# Factorial Congruence, Complete Inverse Pairing and the Modulus-Two Exception

This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.

This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.

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

This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.

Field order uses a source-bound presentation supplement; canonical JSONL and mathematical answers are unchanged.

[Field-order and source-binding data](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=2>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(2, Yes, [[1, 1]], 1, 1, Yes, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (2, Yes, [[1, 1]], 1, 1, Yes, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0001/)

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





## 2. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=3>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(3, Yes, [[1, 1], [2, 2]], 2, 2, Yes, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (3, Yes, [[1, 1], [2, 2]], 2, 2, Yes, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0002/)

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





## 3. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=4>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(4, No, [[1, 1], [2, 2], [3, 2]], 2, 3, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (4, No, [[1, 1], [2, 2], [3, 2]], 2, 3, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0003/)

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





## 4. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=5>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(5, Yes, [[1, 1], [2, 2], [3, 1], [4, 4]], 4, 4, Yes, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (5, Yes, [[1, 1], [2, 2], [3, 1], [4, 4]], 4, 4, Yes, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0004/)

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





## 5. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=6>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(6, No, [[1, 1], [2, 2], [3, 0], [4, 0], [5, 0]], 0, 5, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (6, No, [[1, 1], [2, 2], [3, 0], [4, 0], [5, 0]], 0, 5, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0005/)

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





## 6. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=7>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(7, Yes, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 1], [6, 6]], 6, 6, Yes, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (7, Yes, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 1], [6, 6]], 6, 6, Yes, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0006/)

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





## 7. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=8>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(8, No, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0]], 0, 7, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (8, No, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0]], 0, 7, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0007/)

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





## 8. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=9>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(9, No, [[1, 1], [2, 2], [3, 6], [4, 6], [5, 3], [6, 0], [7, 0], [8, 0]], 0, 8, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (9, No, [[1, 1], [2, 2], [3, 6], [4, 6], [5, 3], [6, 0], [7, 0], [8, 0]], 0, 8, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0008/)

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





## 9. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=10>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(10, No, [[1, 1], [2, 2], [3, 6], [4, 4], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0]], 0, 9, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (10, No, [[1, 1], [2, 2], [3, 6], [4, 4], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0]], 0, 9, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0009/)

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





## 10. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=11>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(11, Yes, [[1, 1], [2, 2], [3, 6], [4, 2], [5, 10], [6, 5], [7, 2], [8, 5], [9, 1], [10, 10]], 10, 10, Yes, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (11, Yes, [[1, 1], [2, 2], [3, 6], [4, 2], [5, 10], [6, 5], [7, 2], [8, 5], [9, 1], [10, 10]], 10, 10, Yes, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0010/)

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





## 11. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=12>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(12, No, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0]], 0, 11, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (12, No, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0]], 0, 11, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0011/)

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





## 12. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=13>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(13, Yes, [[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, Yes, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (13, Yes, [[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, Yes, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0012/)

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





## 13. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=14>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(14, No, [[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, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (14, No, [[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, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0013/)

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





## 14. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=15>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(15, No, [[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, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (15, No, [[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, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0014/)

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





## 15. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=16>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(16, No, [[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, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (16, No, [[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, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0015/)

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





## 16. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=17>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(17, Yes, [[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, Yes, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (17, Yes, [[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, Yes, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0016/)

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





## 17. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=18>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(18, No, [[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, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (18, No, [[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, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0017/)

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





## 18. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=19>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(19, Yes, [[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, Yes, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (19, Yes, [[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, Yes, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0018/)

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





## 19. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=20>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(20, No, [[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, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (20, No, [[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, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0019/)

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





## 20. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=21>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(21, No, [[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, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (21, No, [[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, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0020/)

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





## 21. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=22>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(22, No, [[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, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (22, No, [[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, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0021/)

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





## 22. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=23>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(23, Yes, [[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, Yes, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (23, Yes, [[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, Yes, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0022/)

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





## 23. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=24>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(24, No, [[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, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (24, No, [[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, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0023/)

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





## 24. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=25>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(25, No, [[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, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (25, No, [[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, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0024/)

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





## 25. Wilson Congruence and Primality Agree at the Specified Modulus

Use exact integer n=26>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).

instance_equivalence can be Yes at a composite n because both sides are false; it does not assert primality.

