, 1], [3, 24, 1], [4, 18, 1], [5, 57, 1], [6, 12, 1], [7, 61, 1], [8, 9, 1], [10, 64, 1], [11, 13, 1], [14, 66, 1], [15, 19, 1], [16, 40, 1], [17, 46, 1], [20, 32, 1], [21, 44, 1], [22, 42, 1], [23, 34, 1], [25, 54, 1], [26, 41, 1], [27, 50, 1], [28, 33, 1], [29, 49, 1], [30, 45, 1], [31, 55, 1], [35, 69, 1], [37, 48, 1], [38, 43, 1], [39, 51, 1], [47, 68, 1], [52, 56, 1], [53, 67, 1], [58, 60, 1], [59, 65, 1], [62, 63, 1]], self_inverse_residues_1_and_p_minus_1, 70, 是)

Wave6第十二層核對=(71, [[1, 1], [2, 36], [3, 24], [4, 18], [5, 57], [6, 12], [7, 61], [8, 9], [9, 8], [10, 64], [11, 13], [12, 6], [13, 11], [14, 66], [15, 19], [16, 40], [17, 46], [18, 4], [19, 15], [20, 32], [21, 44], [22, 42], [23, 34], [24, 3], [25, 54], [26, 41], [27, 50], [28, 33], [29, 49], [30, 45], [31, 55], [32, 20], [33, 28], [34, 23], [35, 69], [36, 2], [37, 48], [38, 43], [39, 51], [40, 16], [41, 26], [42, 22], [43, 38], [44, 21], [45, 30], [46, 17], [47, 68], [48, 37], [49, 29], [50, 27], [51, 39], [52, 56], [53, 67], [54, 25], [55, 31], [56, 52], [57, 5], [58, 60], [59, 65], [60, 58], [61, 7], [62, 63], [63, 62], [64, 10], [65, 59], [66, 14], [67, 53], [68, 47], [69, 35], [70, 70]], [1, 70], [[2, 36, 1], [3, 24, 1], [4, 18, 1], [5, 57, 1], [6, 12, 1], [7, 61, 1], [8, 9, 1], [10, 64, 1], [11, 13, 1], [14, 66, 1], [15, 19, 1], [16, 40, 1], [17, 46, 1], [20, 32, 1], [21, 44, 1], [22, 42, 1], [23, 34, 1], [25, 54, 1], [26, 41, 1], [27, 50, 1], [28, 33, 1], [29, 49, 1], [30, 45, 1], [31, 55, 1], [35, 69, 1], [37, 48, 1], [38, 43, 1], [39, 51, 1], [47, 68, 1], [52, 56, 1], [53, 67, 1], [58, 60, 1], [59, 65, 1], [62, 63, 1]], self_inverse_residues_1_and_p_minus_1, 70, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

46. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":73}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(73, [[1, 1], [2, 37], [3, 49], [4, 55], [5, 44], [6, 61], [7, 21], [8, 64], [9, 65], [10, 22], [11, 20], [12, 67], [13, 45], [14, 47], [15, 39], [16, 32], [17, 43], [18, 69], [19, 50], [20, 11], [21, 7], [22, 10], [23, 54], [24, 70], [25, 38], [26, 59], [27, 46], [28, 60], [29, 68], [30, 56], [31, 33], [32, 16], [33, 31], [34, 58], [35, 48], [36, 71], [37, 2], [38, 25], [39, 15], [40, 42], [41, 57], [42, 40], [43, 17], [44, 5], [45, 13], [46, 27], [47, 14], [48, 35], [49, 3], [50, 19], [51, 63], [52, 66], [53, 62], [54, 23], [55, 4], [56, 30], [57, 41], [58, 34], [59, 26], [60, 28], [61, 6], [62, 53], [63, 51], [64, 8], [65, 9], [66, 52], [67, 12], [68, 29], [69, 18], [70, 24], [71, 36], [72, 72]], [1, 72], [[2, 37, 1], [3, 49, 1], [4, 55, 1], [5, 44, 1], [6, 61, 1], [7, 21, 1], [8, 64, 1], [9, 65, 1], [10, 22, 1], [11, 20, 1], [12, 67, 1], [13, 45, 1], [14, 47, 1], [15, 39, 1], [16, 32, 1], [17, 43, 1], [18, 69, 1], [19, 50, 1], [23, 54, 1], [24, 70, 1], [25, 38, 1], [26, 59, 1], [27, 46, 1], [28, 60, 1], [29, 68, 1], [30, 56, 1], [31, 33, 1], [34, 58, 1], [35, 48, 1], [36, 71, 1], [40, 42, 1], [41, 57, 1], [51, 63, 1], [52, 66, 1], [53, 62, 1]], self_inverse_residues_1_and_p_minus_1, 72, 是)

Wave6第十二層核對=(73, [[1, 1], [2, 37], [3, 49], [4, 55], [5, 44], [6, 61], [7, 21], [8, 64], [9, 65], [10, 22], [11, 20], [12, 67], [13, 45], [14, 47], [15, 39], [16, 32], [17, 43], [18, 69], [19, 50], [20, 11], [21, 7], [22, 10], [23, 54], [24, 70], [25, 38], [26, 59], [27, 46], [28, 60], [29, 68], [30, 56], [31, 33], [32, 16], [33, 31], [34, 58], [35, 48], [36, 71], [37, 2], [38, 25], [39, 15], [40, 42], [41, 57], [42, 40], [43, 17], [44, 5], [45, 13], [46, 27], [47, 14], [48, 35], [49, 3], [50, 19], [51, 63], [52, 66], [53, 62], [54, 23], [55, 4], [56, 30], [57, 41], [58, 34], [59, 26], [60, 28], [61, 6], [62, 53], [63, 51], [64, 8], [65, 9], [66, 52], [67, 12], [68, 29], [69, 18], [70, 24], [71, 36], [72, 72]], [1, 72], [[2, 37, 1], [3, 49, 1], [4, 55, 1], [5, 44, 1], [6, 61, 1], [7, 21, 1], [8, 64, 1], [9, 65, 1], [10, 22, 1], [11, 20, 1], [12, 67, 1], [13, 45, 1], [14, 47, 1], [15, 39, 1], [16, 32, 1], [17, 43, 1], [18, 69, 1], [19, 50, 1], [23, 54, 1], [24, 70, 1], [25, 38, 1], [26, 59, 1], [27, 46, 1], [28, 60, 1], [29, 68, 1], [30, 56, 1], [31, 33, 1], [34, 58, 1], [35, 48, 1], [36, 71, 1], [40, 42, 1], [41, 57, 1], [51, 63, 1], [52, 66, 1], [53, 62, 1]], self_inverse_residues_1_and_p_minus_1, 72, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

47. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":79}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(79, [[1, 1], [2, 40], [3, 53], [4, 20], [5, 16], [6, 66], [7, 34], [8, 10], [9, 44], [10, 8], [11, 36], [12, 33], [13, 73], [14, 17], [15, 58], [16, 5], [17, 14], [18, 22], [19, 25], [20, 4], [21, 64], [22, 18], [23, 55], [24, 56], [25, 19], [26, 76], [27, 41], [28, 48], [29, 30], [30, 29], [31, 51], [32, 42], [33, 12], [34, 7], [35, 70], [36, 11], [37, 47], [38, 52], [39, 77], [40, 2], [41, 27], [42, 32], [43, 68], [44, 9], [45, 72], [46, 67], [47, 37], [48, 28], [49, 50], [50, 49], [51, 31], [52, 38], [53, 3], [54, 60], [55, 23], [56, 24], [57, 61], [58, 15], [59, 75], [60, 54], [61, 57], [62, 65], [63, 74], [64, 21], [65, 62], [66, 6], [67, 46], [68, 43], [69, 71], [70, 35], [71, 69], [72, 45], [73, 13], [74, 63], [75, 59], [76, 26], [77, 39], [78, 78]], [1, 78], [[2, 40, 1], [3, 53, 1], [4, 20, 1], [5, 16, 1], [6, 66, 1], [7, 34, 1], [8, 10, 1], [9, 44, 1], [11, 36, 1], [12, 33, 1], [13, 73, 1], [14, 17, 1], [15, 58, 1], [18, 22, 1], [19, 25, 1], [21, 64, 1], [23, 55, 1], [24, 56, 1], [26, 76, 1], [27, 41, 1], [28, 48, 1], [29, 30, 1], [31, 51, 1], [32, 42, 1], [35, 70, 1], [37, 47, 1], [38, 52, 1], [39, 77, 1], [43, 68, 1], [45, 72, 1], [46, 67, 1], [49, 50, 1], [54, 60, 1], [57, 61, 1], [59, 75, 1], [62, 65, 1], [63, 74, 1], [69, 71, 1]], self_inverse_residues_1_and_p_minus_1, 78, 是)

Wave6第十二層核對=(79, [[1, 1], [2, 40], [3, 53], [4, 20], [5, 16], [6, 66], [7, 34], [8, 10], [9, 44], [10, 8], [11, 36], [12, 33], [13, 73], [14, 17], [15, 58], [16, 5], [17, 14], [18, 22], [19, 25], [20, 4], [21, 64], [22, 18], [23, 55], [24, 56], [25, 19], [26, 76], [27, 41], [28, 48], [29, 30], [30, 29], [31, 51], [32, 42], [33, 12], [34, 7], [35, 70], [36, 11], [37, 47], [38, 52], [39, 77], [40, 2], [41, 27], [42, 32], [43, 68], [44, 9], [45, 72], [46, 67], [47, 37], [48, 28], [49, 50], [50, 49], [51, 31], [52, 38], [53, 3], [54, 60], [55, 23], [56, 24], [57, 61], [58, 15], [59, 75], [60, 54], [61, 57], [62, 65], [63, 74], [64, 21], [65, 62], [66, 6], [67, 46], [68, 43], [69, 71], [70, 35], [71, 69], [72, 45], [73, 13], [74, 63], [75, 59], [76, 26], [77, 39], [78, 78]], [1, 78], [[2, 40, 1], [3, 53, 1], [4, 20, 1], [5, 16, 1], [6, 66, 1], [7, 34, 1], [8, 10, 1], [9, 44, 1], [11, 36, 1], [12, 33, 1], [13, 73, 1], [14, 17, 1], [15, 58, 1], [18, 22, 1], [19, 25, 1], [21, 64, 1], [23, 55, 1], [24, 56, 1], [26, 76, 1], [27, 41, 1], [28, 48, 1], [29, 30, 1], [31, 51, 1], [32, 42, 1], [35, 70, 1], [37, 47, 1], [38, 52, 1], [39, 77, 1], [43, 68, 1], [45, 72, 1], [46, 67, 1], [49, 50, 1], [54, 60, 1], [57, 61, 1], [59, 75, 1], [62, 65, 1], [63, 74, 1], [69, 71, 1]], self_inverse_residues_1_and_p_minus_1, 78, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

48. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":83}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(83, [[1, 1], [2, 42], [3, 28], [4, 21], [5, 50], [6, 14], [7, 12], [8, 52], [9, 37], [10, 25], [11, 68], [12, 7], [13, 32], [14, 6], [15, 72], [16, 26], [17, 44], [18, 60], [19, 35], [20, 54], [21, 4], [22, 34], [23, 65], [24, 45], [25, 10], [26, 16], [27, 40], [28, 3], [29, 63], [30, 36], [31, 75], [32, 13], [33, 78], [34, 22], [35, 19], [36, 30], [37, 9], [38, 59], [39, 66], [40, 27], [41, 81], [42, 2], [43, 56], [44, 17], [45, 24], [46, 74], [47, 53], [48, 64], [49, 61], [50, 5], [51, 70], [52, 8], [53, 47], [54, 20], [55, 80], [56, 43], [57, 67], [58, 73], [59, 38], [60, 18], [61, 49], [62, 79], [63, 29], [64, 48], [65, 23], [66, 39], [67, 57], [68, 11], [69, 77], [70, 51], [71, 76], [72, 15], [73, 58], [74, 46], [75, 31], [76, 71], [77, 69], [78, 33], [79, 62], [80, 55], [81, 41], [82, 82]], [1, 82], [[2, 42, 1], [3, 28, 1], [4, 21, 1], [5, 50, 1], [6, 14, 1], [7, 12, 1], [8, 52, 1], [9, 37, 1], [10, 25, 1], [11, 68, 1], [13, 32, 1], [15, 72, 1], [16, 26, 1], [17, 44, 1], [18, 60, 1], [19, 35, 1], [20, 54, 1], [22, 34, 1], [23, 65, 1], [24, 45, 1], [27, 40, 1], [29, 63, 1], [30, 36, 1], [31, 75, 1], [33, 78, 1], [38, 59, 1], [39, 66, 1], [41, 81, 1], [43, 56, 1], [46, 74, 1], [47, 53, 1], [48, 64, 1], [49, 61, 1], [51, 70, 1], [55, 80, 1], [57, 67, 1], [58, 73, 1], [62, 79, 1], [69, 77, 1], [71, 76, 1]], self_inverse_residues_1_and_p_minus_1, 82, 是)

