{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261111-r06-quadratic-residue-0001","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"627cbf63493ee9549aea9c97291b71f5cfe517109d05640b3776d622db894c43","family_id":"qf-mod42-quadratic-square-table","seed":202611116001,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":3}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=3 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(3, [[1, 1], [2, 1]], [1], 1, 1, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (3, [[1, 1], [2, 1]], [1], 1, 1, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0002","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e4271a735d700c32edc4406caecf1642b0cc5c3394be3f955f2fbfab94aff06c","family_id":"qf-mod42-quadratic-square-table","seed":202611116002,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":5}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=5 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(5, [[1, 1], [2, 4], [3, 4], [4, 1]], [1, 4], 2, 2, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (5, [[1, 1], [2, 4], [3, 4], [4, 1]], [1, 4], 2, 2, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0003","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81f18fa169e84f25df9dfc9ad82c55351b52f2bc060938d916c30a1122d2739b","family_id":"qf-mod42-quadratic-square-table","seed":202611116003,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":7}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=7 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(7, [[1, 1], [2, 4], [3, 2], [4, 2], [5, 4], [6, 1]], [1, 2, 4], 3, 3, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (7, [[1, 1], [2, 4], [3, 2], [4, 2], [5, 4], [6, 1]], [1, 2, 4], 3, 3, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0004","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"167452fa495cfd8b5e055e2d377d249fe46721380a21d233ede1e584f8f14544","family_id":"qf-mod42-quadratic-square-table","seed":202611116004,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":11}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=11 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(11, [[1, 1], [2, 4], [3, 9], [4, 5], [5, 3], [6, 3], [7, 5], [8, 9], [9, 4], [10, 1]], [1, 3, 4, 5, 9], 5, 5, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (11, [[1, 1], [2, 4], [3, 9], [4, 5], [5, 3], [6, 3], [7, 5], [8, 9], [9, 4], [10, 1]], [1, 3, 4, 5, 9], 5, 5, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0005","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2571acb0e7af60a199c89aa25451615688affb650f4fcff3e5240d3143f0167d","family_id":"qf-mod42-quadratic-square-table","seed":202611116005,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":13}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=13 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(13, [[1, 1], [2, 4], [3, 9], [4, 3], [5, 12], [6, 10], [7, 10], [8, 12], [9, 3], [10, 9], [11, 4], [12, 1]], [1, 3, 4, 9, 10, 12], 6, 6, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (13, [[1, 1], [2, 4], [3, 9], [4, 3], [5, 12], [6, 10], [7, 10], [8, 12], [9, 3], [10, 9], [11, 4], [12, 1]], [1, 3, 4, 9, 10, 12], 6, 6, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0006","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"408fe49c36e3624983e0a6493be9b3917f876d410cf195a945dd4cbf1a207daa","family_id":"qf-mod42-quadratic-square-table","seed":202611116006,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":17}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=17 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(17, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 8], [6, 2], [7, 15], [8, 13], [9, 13], [10, 15], [11, 2], [12, 8], [13, 16], [14, 9], [15, 4], [16, 1]], [1, 2, 4, 8, 9, 13, 15, 16], 8, 8, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (17, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 8], [6, 2], [7, 15], [8, 13], [9, 13], [10, 15], [11, 2], [12, 8], [13, 16], [14, 9], [15, 4], [16, 1]], [1, 2, 4, 8, 9, 13, 15, 16], 8, 8, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0007","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a41a80d7eeb73a3f2f956bef802339d3bdccd078e063f8b55cbf26a3bdcfab6b","family_id":"qf-mod42-quadratic-square-table","seed":202611116007,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":19}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=19 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(19, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 6], [6, 17], [7, 11], [8, 7], [9, 5], [10, 5], [11, 7], [12, 11], [13, 17], [14, 6], [15, 16], [16, 9], [17, 4], [18, 1]], [1, 4, 5, 6, 7, 9, 11, 16, 17], 9, 9, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (19, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 6], [6, 17], [7, 11], [8, 7], [9, 5], [10, 5], [11, 7], [12, 11], [13, 17], [14, 6], [15, 16], [16, 9], [17, 4], [18, 1]], [1, 4, 5, 6, 7, 9, 11, 16, 17], 9, 9, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0008","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1143d29a194bb583dcc728750e399a20d6b193cd76470680dc79325a695b7a5e","family_id":"qf-mod42-quadratic-square-table","seed":202611116008,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":23}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=23 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(23, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 2], [6, 13], [7, 3], [8, 18], [9, 12], [10, 8], [11, 6], [12, 6], [13, 8], [14, 12], [15, 18], [16, 3], [17, 13], [18, 2], [19, 16], [20, 9], [21, 4], [22, 1]], [1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18], 11, 11, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (23, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 2], [6, 13], [7, 3], [8, 18], [9, 12], [10, 8], [11, 6], [12, 6], [13, 8], [14, 12], [15, 18], [16, 3], [17, 13], [18, 2], [19, 16], [20, 9], [21, 4], [22, 1]], [1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18], 11, 11, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0009","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a72f8b82dd154400f3141bc5647b95b9b310fc1c43ed8ff3e99efb525aba297","family_id":"qf-mod42-quadratic-square-table","seed":202611116009,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":29}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=29 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(29, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 7], [7, 20], [8, 6], [9, 23], [10, 13], [11, 5], [12, 28], [13, 24], [14, 22], [15, 22], [16, 24], [17, 28], [18, 5], [19, 13], [20, 23], [21, 6], [22, 20], [23, 7], [24, 25], [25, 16], [26, 9], [27, 4], [28, 1]], [1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28], 14, 14, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (29, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 7], [7, 20], [8, 6], [9, 23], [10, 13], [11, 5], [12, 28], [13, 24], [14, 22], [15, 22], [16, 24], [17, 28], [18, 5], [19, 13], [20, 23], [21, 6], [22, 20], [23, 7], [24, 25], [25, 16], [26, 9], [27, 4], [28, 1]], [1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28], 14, 14, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0010","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e4809a7ca420c05d2996885db1788426cd604cba0b73d6aab72508d33604fb3","family_id":"qf-mod42-quadratic-square-table","seed":202611116010,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":31}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=31 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(31, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 5], [7, 18], [8, 2], [9, 19], [10, 7], [11, 28], [12, 20], [13, 14], [14, 10], [15, 8], [16, 8], [17, 10], [18, 14], [19, 20], [20, 28], [21, 7], [22, 19], [23, 2], [24, 18], [25, 5], [26, 25], [27, 16], [28, 9], [29, 4], [30, 1]], [1, 2, 4, 5, 7, 8, 9, 10, 14, 16, 18, 19, 20, 25, 28], 15, 15, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (31, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 5], [7, 18], [8, 2], [9, 19], [10, 7], [11, 28], [12, 20], [13, 14], [14, 10], [15, 8], [16, 8], [17, 10], [18, 14], [19, 20], [20, 28], [21, 7], [22, 19], [23, 2], [24, 18], [25, 5], [26, 25], [27, 16], [28, 9], [29, 4], [30, 1]], [1, 2, 4, 5, 7, 8, 9, 10, 14, 16, 18, 19, 20, 25, 28], 15, 15, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0011","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87e1e2861267db9f9e402df5cc0683fcbf27e2d9fde900d0cb1c691a5b23fd94","family_id":"qf-mod42-quadratic-square-table","seed":202611116011,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":37}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=37 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(37, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 12], [8, 27], [9, 7], [10, 26], [11, 10], [12, 33], [13, 21], [14, 11], [15, 3], [16, 34], [17, 30], [18, 28], [19, 28], [20, 30], [21, 34], [22, 3], [23, 11], [24, 21], [25, 33], [26, 10], [27, 26], [28, 7], [29, 27], [30, 12], [31, 36], [32, 25], [33, 16], [34, 9], [35, 4], [36, 1]], [1, 3, 4, 7, 9, 10, 11, 12, 16, 21, 25, 26, 27, 28, 30, 33, 34, 36], 18, 18, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (37, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 12], [8, 27], [9, 7], [10, 26], [11, 10], [12, 33], [13, 21], [14, 11], [15, 3], [16, 34], [17, 30], [18, 28], [19, 28], [20, 30], [21, 34], [22, 3], [23, 11], [24, 21], [25, 33], [26, 10], [27, 26], [28, 7], [29, 27], [30, 12], [31, 36], [32, 25], [33, 16], [34, 9], [35, 4], [36, 1]], [1, 3, 4, 7, 9, 10, 11, 12, 16, 21, 25, 26, 27, 28, 30, 33, 34, 36], 18, 18, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0012","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91b6b605786cdba205d779f276e5ea5278db2042f2016ff2ad8e58662d304068","family_id":"qf-mod42-quadratic-square-table","seed":202611116012,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":41}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=41 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(41, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 8], [8, 23], [9, 40], [10, 18], [11, 39], [12, 21], [13, 5], [14, 32], [15, 20], [16, 10], [17, 2], [18, 37], [19, 33], [20, 31], [21, 31], [22, 33], [23, 37], [24, 2], [25, 10], [26, 20], [27, 32], [28, 5], [29, 21], [30, 39], [31, 18], [32, 40], [33, 23], [34, 8], [35, 36], [36, 25], [37, 16], [38, 9], [39, 4], [40, 1]], [1, 2, 4, 5, 8, 9, 10, 16, 18, 20, 21, 23, 25, 31, 32, 33, 36, 37, 