View answer and explanation(26, No, [[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, No, Yes, Wilson_equivalence_instance)

Compute each running factorial residue through (n−1)! and separately classify primality. Compare the final residue with n−1. The equivalence flag can be Yes for a composite n when both primality and congruence are No. A single instance does not prove the universal equivalence. The exact result, retaining the requested order, is (26, No, [[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, No, Yes, Wilson_equivalence_instance).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0025/)

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





## 26. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=2. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(2, [[1, 1]], [1], [], single_self_inverse_residue_1_equals_minus_1, 1, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (2, [[1, 1]], [1], [], single_self_inverse_residue_1_equals_minus_1, 1, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0026/)

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





## 27. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=3. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(3, [[1, 1], [2, 2]], [1, 2], [], self_inverse_residues_1_and_p_minus_1, 2, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (3, [[1, 1], [2, 2]], [1, 2], [], self_inverse_residues_1_and_p_minus_1, 2, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0027/)

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





## 28. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=5. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(5, [[1, 1], [2, 3], [3, 2], [4, 4]], [1, 4], [[2, 3, 1]], self_inverse_residues_1_and_p_minus_1, 4, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (5, [[1, 1], [2, 3], [3, 2], [4, 4]], [1, 4], [[2, 3, 1]], self_inverse_residues_1_and_p_minus_1, 4, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0028/)

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





## 29. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=7. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0029/)

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





## 30. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=11. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0030/)

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





## 31. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=13. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0031/)

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





## 32. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=17. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0032/)

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





## 33. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=19. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0033/)

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





## 34. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=23. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0034/)

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





## 35. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=29. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0035/)

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





## 36. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=31. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0036/)

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





## 37. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=37. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0037/)

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





## 38. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=41. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0038/)

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





## 39. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=43. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0039/)

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





## 40. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=47. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0040/)

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





## 41. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=53. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0041/)

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





## 42. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=59. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0042/)

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





## 43. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=61. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0043/)

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





## 44. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=67. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0044/)

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





## 45. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=71. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0045/)

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





## 46. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=73. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0046/)

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





## 47. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=79. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0047/)

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





## 48. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=83. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0048/)

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





## 49. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=89. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0049/)

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





## 50. All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case

Use prime modulus p=97. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).

View answer and explanation(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, Yes)

List every nonzero residue and its inverse, identify self-inverse residues and pair each remaining unordered inverse pair once. For p=2,1 and−1 are the same residue, not two distinct ones. Retain the complete factorial congruence for this modulus. The exact result, retaining the requested order, is (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, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0050/)

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





## 51. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=4. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(4, No, [[1, 1], [2, 2], [3, 2]], 2, 3, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (4, No, [[1, 1], [2, 2], [3, 2]], 2, 3, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0051/)

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





## 52. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=6. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(6, No, [[1, 1], [2, 2], [3, 0], [4, 0], [5, 0]], 0, 5, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (6, No, [[1, 1], [2, 2], [3, 0], [4, 0], [5, 0]], 0, 5, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0052/)

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





## 53. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=8. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(8, No, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0]], 0, 7, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (8, No, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0]], 0, 7, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0053/)

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





## 54. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=9. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(9, No, [[1, 1], [2, 2], [3, 6], [4, 6], [5, 3], [6, 0], [7, 0], [8, 0]], 0, 8, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (9, No, [[1, 1], [2, 2], [3, 6], [4, 6], [5, 3], [6, 0], [7, 0], [8, 0]], 0, 8, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0054/)

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





## 55. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=10. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(10, No, [[1, 1], [2, 2], [3, 6], [4, 4], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0]], 0, 9, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (10, No, [[1, 1], [2, 2], [3, 6], [4, 4], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0]], 0, 9, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0055/)

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





## 56. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=12. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(12, No, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0]], 0, 11, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (12, No, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0], [8, 0], [9, 0], [10, 0], [11, 0]], 0, 11, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0056/)

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





## 57. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=14. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(14, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (14, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0057/)

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





## 58. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=15. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(15, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (15, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0058/)

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





## 59. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=16. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(16, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (16, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0059/)

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





## 60. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=18. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(18, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (18, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0060/)

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





## 61. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=20. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(20, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (20, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0061/)

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





## 62. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=21. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(21, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (21, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0062/)

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





## 63. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=22. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(22, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (22, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0063/)

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





## 64. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=24. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(24, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (24, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0064/)

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





## 65. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=25. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(25, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (25, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0065/)

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





## 66. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=26. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(26, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (26, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0066/)

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





## 67. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=27. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(27, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (27, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0067/)

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





## 68. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=28. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(28, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (28, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0068/)

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





## 69. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=30. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(30, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (30, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0069/)

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





## 70. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=32. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(32, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (32, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0070/)

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





## 71. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=33. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(33, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (33, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0071/)

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





## 72. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=34. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(34, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (34, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0072/)

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





## 73. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=35. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(35, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (35, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0073/)

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





## 74. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=36. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(36, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (36, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0074/)

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





## 75. A Composite Factorial Computation Is an Instance, Not the Converse Proof

Use composite integer n=38. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).

