{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261106-r05-fermat-little-theorem-0001","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"330d11f0c17f5a009c69143880e8527a9501b668623151219f96d82d6748285a","family_id":"qf-mod37-flt-prime-coprime","seed":202611065001,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":3,\"p\":2}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=3 and modulus=2, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(2, 1, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (2, 1, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0002","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ab962223d85fc37f8575cd3aca23a87955881152e2a796bab11278b06416342f","family_id":"qf-mod37-flt-prime-coprime","seed":202611065002,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":-4,\"p\":3}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=-4 and modulus=3, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(3, 2, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (3, 2, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0003","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa9985472cf25d602e7785b886e8faf05a14cf7af4ea4a092b46d062fa9a1fbc","family_id":"qf-mod37-flt-prime-coprime","seed":202611065003,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":4,\"p\":5}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=4 and modulus=5, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(5, 4, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (5, 4, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0004","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b158439dbead6e30e5607b23ce9378ebeb040c392df9a553162e43fc1398c4a","family_id":"qf-mod37-flt-prime-coprime","seed":202611065004,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":-5,\"p\":7}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=-5 and modulus=7, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(7, 2, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (7, 2, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0005","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db5574997ae7f7e1db254eeacebb450b3ac9de7bb202d19e7a20f5c1d1163777","family_id":"qf-mod37-flt-prime-coprime","seed":202611065005,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":6,\"p\":11}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=6 and modulus=11, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(11, 6, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (11, 6, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0006","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13c333ceccf4d2acef56b3e09d19b6e5de2bc448e6fcb7c2bff8e64c27be8976","family_id":"qf-mod37-flt-prime-coprime","seed":202611065006,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":-7,\"p\":13}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=-7 and modulus=13, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(13, 6, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (13, 6, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0007","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc9849b1c5531c5b2b1dd4915cdfb6ca7e5bfb07b9d85c7b01494c5ac4152ec2","family_id":"qf-mod37-flt-prime-coprime","seed":202611065007,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":8,\"p\":17}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=8 and modulus=17, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(17, 8, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (17, 8, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0008","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2063be9a996be3c71fb9a33f4e167e1ccb926c4c845ec7ed3dfe1b7cd6665ab0","family_id":"qf-mod37-flt-prime-coprime","seed":202611065008,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":-9,\"p\":19}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=-9 and modulus=19, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(19, 10, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (19, 10, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0009","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ffcb8c044339e70315dfaeb1317c6dd83ac549a8240cc70779f24fb21be4502f","family_id":"qf-mod37-flt-prime-coprime","seed":202611065009,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":10,\"p\":23}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=10 and modulus=23, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(23, 10, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (23, 10, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0010","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc1b33e985c21e27e7878a94dac814cc7a763a9ce84cbc6a840bea96339d6174","family_id":"qf-mod37-flt-prime-coprime","seed":202611065010,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":-11,\"p\":29}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=-11 and modulus=29, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(29, 18, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (29, 18, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0011","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15f891884a1f5c49d837911c1300925fa4dd55bc194b9804ea0eb5809dc695d2","family_id":"qf-mod37-flt-prime-coprime","seed":202611065011,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":12,\"p\":31}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=12 and modulus=31, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(31, 12, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (31, 12, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0012","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be2af5158bd79dfe72009a5c0eb30390570b3838edd0dea4a77526744f1a0239","family_id":"qf-mod37-flt-prime-coprime","seed":202611065012,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":-13,\"p\":37}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=-13 and modulus=37, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(37, 24, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (37, 24, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0013","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91e456a0880400d1c26760362f6e257216275f163b5a0784349fe5040ec7a424","family_id":"qf-mod37-flt-prime-coprime","seed":202611065013,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":14,\"p\":41}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=14 and modulus=41, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(41, 14, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (41, 14, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0014","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0aa2c1548518de10bb9054423d4370ebd381c088232fdfcd94b29e3411fea967","family_id":"qf-mod37-flt-prime-coprime","seed":202611065014,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":-15,\"p\":43}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=-15 and modulus=43, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(43, 28, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (43, 28, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0015","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"051fae0dd1f0a35ddfaa3e71562635f74bfde7f666915569d36d958127043b25","family_id":"qf-mod37-flt-prime-coprime","seed":202611065015,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":16,\"p\":47}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=16 and modulus=47, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(47, 16, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (47, 16, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0016","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"055bcce97128ba9851d4dc416b2e76afa98dd8d991d4e0917f02856166295d9d","family_id":"qf-mod37-flt-prime-coprime","seed":202611065016,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":-17,\"p\":53}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=-17 and modulus=53, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(53, 36, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (53, 36, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0017","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b954fc1199d8466076691d5f47157a82ec12b7444fc55de3b798595cc4f13b03","family_id":"qf-mod37-flt-prime-coprime","seed":202611065017,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":18,\"p\":59}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=18 and modulus=59, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(59, 18, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (59, 18, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0018","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b204dd7f45bba107c5fa11bce92ff9903260d5b91f41556d8bcd7961ffad6e3","family_id":"qf-mod37-flt-prime-coprime","seed":202611065018,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":-19,\"p\":61}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=-19 and modulus=61, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(61, 42, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (61, 42, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0019","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a40b795cd816900d3ab65bfc2d66c87bac8a3ceff7b5d83b5f20fb6738b5c720","family_id":"qf-mod37-flt-prime-coprime","seed":202611065019,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":20,\"p\":67}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=20 and modulus=67, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(67, 20, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (67, 20, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0020","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0720eee073b90c220192dc9d15d78a17a776f822b0d239c262fa771d62583c41","family_id":"qf-mod37-flt-prime-coprime","seed":202611065020,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":-21,\"p\":71}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=-21 and modulus=71, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(71, 50, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (71, 50, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0021","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a55f64a0ced113da501ef01af6bb23ba5a03a3b72253824aa9099a0bafcf702b","family_id":"qf-mod37-flt-prime-coprime","seed":202611065021,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":22,\"p\":73}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=22 and modulus=73, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(73, 22, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (73, 22, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0022","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af84f1aad6bd7327a81142129fce7af39ac08cdeb8dcfb566b346b73fac4984f","family_id":"qf-mod37-flt-prime-coprime","seed":202611065022,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":-23,\"p\":79}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=-23 and modulus=79, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(79, 56, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (79, 56, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0023","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b927da0e0774160f0be1e76beee236f9898ba7be82e283ec5f329c9cd768b61","family_id":"qf-mod37-flt-prime-coprime","seed":202611065023,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":24,\"p\":83}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=24 and modulus=83, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(83, 24, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (83, 24, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0024","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"347ce627d33090206064d66f3351fec33a02c0f4698957824fbaf832f136dd12","family_id":"qf-mod37-flt-prime-coprime","seed":202611065024,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":-25,\"p\":89}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=-25 and modulus=89, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(89, 64, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (89, 64, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0025","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"28d1d9f7a0c31ed763e446019ea71963ccf27b031ea0e67c82407234ebbbb198","family_id":"qf-mod37-flt-prime-coprime","seed":202611065025,"semantic_template":"qf-mod37-flt-prime-coprime/{\"a\":26,\"p\":97}"},"fields":{"title":"Use the Prime-Coprime Fermat Form Only under Both Hypotheses","stem":"Use integer base a=26 and modulus=97, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p_minus_1, congruence_certificate, boundary).","answer":{"kind":"expression","value":"(97, 26, 1, Yes, prime_coprime_form)","explanation":"Require prime p and gcd(a,p)=1, then report a mod p and a^(p−1) mod p with its congruence certificate. Passing one modular power is not a converse primality proof. The exact result, retaining the requested order, is (97, 26, 1, Yes, prime_coprime_form)."},"skills":["Use the Prime-Coprime Fermat Form Only under Both Hypotheses"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0026","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"839ad0ee3b69afee6cec3c3ebaf93eaca2a140a9688b88ca3b8f506d73f5a7b0","family_id":"qf-mod37-flt-all-integer","seed":202611065026,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":2,\"p\":2}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=2 and modulus=2, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(2, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (2, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0027","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa6db4956ef6c8d2a6cc321d768a11dd13ec396bd0bc0ff03be1827bea164f92","family_id":"qf-mod37-flt-all-integer","seed":202611065027,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":-6,\"p\":3}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=-6 and modulus=3, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(3, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (3, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0028","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2c7aa262ae99766c7af1ff2935f0e3d34cee03151fa4e8526f83ee61589952fe","family_id":"qf-mod37-flt-all-integer","seed":202611065028,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":15,\"p\":5}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=15 and modulus=5, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(5, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (5, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0029","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96a18ddd6015c6fcd6f87999023a5907a01ef55d36d6351f4bb4f850840beaac","family_id":"qf-mod37-flt-all-integer","seed":202611065029,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":-28,\"p\":7}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=-28 and