Wave6第十二層核對=(83, [[1, 1], [2, 42], [3, 28], [4, 21], [5, 50], [6, 14], [7, 12], [8, 52], [9, 37], [10, 25], [11, 68], [12, 7], [13, 32], [14, 6], [15, 72], [16, 26], [17, 44], [18, 60], [19, 35], [20, 54], [21, 4], [22, 34], [23, 65], [24, 45], [25, 10], [26, 16], [27, 40], [28, 3], [29, 63], [30, 36], [31, 75], [32, 13], [33, 78], [34, 22], [35, 19], [36, 30], [37, 9], [38, 59], [39, 66], [40, 27], [41, 81], [42, 2], [43, 56], [44, 17], [45, 24], [46, 74], [47, 53], [48, 64], [49, 61], [50, 5], [51, 70], [52, 8], [53, 47], [54, 20], [55, 80], [56, 43], [57, 67], [58, 73], [59, 38], [60, 18], [61, 49], [62, 79], [63, 29], [64, 48], [65, 23], [66, 39], [67, 57], [68, 11], [69, 77], [70, 51], [71, 76], [72, 15], [73, 58], [74, 46], [75, 31], [76, 71], [77, 69], [78, 33], [79, 62], [80, 55], [81, 41], [82, 82]], [1, 82], [[2, 42, 1], [3, 28, 1], [4, 21, 1], [5, 50, 1], [6, 14, 1], [7, 12, 1], [8, 52, 1], [9, 37, 1], [10, 25, 1], [11, 68, 1], [13, 32, 1], [15, 72, 1], [16, 26, 1], [17, 44, 1], [18, 60, 1], [19, 35, 1], [20, 54, 1], [22, 34, 1], [23, 65, 1], [24, 45, 1], [27, 40, 1], [29, 63, 1], [30, 36, 1], [31, 75, 1], [33, 78, 1], [38, 59, 1], [39, 66, 1], [41, 81, 1], [43, 56, 1], [46, 74, 1], [47, 53, 1], [48, 64, 1], [49, 61, 1], [51, 70, 1], [55, 80, 1], [57, 67, 1], [58, 73, 1], [62, 79, 1], [69, 77, 1], [71, 76, 1]], self_inverse_residues_1_and_p_minus_1, 82, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

49. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":89}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(89, [[1, 1], [2, 45], [3, 30], [4, 67], [5, 18], [6, 15], [7, 51], [8, 78], [9, 10], [10, 9], [11, 81], [12, 52], [13, 48], [14, 70], [15, 6], [16, 39], [17, 21], [18, 5], [19, 75], [20, 49], [21, 17], [22, 85], [23, 31], [24, 26], [25, 57], [26, 24], [27, 33], [28, 35], [29, 43], [30, 3], [31, 23], [32, 64], [33, 27], [34, 55], [35, 28], [36, 47], [37, 77], [38, 82], [39, 16], [40, 69], [41, 76], [42, 53], [43, 29], [44, 87], [45, 2], [46, 60], [47, 36], [48, 13], [49, 20], [50, 73], [51, 7], [52, 12], [53, 42], [54, 61], [55, 34], [56, 62], [57, 25], [58, 66], [59, 86], [60, 46], [61, 54], [62, 56], [63, 65], [64, 32], [65, 63], [66, 58], [67, 4], [68, 72], [69, 40], [70, 14], [71, 84], [72, 68], [73, 50], [74, 83], [75, 19], [76, 41], [77, 37], [78, 8], [79, 80], [80, 79], [81, 11], [82, 38], [83, 74], [84, 71], [85, 22], [86, 59], [87, 44], [88, 88]], [1, 88], [[2, 45, 1], [3, 30, 1], [4, 67, 1], [5, 18, 1], [6, 15, 1], [7, 51, 1], [8, 78, 1], [9, 10, 1], [11, 81, 1], [12, 52, 1], [13, 48, 1], [14, 70, 1], [16, 39, 1], [17, 21, 1], [19, 75, 1], [20, 49, 1], [22, 85, 1], [23, 31, 1], [24, 26, 1], [25, 57, 1], [27, 33, 1], [28, 35, 1], [29, 43, 1], [32, 64, 1], [34, 55, 1], [36, 47, 1], [37, 77, 1], [38, 82, 1], [40, 69, 1], [41, 76, 1], [42, 53, 1], [44, 87, 1], [46, 60, 1], [50, 73, 1], [54, 61, 1], [56, 62, 1], [58, 66, 1], [59, 86, 1], [63, 65, 1], [68, 72, 1], [71, 84, 1], [74, 83, 1], [79, 80, 1]], self_inverse_residues_1_and_p_minus_1, 88, 是)

Wave6第十二層核對=(89, [[1, 1], [2, 45], [3, 30], [4, 67], [5, 18], [6, 15], [7, 51], [8, 78], [9, 10], [10, 9], [11, 81], [12, 52], [13, 48], [14, 70], [15, 6], [16, 39], [17, 21], [18, 5], [19, 75], [20, 49], [21, 17], [22, 85], [23, 31], [24, 26], [25, 57], [26, 24], [27, 33], [28, 35], [29, 43], [30, 3], [31, 23], [32, 64], [33, 27], [34, 55], [35, 28], [36, 47], [37, 77], [38, 82], [39, 16], [40, 69], [41, 76], [42, 53], [43, 29], [44, 87], [45, 2], [46, 60], [47, 36], [48, 13], [49, 20], [50, 73], [51, 7], [52, 12], [53, 42], [54, 61], [55, 34], [56, 62], [57, 25], [58, 66], [59, 86], [60, 46], [61, 54], [62, 56], [63, 65], [64, 32], [65, 63], [66, 58], [67, 4], [68, 72], [69, 40], [70, 14], [71, 84], [72, 68], [73, 50], [74, 83], [75, 19], [76, 41], [77, 37], [78, 8], [79, 80], [80, 79], [81, 11], [82, 38], [83, 74], [84, 71], [85, 22], [86, 59], [87, 44], [88, 88]], [1, 88], [[2, 45, 1], [3, 30, 1], [4, 67, 1], [5, 18, 1], [6, 15, 1], [7, 51, 1], [8, 78, 1], [9, 10, 1], [11, 81, 1], [12, 52, 1], [13, 48, 1], [14, 70, 1], [16, 39, 1], [17, 21, 1], [19, 75, 1], [20, 49, 1], [22, 85, 1], [23, 31, 1], [24, 26, 1], [25, 57, 1], [27, 33, 1], [28, 35, 1], [29, 43, 1], [32, 64, 1], [34, 55, 1], [36, 47, 1], [37, 77, 1], [38, 82, 1], [40, 69, 1], [41, 76, 1], [42, 53, 1], [44, 87, 1], [46, 60, 1], [50, 73, 1], [54, 61, 1], [56, 62, 1], [58, 66, 1], [59, 86, 1], [63, 65, 1], [68, 72, 1], [71, 84, 1], [74, 83, 1], [79, 80, 1]], self_inverse_residues_1_and_p_minus_1, 88, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

50. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-prime-inverse-pairing;資料={"p":97}。請依 [p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence] 欄序回答。

查看答案與解釋

(97, [[1, 1], [2, 49], [3, 65], [4, 73], [5, 39], [6, 81], [7, 14], [8, 85], [9, 54], [10, 68], [11, 53], [12, 89], [13, 15], [14, 7], [15, 13], [16, 91], [17, 40], [18, 27], [19, 46], [20, 34], [21, 37], [22, 75], [23, 38], [24, 93], [25, 66], [26, 56], [27, 18], [28, 52], [29, 87], [30, 55], [31, 72], [32, 94], [33, 50], [34, 20], [35, 61], [36, 62], [37, 21], [38, 23], [39, 5], [40, 17], [41, 71], [42, 67], [43, 88], [44, 86], [45, 69], [46, 19], [47, 64], [48, 95], [49, 2], [50, 33], [51, 78], [52, 28], [53, 11], [54, 9], [55, 30], [56, 26], [57, 80], [58, 92], [59, 74], [60, 76], [61, 35], [62, 36], [63, 77], [64, 47], [65, 3], [66, 25], [67, 42], [68, 10], [69, 45], [70, 79], [71, 41], [72, 31], [73, 4], [74, 59], [75, 22], [76, 60], [77, 63], [78, 51], [79, 70], [80, 57], [81, 6], [82, 84], [83, 90], [84, 82], [85, 8], [86, 44], [87, 29], [88, 43], [89, 12], [90, 83], [91, 16], [92, 58], [93, 24], [94, 32], [95, 48], [96, 96]], [1, 96], [[2, 49, 1], [3, 65, 1], [4, 73, 1], [5, 39, 1], [6, 81, 1], [7, 14, 1], [8, 85, 1], [9, 54, 1], [10, 68, 1], [11, 53, 1], [12, 89, 1], [13, 15, 1], [16, 91, 1], [17, 40, 1], [18, 27, 1], [19, 46, 1], [20, 34, 1], [21, 37, 1], [22, 75, 1], [23, 38, 1], [24, 93, 1], [25, 66, 1], [26, 56, 1], [28, 52, 1], [29, 87, 1], [30, 55, 1], [31, 72, 1], [32, 94, 1], [33, 50, 1], [35, 61, 1], [36, 62, 1], [41, 71, 1], [42, 67, 1], [43, 88, 1], [44, 86, 1], [45, 69, 1], [47, 64, 1], [48, 95, 1], [51, 78, 1], [57, 80, 1], [58, 92, 1], [59, 74, 1], [60, 76, 1], [63, 77, 1], [70, 79, 1], [82, 84, 1], [83, 90, 1]], self_inverse_residues_1_and_p_minus_1, 96, 是)

Wave6第十二層核對=(97, [[1, 1], [2, 49], [3, 65], [4, 73], [5, 39], [6, 81], [7, 14], [8, 85], [9, 54], [10, 68], [11, 53], [12, 89], [13, 15], [14, 7], [15, 13], [16, 91], [17, 40], [18, 27], [19, 46], [20, 34], [21, 37], [22, 75], [23, 38], [24, 93], [25, 66], [26, 56], [27, 18], [28, 52], [29, 87], [30, 55], [31, 72], [32, 94], [33, 50], [34, 20], [35, 61], [36, 62], [37, 21], [38, 23], [39, 5], [40, 17], [41, 71], [42, 67], [43, 88], [44, 86], [45, 69], [46, 19], [47, 64], [48, 95], [49, 2], [50, 33], [51, 78], [52, 28], [53, 11], [54, 9], [55, 30], [56, 26], [57, 80], [58, 92], [59, 74], [60, 76], [61, 35], [62, 36], [63, 77], [64, 47], [65, 3], [66, 25], [67, 42], [68, 10], [69, 45], [70, 79], [71, 41], [72, 31], [73, 4], [74, 59], [75, 22], [76, 60], [77, 63], [78, 51], [79, 70], [80, 57], [81, 6], [82, 84], [83, 90], [84, 82], [85, 8], [86, 44], [87, 29], [88, 43], [89, 12], [90, 83], [91, 16], [92, 58], [93, 24], [94, 32], [95, 48], [96, 96]], [1, 96], [[2, 49, 1], [3, 65, 1], [4, 73, 1], [5, 39, 1], [6, 81, 1], [7, 14, 1], [8, 85, 1], [9, 54, 1], [10, 68, 1], [11, 53, 1], [12, 89, 1], [13, 15, 1], [16, 91, 1], [17, 40, 1], [18, 27, 1], [19, 46, 1], [20, 34, 1], [21, 37, 1], [22, 75, 1], [23, 38, 1], [24, 93, 1], [25, 66, 1], [26, 56, 1], [28, 52, 1], [29, 87, 1], [30, 55, 1], [31, 72, 1], [32, 94, 1], [33, 50, 1], [35, 61, 1], [36, 62, 1], [41, 71, 1], [42, 67, 1], [43, 88, 1], [44, 86, 1], [45, 69, 1], [47, 64, 1], [48, 95, 1], [51, 78, 1], [57, 80, 1], [58, 92, 1], [59, 74, 1], [60, 76, 1], [63, 77, 1], [70, 79, 1], [82, 84, 1], [83, 90, 1]], self_inverse_residues_1_and_p_minus_1, 96, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-prime-inverse-pairing · candidate

51. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":4}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

查看答案與解釋

(4, 否, [[1, 1], [2, 2], [3, 2]], 2, 3, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(4, 否, [[1, 1], [2, 2], [3, 2]], 2, 3, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-composite-instance · candidate

52. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":6}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

查看答案與解釋

(6, 否, [[1, 1], [2, 2], [3, 0], [4, 0], [5, 0]], 0, 5, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(6, 否, [[1, 1], [2, 2], [3, 0], [4, 0], [5, 0]], 0, 5, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-composite-instance · candidate

53. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":8}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

查看答案與解釋

(8, 否, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0]], 0, 7, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(8, 否, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0]], 0, 7, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-composite-instance · candidate

54. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":9}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

查看答案與解釋

(9, 否, [[1, 1], [2, 2], [3, 6], [4, 6], [5, 3], [6, 0], [7, 0], [8, 0]], 0, 8, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(9, 否, [[1, 1], [2, 2], [3, 6], [4, 6], [5, 3], [6, 0], [7, 0], [8, 0]], 0, 8, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-composite-instance · candidate

55. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":10}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

查看答案與解釋

(10, 否, [[1, 1], [2, 2], [3, 6], [4, 4], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0]], 0, 9, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(10, 否, [[1, 1], [2, 2], [3, 6], [4, 4], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0]], 0, 9, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-composite-instance · candidate

56. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":12}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

查看答案與解釋

(12, 否, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0]], 0, 11, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(12, 否, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0]], 0, 11, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-composite-instance · candidate

57. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":14}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

查看答案與解釋

(14, 否, [[1, 1], [2, 2], [3, 6], [4, 10], [5, 8], [6, 6], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0]], 0, 13, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(14, 否, [[1, 1], [2, 2], [3, 6], [4, 10], [5, 8], [6, 6], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0]], 0, 13, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-composite-instance · candidate

58. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":15}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

查看答案與解釋

(15, 否, [[1, 1], [2, 2], [3, 6], [4, 9], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0]], 0, 14, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(15, 否, [[1, 1], [2, 2], [3, 6], [4, 9], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0]], 0, 14, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

59. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":16}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(16, 否, [[1, 1], [2, 2], [3, 6], [4, 8], [5, 8], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0]], 0, 15, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(16, 否, [[1, 1], [2, 2], [3, 6], [4, 8], [5, 8], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0]], 0, 15, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

60. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":18}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(18, 否, [[1, 1], [2, 2], [3, 6], [4, 6], [5, 12], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0]], 0, 17, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(18, 否, [[1, 1], [2, 2], [3, 6], [4, 6], [5, 12], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0]], 0, 17, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

61. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":20}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(20, 否, [[1, 1], [2, 2], [3, 6], [4, 4], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0]], 0, 19, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(20, 否, [[1, 1], [2, 2], [3, 6], [4, 4], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0]], 0, 19, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

62. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":21}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(21, 否, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 15], [6, 6], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0]], 0, 20, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(21, 否, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 15], [6, 6], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0]], 0, 20, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

63. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":22}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(22, 否, [[1, 1], [2, 2], [3, 6], [4, 2], [5, 10], [6, 16], [7, 2], [8, 16], [9, 12], [10, 10], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0]], 0, 21, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(22, 否, [[1, 1], [2, 2], [3, 6], [4, 2], [5, 10], [6, 16], [7, 2], [8, 16], [9, 12], [10, 10], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0]], 0, 21, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

64. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":24}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(24, 否, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0]], 0, 23, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(24, 否, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0]], 0, 23, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

65. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":25}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(25, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 20], [6, 20], [7, 15], [8, 20], [9, 5], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0]], 0, 24, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(25, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 20], [6, 20], [7, 15], [8, 20], [9, 5], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0]], 0, 24, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

66. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":26}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(26, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 16], [6, 18], [7, 22], [8, 20], [9, 24], [10, 6], [11, 14], [12, 12], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0]], 0, 25, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(26, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 16], [6, 18], [7, 22], [8, 20], [9, 24], [10, 6], [11, 14], [12, 12], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0]], 0, 25, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

67. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":27}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(27, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 12], [6, 18], [7, 18], [8, 9], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0]], 0, 26, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(27, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 12], [6, 18], [7, 18], [8, 9], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0]], 0, 26, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

68. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":28}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

查看答案與解釋

(28, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 8], [6, 20], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0]], 0, 27, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(28, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 8], [6, 20], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0]], 0, 27, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

69. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":30}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(30, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0]], 0, 29, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(30, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0]], 0, 29, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

70. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":32}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(32, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 24], [6, 16], [7, 16], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0]], 0, 31, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(32, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 24], [6, 16], [7, 16], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0]], 0, 31, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

71. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":33}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(33, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 21], [6, 27], [7, 24], [8, 27], [9, 12], [10, 21], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0]], 0, 32, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(33, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 21], [6, 27], [7, 24], [8, 27], [9, 12], [10, 21], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0]], 0, 32, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

72. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":34}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(34, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 18], [6, 6], [7, 8], [8, 30], [9, 32], [10, 14], [11, 18], [12, 12], [13, 20], [14, 8], [15, 18], [16, 16], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0]], 0, 33, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(34, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 18], [6, 6], [7, 8], [8, 30], [9, 32], [10, 14], [11, 18], [12, 12], [13, 20], [14, 8], [15, 18], [16, 16], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0]], 0, 33, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

73. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":35}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(35, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 15], [6, 20], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0], [34, 0]], 0, 34, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(35, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 15], [6, 20], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0], [34, 0]], 0, 34, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

74. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":36}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(36, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 12], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0], [34, 0], [35, 0]], 0, 35, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(36, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 12], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0], [12, 0], [13, 0], [14, 0], [15, 0], [16, 0], [17, 0], [18, 0], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0], [34, 0], [35, 0]], 0, 35, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

75. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-composite-instance;資料={"n":38}。請依 [n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope] 欄序回答。

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(38, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 6], [6, 36], [7, 24], [8, 2], [9, 18], [10, 28], [11, 4], [12, 10], [13, 16], [14, 34], [15, 16], [16, 28], [17, 20], [18, 18], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0], [34, 0], [35, 0], [36, 0], [37, 0]], 0, 37, 是, finite_composite_instance_not_converse_proof)