View answer and explanation(38, No, [[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, Yes, finite_composite_instance_not_converse_proof)

Compute all running factorial residues and show the final residue differs from n−1. This exact finite instance does not replace the converse proof for all composite integers, and it is not Fermat little theorem. The exact result, retaining the requested order, is (38, No, [[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, Yes, finite_composite_instance_not_converse_proof).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0075/)

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





## 76. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=p2_self_inverse and exact integer n=2. Give (mode, p, self_inverse_residues, modular_identity, distinct_residue_scope, exception_qualified).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(p2_self_inverse, 2, [1], 1_equals_minus_1_mod_2, one_not_two_distinct_self_inverse_residues, Yes)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (p2_self_inverse, 2, [1], 1_equals_minus_1_mod_2, one_not_two_distinct_self_inverse_residues, Yes).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0076/)

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





## 77. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=invalid_n and exact integer n=-1. Give (mode, n, Wilson_domain_qualified, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(invalid_n, -1, No, Wilson_domain_requires_n>1)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (invalid_n, -1, No, Wilson_domain_requires_n>1).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0077/)

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





## 78. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=finite_testing and exact integer n=4. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(finite_testing, 4, 3, checked_moduli_2_through_4, No, finite_cutoff_does_not_prove_the_universal_equivalence)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (finite_testing, 4, 3, checked_moduli_2_through_4, No, finite_cutoff_does_not_prove_the_universal_equivalence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0078/)

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





## 79. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=fermat_confusion and exact integer n=7. Give (mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(fermat_confusion, 7, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 1], [6, 6]], 6, 6, Yes, Wilson:(n-1)!==-1_mod_7, Fermat:a^(7-1)==1_mod_7_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (fermat_confusion, 7, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 1], [6, 6]], 6, 6, Yes, Wilson:(n-1)!==-1_mod_7, Fermat:a^(7-1)==1_mod_7_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0079/)

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





## 80. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=invalid_n and exact integer n=-4. Give (mode, n, Wilson_domain_qualified, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(invalid_n, -4, No, Wilson_domain_requires_n>1)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (invalid_n, -4, No, Wilson_domain_requires_n>1).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0080/)

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





## 81. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=finite_testing and exact integer n=7. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(finite_testing, 7, 6, checked_moduli_2_through_7, No, finite_cutoff_does_not_prove_the_universal_equivalence)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (finite_testing, 7, 6, checked_moduli_2_through_7, No, finite_cutoff_does_not_prove_the_universal_equivalence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0081/)

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





## 82. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=fermat_confusion and exact integer n=17. Give (mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(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, Yes, Wilson:(n-1)!==-1_mod_17, Fermat:a^(17-1)==1_mod_17_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (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, Yes, Wilson:(n-1)!==-1_mod_17, Fermat:a^(17-1)==1_mod_17_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0082/)

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





## 83. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=invalid_n and exact integer n=-7. Give (mode, n, Wilson_domain_qualified, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(invalid_n, -7, No, Wilson_domain_requires_n>1)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (invalid_n, -7, No, Wilson_domain_requires_n>1).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0083/)

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





## 84. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=finite_testing and exact integer n=10. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(finite_testing, 10, 9, checked_moduli_2_through_10, No, finite_cutoff_does_not_prove_the_universal_equivalence)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (finite_testing, 10, 9, checked_moduli_2_through_10, No, finite_cutoff_does_not_prove_the_universal_equivalence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0084/)

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





## 85. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=fermat_confusion and exact integer n=29. Give (mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(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, Yes, Wilson:(n-1)!==-1_mod_29, Fermat:a^(29-1)==1_mod_29_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (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, Yes, Wilson:(n-1)!==-1_mod_29, Fermat:a^(29-1)==1_mod_29_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0085/)

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





## 86. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=invalid_n and exact integer n=-10. Give (mode, n, Wilson_domain_qualified, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(invalid_n, -10, No, Wilson_domain_requires_n>1)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (invalid_n, -10, No, Wilson_domain_requires_n>1).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0086/)

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





## 87. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=finite_testing and exact integer n=13. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(finite_testing, 13, 12, checked_moduli_2_through_13, No, finite_cutoff_does_not_prove_the_universal_equivalence)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (finite_testing, 13, 12, checked_moduli_2_through_13, No, finite_cutoff_does_not_prove_the_universal_equivalence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0087/)

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





## 88. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=fermat_confusion and exact integer n=41. Give (mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(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, Yes, Wilson:(n-1)!==-1_mod_41, Fermat:a^(41-1)==1_mod_41_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (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, Yes, Wilson:(n-1)!==-1_mod_41, Fermat:a^(41-1)==1_mod_41_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0088/)

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





## 89. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=invalid_n and exact integer n=-13. Give (mode, n, Wilson_domain_qualified, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(invalid_n, -13, No, Wilson_domain_requires_n>1)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (invalid_n, -13, No, Wilson_domain_requires_n>1).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0089/)