39, 40], 20, 20, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (41, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 8], [8, 23], [9, 40], [10, 18], [11, 39], [12, 21], [13, 5], [14, 32], [15, 20], [16, 10], [17, 2], [18, 37], [19, 33], [20, 31], [21, 31], [22, 33], [23, 37], [24, 2], [25, 10], [26, 20], [27, 32], [28, 5], [29, 21], [30, 39], [31, 18], [32, 40], [33, 23], [34, 8], [35, 36], [36, 25], [37, 16], [38, 9], [39, 4], [40, 1]], [1, 2, 4, 5, 8, 9, 10, 16, 18, 20, 21, 23, 25, 31, 32, 33, 36, 37, 39, 40], 20, 20, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0013","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3dd771e9d323c11bf7f71f08f9c7d023bc5cc463c40598eb048a4e0263e1166b","family_id":"qf-mod42-quadratic-square-table","seed":202611116013,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":43}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=43 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(43, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 6], [8, 21], [9, 38], [10, 14], [11, 35], [12, 15], [13, 40], [14, 24], [15, 10], [16, 41], [17, 31], [18, 23], [19, 17], [20, 13], [21, 11], [22, 11], [23, 13], [24, 17], [25, 23], [26, 31], [27, 41], [28, 10], [29, 24], [30, 40], [31, 15], [32, 35], [33, 14], [34, 38], [35, 21], [36, 6], [37, 36], [38, 25], [39, 16], [40, 9], [41, 4], [42, 1]], [1, 4, 6, 9, 10, 11, 13, 14, 15, 16, 17, 21, 23, 24, 25, 31, 35, 36, 38, 40, 41], 21, 21, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (43, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 6], [8, 21], [9, 38], [10, 14], [11, 35], [12, 15], [13, 40], [14, 24], [15, 10], [16, 41], [17, 31], [18, 23], [19, 17], [20, 13], [21, 11], [22, 11], [23, 13], [24, 17], [25, 23], [26, 31], [27, 41], [28, 10], [29, 24], [30, 40], [31, 15], [32, 35], [33, 14], [34, 38], [35, 21], [36, 6], [37, 36], [38, 25], [39, 16], [40, 9], [41, 4], [42, 1]], [1, 4, 6, 9, 10, 11, 13, 14, 15, 16, 17, 21, 23, 24, 25, 31, 35, 36, 38, 40, 41], 21, 21, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0014","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"008d7906d05cef29d80bbb7bab0f44be17e5c1863335bd340c37b8299575861b","family_id":"qf-mod42-quadratic-square-table","seed":202611116014,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":47}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=47 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(47, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 2], [8, 17], [9, 34], [10, 6], [11, 27], [12, 3], [13, 28], [14, 8], [15, 37], [16, 21], [17, 7], [18, 42], [19, 32], [20, 24], [21, 18], [22, 14], [23, 12], [24, 12], [25, 14], [26, 18], [27, 24], [28, 32], [29, 42], [30, 7], [31, 21], [32, 37], [33, 8], [34, 28], [35, 3], [36, 27], [37, 6], [38, 34], [39, 17], [40, 2], [41, 36], [42, 25], [43, 16], [44, 9], [45, 4], [46, 1]], [1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 17, 18, 21, 24, 25, 27, 28, 32, 34, 36, 37, 42], 23, 23, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (47, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 2], [8, 17], [9, 34], [10, 6], [11, 27], [12, 3], [13, 28], [14, 8], [15, 37], [16, 21], [17, 7], [18, 42], [19, 32], [20, 24], [21, 18], [22, 14], [23, 12], [24, 12], [25, 14], [26, 18], [27, 24], [28, 32], [29, 42], [30, 7], [31, 21], [32, 37], [33, 8], [34, 28], [35, 3], [36, 27], [37, 6], [38, 34], [39, 17], [40, 2], [41, 36], [42, 25], [43, 16], [44, 9], [45, 4], [46, 1]], [1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 17, 18, 21, 24, 25, 27, 28, 32, 34, 36, 37, 42], 23, 23, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0015","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fea6e68df4ad200862de3db67519effd7c9cecdfd6d2d667ec4aa8f4653bd223","family_id":"qf-mod42-quadratic-square-table","seed":202611116015,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":53}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=53 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(53, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 11], [9, 28], [10, 47], [11, 15], [12, 38], [13, 10], [14, 37], [15, 13], [16, 44], [17, 24], [18, 6], [19, 43], [20, 29], [21, 17], [22, 7], [23, 52], [24, 46], [25, 42], [26, 40], [27, 40], [28, 42], [29, 46], [30, 52], [31, 7], [32, 17], [33, 29], [34, 43], [35, 6], [36, 24], [37, 44], [38, 13], [39, 37], [40, 10], [41, 38], [42, 15], [43, 47], [44, 28], [45, 11], [46, 49], [47, 36], [48, 25], [49, 16], [50, 9], [51, 4], [52, 1]], [1, 4, 6, 7, 9, 10, 11, 13, 15, 16, 17, 24, 25, 28, 29, 36, 37, 38, 40, 42, 43, 44, 46, 47, 49, 52], 26, 26, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (53, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 11], [9, 28], [10, 47], [11, 15], [12, 38], [13, 10], [14, 37], [15, 13], [16, 44], [17, 24], [18, 6], [19, 43], [20, 29], [21, 17], [22, 7], [23, 52], [24, 46], [25, 42], [26, 40], [27, 40], [28, 42], [29, 46], [30, 52], [31, 7], [32, 17], [33, 29], [34, 43], [35, 6], [36, 24], [37, 44], [38, 13], [39, 37], [40, 10], [41, 38], [42, 15], [43, 47], [44, 28], [45, 11], [46, 49], [47, 36], [48, 25], [49, 16], [50, 9], [51, 4], [52, 1]], [1, 4, 6, 7, 9, 10, 11, 13, 15, 16, 17, 24, 25, 28, 29, 36, 37, 38, 40, 42, 43, 44, 46, 47, 49, 52], 26, 26, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0016","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1d0c1fd00b880e93426b0fd1198e02ae691977071aca9acdb024132ca6a3577","family_id":"qf-mod42-quadratic-square-table","seed":202611116016,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":59}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=59 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(59, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 5], [9, 22], [10, 41], [11, 3], [12, 26], [13, 51], [14, 19], [15, 48], [16, 20], [17, 53], [18, 29], [19, 7], [20, 46], [21, 28], [22, 12], [23, 57], [24, 45], [25, 35], [26, 27], [27, 21], [28, 17], [29, 15], [30, 15], [31, 17], [32, 21], [33, 27], [34, 35], [35, 45], [36, 57], [37, 12], [38, 28], [39, 46], [40, 7], [41, 29], [42, 53], [43, 20], [44, 48], [45, 19], [46, 51], [47, 26], [48, 3], [49, 41], [50, 22], [51, 5], [52, 49], [53, 36], [54, 25], [55, 16], [56, 9], [57, 4], [58, 1]], [1, 3, 4, 5, 7, 9, 12, 15, 16, 17, 19, 20, 21, 22, 25, 26, 27, 28, 29, 35, 36, 41, 45, 46, 48, 49, 51, 53, 57], 29, 29, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (59, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 5], [9, 22], [10, 41], [11, 3], [12, 26], [13, 51], [14, 19], [15, 48], [16, 20], [17, 53], [18, 29], [19, 7], [20, 46], [21, 28], [22, 12], [23, 57], [24, 45], [25, 35], [26, 27], [27, 21], [28, 17], [29, 15], [30, 15], [31, 17], [32, 21], [33, 27], [34, 35], [35, 45], [36, 57], [37, 12], [38, 28], [39, 46], [40, 7], [41, 29], [42, 53], [43, 20], [44, 48], [45, 19], [46, 51], [47, 26], [48, 3], [49, 41], [50, 22], [51, 5], [52, 49], [53, 36], [54, 25], [55, 16], [56, 9], [57, 4], [58, 1]], [1, 3, 4, 5, 7, 9, 12, 15, 16, 17, 19, 20, 21, 22, 25, 26, 27, 28, 29, 35, 36, 41, 45, 46, 48, 49, 51, 53, 57], 29, 29, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0017","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"21c0293091360409df807b8f0bb856845b2904ad6095b1f8977acba0cd987512","family_id":"qf-mod42-quadratic-square-table","seed":202611116017,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":61}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=61 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(61, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 3], [9, 20], [10, 39], [11, 60], [12, 22], [13, 47], [14, 13], [15, 42], [16, 12], [17, 45], [18, 19], [19, 56], [20, 34], [21, 14], [22, 57], [23, 41], [24, 27], [25, 15], [26, 5], [27, 58], [28, 52], [29, 48], [30, 46], [31, 46], [32, 48], [33, 52], [34, 58], [35, 5], [36, 15], [37, 27], [38, 41], [39, 57], [40, 14], [41, 34], [42, 56], [43, 19], [44, 45], [45, 12], [46, 42], [47, 13], [48, 47], [49, 22], [50, 60], [51, 39], [52, 20], [53, 3], [54, 49], [55, 36], [56, 25], [57, 16], [58, 9], [59, 4], [60, 1]], [1, 3, 4, 5, 9, 12, 13, 14, 15, 16, 19, 20, 22, 25, 27, 34, 36, 39, 41, 42, 45, 46, 47, 48, 49, 52, 56, 57, 58, 60], 30, 30, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (61, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 3], [9, 20], [10, 39], [11, 60], [12, 22], [13, 47], [14, 13], [15, 42], [16, 12], [17, 45], [18, 19], [19, 56], [20, 34], [21, 14], [22, 57], [23, 41], [24, 27], [25, 15], [26, 5], [27, 58], [28, 52], [29, 48], [30, 46], [31, 46], [32, 48], [33, 52], [34, 58], [35, 5], [36, 15], [37, 27], [38, 41], [39, 57], [40, 14], [41, 34], [42, 56], [43, 19], [44, 45], [45, 12], [46, 42], [47, 13], [48, 47], [49, 22], [50, 60], [51, 39], [52, 20], [53, 3], [54, 49], [55, 36], [56, 25], [57, 16], [58, 9], [59, 4], [60, 1]], [1, 3, 4, 5, 9, 12, 13, 14, 15, 16, 19, 20, 22, 25, 27, 34, 36, 39, 41, 42, 45, 46, 47, 48, 49, 52, 56, 57, 58, 60], 30, 30, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0018","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8729f4a4ab3af20e652bc0bd5924b15b6da87ab3ce0fc9b372966af8de8ff932","family_id":"qf-mod42-quadratic-square-table","seed":202611116018,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":67}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=67 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(67, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 14], [10, 33], [11, 54], [12, 10], [13, 35], [14, 62], [15, 24], [16, 55], [17, 21], [18, 56], [19, 26], [20, 65], [21, 39], [22, 15], [23, 60], [24, 40], [25, 22], [26, 6], [27, 59], [28, 47], [29, 37], [30, 29], [31, 23], [32, 19], [33, 17], [34, 17], [35, 19], [36, 23], [37, 29], [38, 37], [39, 47], [40, 59], [41, 6], [42, 22], [43, 40], [44, 60], [45, 15], [46, 39], [47, 65], [48, 26], [49, 56], [50, 21], [51, 55], [52, 24], [53, 62], [54, 35], [55, 10], [56, 54], [57, 33], [58, 14], [59, 64], [60, 49], [61, 36], [62, 25], [63, 16], [64, 9], [65, 4], [66, 1]], [1, 4, 6, 9, 10, 14, 15, 16, 17, 19, 21, 22, 23, 24, 25, 26, 29, 33, 35, 36, 37, 39, 40, 47, 49, 54, 55, 56, 59, 60, 62, 64, 65], 33, 33, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (67, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 14], [10, 33], [11, 54], [12, 10], [13, 35], [14, 62], [15, 24], [16, 55], [17, 21], [18, 56], [19, 26], [20, 65], [21, 39], [22, 15], [23, 60], [24, 40], [25, 22], [26, 6], [27, 59], [28, 47], [29, 37], [30, 29], [31, 23], [32, 19], [33, 