{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0001","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a77eb15eeda6655eb214643f43bd3b26c4491fd452d168e391e9968f06f018c","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184001,"semantic_template":"wilson-domain-factorial|{n:2}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=2>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(2, Yes, [[1, 1]], 1, 1, Yes, Yes, Wilson_equivalence_instance)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0002","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d17f04eaff5a81c2ae29fc296476588904b59b35639c857cd74ffaebf4404fb","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184002,"semantic_template":"wilson-domain-factorial|{n:3}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=3>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(3, Yes, [[1, 1], [2, 2]], 2, 2, Yes, Yes, Wilson_equivalence_instance)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0003","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"210c94cda6de4ab24f4b858a60b8a6c9724b4bd959f7feb22afa0fcdea055a46","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184003,"semantic_template":"wilson-domain-factorial|{n:4}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=4>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(4, No, [[1, 1], [2, 2], [3, 2]], 2, 3, No, Yes, Wilson_equivalence_instance)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0004","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"199f4057791a1cb77b1ce708cf9a31bc576b362dafbbc776c37005a78f16cba9","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184004,"semantic_template":"wilson-domain-factorial|{n:5}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=5>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(5, Yes, [[1, 1], [2, 2], [3, 1], [4, 4]], 4, 4, Yes, Yes, Wilson_equivalence_instance)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0005","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cd87aaebaeb910e9f17c26e11803742fa870ebe9ae03c01002810df54476394","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184005,"semantic_template":"wilson-domain-factorial|{n:6}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=6>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(6, No, [[1, 1], [2, 2], [3, 0], [4, 0], [5, 0]], 0, 5, No, Yes, Wilson_equivalence_instance)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0006","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a53e45840b9a07600dbf3df01623673904c8a0418dfbf2d7ae093b06d31fa71","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184006,"semantic_template":"wilson-domain-factorial|{n:7}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=7>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(7, Yes, [[1, 1], [2, 2], [3, 6], [4, 3], [5, 1], [6, 6]], 6, 6, Yes, Yes, Wilson_equivalence_instance)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0007","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae0750d5b9f1e3272f6d09f90c67a56144ab1fda1a368b93b23f6515520f884f","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184007,"semantic_template":"wilson-domain-factorial|{n:8}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=8>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(8, No, [[1, 1], [2, 2], [3, 6], [4, 0], [5, 0], [6, 0], [7, 0]], 0, 7, No, Yes, Wilson_equivalence_instance)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0008","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0756e571bd36f0c8e6392131f58f95faebe568e3c4acb59af14bc5b8770987a4","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184008,"semantic_template":"wilson-domain-factorial|{n:9}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=9>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0009","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4e169e7eb548e9a2f3e8989aad53412e8ffc939788a2c541165c5a0fb76e64ae","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184009,"semantic_template":"wilson-domain-factorial|{n:10}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=10>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0010","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3afbc234bf88d3b2d5a2f496f3941b9ec0b33ce1af45dd59682b18d3cad433a5","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184010,"semantic_template":"wilson-domain-factorial|{n:11}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=11>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0011","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17933c16724cdabf33237c68b8aef3c5758d0636f38349951d1c7e41eef589f9","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184011,"semantic_template":"wilson-domain-factorial|{n:12}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=12>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0012","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80b54085d9b4a1127621d67b6e5566eab176276390effe1bbeb6cb79651d955d","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184012,"semantic_template":"wilson-domain-factorial|{n:13}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=13>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0013","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"42c7be71dceabf050015debf4e746c9cf18ad2a391d14b0c7bd7cc01bb62e629","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184013,"semantic_template":"wilson-domain-factorial|{n:14}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=14>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0014","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e167fc449e02803b2912d7cb5a2603120433abcfd4e91769bf49aa47f29820d6","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184014,"semantic_template":"wilson-domain-factorial|{n:15}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=15>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0015","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f595b68087e9c5908c90d5c5134052b0a472359c001cb28615726a051de02091","