{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-0001","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"67405ae92d17057b3ffcd864fbd36cb029afbbc8b430fc619815473050c452e3","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134001,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":1,\"prime\":3}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=3, original a=1. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(3, 1, 1, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (3, 1, 1, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0002","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a25d5e25cdf26aa86500da951187fc579591bb4f95033ac8daa729d507362496","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134002,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":2,\"prime\":5}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=5, original a=2. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(5, 2, 2, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (5, 2, 2, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0003","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"882843cc05ca843ca303e8ee93488c66daae5eb98083df0d302c25ca7ad0046d","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134003,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":3,\"prime\":7}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=7, original a=3. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(7, 3, 3, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (7, 3, 3, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0004","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff8281e9b3edc37bc87bd71401cd204246cd35ed916bc69237e94ac0f889931a","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134004,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":4,\"prime\":11}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=11, original a=4. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(11, 4, 4, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (11, 4, 4, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0005","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25a2ef857114f1eacc9096798e8a1864255d91cb97be869037b83e2de2f7dc57","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134005,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":5,\"prime\":13}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=13, original a=5. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(13, 5, 5, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (13, 5, 5, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0006","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b11a9f58316cf92700ea636fb6d6166932fc1310cb17de42ba112fa772c1d460","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134006,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":6,\"prime\":17}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=17, original a=6. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(17, 6, 6, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (17, 6, 6, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0007","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef1d967cc138666f29f5bbd097e824bbafc20c56fc4af19817a0b76a99446bbc","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134007,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":7,\"prime\":19}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=19, original a=7. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(19, 7, 7, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (19, 7, 7, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0008","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"019e783b3a072266e6f89baf2cb43d535dea574c9a321fd3be4933471e9c65d6","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134008,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":8,\"prime\":23}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=23, original a=8. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(23, 8, 8, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (23, 8, 8, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0009","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"762bbfc5b68f5c55b1b35b6e6c3c6c2cdec6591c2d111bcfae9f7a419f630eee","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134009,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":9,\"prime\":29}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=29, original a=9. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(29, 9, 9, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (29, 9, 9, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0010","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"08e76c5f19cd4ab19fc049225f731b55e9142a5d98ca015546ae061b091ead61","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134010,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":10,\"prime\":31}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=31, original a=10. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(31, 10, 10, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (31, 10, 10, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0011","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2206c3f8dddd4a787e7ec5c73cc694d26871cd68ec59c345b9cc17df4cdac54c","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134011,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":11,\"prime\":37}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=37, original a=11. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(37, 11, 11, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (37, 11, 11, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0012","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6735178bfc1d295f8df58733d847eeb17b967360e73bcbe6b4605e1817bb38f4","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134012,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":12,\"prime\":41}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=41, original a=12. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(41, 12, 12, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (41, 12, 12, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0013","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e836bfb69dc95d3406369202a30fac2d91be2ff938e0926fbe7b7888047923a","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134013,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":13,\"prime\":43}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=43, original a=13. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(43, 13, 13, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (43, 13, 13, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0014","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fc649d814ce8f1bdc5a3d26adcb1affa5eb98c7046f4081403f3e187f3baefe8","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134014,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":14,\"prime\":47}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=47, original a=14. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(47, 14, 14, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (47, 14, 14, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0015","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"38e115ec5cb76c8b5e621419dcb8b74563e5999b1082c368f89077c9cb16dfa3","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134015,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":15,\"prime\":53}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=53, original a=15. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(53, 15, 15, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (53, 15, 15, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0016","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cca6eec5b49e4881fd8d6df04455629023627111c011c6e31c637dd58cd91583","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134016,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":16,\"prime\":59}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=59, original a=16. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(59, 16, 16, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (59, 16, 16, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0017","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c9a5df6657465e6220fe490816743eea55e8aad9f3cb62e3c7599125509d4ef","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134017,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":17,\"prime\":61}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=61, original a=17. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(61, 17, 17, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (61, 17, 17, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0018","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cab35043a1ebef12803dbb17dae8a53860d0222d49513f2f9dbd3eacd0964b85","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134018,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":18,\"prime\":67}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=67, original a=18. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(67, 18, 18, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (67, 18, 18, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0019","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea0a2d06464ecca7bf581b9d72ed189fbe8183cb02715e16ab32edd1b2216d29","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134019,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":19,\"prime\":71}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=71, original a=19. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(71, 19, 19, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (71, 19, 19, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0020","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b3a12d3f56daacc110cf15096f47acf3fea906da1883e3e8d41f8d9a7bb24ab5","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134020,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":20,\"prime\":73}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=73, original a=20. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(73, 20, 20, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (73, 20, 20, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0021","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d001b070fe5dbce58078d645a96009d72ae206694351399cd707e4a3a115b459","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134021,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":21,\"prime\":79}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=79, original a=21. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(79, 21, 21, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (79, 21, 21, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0022","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8b59158a22f67615ca4e56755a7afbc57d5039b1635054550115f020c328333","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134022,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":22,\"prime\":83}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=83, original a=22. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(83, 22, 22, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (83, 22, 22, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0023","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"40ba5129c1ef1c95fffa32b90c9bbdac92b89656ebb90a753593c052086c29bb","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134023,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":23,\"prime\":89}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=89, original a=23. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(89, 23, 23, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (89, 23, 23, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0024","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"88b4c426f783cead29d6dd7903697131436b13b9cc392239d6de139c477f2312","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134024,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":24,\"prime\":97}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=97, original a=24. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(97, 24, 24, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (97, 24, 24, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0025","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9070fbfc80501571100e98c67e901ca17ed784375c894fcc278f0fb3d074a2f7","family_id":"qf-mod44-eulercrit-domain-qualification","seed":202611134025,"semantic_template":"qf-mod44-eulercrit-domain-qualification/{\"a\":25,\"prime\":101}"},"fields":{"title":"Qualify Odd Primality and the Normalized Nonzero Class First","stem":"Use prime=101, original a=25. Give (prime, a, normalized_class, odd_prime, nonzero_class, certificate).","answer":{"kind":"expression","value":"(101, 25, 25, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)","explanation":"Require a bounded odd prime and a mod p≠0 before this criterion. The Legendre symbol is already defined by square classification, not by assuming this theorem. The exact result, retaining the requested order, is (101, 25, 25, Yes, Yes, odd_prime_and_nonzero_modular_class_qualified)."},"skills":["Qualify Odd Primality and the Normalized Nonzero Class First"],"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-20261113-r04-euler-criterion-quadratic-residue-0026","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5486415e2acda828bb4ade55cad7cbe359a5a40e69715901d1e3c81a5b2adba6","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134026,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":1,\"prime\":3}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=3, original a=1. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(3, 1, [1, 2], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (3, 1, [1, 2], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0027","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4c37515e0675caa33f0ba8ce3aba36ae062cf445cbcd8312f3477427300f0da9","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134027,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":2,\"prime\":5}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=5, original a=2. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(5, 2, [], -1, 4, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (5, 2, [], -1, 4, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0028","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3bcba59f6fea25bfe6ef16ef7cf061b56a72309723505cea1a1f7c00b4cfc0b0","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134028,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":3,\"prime\":7}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=7, original a=3. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(7, 3, [], -1, 6, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (7, 3, [], -1, 6, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0029","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b74da33ad5d0dd2fdfc80d9d8b472d86061be75612b0a78ce4c0ff7f6e587cb4","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134029,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":4,\"prime\":11}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=11, original a=4. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(11, 4, [2, 9], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (11, 4, [2, 9], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0030","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7ad5b81fea5383217af8afc01c0ec13fb5639fe9a25897624c38ec16afac9f95","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134030,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":5,\"prime\":13}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=13, original a=5. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(13, 5, [], -1, 12, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (13, 5, [], -1, 12, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0031","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"519cfe01aabf5db5105a552e5e3f551fe253b22e76593e8740faabd63146b7d3","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134031,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":6,\"prime\":17}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=17, original a=6. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(17, 6, [], -1, 16, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (17, 6, [], -1, 16, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0032","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e30b91753af1e8fcbe660791937e6fe362c2d110acb8a750b4565fda5bbea1f6","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134032,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":7,\"prime\":19}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=19, original a=7. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(19, 7, [8, 11], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (19, 7, [8, 11], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0033","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a0397e6d6652673b92c63574adcf510501c50e7faa6fb741fa1adfc58aa0903","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134033,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":8,\"prime\":23}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=23, original a=8. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(23, 8, [10, 13], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (23, 8, [10, 13], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0034","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5ffc45ac0d1fd157074f2da9966c4f62d0d7c7c64788c401e981b567c648e957","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134034,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":9,\"prime\":29}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=29, original a=9. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(29, 9, [3, 26], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (29, 9, [3, 26], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0035","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9a00068cfe8c4e8cacb7a6512b2821de358a302c4abf90adf2059ed3f6dc7f4","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134035,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":10,\"prime\":31}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=31, original a=10. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(31, 10, [14, 17], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (31, 10, [14, 17], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0036","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9870745e4c0bc6e571f578844fee6e0f3367c1e93b2c0567c5526c705f940ac0","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134036,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":11,\"prime\":37}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=37, original a=11. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(37, 11, [14, 23], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (37, 11, [14, 23], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0037","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"972a55f3b6456ed9f2fc4cd50b76daa0c6452357ea9a9fb84690061c91247ab3","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134037,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":12,\"prime\":41}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=41, original a=12. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(41, 12, [], -1, 40, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (41, 12, [], -1, 40, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0038","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15cb5a65aa16d4e602605e96c43ca5a2259488f930ca2bb8b7ab762971e3b7e3","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134038,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":13,\"prime\":43}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=43, original a=13. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(43, 13, [20, 23], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (43, 13, [20, 23], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0039","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c88697312ec9128c289414f0e707dfcecdbf38afe49b77bbc49afcc5cba670ee","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134039,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":14,\"prime\":47}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=47, original a=14. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(47, 14, [22, 25], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (47, 14, [22, 25], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0040","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cdb9ccd4ca797964fd1e1091ce801fa4e7a57175d8866d34d75d7cef36446fe","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134040,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":15,\"prime\":53}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=53, original a=15. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(53, 15, [11, 42], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (53, 15, [11, 42], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0041","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e0f8b636d549094912bf6f22bc35fb77018748ee9b70d053cdadcaece44d3a3e","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134041,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":16,\"prime\":59}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=59, original a=16. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(59, 16, [4, 55], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (59, 16, [4, 55], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0042","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3191fdcc4df77cf69caefe477b403b591a76c4b3afb0d4734f570cd6ffc4c19f","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134042,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":17,\"prime\":61}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=61, original a=17. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(61, 17, [], -1, 60, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (61, 17, [], -1, 60, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0043","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"34a6eb053a61d579db9f3087e98739e73e33cba36ee0f593cb2af61cdd7f6495","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134043,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":18,\"prime\":67}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=67, original a=18. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(67, 18, [], -1, 66, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (67, 18, [], -1, 66, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0044","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90681c1d8779e94d7ec7c0c48e5f2c1808b550708e8ed49a452f67aace1e3e9c","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134044,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":19,\"prime\":71}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=71, original a=19. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(71, 19, [27, 44], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (71, 19, [27, 44], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0045","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"42cc63cd54a407d0712d7a5aef8fb0400d6d86417bc98e7cd835b7de0b450880","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134045,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":20,\"prime\":73}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=73, original a=20. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(73, 20, [], -1, 72, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (73, 20, [], -1, 72, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0046","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36442d4d08c81edfaacd0af08a0dd5166b124e2bcc921f91e21416670a47871a","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134046,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":21,\"prime\":79}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=79, original a=21. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(79, 21, [10, 69], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (79, 21, [10, 69], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0047","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3df45c2472e67657795a2968242bf35c257d82afb02eeb86329ac89b0be9da1","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134047,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":22,\"prime\":83}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=83, original a=22. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(83, 22, [], -1, 82, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (83, 22, [], -1, 82, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0048","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cbe38ce0ad0e5426af2af5ab2c775672f56321d6fbd16e8d3399c4d698eee863","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134048,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":23,\"prime\":89}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=89, original a=23. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(89, 23, [], -1, 88, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (89, 23, [], -1, 88, -1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0049","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"27c70dd21179a6ef06b0eef1eccc9afcd33f6a35bd5cd67c2d786892eee7c2b2","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134049,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":24,\"prime\":97}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=97, original a=24. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(97, 24, [11, 86], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (97, 24, [11, 86], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0050","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19f2fed97a6a99f97c0d377c6ec538eab6b784b0c4d3c938224bb9ddedf150a5","family_id":"qf-mod44-eulercrit-independent-comparison","seed":202611134050,"semantic_template":"qf-mod44-eulercrit-independent-comparison/{\"a\":25,\"prime\":101}"},"fields":{"title":"Compare a Square-Table Legendre Value with a Separately Computed Half Power","stem":"Use odd prime p=101, original a=25. Give (prime, normalized_class, nonzero_roots, Legendre_symbol, modular_power, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(101, 25, [5, 96], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)","explanation":"Normalize a and obtain all nonzero roots from the complete square table, giving Legendre±1 independently. Compute a^((p−1)/2) modulo p by repeated squaring, then interpret raw residue p−1 as signed−1. Compare signed values; one finite instance is not the general theorem proof. The exact result, retaining the requested order, is (101, 25, [5, 96], 1, 1, 1, Yes, square_table_and_repeated_squaring_are_independent_computations)."},"skills":["Compare a Square-Table Legendre Value with a Separately Computed Half Power"],"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-20261113-r04-euler-criterion-quadratic-residue-0051","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"26c2dabfcaea2ca196c84f66aa019daa3d0ba714be73efc97b69623e1868fa32","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134051,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":1,\"prime\":3}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=3, original a=1, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(3, 1, quadratic_residue, [1, 2], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (3, 1, quadratic_residue, [1, 2], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0052","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"500f2e399b8dbbd6c5dc97b2326e4a2d615e2b5b33be1411cfbae831ef07c4e3","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134052,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":2,\"prime\":5}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=5, original a=2, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(5, 2, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (5, 2, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0053","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d333f53ef4c5b55e945ff808e34a6726a6b6ebbb59c9481d876b43d8c3674d5f","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134053,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":2,\"prime\":7}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=7, original a=2, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(7, 2, quadratic_residue, [3, 4], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (7, 2, quadratic_residue, [3, 4], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0054","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3ac9df813c269259085f4d68c0790311f2aed0e213e62e8eb38041b8e436cd55","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134054,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":2,\"prime\":11}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=11, original a=2, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(11, 2, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (11, 2, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0055","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6c4759c1213ca82e728ca3acf61e241934d4b09dfa6efd4cb573cfbf5eb59585","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134055,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":12,\"prime\":13}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=13, original a=12, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(13, 12, quadratic_residue, [5, 8], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (13, 12, quadratic_residue, [5, 8], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0056","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5c8714a92f325232aad481405abd3c06d4b4cbbecd4b5dee56c0770bc29fd16","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134056,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":3,\"prime\":17}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=17, original a=3, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(17, 3, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (17, 3, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0057","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e684c15d7c516a6b0e8e0b8f074760e36b5f4ef61a4b37ddcc2f9de0de440b84","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134057,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":11,\"prime\":19}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=19, original a=11, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(19, 11, quadratic_residue, [7, 12], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (19, 11, quadratic_residue, [7, 12], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0058","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bc625c12fc1fe91c05785d5f18098e157be86d8ee1c04d7a9c77a19e63feb1b8","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134058,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":5,\"prime\":23}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=23, original a=5, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(23, 5, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (23, 5, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0059","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3af25379402d1d990380c5f7a1869086cccceece06bf1921a34f0e6014efe8f0","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134059,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":23,\"prime\":29}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=29, original