17], [34, 17], [35, 19], [36, 23], [37, 29], [38, 37], [39, 47], [40, 59], [41, 6], [42, 22], [43, 40], [44, 60], [45, 15], [46, 39], [47, 65], [48, 26], [49, 56], [50, 21], [51, 55], [52, 24], [53, 62], [54, 35], [55, 10], [56, 54], [57, 33], [58, 14], [59, 64], [60, 49], [61, 36], [62, 25], [63, 16], [64, 9], [65, 4], [66, 1]], [1, 4, 6, 9, 10, 14, 15, 16, 17, 19, 21, 22, 23, 24, 25, 26, 29, 33, 35, 36, 37, 39, 40, 47, 49, 54, 55, 56, 59, 60, 62, 64, 65], 33, 33, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0019","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d4dd601ca3f6e8008613cf7c4ca9e4f77506f299e6a00fc219e159589cde65ff","family_id":"qf-mod42-quadratic-square-table","seed":202611116019,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":71}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=71 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(71, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 10], [10, 29], [11, 50], [12, 2], [13, 27], [14, 54], [15, 12], [16, 43], [17, 5], [18, 40], [19, 6], [20, 45], [21, 15], [22, 58], [23, 32], [24, 8], [25, 57], [26, 37], [27, 19], [28, 3], [29, 60], [30, 48], [31, 38], [32, 30], [33, 24], [34, 20], [35, 18], [36, 18], [37, 20], [38, 24], [39, 30], [40, 38], [41, 48], [42, 60], [43, 3], [44, 19], [45, 37], [46, 57], [47, 8], [48, 32], [49, 58], [50, 15], [51, 45], [52, 6], [53, 40], [54, 5], [55, 43], [56, 12], [57, 54], [58, 27], [59, 2], [60, 50], [61, 29], [62, 10], [63, 64], [64, 49], [65, 36], [66, 25], [67, 16], [68, 9], [69, 4], [70, 1]], [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 19, 20, 24, 25, 27, 29, 30, 32, 36, 37, 38, 40, 43, 45, 48, 49, 50, 54, 57, 58, 60, 64], 35, 35, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (71, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 10], [10, 29], [11, 50], [12, 2], [13, 27], [14, 54], [15, 12], [16, 43], [17, 5], [18, 40], [19, 6], [20, 45], [21, 15], [22, 58], [23, 32], [24, 8], [25, 57], [26, 37], [27, 19], [28, 3], [29, 60], [30, 48], [31, 38], [32, 30], [33, 24], [34, 20], [35, 18], [36, 18], [37, 20], [38, 24], [39, 30], [40, 38], [41, 48], [42, 60], [43, 3], [44, 19], [45, 37], [46, 57], [47, 8], [48, 32], [49, 58], [50, 15], [51, 45], [52, 6], [53, 40], [54, 5], [55, 43], [56, 12], [57, 54], [58, 27], [59, 2], [60, 50], [61, 29], [62, 10], [63, 64], [64, 49], [65, 36], [66, 25], [67, 16], [68, 9], [69, 4], [70, 1]], [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 19, 20, 24, 25, 27, 29, 30, 32, 36, 37, 38, 40, 43, 45, 48, 49, 50, 54, 57, 58, 60, 64], 35, 35, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0020","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2e34ff4fecbd00af4ea1309a575aebed3c35c66b8cdfb6c9a50006992e1cf0cc","family_id":"qf-mod42-quadratic-square-table","seed":202611116020,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":73}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=73 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(73, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 8], [10, 27], [11, 48], [12, 71], [13, 23], [14, 50], [15, 6], [16, 37], [17, 70], [18, 32], [19, 69], [20, 35], [21, 3], [22, 46], [23, 18], [24, 65], [25, 41], [26, 19], [27, 72], [28, 54], [29, 38], [30, 24], [31, 12], [32, 2], [33, 67], [34, 61], [35, 57], [36, 55], [37, 55], [38, 57], [39, 61], [40, 67], [41, 2], [42, 12], [43, 24], [44, 38], [45, 54], [46, 72], [47, 19], [48, 41], [49, 65], [50, 18], [51, 46], [52, 3], [53, 35], [54, 69], [55, 32], [56, 70], [57, 37], [58, 6], [59, 50], [60, 23], [61, 71], [62, 48], [63, 27], [64, 8], [65, 64], [66, 49], [67, 36], [68, 25], [69, 16], [70, 9], [71, 4], [72, 1]], [1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 19, 23, 24, 25, 27, 32, 35, 36, 37, 38, 41, 46, 48, 49, 50, 54, 55, 57, 61, 64, 65, 67, 69, 70, 71, 72], 36, 36, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (73, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 8], [10, 27], [11, 48], [12, 71], [13, 23], [14, 50], [15, 6], [16, 37], [17, 70], [18, 32], [19, 69], [20, 35], [21, 3], [22, 46], [23, 18], [24, 65], [25, 41], [26, 19], [27, 72], [28, 54], [29, 38], [30, 24], [31, 12], [32, 2], [33, 67], [34, 61], [35, 57], [36, 55], [37, 55], [38, 57], [39, 61], [40, 67], [41, 2], [42, 12], [43, 24], [44, 38], [45, 54], [46, 72], [47, 19], [48, 41], [49, 65], [50, 18], [51, 46], [52, 3], [53, 35], [54, 69], [55, 32], [56, 70], [57, 37], [58, 6], [59, 50], [60, 23], [61, 71], [62, 48], [63, 27], [64, 8], [65, 64], [66, 49], [67, 36], [68, 25], [69, 16], [70, 9], [71, 4], [72, 1]], [1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 19, 23, 24, 25, 27, 32, 35, 36, 37, 38, 41, 46, 48, 49, 50, 54, 55, 57, 61, 64, 65, 67, 69, 70, 71, 72], 36, 36, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0021","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f025b80ec39aa361c555492910e765d6b041840adcbd0c99bafd411361256540","family_id":"qf-mod42-quadratic-square-table","seed":202611116021,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":79}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=79 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(79, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 2], [10, 21], [11, 42], [12, 65], [13, 11], [14, 38], [15, 67], [16, 19], [17, 52], [18, 8], [19, 45], [20, 5], [21, 46], [22, 10], [23, 55], [24, 23], [25, 72], [26, 44], [27, 18], [28, 73], [29, 51], [30, 31], [31, 13], [32, 76], [33, 62], [34, 50], [35, 40], [36, 32], [37, 26], [38, 22], [39, 20], [40, 20], [41, 22], [42, 26], [43, 32], [44, 40], [45, 50], [46, 62], [47, 76], [48, 13], [49, 31], [50, 51], [51, 73], [52, 18], [53, 44], [54, 72], [55, 23], [56, 55], [57, 10], [58, 46], [59, 5], [60, 45], [61, 8], [62, 52], [63, 19], [64, 67], [65, 38], [66, 11], [67, 65], [68, 42], [69, 21], [70, 2], [71, 64], [72, 49], [73, 36], [74, 25], [75, 16], [76, 9], [77, 4], [78, 1]], [1, 2, 4, 5, 8, 9, 10, 11, 13, 16, 18, 19, 20, 21, 22, 23, 25, 26, 31, 32, 36, 38, 40, 42, 44, 45, 46, 49, 50, 51, 52, 55, 62, 64, 65, 67, 72, 73, 76], 39, 39, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (79, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 2], [10, 21], [11, 42], [12, 65], [13, 11], [14, 38], [15, 67], [16, 19], [17, 52], [18, 8], [19, 45], [20, 5], [21, 46], [22, 10], [23, 55], [24, 23], [25, 72], [26, 44], [27, 18], [28, 73], [29, 51], [30, 31], [31, 13], [32, 76], [33, 62], [34, 50], [35, 40], [36, 32], [37, 26], [38, 22], [39, 20], [40, 20], [41, 22], [42, 26], [43, 32], [44, 40], [45, 50], [46, 62], [47, 76], [48, 13], [49, 31], [50, 51], [51, 73], [52, 18], [53, 44], [54, 72], [55, 23], [56, 55], [57, 10], [58, 46], [59, 5], [60, 45], [61, 8], [62, 52], [63, 19], [64, 67], [65, 38], [66, 11], [67, 65], [68, 42], [69, 21], [70, 2], [71, 64], [72, 49], [73, 36], [74, 25], [75, 16], [76, 9], [77, 4], [78, 1]], [1, 2, 4, 5, 8, 9, 10, 11, 13, 16, 18, 19, 20, 21, 22, 23, 25, 26, 31, 32, 36, 38, 40, 42, 44, 45, 46, 49, 50, 51, 52, 55, 62, 64, 65, 67, 72, 73, 76], 39, 39, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0022","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c257360f6ca79452e72d98b1b9a35188a35c9b2b26a4888d659d94297a1128fc","family_id":"qf-mod42-quadratic-square-table","seed":202611116022,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":83}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=83 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(83, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 17], [11, 38], [12, 61], [13, 3], [14, 30], [15, 59], [16, 7], [17, 40], [18, 75], [19, 29], [20, 68], [21, 26], [22, 69], [23, 31], [24, 78], [25, 44], [26, 12], [27, 65], [28, 37], [29, 11], [30, 70], [31, 48], [32, 28], [33, 10], [34, 77], [35, 63], [36, 51], [37, 41], [38, 33], [39, 27], [40, 23], [41, 21], [42, 21], [43, 23], [44, 27], [45, 33], [46, 41], [47, 51], [48, 63], [49, 77], [50, 10], [51, 28], [52, 48], [53, 70], [54, 11], [55, 37], [56, 65], [57, 12], [58, 44], [59, 78], [60, 31], [61, 69], [62, 26], [63, 68], [64, 29], [65, 75], [66, 40], [67, 7], [68, 59], [69, 30], [70, 3], [71, 61], [72, 38], [73, 17], [74, 81], [75, 64], [76, 49], [77, 36], [78, 25], [79, 16], [80, 9], [81, 4], [82, 1]], [1, 3, 4, 7, 9, 10, 11, 12, 16, 17, 21, 23, 25, 26, 27, 28, 29, 30, 31, 33, 36, 37, 38, 40, 41, 44, 48, 49, 51, 59, 61, 63, 64, 65, 68, 69, 70, 75, 77, 78, 81], 41, 41, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (83, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 17], [11, 38], [12, 61], [13, 3], [14, 30], [15, 59], [16, 7], [17, 40], [18, 75], [19, 29], [20, 68], [21, 26], [22, 69], [23, 31], [24, 78], [25, 44], [26, 12], [27, 65], [28, 37], [29, 11], [30, 70], [31, 48], [32, 28], [33, 10], [34, 77], [35, 63], [36, 51], [37, 41], [38, 33], [39, 27], [40, 23], [41, 21], [42, 21], [43, 23], [44, 27], [45, 33], [46, 41], [47, 51], [48, 63], [49, 77], [50, 10], [51, 28], [52, 48], [53, 70], [54, 11], [55, 37], [56, 65], [57, 12], [58, 44], [59, 78], [60, 31], [61, 69], [62, 26], [63, 68], [64, 29], [65, 75], [66, 40], [67, 7], [68, 59], [69, 30], [70, 3], [71, 61], [72, 38], [73, 17], [74, 81], [75, 64], [76, 49], [77, 36], [78, 25], [79, 16], [80, 9], [81, 4], [82, 1]], [1, 3, 4, 7, 9, 10, 11, 12, 16, 17, 21, 23, 25, 26, 27, 28, 29, 30, 31, 33, 36, 37, 38, 40, 41, 44, 48, 49, 51, 59, 61, 63, 64, 65, 68, 69, 70, 75, 77, 78, 81], 41, 41, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0023","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"218190b6c66764547ba123f2649152e863b9576e3f34f3cbcc3a5a38e8c569bc","family_id":"qf-mod42-quadratic-square-table","seed":202611116023,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":89}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=89 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(89, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 11], [11, 32], [12, 55], [13, 80], [14, 18], [15, 47], [16, 78], [17, 22], [18, 57], [19, 5], [20, 44], [21, 85], [22, 39], [23, 84], [24, 42], [25, 2], [26, 53], [27, 17], [28, 72], [29, 40], [30, 10], [31, 71], [32, 45], [33, 21], [34, 88], [35, 68], [36, 50], [37, 34], [38, 20], [39, 8], [40, 87], [41, 79], [42, 73], [43, 69], [44, 67], [45, 67], [46, 69], [47, 73], [48, 79], [49, 87], [50, 8], [51, 20], [52, 34], [53, 50], [54, 68], [55, 88], [56, 21], [57, 45], [58, 71], [59, 10], [60, 40], [61, 72], [62, 17], [63, 53], [64, 2], [65, 42], [66, 84], [67, 39], [68, 85], [69, 44], [70, 5], [71, 57], [72, 