Wave6第十二層核對=(38, 否, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 6], [6, 36], [7, 24], [8, 2], [9, 18], [10, 28], [11, 4], [12, 10], [13, 16], [14, 34], [15, 16], [16, 28], [17, 20], [18, 18], [19, 0], [20, 0], [21, 0], [22, 0], [23, 0], [24, 0], [25, 0], [26, 0], [27, 0], [28, 0], [29, 0], [30, 0], [31, 0], [32, 0], [33, 0], [34, 0], [35, 0], [36, 0], [37, 0]], 0, 37, 是, finite_composite_instance_not_converse_proof);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-composite-instance · candidate

76. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"p2_self_inverse","n":2}。請依 [mode, p, self_inverse_residues, modular_identity, distinct_residue_scope, exception_qualified] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(p2_self_inverse, 2, [1], 1_equals_minus_1_mod_2, one_not_two_distinct_self_inverse_residues, 是)

Wave6第十二層核對=(p2_self_inverse, 2, [1], 1_equals_minus_1_mod_2, one_not_two_distinct_self_inverse_residues, 是);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

77. Wilson定理

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

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(invalid_n, -1, 否, Wilson_domain_requires_n>1)

Wave6第十二層核對=(invalid_n, -1, 否, Wilson_domain_requires_n>1);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

78. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"finite_testing","n":4}。請依 [mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(finite_testing, 4, 3, checked_moduli_2_through_4, 否, finite_cutoff_does_not_prove_the_universal_equivalence)

Wave6第十二層核對=(finite_testing, 4, 3, checked_moduli_2_through_4, 否, finite_cutoff_does_not_prove_the_universal_equivalence);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

79. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"fermat_confusion","n":7}。請依 [mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(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)

Wave6第十二層核對=(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);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

80. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"invalid_n","n":-4}。請依 [mode, n, Wilson_domain_qualified, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(invalid_n, -4, 否, Wilson_domain_requires_n>1)

Wave6第十二層核對=(invalid_n, -4, 否, Wilson_domain_requires_n>1);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

81. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"finite_testing","n":7}。請依 [mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(finite_testing, 7, 6, checked_moduli_2_through_7, 否, finite_cutoff_does_not_prove_the_universal_equivalence)

Wave6第十二層核對=(finite_testing, 7, 6, checked_moduli_2_through_7, 否, finite_cutoff_does_not_prove_the_universal_equivalence);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

82. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"fermat_confusion","n":17}。請依 [mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(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)

Wave6第十二層核對=(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);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

83. Wilson定理

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

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(invalid_n, -7, 否, Wilson_domain_requires_n>1)

Wave6第十二層核對=(invalid_n, -7, 否, Wilson_domain_requires_n>1);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

84. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"finite_testing","n":10}。請依 [mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(finite_testing, 10, 9, checked_moduli_2_through_10, 否, finite_cutoff_does_not_prove_the_universal_equivalence)

Wave6第十二層核對=(finite_testing, 10, 9, checked_moduli_2_through_10, 否, finite_cutoff_does_not_prove_the_universal_equivalence);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

85. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"fermat_confusion","n":29}。請依 [mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(fermat_confusion, 29, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 4], [6, 24], [7, 23], [8, 10], [9, 3], [10, 1], [11, 11], [12, 16], [13, 5], [14, 12], [15, 6], [16, 9], [17, 8], [18, 28], [19, 10], [20, 26], [21, 24], [22, 6], [23, 22], [24, 6], [25, 5], [26, 14], [27, 1], [28, 28]], 28, 28, 是, Wilson:(n-1)!==-1_mod_29, Fermat:a^(29-1)==1_mod_29_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed)

Wave6第十二層核對=(fermat_confusion, 29, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 4], [6, 24], [7, 23], [8, 10], [9, 3], [10, 1], [11, 11], [12, 16], [13, 5], [14, 12], [15, 6], [16, 9], [17, 8], [18, 28], [19, 10], [20, 26], [21, 24], [22, 6], [23, 22], [24, 6], [25, 5], [26, 14], [27, 1], [28, 28]], 28, 28, 是, Wilson:(n-1)!==-1_mod_29, Fermat:a^(29-1)==1_mod_29_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

86. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"invalid_n","n":-10}。請依 [mode, n, Wilson_domain_qualified, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(invalid_n, -10, 否, Wilson_domain_requires_n>1)

Wave6第十二層核對=(invalid_n, -10, 否, Wilson_domain_requires_n>1);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

87. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"finite_testing","n":13}。請依 [mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(finite_testing, 13, 12, checked_moduli_2_through_13, 否, finite_cutoff_does_not_prove_the_universal_equivalence)

Wave6第十二層核對=(finite_testing, 13, 12, checked_moduli_2_through_13, 否, finite_cutoff_does_not_prove_the_universal_equivalence);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

88. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"fermat_confusion","n":41}。請依 [mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(fermat_confusion, 41, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 38], [6, 23], [7, 38], [8, 17], [9, 30], [10, 13], [11, 20], [12, 35], [13, 4], [14, 15], [15, 20], [16, 33], [17, 28], [18, 12], [19, 23], [20, 9], [21, 25], [22, 17], [23, 22], [24, 36], [25, 39], [26, 30], [27, 31], [28, 7], [29, 39], [30, 22], [31, 26], [32, 12], [33, 27], [34, 16], [35, 27], [36, 29], [37, 7], [38, 20], [39, 1], [40, 40]], 40, 40, 是, Wilson:(n-1)!==-1_mod_41, Fermat:a^(41-1)==1_mod_41_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed)

Wave6第十二層核對=(fermat_confusion, 41, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 38], [6, 23], [7, 38], [8, 17], [9, 30], [10, 13], [11, 20], [12, 35], [13, 4], [14, 15], [15, 20], [16, 33], [17, 28], [18, 12], [19, 23], [20, 9], [21, 25], [22, 17], [23, 22], [24, 36], [25, 39], [26, 30], [27, 31], [28, 7], [29, 39], [30, 22], [31, 26], [32, 12], [33, 27], [34, 16], [35, 27], [36, 29], [37, 7], [38, 20], [39, 1], [40, 40]], 40, 40, 是, Wilson:(n-1)!==-1_mod_41, Fermat:a^(41-1)==1_mod_41_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

89. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"invalid_n","n":-13}。請依 [mode, n, Wilson_domain_qualified, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(invalid_n, -13, 否, Wilson_domain_requires_n>1)

Wave6第十二層核對=(invalid_n, -13, 否, Wilson_domain_requires_n>1);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

90. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"finite_testing","n":16}。請依 [mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(finite_testing, 16, 15, checked_moduli_2_through_16, 否, finite_cutoff_does_not_prove_the_universal_equivalence)