a=23, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(29, 23, quadratic_residue, [9, 20], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (29, 23, quadratic_residue, [9, 20], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0060","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd42409ad85504bfe91a8a962aa40aeb84935604aecb6dd0a755262ac154e373","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134060,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":3,\"prime\":31}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=31, original a=3, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(31, 3, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (31, 3, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0061","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7687e5005d3ec89b2b5a4ea0e00d584eff388d54bf10b23e3a14219485d1a70c","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134061,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":10,\"prime\":37}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=37, original a=10, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(37, 10, quadratic_residue, [11, 26], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (37, 10, quadratic_residue, [11, 26], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0062","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e69d2a90ca93fe095f5aa3182c9213626c9ba3b3b25d160c977f562aa8ea207","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134062,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":3,\"prime\":41}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=41, original a=3, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(41, 3, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (41, 3, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0063","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"086d42e11349ab22f4d9aa827a9e4181d7af87f22bd7c5f571542b0591484675","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134063,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":40,\"prime\":43}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=43, original a=40, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(43, 40, quadratic_residue, [13, 30], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (43, 40, quadratic_residue, [13, 30], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0064","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9463dac9351e98e3b6616918b39f72dc56f19853bb822f69fc6b722ed886bcc6","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134064,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":5,\"prime\":47}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=47, original a=5, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(47, 5, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (47, 5, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0065","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49db04bbc826e0a9f6dfefbfe306afb7a8c3b0a1473dc8dbc296bcf6ecdb4435","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134065,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":13,\"prime\":53}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=53, original a=13, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(53, 13, quadratic_residue, [15, 38], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (53, 13, quadratic_residue, [15, 38], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0066","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ab00abb71a1219e7d788b552be95da97b5e52d80ca6a3c844030e8003538628","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134066,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":2,\"prime\":59}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=59, original a=2, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(59, 2, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (59, 2, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0067","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5165d4050fcbc1e7de319c5dcedc7a77ca47d10f9999bb85a61dff31fef01044","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134067,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":45,\"prime\":61}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=61, original a=45, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(61, 45, quadratic_residue, [17, 44], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (61, 45, quadratic_residue, [17, 44], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0068","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"721ea15bdf7489b4f82165306d3324aa897bf3e23077460e082f075ed425ab50","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134068,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":2,\"prime\":67}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=67, original a=2, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(67, 2, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (67, 2, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0069","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc3592919516091a08d733e0d52fb4fe42439dee2f4db8c57df7658b07fafe0f","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134069,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":6,\"prime\":71}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=71, original a=6, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(71, 6, quadratic_residue, [19, 52], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (71, 6, quadratic_residue, [19, 52], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0070","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2919dfbe64fef94c833f328d1f4d2471be010f00d2b9badc9b944859665f35a0","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134070,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":5,\"prime\":73}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=73, original a=5, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(73, 5, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (73, 5, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0071","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a890366688987bd6343986485a905232006fb554e4af3e2193b02d2a15f5419f","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134071,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":46,\"prime\":79}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=79, original a=46, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(79, 46, quadratic_residue, [21, 58], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (79, 46, quadratic_residue, [21, 58], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0072","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7c90fa0ea6695ba041a45d5716f39134070fcecab1a792ef900b7b58c3e2976","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134072,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":2,\"prime\":83}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=83, original a=2, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(83, 2, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (83, 2, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0073","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15829f747ecdb677397678ea65eff01c49b0c97bd9479728e76134231334c3a0","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134073,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":84,\"prime\":89}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=89, original a=84, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(89, 84, quadratic_residue, [23, 66], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (89, 84, quadratic_residue, [23, 66], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0074","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef141809baa9253ca926a5b6acb06b1f3f8441dcaf3601c1105146267399d03a","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134074,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":5,\"prime\":97}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=97, original