modulus=7, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(7, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (7, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0030","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"687e819e038833d40ea933c742461d9df8a0e590ff43540ba0f96b8b6ebc0a4c","family_id":"qf-mod37-flt-all-integer","seed":202611065030,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":55,\"p\":11}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=55 and modulus=11, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(11, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (11, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0031","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"30293ac5bd186ecec07709d7ed245d3b66f4e9e573c1184124d307ec0bc575d0","family_id":"qf-mod37-flt-all-integer","seed":202611065031,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":-78,\"p\":13}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=-78 and modulus=13, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(13, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (13, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0032","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cdd1f0cb65ebee8328b4d4f57e546075ba77a6a7317c38b60b32c385e8e7486d","family_id":"qf-mod37-flt-all-integer","seed":202611065032,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":119,\"p\":17}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=119 and modulus=17, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(17, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (17, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0033","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65e0f3314285ee40d9ceb7e34258a39f71c1f6f66ed479cf29398cc589954d90","family_id":"qf-mod37-flt-all-integer","seed":202611065033,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":-152,\"p\":19}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=-152 and modulus=19, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(19, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (19, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0034","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fbfc026185b205fde41c6419518c544cb0d365173e5bd710931e7d534b7d2e8e","family_id":"qf-mod37-flt-all-integer","seed":202611065034,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":207,\"p\":23}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=207 and modulus=23, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(23, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (23, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0035","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b8a665e4b6666c8291d86dc9db52eea33f70edf865f50a5c5cedcd486ac7f94f","family_id":"qf-mod37-flt-all-integer","seed":202611065035,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":-290,\"p\":29}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=-290 and modulus=29, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(29, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (29, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0036","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"847e9c006f4ffd152881cb47e8c4ba34ff037ccb862398780a680b835c84e016","family_id":"qf-mod37-flt-all-integer","seed":202611065036,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":341,\"p\":31}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=341 and modulus=31, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(31, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (31, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0037","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"be50d84eb4f2186fda88837701bed6bf2a9c0883050515c4f19a134709a0f851","family_id":"qf-mod37-flt-all-integer","seed":202611065037,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":-444,\"p\":37}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=-444 and modulus=37, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(37, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (37, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0038","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e1fb6d5e0469292775aeec4626a0c27b5459c218bb5136478f590dcfc7e76d1d","family_id":"qf-mod37-flt-all-integer","seed":202611065038,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":533,\"p\":41}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=533 and modulus=41, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(41, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (41, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0039","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a5df886f057aa2b1072084ab8a5fb2444c46802ad7e9bc6b6633601d1b15345","family_id":"qf-mod37-flt-all-integer","seed":202611065039,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":-602,\"p\":43}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=-602 and modulus=43, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(43, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (43, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0040","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8c71586767d65ef20f235b00d529ba5ef7f29fb05b191e10735fc37c9165fbc0","family_id":"qf-mod37-flt-all-integer","seed":202611065040,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":705,\"p\":47}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=705 and modulus=47, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(47, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (47, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0041","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e54c317159d930ad07d7fc3ee3956c19e35ad1bdbc108461bb5df3a428c8c5a","family_id":"qf-mod37-flt-all-integer","seed":202611065041,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":-848,\"p\":53}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=-848 and modulus=53, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(53, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (53, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0042","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b3d5e6d4d829da47e9ed0db3da6461a1fff7cea708d20a353b4eda6af51b67e3","family_id":"qf-mod37-flt-all-integer","seed":202611065042,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":1003,\"p\":59}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=1003 and modulus=59, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(59, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (59, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0043","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a9f40e89b566f87c8b0bd4d0e3b338756bce124236560478cfa709e904d4f6f","family_id":"qf-mod37-flt-all-integer","seed":202611065043,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":-1098,\"p\":61}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=-1098 and modulus=61, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(61, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (61, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0044","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"73ed6d59c1c648e0d45ccb4e582f7e0acfb836fcdc870e3159adad82852efdc9","family_id":"qf-mod37-flt-all-integer","seed":202611065044,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":1273,\"p\":67}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=1273 and modulus=67, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(67, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (67, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0045","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"eb83dc06ec1c95ecf961693d78a20b3fbab0e304463b242ca89014a48f8a0d05","family_id":"qf-mod37-flt-all-integer","seed":202611065045,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":-1420,\"p\":71}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=-1420 