22], [73, 78], [74, 47], [75, 18], [76, 80], [77, 55], [78, 32], [79, 11], [80, 81], [81, 64], [82, 49], [83, 36], [84, 25], [85, 16], [86, 9], [87, 4], [88, 1]], [1, 2, 4, 5, 8, 9, 10, 11, 16, 17, 18, 20, 21, 22, 25, 32, 34, 36, 39, 40, 42, 44, 45, 47, 49, 50, 53, 55, 57, 64, 67, 68, 69, 71, 72, 73, 78, 79, 80, 81, 84, 85, 87, 88], 44, 44, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (89, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 11], [11, 32], [12, 55], [13, 80], [14, 18], [15, 47], [16, 78], [17, 22], [18, 57], [19, 5], [20, 44], [21, 85], [22, 39], [23, 84], [24, 42], [25, 2], [26, 53], [27, 17], [28, 72], [29, 40], [30, 10], [31, 71], [32, 45], [33, 21], [34, 88], [35, 68], [36, 50], [37, 34], [38, 20], [39, 8], [40, 87], [41, 79], [42, 73], [43, 69], [44, 67], [45, 67], [46, 69], [47, 73], [48, 79], [49, 87], [50, 8], [51, 20], [52, 34], [53, 50], [54, 68], [55, 88], [56, 21], [57, 45], [58, 71], [59, 10], [60, 40], [61, 72], [62, 17], [63, 53], [64, 2], [65, 42], [66, 84], [67, 39], [68, 85], [69, 44], [70, 5], [71, 57], [72, 22], [73, 78], [74, 47], [75, 18], [76, 80], [77, 55], [78, 32], [79, 11], [80, 81], [81, 64], [82, 49], [83, 36], [84, 25], [85, 16], [86, 9], [87, 4], [88, 1]], [1, 2, 4, 5, 8, 9, 10, 11, 16, 17, 18, 20, 21, 22, 25, 32, 34, 36, 39, 40, 42, 44, 45, 47, 49, 50, 53, 55, 57, 64, 67, 68, 69, 71, 72, 73, 78, 79, 80, 81, 84, 85, 87, 88], 44, 44, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0024","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9d320d1ffe16b8d1a4ccfddc9487a03e6c3be3a0db7902e18700d1f7e7d71c18","family_id":"qf-mod42-quadratic-square-table","seed":202611116024,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":97}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=97 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(97, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 3], [11, 24], [12, 47], [13, 72], [14, 2], [15, 31], [16, 62], [17, 95], [18, 33], [19, 70], [20, 12], [21, 53], [22, 96], [23, 44], [24, 91], [25, 43], [26, 94], [27, 50], [28, 8], [29, 65], [30, 27], [31, 88], [32, 54], [33, 22], [34, 89], [35, 61], [36, 35], [37, 11], [38, 86], [39, 66], [40, 48], [41, 32], [42, 18], [43, 6], [44, 93], [45, 85], [46, 79], [47, 75], [48, 73], [49, 73], [50, 75], [51, 79], [52, 85], [53, 93], [54, 6], [55, 18], [56, 32], [57, 48], [58, 66], [59, 86], [60, 11], [61, 35], [62, 61], [63, 89], [64, 22], [65, 54], [66, 88], [67, 27], [68, 65], [69, 8], [70, 50], [71, 94], [72, 43], [73, 91], [74, 44], [75, 96], [76, 53], [77, 12], [78, 70], [79, 33], [80, 95], [81, 62], [82, 31], [83, 2], [84, 72], [85, 47], [86, 24], [87, 3], [88, 81], [89, 64], [90, 49], [91, 36], [92, 25], [93, 16], [94, 9], [95, 4], [96, 1]], [1, 2, 3, 4, 6, 8, 9, 11, 12, 16, 18, 22, 24, 25, 27, 31, 32, 33, 35, 36, 43, 44, 47, 48, 49, 50, 53, 54, 61, 62, 64, 65, 66, 70, 72, 73, 75, 79, 81, 85, 86, 88, 89, 91, 93, 94, 95, 96], 48, 48, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (97, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 3], [11, 24], [12, 47], [13, 72], [14, 2], [15, 31], [16, 62], [17, 95], [18, 33], [19, 70], [20, 12], [21, 53], [22, 96], [23, 44], [24, 91], [25, 43], [26, 94], [27, 50], [28, 8], [29, 65], [30, 27], [31, 88], [32, 54], [33, 22], [34, 89], [35, 61], [36, 35], [37, 11], [38, 86], [39, 66], [40, 48], [41, 32], [42, 18], [43, 6], [44, 93], [45, 85], [46, 79], [47, 75], [48, 73], [49, 73], [50, 75], [51, 79], [52, 85], [53, 93], [54, 6], [55, 18], [56, 32], [57, 48], [58, 66], [59, 86], [60, 11], [61, 35], [62, 61], [63, 89], [64, 22], [65, 54], [66, 88], [67, 27], [68, 65], [69, 8], [70, 50], [71, 94], [72, 43], [73, 91], [74, 44], [75, 96], [76, 53], [77, 12], [78, 70], [79, 33], [80, 95], [81, 62], [82, 31], [83, 2], [84, 72], [85, 47], [86, 24], [87, 3], [88, 81], [89, 64], [90, 49], [91, 36], [92, 25], [93, 16], [94, 9], [95, 4], [96, 1]], [1, 2, 3, 4, 6, 8, 9, 11, 12, 16, 18, 22, 24, 25, 27, 31, 32, 33, 35, 36, 43, 44, 47, 48, 49, 50, 53, 54, 61, 62, 64, 65, 66, 70, 72, 73, 75, 79, 81, 85, 86, 88, 89, 91, 93, 94, 95, 96], 48, 48, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0025","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"831c7dc9730e28dfb5a264c5c0072a14fd27584d10cb1ceadb8b2fcc6a32a8a2","family_id":"qf-mod42-quadratic-square-table","seed":202611116025,"semantic_template":"qf-mod42-quadratic-square-table/{\"p\":101}"},"fields":{"title":"The Complete Nonzero Square Table Has Half as Many Distinct Residues","stem":"Use bounded odd prime p=101 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Give (p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal).","answer":{"kind":"expression","value":"(101, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 100], [11, 20], [12, 43], [13, 68], [14, 95], [15, 23], [16, 54], [17, 87], [18, 21], [19, 58], [20, 97], [21, 37], [22, 80], [23, 24], [24, 71], [25, 19], [26, 70], [27, 22], [28, 77], [29, 33], [30, 92], [31, 52], [32, 14], [33, 79], [34, 45], [35, 13], [36, 84], [37, 56], [38, 30], [39, 6], [40, 85], [41, 65], [42, 47], [43, 31], [44, 17], [45, 5], [46, 96], [47, 88], [48, 82], [49, 78], [50, 76], [51, 76], [52, 78], [53, 82], [54, 88], [55, 96], [56, 5], [57, 17], [58, 31], [59, 47], [60, 65], [61, 85], [62, 6], [63, 30], [64, 56], [65, 84], [66, 13], [67, 45], [68, 79], [69, 14], [70, 52], [71, 92], [72, 33], [73, 77], [74, 22], [75, 70], [76, 19], [77, 71], [78, 24], [79, 80], [80, 37], [81, 97], [82, 58], [83, 21], [84, 87], [85, 54], [86, 23], [87, 95], [88, 68], [89, 43], [90, 20], [91, 100], [92, 81], [93, 64], [94, 49], [95, 36], [96, 25], [97, 16], [98, 9], [99, 4], [100, 1]], [1, 4, 5, 6, 9, 13, 14, 16, 17, 19, 20, 21, 22, 23, 24, 25, 30, 31, 33, 36, 37, 43, 45, 47, 49, 52, 54, 56, 58, 64, 65, 68, 70, 71, 76, 77, 78, 79, 80, 81, 82, 84, 85, 87, 88, 92, 95, 96, 97, 100], 50, 50, Yes)","explanation":"List [root,square class] in increasing nonzero-root order, deduplicate square classes increasingly, and compare their count with (p−1)/2. The zero class is not added. The exact result, retaining the requested order, is (101, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 100], [11, 20], [12, 43], [13, 68], [14, 95], [15, 23], [16, 54], [17, 87], [18, 21], [19, 58], [20, 97], [21, 37], [22, 80], [23, 24], [24, 71], [25, 19], [26, 70], [27, 22], [28, 77], [29, 33], [30, 92], [31, 52], [32, 14], [33, 79], [34, 45], [35, 13], [36, 84], [37, 56], [38, 30], [39, 6], [40, 85], [41, 65], [42, 47], [43, 31], [44, 17], [45, 5], [46, 96], [47, 88], [48, 82], [49, 78], [50, 76], [51, 76], [52, 78], [53, 82], [54, 88], [55, 96], [56, 5], [57, 17], [58, 31], [59, 47], [60, 65], [61, 85], [62, 6], [63, 30], [64, 56], [65, 84], [66, 13], [67, 45], [68, 79], [69, 14], [70, 52], [71, 92], [72, 33], [73, 77], [74, 22], [75, 70], [76, 19], [77, 71], [78, 24], [79, 80], [80, 37], [81, 97], [82, 58], [83, 21], [84, 87], [85, 54], [86, 23], [87, 95], [88, 68], [89, 43], [90, 20], [91, 100], [92, 81], [93, 64], [94, 49], [95, 36], [96, 25], [97, 16], [98, 9], [99, 4], [100, 1]], [1, 4, 5, 6, 9, 13, 14, 16, 17, 19, 20, 21, 22, 23, 24, 25, 30, 31, 33, 36, 37, 43, 45, 47, 49, 52, 54, 56, 58, 64, 65, 68, 70, 71, 76, 77, 78, 79, 80, 81, 82, 84, 85, 87, 88, 92, 95, 96, 97, 100], 50, 50, Yes)."},"skills":["The Complete Nonzero Square Table Has Half as Many Distinct Residues"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0026","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"decf97c404fd3d8fa7eeb440004fd431b54d1450d372188a0c0023e2d9bdc9e4","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116026,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":1,\"p\":3}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=3 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=1. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(3, 1, 1, [1, 2], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (3, 1, 1, [1, 2], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0027","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6343b48ebee271d8fb5816eff30e3837c01637b659e20643a9d324afe3a43360","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116027,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":2,\"p\":5}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=5 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=2. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(5, 2, 2, [], No, complete_nonzero_square_table_has_no_root)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (5, 2, 2, [], No, complete_nonzero_square_table_has_no_root)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0028","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1b57bc178d22e5207836836d885a56796c70ab2a5b50a29566a236748c1e5936","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116028,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":3,\"p\":7}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=7 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=3. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(7, 3, 3, [], No, complete_nonzero_square_table_has_no_root)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (7, 3, 3, [], No, complete_nonzero_square_table_has_no_root)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0029","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"52feb03b61b53004b41a724b8f836d31e75736e8846b5a32e7673f7a8175d31d","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116029,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":4,\"p\":11}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=11 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=4. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(11, 4, 4, [2, 9], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (11, 4, 4, [2, 9], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0030","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7a0665971a89f799ccae29b670347898db65493cee8a0dc4c9e19b05daf0373","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116030,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":5,\"p\":13}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=13 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=5. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(13, 5, 5, [], No, complete_nonzero_square_table_has_no_root)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (13, 5, 5, [], No, complete_nonzero_square_table_has_no_root)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0031","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9467865ed5d4b5768a934203ef48143aff31da7b110ac08cce5bc74c0c5db548","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116031,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":6,\"p\":17}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=17 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=6. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(17, 6, 6, [], No, complete_nonzero_square_table_has_no_root)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (17, 6, 6, [], No, complete_nonzero_square_table_has_no_root)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0032","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"034c2ed9695944c364c02aa06c57cf174921e3188fc06aad7f447aa13643b88a","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116032,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":7,\"p\":19}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=19 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=7. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(19, 7, 7, [8, 11], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (19, 7, 7, [8, 11], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0033","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e9197395b7f756f14a7347c5187fba066db231630c8e2193632d399612c3d57b","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116033,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":8,\"p\":23}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=23 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=8. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(23, 8, 8, [10, 13], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (23, 8, 8, [10, 13], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0034","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91acf6048c61fef1fa996909e92c0cdbed1c725b36ad64f12b781ae603a4d8e2","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116034,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":9,\"p\":29}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=29 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=9. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(29, 9, 9, [3, 26], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (29, 9, 9, [3, 26], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0035","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c2fa37697fe6ab0733ae933684f2a6093657c2134a4ed3603a0eea19ab293ee4","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116035,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":10,\"p\":31}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=31 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=10. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(31, 10, 10, [14, 17], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (31, 10, 10, [14, 17], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0036","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6627f743584af2c904858a347ab246668193ebc51d0fb06b472a97bf4ed36dce","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116036,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":11,\"p\":37}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=37 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=11. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(37, 11, 11, [14, 23], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (37, 11, 11, [14, 23], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0037","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d3bdda5403bf7e4f16b7ef85644af453435ed3e1bdb40fda942055c04468356a","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116037,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":12,\"p\":41}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=41 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=12. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(41, 12, 12, [], No, complete_nonzero_square_table_has_no_root)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (41, 12, 12, [], No, complete_nonzero_square_table_has_no_root)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0038","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b9d27199879cc672aa7ff929b74c23a6773fe480f02a376940ac313cea23fb6","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116038,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":13,\"p\":43}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=43 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=13. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(43, 13, 13, [20, 23], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (43, 13, 13, [20, 23], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0039","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"036e43dfa3631e349e9cf40791e007ddad0469682763b6ab8afe6e854183e3b0","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116039,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":14,\"p\":47}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=47 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=14. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(47, 14, 14, [22, 25], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (47, 14, 14, [22, 25], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0040","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d933463c5f2f0d58e88e28b4045fced5b9ab9eebf5a635ec2242527e7cb9edda","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116040,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":15,\"p\":53}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=53 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=15. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(53, 15, 15, [11, 42], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (53, 15, 15, [11, 42], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0041","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca22470b98ddbba92ca113fc3e4e7a7948eede66d8afca6a43be5b9b954e3cbe","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116041,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":16,\"p\":59}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=59 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=16. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(59, 16, 16, [4, 55], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (59, 16, 16, [4, 55], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0042","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"93a35e8cda24c21f19f8afb14ff84bb86afcf653c50bb421f0e35fda547bdfc4","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116042,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":17,\"p\":61}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=61 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=17. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(61, 17, 17, [], No, complete_nonzero_square_table_has_no_root)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (61, 17, 17, [], No, complete_nonzero_square_table_has_no_root)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0043","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6ae624fe7a51e00c497b536adc4b06c30b5254ca0bd9f60a7eee10dc8282f611","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116043,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":18,\"p\":67}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=67 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=18. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(67, 18, 18, [], No, complete_nonzero_square_table_has_no_root)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (67, 18, 18, [], No, complete_nonzero_square_table_has_no_root)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0044","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"19ed1d4ca0b6846e1599b67b08ac828dec8cb7a218520752c57dbcae21adb05a","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116044,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":19,\"p\":71}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=71 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=19. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(71, 19, 19, [27, 44], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (71, 19, 19, [27, 44], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0045","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cdde1a58114cfdc4cbb6d3caa1e0dbabed399a967c980fd556095af97e5c78a3","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116045,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":20,\"p\":73}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=73 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=20. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(73, 20, 20, [], No, complete_nonzero_square_table_has_no_root)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (73, 20, 20, [], No, complete_nonzero_square_table_has_no_root)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0046","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b665a72be8d409648df3a15f489274acc1f63a14841cc6bbb83bf70d088e5bda","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116046,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":21,\"p\":79}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=79 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=21. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(79, 21, 21, [10, 69], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (79, 21, 21, [10, 69], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0047","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1f9b95a5c7f09a493801499e5d7098249d4486c098e0a3f1b1a5022591db043","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116047,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":22,\"p\":83}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=83 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=22. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(83, 22, 22, [], No, complete_nonzero_square_table_has_no_root)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (83, 22, 22, [], No, complete_nonzero_square_table_has_no_root)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0048","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ae41d31030a5ec6ecfb5488f959a7c296474bd4ea57d97e6021f416e0219b976","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116048,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":23,\"p\":89}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=89 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=23. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(89, 23, 23, [], No, complete_nonzero_square_table_has_no_root)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (89, 23, 23, [], No, complete_nonzero_square_table_has_no_root)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0049","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a85450e5c93ac9f9e674b477d8babd3583d7af74f4cc16c76bab1d7b1f59cb4d","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116049,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":24,\"p\":97}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=97 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=24. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(97, 24, 24, [11, 86], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (97, 24, 24, [11, 86], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0050","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a12c3eb8873652866b27fe2dd00a3c836e39b47c719dbf63797f91a84710ba4c","family_id":"qf-mod42-quadratic-membership-witness","seed":202611116050,"semantic_template":"qf-mod42-quadratic-membership-witness/{\"a\":25,\"p\":101}"},"fields":{"title":"Normalize the Candidate before Checking Nonzero Square Roots","stem":"Use bounded odd prime p=101 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=25. Give (p, a, normalized_class, nonzero_roots, nonzero_residue, certificate).","answer":{"kind":"expression","value":"(101, 25, 25, [5, 96], Yes, nonzero_square_root_witness)","explanation":"Retain original a and its normalized class. List every nonzero root in order. A nonzero class with a root qualifies; zero is excluded by this convention, and another rootless class is refuted by the complete table. The exact result, retaining the requested order, is (101, 25, 25, [5, 96], Yes, nonzero_square_root_witness)."},"skills":["Normalize the Candidate before Checking Nonzero Square Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0051","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1ce571b41083f20b3d00713e0d158e37609133ea0c9c23c71993cef244a68881","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116051,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":1,\"p\":3}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=3 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=1. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(3, 1, [1, 2], 2, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (3, 1, [1, 2], 2, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0052","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"885e693341f2da1b7e4ac609ddeebc9f4857d0a75aa191aa8c21c31f41e9ef18","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116052,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":4,\"p\":5}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=5 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=4. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(5, 4, [2, 3], 3, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (5, 4, [2, 3], 3, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0053","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e46bf80d3e19886ca972f4de26f523e53b0993da0c40808f44ce996fd27ec7d5","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116053,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":2,\"p\":7}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=7 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=2. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(7, 2, [3, 4], 4, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (7, 2, [3, 4], 4, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0054","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4002d38d123b0186bfeda778ee35f63a49825bf3bec4c11f566d3192fd436e30","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116054,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":5,\"p\":11}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=11 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=5. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(11, 5, [4, 7], 7, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (11, 5, [4, 7], 7, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0055","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2cdde5155c7f599256554f06838ab77eab411629d01bd4326843545b94bab208","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116055,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":12,\"p\":13}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=13 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=12. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(13, 12, [5, 8], 8, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (13, 12, [5, 8], 8, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0056","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf57df782c2c6bfa36732647829fa386c9c2270c89c705deba7e084d7c1ed990","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116056,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":2,\"p\":17}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=17 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=2. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(17, 2, [6, 11], 11, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (17, 2, [6, 11], 11, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0057","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b3aa90be5fa543b4485629a53e78cbe5c4018e9e3165cd0f6cba8e330091f93","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116057,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":11,\"p\":19}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=19 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=11. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(19, 11, [7, 12], 12, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (19, 11, [7, 12], 12, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0058","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b058468aa89dc142bf420a4c78d17516224e261881130adebe77fac070a72a67","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116058,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":18,\"p\":23}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=23 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=18. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(23, 18, [8, 15], 15, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (23, 18, [8, 15], 15, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0059","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4339b7d424967f20506f6aa01e6ed905317440438133687c55220bcb6ffd532f","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116059,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":23,\"p\":29}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=29 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=23. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(29, 23, [9, 20], 20, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (29, 23, [9, 20], 20, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0060","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa8e740ce70e805416b4db3cba2b4c8c0c6a2decb53323212a90393f3dd4cb84","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116060,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":7,\"p\":31}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=31 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=7. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(31, 7, [10, 21], 21, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (31, 7, [10, 21], 21, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0061","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1384d7348f210b5ac07a79418c8128c4e4ad355b1c8876e95514b50a78c56666","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116061,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":10,\"p\":37}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=37 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=10. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(37, 10, [11, 26], 26, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (37, 10, [11, 26], 26, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0062","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ca652c5e68e24fc6934f481fea146330bd97008e85af1714d7fc0678cbe759c","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116062,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":21,\"p\":41}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=41 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=21. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(41, 21, [12, 29], 29, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (41, 21, [12, 29], 29, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0063","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"70fcc6798ed873445ba147de4c344da8ab3baf1321c4786dc082c7159756639e","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116063,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":40,\"p\":43}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=43 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=40. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(43, 40, [13, 30], 30, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (43, 40, [13, 30], 30, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0064","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a382c8ebf688462b034fa55188c0e202df4c10e0c346b7bf0326823d6afd8c1","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116064,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":8,\"p\":47}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=47 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=8. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(47, 8, [14, 33], 33, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (47, 8, [14, 33], 33, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0065","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"65cc17f2936a5d001de0ca41f1ac9f890e1091f5368b4318402fc21afbb0d5e8","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116065,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":13,\"p\":53}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=53 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=13. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(53, 13, [15, 38], 38, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (53, 13, [15, 38], 38, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0066","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c4b5ba225d5d25617449f82aab5e8763314e5949a4726382ea0e74de6027057","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116066,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":20,\"p\":59}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=59 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=20. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(59, 20, [16, 43], 43, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (59, 20, [16, 43], 43, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0067","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"95e032dbc19d2d706068fe9f28a1c6fb3f0073aede4b22132d0a284cbcbebdab","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116067,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":45,\"p\":61}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=61 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=45. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(61, 45, [17, 44], 44, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (61, 45, [17, 44], 44, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0068","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0325918ae77d2392f52c1f91fe28a6b3c483ac3541c057be658e4b6d21314eda","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116068,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":56,\"p\":67}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=67 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=56. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(67, 56, [18, 49], 49, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (67, 56, [18, 49], 49, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0069","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16af94587954f914bbded268b9ab2fddb29f5a48717822c41dee96e12aafc009","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116069,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":6,\"p\":71}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=71 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=6. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(71, 6, [19, 52], 52, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (71, 6, [19, 52], 52, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0070","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a1c3c5bf0180972accfef2b7f3bfdef5f5f6c66342f8b9e70efdea3a67fc044b","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116070,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":35,\"p\":73}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=73 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=35. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(73, 35, [20, 53], 53, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (73, 35, [20, 53], 53, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0071","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c2cac68c3aa749555b136b917e0cc89a86cbeef1ea809ce819b30d657108f56","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116071,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":46,\"p\":79}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=79 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=46. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(79, 46, [21, 58], 58, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (79, 46, [21, 58], 58, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0072","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dc62d8f6d4b69302e237990cdbc6aaee592dfc065982fbb53dcf5f59274880cc","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116072,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":69,\"p\":83}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=83 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=69. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(83, 69, [22, 61], 61, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (83, 69, [22, 61], 61, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0073","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b05e4115094a3ab6ae3187eab823afdc2b6495c0e2d4ec38d5e49cd947d59c6e","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116073,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":84,\"p\":89}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=89 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=84. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(89, 84, [23, 66], 66, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (89, 84, [23, 66], 66, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0074","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80926a082215c8d11f1e972051c1b252a7927d8ec8a098523a8435707d4952b0","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116074,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":91,\"p\":97}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=97 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=91. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(97, 91, [24, 73], 73, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (97, 91, [24, 73], 73, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0075","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6be02f17350603aac9738a3773129be5ec49677d41c319b8a40e48c587e689b3","family_id":"qf-mod42-quadratic-paired-roots","seed":202611116075,"semantic_template":"qf-mod42-quadratic-paired-roots/{\"a\":19,\"p\":101}"},"fields":{"title":"A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots","stem":"Use bounded odd prime p=101 and the declared nonzero quadratic-residue convention. Enumerate every x=1..p−1 and x^2 modulo p; normalize input classes into0..p−1. Zero is excluded from nonzero residues. Do not assume an Euler-criterion or Legendre-symbol result. Original a=19. Give (p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate).","answer":{"kind":"expression","value":"(101, 19, [25, 76], 76, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)","explanation":"Require actual nonzero residue membership and exactly two roots. Report the normalized negative of the first root, its presence and distinctness. Do not transfer the two-root conclusion to zero, modulus2 or composite moduli. The exact result, retaining the requested order, is (101, 19, [25, 76], 76, Yes, Yes, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)."},"skills":["A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0076","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"46c6ab13904af88162a10e532336b3d983c75eb0772e869414ef8b6c4840daa7","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116076,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":0,\"mode\":\"zero_class\",\"modulus\":3}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=zero_class, modulus=3, original a=0. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 3, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (zero_class, 3, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0077","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5302f9b2dcbd8bd1009e2b02976cbc31abad79890f2a4ed479232e1ad65e77ee","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116077,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":2,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=even_prime, modulus=2, original a=2. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (even_prime, 2, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0078","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"85b6a4bf67b6f3915f1a2cfb795ed51453dd31ab80c77b90bd79c5386bc79958","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116078,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":3,\"mode\":\"composite_modulus\",\"modulus\":9}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=composite_modulus, modulus=9, original a=3. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 9, 3, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (composite_modulus, 9, 3, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0079","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9ff70dfc4ffe1021f5f3039cf5306843bd4c1025f388ddcd0c2909917184968","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116079,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":0,\"mode\":\"zero_class\",\"modulus\":11}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=zero_class, modulus=11, original a=0. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 11, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (zero_class, 11, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0080","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e6bb8d7e73865f90f8309c55f4a480c7dcf43a98388fa98ed7c24ce5e5c0c351","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116080,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":5,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=even_prime, modulus=2, original a=5. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 1, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (even_prime, 2, 1, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0081","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"209e2357f519e9b1a69afc4b81e038efe779aba4a16a3952380fb8809a55e1eb","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116081,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":6,\"mode\":\"composite_modulus\",\"modulus\":15}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=composite_modulus, modulus=15, original a=6. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 15, 6, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (composite_modulus, 15, 6, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0082","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ccee2f6bb9c70f60795ecb3dcae6a75c1f3b88bd50404de6011a77dc4dca2ae6","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116082,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":0,\"mode\":\"zero_class\",\"modulus\":19}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=zero_class, modulus=19, original a=0. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 19, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (zero_class, 19, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0083","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"067a927d84bd92333a52cc3c8fb49c3ab539194ed4a57458d113536a22bca7cb","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116083,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":8,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=even_prime, modulus=2, original a=8. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (even_prime, 2, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0084","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f625537aa83c73a65b98737208418f424d560aa8b9c1ca43b9e1e1798d907d4b","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116084,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":9,\"mode\":\"composite_modulus\",\"modulus\":21}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=composite_modulus, modulus=21, original a=9. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 21, 9, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (composite_modulus, 21, 9, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0085","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f4e6c23cf5cc7621e45b4eebb325a3c83eb685bd07a30338e796c6b34072331f","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116085,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":0,\"mode\":\"zero_class\",\"modulus\":31}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=zero_class, modulus=31, original a=0. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 31, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (zero_class, 31, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0086","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3afc5545f2912168b1557816acbe03697b32d00e57633e19f8ce756b389d459b","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116086,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":11,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=even_prime, modulus=2, original a=11. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 1, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (even_prime, 2, 1, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0087","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"57517bf3796d038f38aa4152855317e889990bd3f1ea9538cfc0d3668b64e85f","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116087,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":12,\"mode\":\"composite_modulus\",\"modulus\":27}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=composite_modulus, modulus=27, original a=12. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 27, 12, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (composite_modulus, 27, 12, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0088","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"968507af95355ef35d5677509e257a9802351981a8e9776af2d6048c1beb8f0b","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116088,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":0,\"mode\":\"zero_class\",\"modulus\":43}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=zero_class, modulus=43, original a=0. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 43, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (zero_class, 43, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0089","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"64dc9939054ba794225277667e0cf81c2b3857774bdc7c7c7f41c6de64312c52","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116089,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":14,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=even_prime, modulus=2, original a=14. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (even_prime, 2, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0090","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4bb63b225f649f49b07bcc6ed65057468ad0b9ef7fe96eba089d992daee84e94","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116090,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":15,\"mode\":\"composite_modulus\",\"modulus\":33}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=composite_modulus, modulus=33, original a=15. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 33, 15, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (composite_modulus, 33, 15, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0091","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4d8be2e722cfb7f6e677f59cd5177569db04a3dae1b2d630a107c9d76899eff8","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116091,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":0,\"mode\":\"zero_class\",\"modulus\":59}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=zero_class, modulus=59, original a=0. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 59, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (zero_class, 59, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0092","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d9839ab685f4448fccf59ae1b3160d0af94a444bc59968fd4cba07abe19c36cc","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116092,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":17,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=even_prime, modulus=2, original a=17. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 1, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (even_prime, 2, 1, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0093","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f00a5e87d49d241f325759f8ff7d50bc10c408447cc725bc2c3e78e49d67b3b9","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116093,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":18,\"mode\":\"composite_modulus\",\"modulus\":39}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=composite_modulus, modulus=39, original a=18. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 39, 18, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (composite_modulus, 39, 18, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0094","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa5287e41ca929bd8ae897f5b4eff3f0303444d5a98317ffe464313c76860db9","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116094,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":0,\"mode\":\"zero_class\",\"modulus\":71}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=zero_class, modulus=71, original a=0. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 71, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (zero_class, 71, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0095","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82395d663ba3abdbb20ae8c9f66d156d31be16c1ffc98e1983d540b921abc78b","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116095,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":20,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=even_prime, modulus=2, original a=20. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (even_prime, 2, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0096","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d7f6dea17782d9a6b85f29df126f8c7d37fa23edfce86754fc1b3a27c7267331","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116096,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":21,\"mode\":\"composite_modulus\",\"modulus\":45}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=composite_modulus, modulus=45, original a=21. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 45, 21, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (composite_modulus, 45, 21, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0097","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3430976294cfb0f438f80fa1db8d997123f09d6ce5dc401fff4963cfedc4a188","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116097,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":0,\"mode\":\"zero_class\",\"modulus\":83}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=zero_class, modulus=83, original a=0. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 83, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (zero_class, 83, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0098","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3778eb03ef1d6a49024f0cacfd2e0c96c5f9d755ee59352d8e3180cd9b22a45","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116098,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":23,\"mode\":\"even_prime\",\"modulus\":2}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=even_prime, modulus=2, original a=23. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(even_prime, 2, 1, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (even_prime, 2, 1, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0099","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e0d652c56632d9ecb772355871fb8d369cdc32ccc84fbd5ed1e78089bc453f8a","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116099,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":24,\"mode\":\"composite_modulus\",\"modulus\":51}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=composite_modulus, modulus=51, original a=24. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(composite_modulus, 51, 24, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (composite_modulus, 51, 24, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the 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-20261111-r06-quadratic-residue-0100","batch_id":"batch-20261014-advance-20261111-r06-quadratic-residue-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9556a3b26a2f8f42039921944a4e3318335f827216a9b95b10d9781137e680a2","family_id":"qf-mod42-quadratic-domain-boundary","seed":202611116100,"semantic_template":"qf-mod42-quadratic-domain-boundary/{\"a\":0,\"mode\":\"zero_class\",\"modulus\":101}"},"fields":{"title":"Zero, Modulus Two and Composite Moduli Need Separate Conventions","stem":"Use mode=zero_class, modulus=101, original a=0. Give (mode, modulus, normalized_class, declared_model_qualified, boundary).","answer":{"kind":"expression","value":"(zero_class, 101, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)","explanation":"Normalize a under the declared modulus. The fixed No means this odd-prime/nonzero model is not qualified; it does not say that a has no square root modulo2 or a composite modulus. Validate the named boundary premise instead of using an odd-prime theorem outside its domain. The exact result, retaining the requested order, is (zero_class, 101, 0, No, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)."},"skills":["Zero, Modulus Two and Composite Moduli Need Separate Conventions"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}