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





## 90. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=finite_testing and exact integer n=16. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(finite_testing, 16, 15, checked_moduli_2_through_16, No, finite_cutoff_does_not_prove_the_universal_equivalence)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (finite_testing, 16, 15, checked_moduli_2_through_16, No, finite_cutoff_does_not_prove_the_universal_equivalence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0090/)

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





## 91. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=fermat_confusion and exact integer n=53. Give (mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(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, Yes, Wilson:(n-1)!==-1_mod_53, Fermat:a^(53-1)==1_mod_53_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (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, Yes, Wilson:(n-1)!==-1_mod_53, Fermat:a^(53-1)==1_mod_53_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0091/)

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





## 92. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=invalid_n and exact integer n=-16. Give (mode, n, Wilson_domain_qualified, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(invalid_n, -16, No, Wilson_domain_requires_n>1)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (invalid_n, -16, No, Wilson_domain_requires_n>1).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0092/)

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





## 93. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=finite_testing and exact integer n=19. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(finite_testing, 19, 18, checked_moduli_2_through_19, No, finite_cutoff_does_not_prove_the_universal_equivalence)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (finite_testing, 19, 18, checked_moduli_2_through_19, No, finite_cutoff_does_not_prove_the_universal_equivalence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0093/)

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





## 94. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=fermat_confusion and exact integer n=67. Give (mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(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, Yes, Wilson:(n-1)!==-1_mod_67, Fermat:a^(67-1)==1_mod_67_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (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, Yes, Wilson:(n-1)!==-1_mod_67, Fermat:a^(67-1)==1_mod_67_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0094/)

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





## 95. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=invalid_n and exact integer n=-19. Give (mode, n, Wilson_domain_qualified, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(invalid_n, -19, No, Wilson_domain_requires_n>1)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (invalid_n, -19, No, Wilson_domain_requires_n>1).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0095/)

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





## 96. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=finite_testing and exact integer n=22. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(finite_testing, 22, 21, checked_moduli_2_through_22, No, finite_cutoff_does_not_prove_the_universal_equivalence)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (finite_testing, 22, 21, checked_moduli_2_through_22, No, finite_cutoff_does_not_prove_the_universal_equivalence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0096/)

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





## 97. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=fermat_confusion and exact integer n=79. Give (mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(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, Yes, Wilson:(n-1)!==-1_mod_79, Fermat:a^(79-1)==1_mod_79_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (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, Yes, Wilson:(n-1)!==-1_mod_79, Fermat:a^(79-1)==1_mod_79_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0097/)

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





## 98. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=invalid_n and exact integer n=-22. Give (mode, n, Wilson_domain_qualified, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(invalid_n, -22, No, Wilson_domain_requires_n>1)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (invalid_n, -22, No, Wilson_domain_requires_n>1).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0098/)

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





## 99. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=finite_testing and exact integer n=25. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(finite_testing, 25, 24, checked_moduli_2_through_25, No, finite_cutoff_does_not_prove_the_universal_equivalence)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (finite_testing, 25, 24, checked_moduli_2_through_25, No, finite_cutoff_does_not_prove_the_universal_equivalence).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0099/)

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





## 100. Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate

Use mode=fermat_confusion and exact integer n=97. Give (mode, n, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, Wilson_statement, Fermat_statement, theorem_identification_claim, boundary).

The four modes have4/6/10/6 fields. At p=2,1 and−1 are the same residue. The Fermat string is a theorem comparison; that branch does not recompute powers of a.

View answer and explanation(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, Yes, Wilson:(n-1)!==-1_mod_97, Fermat:a^(97-1)==1_mod_97_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed)

Reject n≤1 from the theorem domain. Testing moduli2 through a cutoff is not a universal proof. The Fermat comparison supplies named statements but does not recompute powers of a. At p=2 there is one self-inverse nonzero residue. Preserve the mode-specific layouts and distinguish a failed inference from a false congruence. The exact result, retaining the requested order, is (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, Yes, Wilson:(n-1)!==-1_mod_97, Fermat:a^(97-1)==1_mod_97_when_n_does_not_divide_a, No, distinct_theorems_no_Fermat_power_claim_recomputed).



[Reviewed step-by-step derivation](https://ocme.evemisslab.com/en/questions/q-20261015-advance-20261118-r04-wilson-theorem-0100/)

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



[Canonical question JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261118-r04-wilson-theorem-foundations/questions.jsonl) · [English JSONL](https://ocme.evemisslab.com/data/i18n/questions/batch-20261015-advance-20261118-r04-wilson-theorem-foundations.en.jsonl) · [Validation report](https://ocme.evemisslab.com/data/questions/batches/batch-20261015-advance-20261118-r04-wilson-theorem-foundations/validation-report.json)