and modulus=71, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(71, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (71, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0046","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7deacb88d9bc2f979813ae12333ab8bf717e61f43ba638839ffc18e5399a91d","family_id":"qf-mod37-flt-all-integer","seed":202611065046,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":1533,\"p\":73}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=1533 and modulus=73, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(73, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (73, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0047","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"05b610bad458fc6e9c1f7b3c25baa48d57e75d63ced99485dbe1f7c7a04f76ec","family_id":"qf-mod37-flt-all-integer","seed":202611065047,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":-1738,\"p\":79}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=-1738 and modulus=79, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(79, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (79, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0048","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b257d9988f3b94ed8e13c66a1945fbf20e078aed659c607d3545b8bd32279952","family_id":"qf-mod37-flt-all-integer","seed":202611065048,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":1909,\"p\":83}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=1909 and modulus=83, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(83, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (83, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0049","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ff33aef51b6059d21d6ef3db13c686cb9734cd3175a5a563b0fce98cf1287a1d","family_id":"qf-mod37-flt-all-integer","seed":202611065049,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":-2136,\"p\":89}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=-2136 and modulus=89, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(89, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (89, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0050","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f6d2a688f691053f0780e6f7f62c2d99ec74ce6c384847caac8e007530faad8","family_id":"qf-mod37-flt-all-integer","seed":202611065050,"semantic_template":"qf-mod37-flt-all-integer/{\"a\":2425,\"p\":97}"},"fields":{"title":"The All-Integer Prime Form Includes Bases Divisible by p","stem":"Use integer base a=2425 and modulus=97, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (p, normalized_base, power_p, comparison_residue, congruent, boundary).","answer":{"kind":"expression","value":"(97, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)","explanation":"Require prime p, compute a^p mod p and compare it with a mod p, retaining the comparison residue separately. The case p dividing a is allowed without canceling the zero residue. The exact result, retaining the requested order, is (97, 0, 0, 0, Yes, all_integer_form_including_prime_divides_a)."},"skills":["The All-Integer Prime Form Includes Bases Divisible by p"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0051","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4270cc634012ed80297786778a82df9e574ca9357f0405cd41fbf7558eb38888","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065051,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":2,\"modulus\":4}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=2 and modulus=4, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(4, 2, No, No, No, 0, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (4, 2, No, No, No, 0, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0052","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"05a922af20d71dfe31a890776161d129fb2bdbd978c64d180f80f73d228add02","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065052,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":-3,\"modulus\":5}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=-3 and modulus=5, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(5, -3, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (5, -3, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0053","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"818ee8eea80612eb0ad6b4aab41a0a4c9ccbeb64b1de3a5d8fe1e1771aaaa266","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065053,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":4,\"modulus\":6}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=4 and modulus=6, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(6, 4, No, No, No, 4, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (6, 4, No, No, No, 4, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0054","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f055b9c5d02a7d8be770e2877f77a0151e092b6f1954d058cdfa67a9669d866","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065054,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":-5,\"modulus\":7}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=-5 and modulus=7, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(7, -5, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (7, -5, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0055","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"47c824cd08529f138b41fc910321255ec021fae2f99c92b95d891d28510746b8","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065055,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":6,\"modulus\":8}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=6 and modulus=8, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(8, 6, No, No, No, 0, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (8, 6, No, No, No, 0, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0056","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13090189f2f6f3bd14f9a1df50c14ecb3469561a86b281f80848c0aff9198b56","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065056,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":-7,\"modulus\":9}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=-7 and modulus=9, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(9, -7, No, Yes, No, 4, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (9, -7, No, Yes, No, 4, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0057","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"64d0c35721b7b7b05f7945c1958e6b21cc3045d652bed99dca97679c0e7d0ec8","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065057,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":8,\"modulus\":10}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=8 and modulus=10, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(10, 8, No, No, No, 8, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (10, 8, No, No, No, 8, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0058","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6cc0817704fd63c055575ff4d3efb48e29962ecb99ce7da340bc2e6f4db37b88","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065058,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":-9,\"modulus\":11}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=-9 and