a=5, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(97, 5, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (97, 5, quadratic_nonresidue, [], -1, -1, Yes, nonresidue_maps_to_minus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0075","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"305580e1f9bc3b77d4e8da3148a131ce02285443b49c4f374df34373d0b99e84","family_id":"qf-mod44-eulercrit-residue-nonresidue-cases","seed":202611134075,"semantic_template":"qf-mod44-eulercrit-residue-nonresidue-cases/{\"a\":19,\"prime\":101}"},"fields":{"title":"Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion","stem":"Use odd prime p=101, original a=19, normalized nonzero class. Give (prime, normalized_class, classification, nonzero_roots, Legendre_symbol, signed_power, equal, certificate).","answer":{"kind":"expression","value":"(101, 19, quadratic_residue, [25, 76], 1, 1, Yes, residue_maps_to_plus_one)","explanation":"Keep square-table classification, roots, Legendre value and signed half-power result separate. Modulo-p representative p−1 means−1, not integer equality of−1 and p−1. No Jacobi or composite-modulus converse is asserted. The exact result, retaining the requested order, is (101, 19, quadratic_residue, [25, 76], 1, 1, Yes, residue_maps_to_plus_one)."},"skills":["Residues Give Plus One and Nonresidues Minus One under the Qualified Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0076","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f0d5d872613b284987e7b86b8098d82737ed263f983850187166f77e1319198","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134076,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":1,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=even_prime, modulus=2, original a=1. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 1, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (even_prime, 2, 1, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0077","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5aedee15430c7590423c32723e5f8c930d893f81c884c97c7ab837237a6d5a53","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134077,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":2,\"mode\":\"composite_modulus\",\"modulus\":9}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=composite_modulus, modulus=9, original a=2. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 9, 2, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (composite_modulus, 9, 2, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0078","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"88e1b5d49865452ee2d95afb082b1c0715aabece515166f069308934864d269b","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134078,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":21,\"mode\":\"zero_class\",\"modulus\":7}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=zero_class, modulus=7, original a=21. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 7, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (zero_class, 7, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0079","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c86e1b696367c94c70f6711b624d3636a771347506ef9201fd291afc1c950223","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134079,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":4,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=even_prime, modulus=2, original a=4. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (even_prime, 2, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0080","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7117c7c8372f71424d41d456fe9c56407e8b7913a5ae427144d8fe97a4606030","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134080,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":5,\"mode\":\"composite_modulus\",\"modulus\":15}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=composite_modulus, modulus=15, original a=5. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 15, 5, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (composite_modulus, 15, 5, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0081","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d6ad10e3f8171ffffcf0ec9eadbb8f47b77b40d8c4a9543d6c9c3d6c491141e","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134081,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":102,\"mode\":\"zero_class\",\"modulus\":17}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=zero_class, modulus=17, original a=102. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 17, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (zero_class, 17, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0082","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2baa5a0ffc34d739ef56e54102f668f89ab929f66e28a264f3389514b5d679a","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134082,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":7,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=even_prime, modulus=2, original a=7. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 1, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (even_prime, 2, 1, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0083","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3fda9fba7912d466ee1d985e1abe1f86ccb7c322d49bc9b520d99a034b30a5a6","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134083,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":8,\"mode\":\"composite_modulus\",\"modulus\":21}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=composite_modulus, modulus=21, original a=8. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 21, 8, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (composite_modulus, 21, 8, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0084","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea110cc37b5091fa55737aa9bfbaa1ff081db998e4594c03c0ea09fd7226da1c","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134084,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":261,\"mode\":\"zero_class\",\"modulus\":29}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=zero_class, modulus=29, original a=261. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 29, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (zero_class, 29, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0085","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6570578bb868a0cc97304f6eef92393ec97cd6348062dc7afe4425729d5a66b5","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134085,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":10,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=even_prime, modulus=2, original a=10. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (even_prime, 2, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0086","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"17b5e7bebb9b598bd1473050d7ba238e6401183b07138b877fab3ef0c8ea8c88","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134086,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":11,\"mode\":\"composite_modulus\",\"modulus\":27}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=composite_modulus, modulus=27, original a=11. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 27, 11, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (composite_modulus, 27, 11, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0087","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"249d2369f2d295dcd7217664617af21d32602828f8ca5859215ac070fb6260d3","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134087,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":492,\"mode\":\"zero_class\",\"modulus\":41}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=zero_class, modulus=41, original a=492. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 41, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (zero_class, 41, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0088","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ce8cbad7d6e13a6a6de3d5daf5b714300e16f2cc046beee716a2973c78b5417c","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134088,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":13,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=even_prime, modulus=2, original a=13. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 1, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (even_prime, 2, 1, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0089","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b56756bedd8f07f2a951cb784bf16b5e35211b6b2dac07afd99d5e2715c1ca82","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134089,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":14,\"mode\":\"composite_modulus\",\"modulus\":33}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=composite_modulus, modulus=33, original a=14. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 33, 14, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (composite_modulus, 33, 14, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0090","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cfeea8b22c5cabd21d7473e9313b888c247c3a762608bea789ad03b7aea055ec","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134090,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":795,\"mode\":\"zero_class\",\"modulus\":53}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=zero_class, modulus=53, original a=795. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 53, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (zero_class, 53, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0091","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a349949ce3785800cbc942700a6aa4367e4a452254b53956f32a24fe33439ec5","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134091,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":16,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=even_prime, modulus=2, original a=16. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (even_prime, 2, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0092","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"116fe577f4e8d2b543bf02703e288bdcbd0d82ee099f25cb5032cac8b6d45eab","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134092,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":17,\"mode\":\"composite_modulus\",\"modulus\":39}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=composite_modulus, modulus=39, original a=17. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 39, 17, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (composite_modulus, 39, 17, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0093","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8dd4e64e554294c1549d7f98ed3b542041509cd5c326e88b5c9cac4a7aa8a211","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134093,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":1206,\"mode\":\"zero_class\",\"modulus\":67}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=zero_class, modulus=67, original a=1206. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 67, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (zero_class, 67, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0094","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf50a3619c01c59df16ec87bac9f74c3e3887e9bbdab7748e507c30440e984a6","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134094,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":19,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=even_prime, modulus=2, original a=19. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 1, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (even_prime, 2, 1, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0095","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87bf73fc89d1424838b9e9223d3bb75c2c4cf204dc59f04239b5dfd308fa7f6b","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134095,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":20,\"mode\":\"composite_modulus\",\"modulus\":45}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=composite_modulus, modulus=45, original a=20. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 45, 20, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (composite_modulus, 45, 20, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0096","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e0113f7302fe8a6302179a6c00dd33afee2592bb8b382e7a9463a2ec8ccb2694","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134096,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":1659,\"mode\":\"zero_class\",\"modulus\":79}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=zero_class, modulus=79, original a=1659. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 79, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (zero_class, 79, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0097","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"30c4efe1c2f3c1049ff785b0b32078992758002de495505bf71766b9493baece","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134097,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":22,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=even_prime, modulus=2, original a=22. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (even_prime, 2, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0098","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6c74ff2eca425c0a9b7ddab9ff496848bbf5f1c89a7abc1d78860d1ff68f1bb6","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134098,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":23,\"mode\":\"composite_modulus\",\"modulus\":51}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=composite_modulus, modulus=51, original a=23. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 51, 23, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (composite_modulus, 51, 23, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0099","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf4e00a13d7c92b08716b1001fd9b23bfea6c34e2e19ee2c3f9d0e6600706b23","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134099,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":2328,\"mode\":\"zero_class\",\"modulus\":97}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=zero_class, modulus=97, original a=2328. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 97, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (zero_class, 97, 0, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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-20261113-r04-euler-criterion-quadratic-residue-0100","batch_id":"batch-20261015-advance-20261113-r04-euler-criterion-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bac759dff90c8235bc55dc5eba139e6a717ab85ff826c7f7ed169621765be786","family_id":"qf-mod44-eulercrit-domain-boundary","seed":202611134100,"semantic_template":"qf-mod44-eulercrit-domain-boundary/{\"a\":25,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion","stem":"Use mode=even_prime, modulus=2, original a=25. Give (mode, modulus, normalized_class, declared_domain_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 1, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)","explanation":"Validate the named failed hypothesis and normalize the class. No means this odd-prime/nonzero domain is not qualified; it does not assert that every out-of-domain numerical equality must fail or redefine the Legendre symbol. The exact result, retaining the requested order, is (even_prime, 2, 1, No, Euler_criterion_requires_an_odd_prime_and_a_nonzero_class_and_does_not_define_the_Legendre_symbol)."},"skills":["Zero, Even Prime and Composite Inputs Are Outside This Declared Criterion"],"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."}}