Wave6第十二層核對=(finite_testing, 16, 15, checked_moduli_2_through_16, 否, finite_cutoff_does_not_prove_the_universal_equivalence);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-boundary · candidate

91. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"fermat_confusion","n":53}。請依 [mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(fermat_confusion, 53, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 14], [6, 31], [7, 5], [8, 40], [9, 42], [10, 49], [11, 9], [12, 2], [13, 26], [14, 46], [15, 1], [16, 16], [17, 7], [18, 20], [19, 9], [20, 21], [21, 17], [22, 3], [23, 16], [24, 13], [25, 7], [26, 23], [27, 38], [28, 4], [29, 10], [30, 35], [31, 25], [32, 5], [33, 6], [34, 45], [35, 38], [36, 43], [37, 1], [38, 38], [39, 51], [40, 26], [41, 6], [42, 40], [43, 24], [44, 49], [45, 32], [46, 41], [47, 19], [48, 11], [49, 9], [50, 26], [51, 1], [52, 52]], 52, 52, 是, Wilson:(n-1)!==-1_mod_53, Fermat:a^(53-1)==1_mod_53_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed)

Wave6第十二層核對=(fermat_confusion, 53, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 14], [6, 31], [7, 5], [8, 40], [9, 42], [10, 49], [11, 9], [12, 2], [13, 26], [14, 46], [15, 1], [16, 16], [17, 7], [18, 20], [19, 9], [20, 21], [21, 17], [22, 3], [23, 16], [24, 13], [25, 7], [26, 23], [27, 38], [28, 4], [29, 10], [30, 35], [31, 25], [32, 5], [33, 6], [34, 45], [35, 38], [36, 43], [37, 1], [38, 38], [39, 51], [40, 26], [41, 6], [42, 40], [43, 24], [44, 49], [45, 32], [46, 41], [47, 19], [48, 11], [49, 9], [50, 26], [51, 1], [52, 52]], 52, 52, 是, Wilson:(n-1)!==-1_mod_53, Fermat:a^(53-1)==1_mod_53_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

92. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"invalid_n","n":-16}。請依 [mode, n, Wilson_domain_qualified, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(invalid_n, -16, 否, Wilson_domain_requires_n>1)

Wave6第十二層核對=(invalid_n, -16, 否, Wilson_domain_requires_n>1);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-boundary · candidate

93. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"finite_testing","n":19}。請依 [mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(finite_testing, 19, 18, checked_moduli_2_through_19, 否, finite_cutoff_does_not_prove_the_universal_equivalence)

Wave6第十二層核對=(finite_testing, 19, 18, checked_moduli_2_through_19, 否, finite_cutoff_does_not_prove_the_universal_equivalence);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-boundary · candidate

94. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"fermat_confusion","n":67}。請依 [mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(fermat_confusion, 67, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 53], [6, 50], [7, 15], [8, 53], [9, 8], [10, 13], [11, 9], [12, 41], [13, 64], [14, 25], [15, 40], [16, 37], [17, 26], [18, 66], [19, 48], [20, 22], [21, 60], [22, 47], [23, 9], [24, 15], [25, 40], [26, 35], [27, 7], [28, 62], [29, 56], [30, 5], [31, 21], [32, 2], [33, 66], [34, 33], [35, 16], [36, 40], [37, 6], [38, 27], [39, 48], [40, 44], [41, 62], [42, 58], [43, 15], [44, 57], [45, 19], [46, 3], [47, 7], [48, 1], [49, 49], [50, 38], [51, 62], [52, 8], [53, 22], [54, 49], [55, 15], [56, 36], [57, 42], [58, 24], [59, 9], [60, 4], [61, 43], [62, 53], [63, 56], [64, 33], [65, 1], [66, 66]], 66, 66, 是, Wilson:(n-1)!==-1_mod_67, Fermat:a^(67-1)==1_mod_67_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed)

Wave6第十二層核對=(fermat_confusion, 67, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 53], [6, 50], [7, 15], [8, 53], [9, 8], [10, 13], [11, 9], [12, 41], [13, 64], [14, 25], [15, 40], [16, 37], [17, 26], [18, 66], [19, 48], [20, 22], [21, 60], [22, 47], [23, 9], [24, 15], [25, 40], [26, 35], [27, 7], [28, 62], [29, 56], [30, 5], [31, 21], [32, 2], [33, 66], [34, 33], [35, 16], [36, 40], [37, 6], [38, 27], [39, 48], [40, 44], [41, 62], [42, 58], [43, 15], [44, 57], [45, 19], [46, 3], [47, 7], [48, 1], [49, 49], [50, 38], [51, 62], [52, 8], [53, 22], [54, 49], [55, 15], [56, 36], [57, 42], [58, 24], [59, 9], [60, 4], [61, 43], [62, 53], [63, 56], [64, 33], [65, 1], [66, 66]], 66, 66, 是, Wilson:(n-1)!==-1_mod_67, Fermat:a^(67-1)==1_mod_67_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

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Wilson定理 · qf-mod49-wilson-boundary · candidate

95. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"invalid_n","n":-19}。請依 [mode, n, Wilson_domain_qualified, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(invalid_n, -19, 否, Wilson_domain_requires_n>1)

Wave6第十二層核對=(invalid_n, -19, 否, Wilson_domain_requires_n>1);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-boundary · candidate

96. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"finite_testing","n":22}。請依 [mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(finite_testing, 22, 21, checked_moduli_2_through_22, 否, finite_cutoff_does_not_prove_the_universal_equivalence)

Wave6第十二層核對=(finite_testing, 22, 21, checked_moduli_2_through_22, 否, finite_cutoff_does_not_prove_the_universal_equivalence);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-boundary · candidate

97. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"fermat_confusion","n":79}。請依 [mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(fermat_confusion, 79, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 41], [6, 9], [7, 63], [8, 30], [9, 33], [10, 14], [11, 75], [12, 31], [13, 8], [14, 33], [15, 21], [16, 20], [17, 24], [18, 37], [19, 71], [20, 77], [21, 37], [22, 24], [23, 78], [24, 55], [25, 32], [26, 42], [27, 28], [28, 73], [29, 63], [30, 73], [31, 51], [32, 52], [33, 57], [34, 42], [35, 48], [36, 69], [37, 25], [38, 2], [39, 78], [40, 39], [41, 19], [42, 8], [43, 28], [44, 47], [45, 61], [46, 41], [47, 31], [48, 66], [49, 74], [50, 66], [51, 48], [52, 47], [53, 42], [54, 56], [55, 78], [56, 23], [57, 47], [58, 40], [59, 69], [60, 32], [61, 56], [62, 75], [63, 64], [64, 67], [65, 10], [66, 28], [67, 59], [68, 62], [69, 12], [70, 50], [71, 74], [72, 35], [73, 27], [74, 23], [75, 66], [76, 39], [77, 1], [78, 78]], 78, 78, 是, Wilson:(n-1)!==-1_mod_79, Fermat:a^(79-1)==1_mod_79_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed)