modulus=11, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(11, -9, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (11, -9, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0059","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dfa52af733450942d36a3035dd989916ecfa500ea85972c3bfc68284bc5a97ba","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065059,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":10,\"modulus\":12}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=10 and modulus=12, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(12, 10, No, No, No, 4, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (12, 10, No, No, No, 4, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0060","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7556ddd70ccfcf0abcc696fad2713b09d2bcd36059eb3ce2346a31cd95e4dd3f","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065060,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":-11,\"modulus\":13}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=-11 and modulus=13, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(13, -11, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (13, -11, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0061","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd37f2b46782e3d485c44e4e3866f6c8e3d652d5df3d2a4d049e8a8d587c1aa5","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065061,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":12,\"modulus\":14}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=12 and modulus=14, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(14, 12, No, No, No, 12, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (14, 12, No, No, No, 12, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0062","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ba4945d91fb18eb0d9f8d34ddb2f50096fe1cf1376ff5ca267ccc8658f94053","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065062,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":-13,\"modulus\":15}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=-13 and modulus=15, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(15, -13, No, Yes, No, 4, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (15, -13, No, Yes, No, 4, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0063","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e9ed0395a84cb3f4ef399e45e4542e9c487408798b81c78d9da78a5757597973","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065063,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":14,\"modulus\":16}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=14 and modulus=16, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(16, 14, No, No, No, 0, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (16, 14, No, No, No, 0, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0064","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f83cb2436277997c63f183284f4c17c55bf91a925d12e7db207ab712cb32e33","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065064,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":-15,\"modulus\":17}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=-15 and modulus=17, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(17, -15, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (17, -15, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0065","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ea6748cf84b381d2d68a5ce12d286ea8e52c725dcd20ba2191058d509b26e595","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065065,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":16,\"modulus\":18}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=16 and modulus=18, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(18, 16, No, No, No, 4, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (18, 16, No, No, No, 4, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0066","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"73d637063dd19be0970bd270ec99d77069aa6ce3a6d9b09a3f3714bbe9846f27","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065066,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":-17,\"modulus\":19}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=-17 and modulus=19, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(19, -17, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (19, -17, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0067","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4552d0692f81c083de5423605d7d9e339e9305e9f9510a9bdd81aac43ccd8ee8","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065067,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":18,\"modulus\":20}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=18 and modulus=20, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(20, 18, No, No, No, 12, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (20, 18, No, No, No, 12, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0068","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"62cd84cf00f8a48e66cc5a9d2505425bb4bad8601ca97e9db423b962c6f3f87c","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065068,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":-19,\"modulus\":21}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=-19 and modulus=21, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(21, -19, No, Yes, No, 4, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (21, -19, No, Yes, No, 4, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0069","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"936c6da6b8efe4cc440cb965141523c6d26b605a169f1028a98d29e9e36b92bb","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065069,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":20,\"modulus\":22}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=20 and modulus=22, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(22, 20, No, No, No, 20, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (22, 20, No, No, No, 20, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0070","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"374675c0c956560991134616cc23225c37ee8880bdd251f50f647fc3548a2fab","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065070,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":-21,\"modulus\":23}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=-21 and modulus=23, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(23, -21, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (23, -21, Yes, Yes, Yes, 1, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0071","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"769d4d2f114838c8d493d4692bf16380b6ae1ba372aa5bbbaadea2d6819e898c","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065071,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":22,\"modulus\":24}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=22 and modulus=24, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(24, 22, No, No, No, 16, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (24, 22, No, No, No, 16, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0072","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7800081c6a8045b776676ff6efc83dd7aa1050ceac8b9d935574a9ddaf37cf1","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065072,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":-23,\"modulus\":25}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=-23 and modulus=25, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(25, -23, No, Yes, No, 16, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (25, -23, No, Yes, No, 16, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0073","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"45e5746a171b07e78d317a69516b46a0c535984f46f75bc28bfcfb0d21df6592","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065073,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":24,\"modulus\":26}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=24 and modulus=26, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(26, 24, No, No, No, 24, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (26, 24, No, No, No, 24, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0074","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16146ecfcf0d96e2bb3d8619804c2b1b803ec72338916443d93422603e950303","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065074,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":-25,\"modulus\":27}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=-25 and modulus=27, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(27, -25, No, Yes, No, 13, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (27, -25, No, Yes, No, 13, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0075","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3d542976a958d19ff536f0a95fa136c782ec9f3a21fe818e16b59faa8538c476","family_id":"qf-mod37-flt-hypothesis-audit","seed":202611065075,"semantic_template":"qf-mod37-flt-hypothesis-audit/{\"a\":26,\"modulus\":28}"},"fields":{"title":"An Inapplicable Theorem Need Not Mean a Failed Modular Test","stem":"Use integer base a=26 and modulus=28, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, original_base, prime, coprime, coprime_form_applicable, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(28, 26, No, No, No, 20, coprime_form_not_applicable_without_both_hypotheses)","explanation":"Retain the original signed base, primality, coprimality, joint applicability and actual a^(modulus−1) residue separately. Failure of either hypothesis blocks the prime-coprime theorem, but does not force that residue to differ from1. The exact result, retaining the requested order, is (28, 26, No, No, No, 20, coprime_form_not_applicable_without_both_hypotheses)."},"skills":["An Inapplicable Theorem Need Not Mean a Failed Modular Test"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0076","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2cc3519a2d1927aae8d498a88315ff2726e6c91e51506b984b08ef30b1e1846d","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065076,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":2,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=2 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 2, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 2, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0077","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"03655592fe1a17ab4af15cc6a7bf6e1263518658db6f488f12adaaac8a5fa5e2","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065077,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":4,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=4 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 4, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 4, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0078","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f2a4fac34f7002b0114b3675809ba4d612e4c0f2ca519cd0322f1cecad428e28","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065078,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":5,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=5 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 5, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 5, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0079","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de67be93c6bb938a5fd731c71fce94741240b2858f4d51099ac0f8cb824447ad","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065079,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":7,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=7 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 7, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 7, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0080","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e674469110602a30d9ede0ef79edcf44871da262fbf34db05ed78da1dd61cb7","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065080,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":8,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=8 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 8, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 8, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0081","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af9418f9edc31d15b99764e3830da9efab04b88753377eb4fc7957abce4de616","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065081,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":10,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=10 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 10, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 10, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0082","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"69c97a8c5d94c34155acdbd6131637d339e39a548525c84e98c34562a583c370","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065082,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":13,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=13 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 13, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 13, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0083","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"534b401b4dd4b878de9b55c3275d7c4342a92c5d7898729048cde92e26b3681d","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065083,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":14,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=14 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 14, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 14, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0084","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8bb8f0e42dd2cb746b84e1f869534eb9839479f4fa63e09e6f4297890c52048f","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065084,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":16,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=16 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 16, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 16, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0085","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6aa0b07010957a85af47d55926058e7cfef11ee31ba9c0bb53ad034ab13b8a1a","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065085,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":19,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=19 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 19, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 19, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0086","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a88ed4140d8d608b9667c78452404e9ef99467e9771e412de81a4b3f7db7f1f5","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065086,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":20,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=20 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 20, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 20, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0087","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af6781c1d8a88fa95d95094e10eee560f1d0bf9f4b4c68e4f7faa6070d4ac761","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065087,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":23,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=23 