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184015,"semantic_template":"wilson-domain-factorial|{n:16}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=16>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0016","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7a2fe979415586996d04a454da0ef46ede45fc0198ce0d7e35e1127faaf65844","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184016,"semantic_template":"wilson-domain-factorial|{n:17}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=17>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0017","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8601a830e482afde3e54a62e3158c9bdb721c49a34f35a525fd06b2eab6100c8","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184017,"semantic_template":"wilson-domain-factorial|{n:18}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=18>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0018","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49fa61366e75e171893924b7b6cf45aac6dbcc42d78671570822b4afa01ec2bb","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184018,"semantic_template":"wilson-domain-factorial|{n:19}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=19>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0019","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9bdb9e941493f4cb2755ba50396624f8d4fbe2ae27a748bde47df1cb0910fa17","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184019,"semantic_template":"wilson-domain-factorial|{n:20}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=20>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0020","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"39e765daa4a86573a8cca9bbcbf7fc1b2546c951a37d9994cd9f23d572fd51ae","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184020,"semantic_template":"wilson-domain-factorial|{n:21}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=21>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0021","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b94cfe35efbe772e46647625c027a69c6df8dda909e3b1720127ada696c7db1","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184021,"semantic_template":"wilson-domain-factorial|{n:22}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=22>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0022","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0031943a204a5f1466490715881e5a45a7526b5ab814ba2632f000ac80f5a646","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184022,"semantic_template":"wilson-domain-factorial|{n:23}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=23>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0023","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eb6a167c8344d5670f68f7a08d5d8d30fb4e8cf55cec91d8c77f938b7d1d4355","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184023,"semantic_template":"wilson-domain-factorial|{n:24}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=24>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0024","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"37e17f3aab4388da62f268182216622948330fd4ae14d83bd1d4fe0b16b0f26c","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184024,"semantic_template":"wilson-domain-factorial|{n:25}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=25>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0025","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dfe52e81ffa2416c46d617bef10567cc3bf97b5afc73791d405f4748113f231a","family_id":"qf-mod49-wilson-domain-factorial","seed":202611184025,"semantic_template":"wilson-domain-factorial|{n:26}"},"fields":{"title":"Wilson Congruence and Primality Agree at the Specified Modulus","stem":"Use exact integer n=26>1. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence, instance_equivalence, instance_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Congruence and Primality Agree at the Specified Modulus"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0026","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7061a0dccc760991e807b903294efd4126de3ec4c57376e7c323728f9d0a3083","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184026,"semantic_template":"wilson-prime-inverse-pairing|{n:2}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=2. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(2, [[1, 1]], [1], [], single_self_inverse_residue_1_equals_minus_1, 1, Yes)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0027","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d12d290327fbe2944f13a8bbb644c3beffcfcd8854001c47c68f97a66b98b2d9","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184027,"semantic_template":"wilson-prime-inverse-pairing|{n:3}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=3. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(3, [[1, 1], [2, 2]], [1, 2], [], self_inverse_residues_1_and_p_minus_1, 2, Yes)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0028","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ad79b0e14b94312f9fae96c93c1928d386965b83dd9c1fc10158457120c5bc79","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184028,"semantic_template":"wilson-prime-inverse-pairing|{n:5}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=5. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(5, [[1, 1], [2, 3], [3, 2], [4, 4]], [1, 4], [[2, 3, 1]], self_inverse_residues_1_and_p_minus_1, 4, Yes)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0029","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ef78b541b9caee28ca0f3c89acbb53e3abf98f7a50bbe00e62cb8c9ef266eea","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184029,"semantic_template":"wilson-prime-inverse-pairing|{n:7}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=7. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0030","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"89ad139f7e9fe25e1052123c77d4ea5fc5641afb22fa6f50466986257ff7a89f","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184030,"semantic_template":"wilson-prime-inverse-pairing|{n:11}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=11. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0031","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"52ee477a1a5464290dbe956741b61b27428616d20da17c9cebe866e69860409c","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184031,"semantic_template":"wilson-prime-inverse-pairing|{n:13}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=13. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0032","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a78ff92645e28b00153ad30d1944c4a234a2dc568c9e86f22f884fc7d247e9a0","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184032,"semantic_template":"wilson-prime-inverse-pairing|{n:17}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=17. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0033","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"624b323d9ea12f8c6de303cc21aa0222b6fa0b4e1fc22f48a5f6b27450f23667","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184033,"semantic_template":"wilson-prime-inverse-pairing|{n:19}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=19. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0034","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe81d427df92e7051986c0117c3fbb12a6726f8a5c8161fbcfe49d74906710fa","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184034,"semantic_template":"wilson-prime-inverse-pairing|{n:23}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=23. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0035","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3da309e8daff7e5a84177ba132f7a7f8efcbf94d7a93b127fecb365ec93f2419","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184035,"semantic_template":"wilson-prime-inverse-pairing|{n:29}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=29. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0036","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e1b314590ae01c803bc9f2d2291bf3c29d0bba5b0be9fa5f23b8234f590d76d","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184036,"semantic_template":"wilson-prime-inverse-pairing|{n:31}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=31. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0037","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a4050cbd599d9557abdddc655e1c2b4513c969bbc7b52ae8661c08330f77618","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184037,"semantic_template":"wilson-prime-inverse-pairing|{n:37}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=37. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0038","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"acf49f42709a1f3f24f6a05a8aa64c782b9730f72ed6bd04843d53b5f0008ee3","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184038,"semantic_template":"wilson-prime-inverse-pairing|{n:41}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=41. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0039","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2d4ade26a7fa4ba9eef57dcdc01fc4fd08ae1cf8fcadc5abd484ffd7f19b70f4","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184039,"semantic_template":"wilson-prime-inverse-pairing|{n:43}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=43. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0040","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b5b3073e6c7c1d219348880adb2e474e78a8ddc3a0c23d0039e071898b309bf","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184040,"semantic_template":"wilson-prime-inverse-pairing|{n:47}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=47. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0041","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9dfee08dc8ecd5d87af35d342fdfab535636c1b5fe61261dc47c25511ab366a6","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184041,"semantic_template":"wilson-prime-inverse-pairing|{n:53}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=53. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0042","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ca379e3adddc107e562b40e2300af40703da2ca3271170f27afcc60c8373a35","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184042,"semantic_template":"wilson-prime-inverse-pairing|{n:59}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=59. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0043","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd37945115ea2b8850f3711fe9df9e4ef05538e70d2ead96e27014e13f81dfde","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184043,"semantic_template":"wilson-prime-inverse-pairing|{n:61}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=61. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0044","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d04541a70790756d135eb291442716bb2d65db0639be3fb64314722bd1a7816a","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184044,"semantic_template":"wilson-prime-inverse-pairing|{n:67}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=67. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0045","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6c46df0f139a763f95e6e918574e61ecb258e6b890066591e9ff75ebcbb92ead","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184045,"semantic_template":"wilson-prime-inverse-pairing|{n:71}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=71. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0046","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db11518e4273b1e3e6135d8e2f00a2c20770532bcead159fdb93ecec6ed90ff1","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184046,"semantic_template":"wilson-prime-inverse-pairing|{n:73}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=73. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0047","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65d3b50d827eb9d5efb9b2a1b2faed4c60703947bc859ba8bd40326e796553bd","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184047,"semantic_template":"wilson-prime-inverse-pairing|{n:79}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=79. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0048","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"247e6b7ca2e1d63aab528bcdf562b7c531f6474216e3b56642c9489ad556f09f","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184048,"semantic_template":"wilson-prime-inverse-pairing|{n:83}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=83. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0049","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef7f015a118f2bcad620859f044de68df6b8271fa6a5c0a9ccd977768ad766e0","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184049,"semantic_template":"wilson-prime-inverse-pairing|{n:89}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=89. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0050","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21f8aa4e77f342131da77155a4304123637fd59f4175367d6b10129b4d669a8c","family_id":"qf-mod49-wilson-prime-inverse-pairing","seed":202611184050,"semantic_template":"wilson-prime-inverse-pairing|{n:97}"},"fields":{"title":"All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case","stem":"Use prime modulus p=97. Give (p, all_inverse_pairs, self_inverse_residues, distinct_cancelling_pairs, self_inverse_scope, factorial_residue, Wilson_congruence).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["All Nonzero Prime Residues Pair with Inverses, with a Special p=2 Case"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0051","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15a4e19f65222aa48923b02c6df67f4f13d0604e3f4b57fb52f20482254cc764","family_id":"qf-mod49-wilson-composite-instance","seed":202611184051,"semantic_template":"wilson-composite-instance|{n:4}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=4. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(4, No, [[1, 1], [2, 2], [3, 2]], 2, 3, Yes, finite_composite_instance_not_converse_proof)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0052","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b41041fcf3489a1c984a5c056ebeae9e33defbbe57291adff13542bb4a31ccf8","family_id":"qf-mod49-wilson-composite-instance","seed":202611184052,"semantic_template":"wilson-composite-instance|{n:6}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=6. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(6, No, [[1, 1], [2, 2], [3, 0], [4, 0], [5, 0]], 0, 5, Yes, finite_composite_instance_not_converse_proof)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0053","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87ba84699d299c52955895cc11cc4cc29a1913b73b1ed40b0ae447c8c68e7899","family_id":"qf-mod49-wilson-composite-instance","seed":202611184053,"semantic_template":"wilson-composite-instance|{n:8}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=8. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0054","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93e9933f502dcbf70671c11f32d92cb2316d21f153521aa9d9cd24183e8966c9","family_id":"qf-mod49-wilson-composite-instance","seed":202611184054,"semantic_template":"wilson-composite-instance|{n:9}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=9. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0055","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e6f7321b02224e73ed614b2f7ca8362ca1ffe40233f95d85efee4f89ddb1ee18","family_id":"qf-mod49-wilson-composite-instance","seed":202611184055,"semantic_template":"wilson-composite-instance|{n:10}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=10. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0056","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de5c8e3172cc9af3ad7612cb5f5dc3353a301f0c37f992820fd10e2b04daad09","family_id":"qf-mod49-wilson-composite-instance","seed":202611184056,"semantic_template":"wilson-composite-instance|{n:12}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=12. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0057","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f7242c980974baea26d2ee79e6642d9bf0371bd27610deada1dff37f3caa70f0","family_id":"qf-mod49-wilson-composite-instance","seed":202611184057,"semantic_template":"wilson-composite-instance|{n:14}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=14. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0058","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"66680e1f50a35b91f3246acdf1604ddaac0ca958e43f84dda27f00cb01e386f1","family_id":"qf-mod49-wilson-composite-instance","seed":202611184058,"semantic_template":"wilson-composite-instance|{n:15}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=15. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0059","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3ecc65b3327982dc89860f68d53a670f1acf75df8bb85c138644463f9f506bfd","family_id":"qf-mod49-wilson-composite-instance","seed":202611184059,"semantic_template":"wilson-composite-instance|{n:16}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=16. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0060","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c50aacf19f644c817f0666095cc195032615d2872e01315c52530165111aad31","family_id":"qf-mod49-wilson-composite-instance","seed":202611184060,"semantic_template":"wilson-composite-instance|{n:18}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=18. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0061","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"676e4ba4055840c91345e27a5cb786d52362094e8c87f1c010aac41ae51e3fca","family_id":"qf-mod49-wilson-composite-instance","seed":202611184061,"semantic_template":"wilson-composite-instance|{n:20}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=20. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0062","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49cfcbef171f6ac640ad4cc8ec0d612eebd0a9305921846c0cc65b8367811530","family_id":"qf-mod49-wilson-composite-instance","seed":202611184062,"semantic_template":"wilson-composite-instance|{n:21}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=21. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0063","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e1c28a41f61ecd180ffd8619ee9f8a06698ba5c025f70bf194df2352f7ac2d16","family_id":"qf-mod49-wilson-composite-instance","seed":202611184063,"semantic_template":"wilson-composite-instance|{n:22}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=22. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0064","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cfa2f12bb6ec15a7641e4e9c5557209591d910ebfaa57d78522c4d8cca74ddb0","family_id":"qf-mod49-wilson-composite-instance","seed":202611184064,"semantic_template":"wilson-composite-instance|{n:24}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=24. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0065","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4c3af05a71706345e0e5def4b4a151170d39c9745e0bf0e7f9293b8bd74c3fd8","family_id":"qf-mod49-wilson-composite-instance","seed":202611184065,"semantic_template":"wilson-composite-instance|{n:25}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=25. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0066","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2fb198e58d85a530a63382cfbb71f6915f2d092881eb288e9992b83693b3bda3","family_id":"qf-mod49-wilson-composite-instance","seed":202611184066,"semantic_template":"wilson-composite-instance|{n:26}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=26. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0067","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"559b71d1d675604e1dd9cd6be7f11a544b2673d288aaa79b816a9a6c57efe020","family_id":"qf-mod49-wilson-composite-instance","seed":202611184067,"semantic_template":"wilson-composite-instance|{n:27}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=27. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0068","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1dcfe19bde64deb4158d9328abe55cd3e2be66c21620c7482b49de3ce1d293d","family_id":"qf-mod49-wilson-composite-instance","seed":202611184068,"semantic_template":"wilson-composite-instance|{n:28}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=28. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0069","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11826a59291479efe50a32d684909827779fb0212b4880458ab1ab9db1de2f18","family_id":"qf-mod49-wilson-composite-instance","seed":202611184069,"semantic_template":"wilson-composite-instance|{n:30}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=30. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0070","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"feff3eca909b3d9721057106c3aa3f5420d12e2e9c3b735513d443be0de02c0f","family_id":"qf-mod49-wilson-composite-instance","seed":202611184070,"semantic_template":"wilson-composite-instance|{n:32}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=32. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0071","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3930e0443003e57326fd3b8a8e0fccbf9a0a6f850bd0f31aa5adb7752fcadd2c","family_id":"qf-mod49-wilson-composite-instance","seed":202611184071,"semantic_template":"wilson-composite-instance|{n:33}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=33. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0072","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f665d9ec85f2afc18444f85f48b9219462de8946fe66cf831e20350ca3cdeded","family_id":"qf-mod49-wilson-composite-instance","seed":202611184072,"semantic_template":"wilson-composite-instance|{n:34}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=34. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0073","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a05c5893d762179f074fbaae998dc73f5313611b5f0b8bc28c5412836fa5c6c","family_id":"qf-mod49-wilson-composite-instance","seed":202611184073,"semantic_template":"wilson-composite-instance|{n:35}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=35. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0074","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83a96f3f7069208c42a829074d32d773e7f6b3e0137cb5d5619017133489a814","family_id":"qf-mod49-wilson-composite-instance","seed":202611184074,"semantic_template":"wilson-composite-instance|{n:36}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=36. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0075","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bead41d032c170735057820518e684d56c0a35c93cbcac7be85b2f71481332ce","family_id":"qf-mod49-wilson-composite-instance","seed":202611184075,"semantic_template":"wilson-composite-instance|{n:38}"},"fields":{"title":"A Composite Factorial Computation Is an Instance, Not the Converse Proof","stem":"Use composite integer n=38. Give (n, prime, running_factorial_residues, factorial_residue, minus_one_residue, Wilson_congruence_fails, finite_scope).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["A Composite Factorial Computation Is an Instance, Not the Converse Proof"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0076","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"104edd4716a16d5d8af3588be25b9e65ac70bd19141faea9e3007e64bd73c330","family_id":"qf-mod49-wilson-boundary","seed":202611184076,"semantic_template":"wilson-boundary|{mode:\"p2_self_inverse\",n:2}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=p2_self_inverse and exact integer n=2. Give (mode, p, self_inverse_residues, modular_identity, distinct_residue_scope, exception_qualified).","answer":{"kind":"expression","value":"(p2_self_inverse, 2, [1], 1_equals_minus_1_mod_2, one_not_two_distinct_self_inverse_residues, Yes)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0077","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"92678dfa2e9d3de734af63066fc7754249f34598cc8ada7672c43cc137a9f6aa","family_id":"qf-mod49-wilson-boundary","seed":202611184077,"semantic_template":"wilson-boundary|{mode:\"invalid_n\",n:-1}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=invalid_n and exact integer n=-1. Give (mode, n, Wilson_domain_qualified, boundary).","answer":{"kind":"expression","value":"(invalid_n, -1, No, Wilson_domain_requires_n>1)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0078","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3784c172e1ca97d4e761728fef01f05530d8dd51428a4701c3a84ca9557a9655","family_id":"qf-mod49-wilson-boundary","seed":202611184078,"semantic_template":"wilson-boundary|{mode:\"finite_testing\",n:4}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=finite_testing and exact integer n=4. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_testing, 4, 3, checked_moduli_2_through_4, No, finite_cutoff_does_not_prove_the_universal_equivalence)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0079","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"731e29e43dd1493684ac042eb668ea77c051dcff0abcf909a462ed96d146de9b","family_id":"qf-mod49-wilson-boundary","seed":202611184079,"semantic_template":"wilson-boundary|{mode:\"fermat_confusion\",n:7}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"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).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0080","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee1e79c918c6eea2ef15b8a3b9145f26d5bc1e843ba6c90b0b4294563b4f6ac7","family_id":"qf-mod49-wilson-boundary","seed":202611184080,"semantic_template":"wilson-boundary|{mode:\"invalid_n\",n:-4}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=invalid_n and exact integer n=-4. Give (mode, n, Wilson_domain_qualified, boundary).","answer":{"kind":"expression","value":"(invalid_n, -4, No, Wilson_domain_requires_n>1)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0081","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b05063093428eb53b676fc54c0fbc62c0315d305ea8923edd8145e11f4d2c1e","family_id":"qf-mod49-wilson-boundary","seed":202611184081,"semantic_template":"wilson-boundary|{mode:\"finite_testing\",n:7}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=finite_testing and exact integer n=7. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_testing, 7, 6, checked_moduli_2_through_7, No, finite_cutoff_does_not_prove_the_universal_equivalence)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0082","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e4b418eaaa838c0f4a9309a948a15e8b164213771fb4d3f1669f7314296c6850","family_id":"qf-mod49-wilson-boundary","seed":202611184082,"semantic_template":"wilson-boundary|{mode:\"fermat_confusion\",n:17}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"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).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0083","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d5bb03e96a805799ed3e9e5e8de9e34ef399f523f451ddf3272e24cf7f86268f","family_id":"qf-mod49-wilson-boundary","seed":202611184083,"semantic_template":"wilson-boundary|{mode:\"invalid_n\",n:-7}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=invalid_n and exact integer n=-7. Give (mode, n, Wilson_domain_qualified, boundary).","answer":{"kind":"expression","value":"(invalid_n, -7, No, Wilson_domain_requires_n>1)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0084","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2beedeb9d56404edd43c1156103cc7e40ac9dbe26ce7483b6cb2a49c281a998a","family_id":"qf-mod49-wilson-boundary","seed":202611184084,"semantic_template":"wilson-boundary|{mode:\"finite_testing\",n:10}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=finite_testing and exact integer n=10. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_testing, 10, 9, checked_moduli_2_through_10, No, finite_cutoff_does_not_prove_the_universal_equivalence)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0085","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"40bf2897de447bf05f7caba565a6a771f1210d700d22d914826ec55049bc2b75","family_id":"qf-mod49-wilson-boundary","seed":202611184085,"semantic_template":"wilson-boundary|{mode:\"fermat_confusion\",n:29}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"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).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0086","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"12e06aaa9a03503b74a0db03f5d48def5a531789e6f6e7a5de06bbab855b9e54","family_id":"qf-mod49-wilson-boundary","seed":202611184086,"semantic_template":"wilson-boundary|{mode:\"invalid_n\",n:-10}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=invalid_n and exact integer n=-10. Give (mode, n, Wilson_domain_qualified, boundary).","answer":{"kind":"expression","value":"(invalid_n, -10, No, Wilson_domain_requires_n>1)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0087","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"721701c1d09cd5997a902d22578360383aaea8c8e108e725fee79c7a2e530f8c","family_id":"qf-mod49-wilson-boundary","seed":202611184087,"semantic_template":"wilson-boundary|{mode:\"finite_testing\",n:13}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=finite_testing and exact integer n=13. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_testing, 13, 12, checked_moduli_2_through_13, No, finite_cutoff_does_not_prove_the_universal_equivalence)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0088","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2941fea468674d19dcf312694b2a8218f03237c14acbbeed85d14ac6a8318165","family_id":"qf-mod49-wilson-boundary","seed":202611184088,"semantic_template":"wilson-boundary|{mode:\"fermat_confusion\",n:41}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"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).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0089","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b6005f23d39fda01e0782e75376dfd2eb8a9d3897e3e4f3c16e300f0713ea89","family_id":"qf-mod49-wilson-boundary","seed":202611184089,"semantic_template":"wilson-boundary|{mode:\"invalid_n\",n:-13}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=invalid_n and exact integer n=-13. Give (mode, n, Wilson_domain_qualified, boundary).","answer":{"kind":"expression","value":"(invalid_n, -13, No, Wilson_domain_requires_n>1)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0090","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3515a37d38d43364afa7d5ca558868f5589ee6404ac4009db33262e6ad870e78","family_id":"qf-mod49-wilson-boundary","seed":202611184090,"semantic_template":"wilson-boundary|{mode:\"finite_testing\",n:16}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=finite_testing and exact integer n=16. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_testing, 16, 15, checked_moduli_2_through_16, No, finite_cutoff_does_not_prove_the_universal_equivalence)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0091","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7f3c728274649d991858070e3bcd2814e163ac096df2f33dbf3f463625272da1","family_id":"qf-mod49-wilson-boundary","seed":202611184091,"semantic_template":"wilson-boundary|{mode:\"fermat_confusion\",n:53}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"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).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0092","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ac71e70eb5cff54d2cbebc194a7430c10628804a3064e7050ee1e801de84a8b","family_id":"qf-mod49-wilson-boundary","seed":202611184092,"semantic_template":"wilson-boundary|{mode:\"invalid_n\",n:-16}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=invalid_n and exact integer n=-16. Give (mode, n, Wilson_domain_qualified, boundary).","answer":{"kind":"expression","value":"(invalid_n, -16, No, Wilson_domain_requires_n>1)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0093","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5add793bd9bd57725ca9ea9a790a8430046b7f2200e6972c26fdc568d868471b","family_id":"qf-mod49-wilson-boundary","seed":202611184093,"semantic_template":"wilson-boundary|{mode:\"finite_testing\",n:19}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=finite_testing and exact integer n=19. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_testing, 19, 18, checked_moduli_2_through_19, No, finite_cutoff_does_not_prove_the_universal_equivalence)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0094","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"88c99669ed29f8922e02d80ec8c9e9cd40fc74d723a1675c75e3db0b148a2279","family_id":"qf-mod49-wilson-boundary","seed":202611184094,"semantic_template":"wilson-boundary|{mode:\"fermat_confusion\",n:67}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"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).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0095","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65be23541b480cad796296589bf68cf5cb4ea84140e8ea95e7c3a0dfb953b6c3","family_id":"qf-mod49-wilson-boundary","seed":202611184095,"semantic_template":"wilson-boundary|{mode:\"invalid_n\",n:-19}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=invalid_n and exact integer n=-19. Give (mode, n, Wilson_domain_qualified, boundary).","answer":{"kind":"expression","value":"(invalid_n, -19, No, Wilson_domain_requires_n>1)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0096","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"04a484305b8333a33e1c88eef79a594f3de79fbd86e489666b52069f531429c7","family_id":"qf-mod49-wilson-boundary","seed":202611184096,"semantic_template":"wilson-boundary|{mode:\"finite_testing\",n:22}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=finite_testing and exact integer n=22. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_testing, 22, 21, checked_moduli_2_through_22, No, finite_cutoff_does_not_prove_the_universal_equivalence)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0097","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9622639eae2bd4b6f4aa1620782b91f35e6c378422ec2121dcc889cc1a09e875","family_id":"qf-mod49-wilson-boundary","seed":202611184097,"semantic_template":"wilson-boundary|{mode:\"fermat_confusion\",n:79}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"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).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0098","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"424ac99855af63af50cf3da276b82fc05031f19d86ef911ebb7ec09c52d2060b","family_id":"qf-mod49-wilson-boundary","seed":202611184098,"semantic_template":"wilson-boundary|{mode:\"invalid_n\",n:-22}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=invalid_n and exact integer n=-22. Give (mode, n, Wilson_domain_qualified, boundary).","answer":{"kind":"expression","value":"(invalid_n, -22, No, Wilson_domain_requires_n>1)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0099","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8374537e183ab9fddbfb656620e4674971746b8fb04d81297cb93a9d166c88c2","family_id":"qf-mod49-wilson-boundary","seed":202611184099,"semantic_template":"wilson-boundary|{mode:\"finite_testing\",n:25}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"Use mode=finite_testing and exact integer n=25. Give (mode, cutoff, tested_modulus_count, tested_range, universal_proof_claim, boundary).","answer":{"kind":"expression","value":"(finite_testing, 25, 24, checked_moduli_2_through_25, No, finite_cutoff_does_not_prove_the_universal_equivalence)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261118-r04-wilson-theorem-0100","batch_id":"batch-20261015-advance-20261118-r04-wilson-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a9934517116a1f6aed3acd34d94fd0275b63ae29f200df0da0735cb2da3125b","family_id":"qf-mod49-wilson-boundary","seed":202611184100,"semantic_template":"wilson-boundary|{mode:\"fermat_confusion\",n:97}"},"fields":{"title":"Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate","stem":"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).","answer":{"kind":"expression","value":"(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)","explanation":"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)."},"skills":["Wilson Domain, Finite Testing, Fermat Comparison and the p=2 Exception Stay Separate"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}