Wave6第十二層核對=(fermat_confusion, 79, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 41], [6, 9], [7, 63], [8, 30], [9, 33], [10, 14], [11, 75], [12, 31], [13, 8], [14, 33], [15, 21], [16, 20], [17, 24], [18, 37], [19, 71], [20, 77], [21, 37], [22, 24], [23, 78], [24, 55], [25, 32], [26, 42], [27, 28], [28, 73], [29, 63], [30, 73], [31, 51], [32, 52], [33, 57], [34, 42], [35, 48], [36, 69], [37, 25], [38, 2], [39, 78], [40, 39], [41, 19], [42, 8], [43, 28], [44, 47], [45, 61], [46, 41], [47, 31], [48, 66], [49, 74], [50, 66], [51, 48], [52, 47], [53, 42], [54, 56], [55, 78], [56, 23], [57, 47], [58, 40], [59, 69], [60, 32], [61, 56], [62, 75], [63, 64], [64, 67], [65, 10], [66, 28], [67, 59], [68, 62], [69, 12], [70, 50], [71, 74], [72, 35], [73, 27], [74, 23], [75, 66], [76, 39], [77, 1], [78, 78]], 78, 78, 是, Wilson:(n-1)!==-1_mod_79, Fermat:a^(79-1)==1_mod_79_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-boundary · candidate

98. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"invalid_n","n":-22}。請依 [mode, n, Wilson_domain_qualified, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(invalid_n, -22, 否, Wilson_domain_requires_n>1)

Wave6第十二層核對=(invalid_n, -22, 否, Wilson_domain_requires_n>1);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-boundary · candidate

99. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"finite_testing","n":25}。請依 [mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(finite_testing, 25, 24, checked_moduli_2_through_25, 否, finite_cutoff_does_not_prove_the_universal_equivalence)

Wave6第十二層核對=(finite_testing, 25, 24, checked_moduli_2_through_25, 否, finite_cutoff_does_not_prove_the_universal_equivalence);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-boundary · candidate

100. Wilson定理

n>1、factorial congruence與prime-residue inverse pairing責任。操作=wilson-boundary;資料={"mode":"fermat_confusion","n":97}。請依 [mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary] 欄序回答。

四種mode為4/6/10/6欄;p=2的1與−1是同一residue。Fermat字串只作定理比較,這一分支沒有重算a的冪。

查看答案與解釋

(fermat_confusion, 97, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 23], [6, 41], [7, 93], [8, 65], [9, 3], [10, 30], [11, 39], [12, 80], [13, 70], [14, 10], [15, 53], [16, 72], [17, 60], [18, 13], [19, 53], [20, 90], [21, 47], [22, 64], [23, 17], [24, 20], [25, 15], [26, 2], [27, 54], [28, 57], [29, 4], [30, 23], [31, 34], [32, 21], [33, 14], [34, 88], [35, 73], [36, 9], [37, 42], [38, 44], [39, 67], [40, 61], [41, 76], [42, 88], [43, 1], [44, 44], [45, 40], [46, 94], [47, 53], [48, 22], [49, 11], [50, 65], [51, 17], [52, 11], [53, 1], [54, 54], [55, 60], [56, 62], [57, 42], [58, 11], [59, 67], [60, 43], [61, 4], [62, 54], [63, 7], [64, 60], [65, 20], [66, 59], [67, 73], [68, 17], [69, 9], [70, 48], [71, 13], [72, 63], [73, 40], [74, 50], [75, 64], [76, 14], [77, 11], [78, 82], [79, 76], [80, 66], [81, 11], [82, 29], [83, 79], [84, 40], [85, 5], [86, 42], [87, 65], [88, 94], [89, 24], [90, 26], [91, 38], [92, 4], [93, 81], [94, 48], [95, 1], [96, 96]], 96, 96, 是, Wilson:(n-1)!==-1_mod_97, Fermat:a^(97-1)==1_mod_97_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed)

Wave6第十二層核對=(fermat_confusion, 97, [[1, 1], [2, 2], [3, 6], [4, 24], [5, 23], [6, 41], [7, 93], [8, 65], [9, 3], [10, 30], [11, 39], [12, 80], [13, 70], [14, 10], [15, 53], [16, 72], [17, 60], [18, 13], [19, 53], [20, 90], [21, 47], [22, 64], [23, 17], [24, 20], [25, 15], [26, 2], [27, 54], [28, 57], [29, 4], [30, 23], [31, 34], [32, 21], [33, 14], [34, 88], [35, 73], [36, 9], [37, 42], [38, 44], [39, 67], [40, 61], [41, 76], [42, 88], [43, 1], [44, 44], [45, 40], [46, 94], [47, 53], [48, 22], [49, 11], [50, 65], [51, 17], [52, 11], [53, 1], [54, 54], [55, 60], [56, 62], [57, 42], [58, 11], [59, 67], [60, 43], [61, 4], [62, 54], [63, 7], [64, 60], [65, 20], [66, 59], [67, 73], [68, 17], [69, 9], [70, 48], [71, 13], [72, 63], [73, 40], [74, 50], [75, 64], [76, 14], [77, 11], [78, 82], [79, 76], [80, 66], [81, 11], [82, 29], [83, 79], [84, 40], [85, 5], [86, 42], [87, 65], [88, 94], [89, 24], [90, 26], [91, 38], [92, 4], [93, 81], [94, 48], [95, 1], [96, 96]], 96, 96, 是, Wilson:(n-1)!==-1_mod_97, Fermat:a^(97-1)==1_mod_97_when_n_does_not_divide_a, 否, distinct_theorems_no_Fermat_power_claim_recomputed);依完整明示資料獨立重算;p=2只有一個self-inverse residue 1=-1;composite instances與finite testing不取代converse proof,Wilson不等於Fermat。

本題逐步詳解(AI 二次覆核)

Wilson定理 · qf-mod49-wilson-boundary · candidate

原始題目 JSONL · 驗收報告