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 23, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 23, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0088","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49dcbc75c4d281e5ef078bcd702eeb942ecdd3885e8a076f3fcb24e8853bd040","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065088,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":25,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=25 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 25, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 25, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0089","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2efd4155801733a816a78e18b0908555adec6d35230d5ea4395e8a634a4472c0","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065089,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":26,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=26 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 26, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 26, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0090","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cd44acba7e246604afe88522bc9f7f49e6e508e38f26628898b1046baeea499d","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065090,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":28,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=28 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 28, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 28, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0091","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25b7915e8a08f626db404e0c46fdee405bbbcf450dc3f17ba3952c6908155c37","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065091,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":29,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=29 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 29, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 29, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0092","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"10402f8cc1613fbb0738903f031cbc3a24f7b763f25afeafd2e43b4efc986ef2","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065092,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":31,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=31 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 31, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 31, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0093","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"394ab6ade644c81ff53786516d4072a0abaeeba31912de495bd9ba2e5ef0d14e","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065093,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":32,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=32 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 32, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 32, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0094","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8cb6f77cf0fe895d29d6defebabb32b03fb62f62faa4b69f5de6fd40c63bbfe1","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065094,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":35,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=35 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 35, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 35, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0095","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f392a0d98f20d5d13f086274b93a9fb7fe33ec8379ec709027ff34e935fa158e","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065095,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":37,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=37 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 37, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 37, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0096","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3fc2ce2349df990a86ea77bbf664e8e6ee977517e262325b18f8b5db2750d9aa","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065096,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":38,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=38 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 38, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 38, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0097","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2acc26eb32541869cacb4fbc7fc8ec96224e59492ba980a0a0aebdfe806a84bb","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065097,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":40,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=40 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 40, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 40, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0098","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e3c254e0209284cfc5aee3175ec93b6faaca6cc2c9e649438afab82e0d1d0e0","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065098,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":41,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=41 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 41, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 41, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0099","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c045fac00504a6e075a38ee759d89d0a969a7f7ed6b3f2852b87340664cedb6d","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065099,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":43,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=43 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 43, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 43, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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-20261014-advance-20261106-r05-fermat-little-theorem-0100","batch_id":"batch-20261014-advance-20261106-r05-fermat-little-theorem-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d50076734db2acbc44bb352362deddfb19900509ce5fb9b4b108ccb18579a86","family_id":"qf-mod37-flt-converse-counterexample","seed":202611065100,"semantic_template":"qf-mod37-flt-converse-counterexample/{\"a\":46,\"modulus\":561}"},"fields":{"title":"A Composite Can Pass a Fermat Test for a Chosen Base","stem":"Use integer base a=46 and modulus=561, normalized residues in [0,modulus). Exact integer square-and-multiply computes powers; trial division checks primality. Every equality of residues here is a congruence modulo the stated modulus, not an unrestricted integer equality. The coprime form requires prime modulus and an invertible base; the all-integer form for prime p also allows p to divide a. Never cancel a noninvertible residue. Give (modulus, base, prime, coprime, power_modulus_minus_1, boundary).","answer":{"kind":"expression","value":"(561, 46, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)","explanation":"Verify the modulus is composite, the base is coprime and the computed (modulus−1)-st power is1 modulo the modulus. That is an explicit counterexample to the converse, not a proof that every base or every composite passes. The exact result, retaining the requested order, is (561, 46, No, Yes, 1, converse_false_composite_can_pass_Fermat_test)."},"skills":["A Composite Can Pass a Fermat Test for a Chosen Base"],"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."}}