{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261014-advance-20261112-r06-legendre-symbol-0060", "pilot_number": "R0046-23785", "number": 23785, "family_id": "qf-mod43-legendre-zero-class", "title": { "zh-Hant": "第 23785 題：Legendre符號", "en": "Question 23785: Divisibility by p Gives Symbol Zero, Not Plus One" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261014-advance-20261112-r06-legendre-symbol-0060", "batch_id": "batch-20261014-advance-20261112-r06-legendre-symbol-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-legendre-symbol" ], "title_zh": "Legendre符號", "stem_zh": "odd prime上的zero/residue/nonresidue三值完整分類。操作=legendre-zero-class；資料={\"prime\":31,\"a\":310}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "answer": { "kind": "expression", "value": "(31, 310, 0, 0, [], p_divides_a_so_Legendre_symbol_is_zero_before_residue_classification)", "explanation_zh": "Wave6第六層核對=(31, 310, 0, 0, [], p_divides_a_so_Legendre_symbol_is_zero_before_residue_classification)；依完整明示資料獨立重算；a先normalize modulo p且zero class輸出0；nonzero classification由完整square table取得；p=2與composite modulus不在domain，且不得提前使用Euler criterion。" }, "difficulty": { "band": "undergraduate_core", "level": 2, "skills": [ "Legendre符號" ] }, "variant": { "family_id": "qf-mod43-legendre-zero-class", "seed": 202611126060, "parameters": { "prime": 31, "a": 310, "domain": "Odd_prime_Legendre_symbol", "operation": "legendre-zero-class", "semantic_template": "qf-mod43-legendre-zero-class/{\"a\":310,\"prime\":31}" } }, "verification": { "status": "candidate", "checks": [ "producer_exact_finite_or_symbolic_model", "producer_source_snapshot_check", "producer_substantive_template_deduplication", "producer_json_schema" ], "warning_zh": "候選須主端重算；不構成人工審定、一般定理的有限枚舉替代、未宣告瑕積分、未明示係數律、Euler criterion循環、有限word等價、Lean或Evidence。" }, "provenance": { "generator_role": "question_producer", "generator_model": "model_unknown", "generated_at": "2026-10-14", "contract_version": "ocme-question-factory-v0.1" } }, "original_english_edition": { "schema_version": "ocme-question-localization-v0.1", "question_id": "q-20261014-advance-20261112-r06-legendre-symbol-0060", "batch_id": "batch-20261014-advance-20261112-r06-legendre-symbol-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "a5dc4943bbbd7522e7b5350b1b9c469a8026a5cc915dc40a7261b25c84cd2a64", "family_id": "qf-mod43-legendre-zero-class", "seed": 202611126060, "semantic_template": "qf-mod43-legendre-zero-class/{\"a\":310,\"prime\":31}" }, "fields": { "title": "Divisibility by p Gives Symbol Zero, Not Plus One", "stem": "Use odd prime p=31, original a=310. Give (prime, a, normalized_class, Legendre_symbol, nonzero_roots, certificate).", "answer": { "kind": "expression", "value": "(31, 310, 0, 0, [], p_divides_a_so_Legendre_symbol_is_zero_before_residue_classification)", "explanation": "Require a mod p=0 and report Legendre symbol0 before residue/nonresidue classification. The nonzero-root list is empty; this does not deny that0 itself squares to0. The exact result, retaining the requested order, is (31, 310, 0, 0, [], p_divides_a_so_Legendre_symbol_is_zero_before_residue_classification)." }, "skills": [ "Divisibility by p Gives Symbol Zero, Not Plus One" ], "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review." } }, "givens": [ { "id": "G001", "text": { "zh-Hant": "odd prime上的zero/residue/nonresidue三值完整分類。操作=legendre-zero-class；資料={\"prime\":31,\"a\":310}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "en": "Use odd prime p=31, original a=310. Give (prime, a, normalized_class, Legendre_symbol, nonzero_roots, certificate)." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "source operation=legendre-zero-class;original snapshot and mathematics retained", "after": "explicit ordered source output fields=[prime, a, normalized_class, Legendre_symbol, nonzero_roots, certificate]", "relation": "implies", "rule_id": "source_ordered_interface_explicit_without_javascript", "conditions": [], "explanation": { "zh-Hant": "保留原題介面快照，本頁直接列來源欄序，不需JavaScript。radius由原mode選6／11欄，不補欄；No與實際否定依來源責任分清。", "en": "Keep the source snapshot while listing ordered fields without JavaScript.Radius uses its original mode to select6 or11 fields without padding;distinguish nonclaims from actual negations." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S002", "uses": [ "G001", "S001" ], "before": "Legendre ORIGINAL odd prime:ORIGINAL modulus=31;ALL possible trial divisors through square root=[2, 3, 4, 5]", "after": "ALL [divisor,remainder] rows=[[2, 1], [3, 1], [4, 3], [5, 1]];prime=是", "relation": "implies", "rule_id": "legendre_complete_prime_trial_not_Euler_classification", "conditions": [], "explanation": { "zh-Hant": "先完整核prime資格，不用Legendre或Euler分類反過來猜prime；這是modulus前提，不是a的平方性。", "en": "Qualify primality over the full trial range,not by guessing from Legendre/Euler classification.This is a modulus premise,not the square status of a." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S003", "uses": [ "G001", "S002" ], "before": "Legendre ORIGINAL nonzero root1 exact square := mul(1, 1)", "after": "Legendre ORIGINAL nonzero root1 exact square = 1", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root1 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root1 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "1", "1" ], "output": "1", "label": "Legendre ORIGINAL nonzero root1 exact square" } }, { "id": "S004", "uses": [ "G001", "S003" ], "before": "ORIGINAL root x=1 in1..30;exact square=1", "after": "square=0*31+1;canonical square class=1;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S005", "uses": [ "G001", "S004" ], "before": "Legendre ORIGINAL nonzero root2 exact square := mul(2, 2)", "after": "Legendre ORIGINAL nonzero root2 exact square = 4", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root2 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root2 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "2", "2" ], "output": "4", "label": "Legendre ORIGINAL nonzero root2 exact square" } }, { "id": "S006", "uses": [ "G001", "S005" ], "before": "ORIGINAL root x=2 in1..30;exact square=4", "after": "square=0*31+4;canonical square class=4;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S007", "uses": [ "G001", "S006" ], "before": "Legendre ORIGINAL nonzero root3 exact square := mul(3, 3)", "after": "Legendre ORIGINAL nonzero root3 exact square = 9", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root3 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root3 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "3", "3" ], "output": "9", "label": "Legendre ORIGINAL nonzero root3 exact square" } }, { "id": "S008", "uses": [ "G001", "S007" ], "before": "ORIGINAL root x=3 in1..30;exact square=9", "after": "square=0*31+9;canonical square class=9;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S009", "uses": [ "G001", "S008" ], "before": "Legendre ORIGINAL nonzero root4 exact square := mul(4, 4)", "after": "Legendre ORIGINAL nonzero root4 exact square = 16", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root4 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root4 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "4", "4" ], "output": "16", "label": "Legendre ORIGINAL nonzero root4 exact square" } }, { "id": "S010", "uses": [ "G001", "S009" ], "before": "ORIGINAL root x=4 in1..30;exact square=16", "after": "square=0*31+16;canonical square class=16;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S011", "uses": [ "G001", "S010" ], "before": "Legendre ORIGINAL nonzero root5 exact square := mul(5, 5)", "after": "Legendre ORIGINAL nonzero root5 exact square = 25", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root5 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root5 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "5", "5" ], "output": "25", "label": "Legendre ORIGINAL nonzero root5 exact square" } }, { "id": "S012", "uses": [ "G001", "S011" ], "before": "ORIGINAL root x=5 in1..30;exact square=25", "after": "square=0*31+25;canonical square class=25;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S013", "uses": [ "G001", "S012" ], "before": "Legendre ORIGINAL nonzero root6 exact square := mul(6, 6)", "after": "Legendre ORIGINAL nonzero root6 exact square = 36", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root6 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root6 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "6", "6" ], "output": "36", "label": "Legendre ORIGINAL nonzero root6 exact square" } }, { "id": "S014", "uses": [ "G001", "S013" ], "before": "ORIGINAL root x=6 in1..30;exact square=36", "after": "square=1*31+5;canonical square class=5;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S015", "uses": [ "G001", "S014" ], "before": "Legendre ORIGINAL nonzero root7 exact square := mul(7, 7)", "after": "Legendre ORIGINAL nonzero root7 exact square = 49", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root7 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root7 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "7", "7" ], "output": "49", "label": "Legendre ORIGINAL nonzero root7 exact square" } }, { "id": "S016", "uses": [ "G001", "S015" ], "before": "ORIGINAL root x=7 in1..30;exact square=49", "after": "square=1*31+18;canonical square class=18;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S017", "uses": [ "G001", "S016" ], "before": "Legendre ORIGINAL nonzero root8 exact square := mul(8, 8)", "after": "Legendre ORIGINAL nonzero root8 exact square = 64", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root8 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root8 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "8", "8" ], "output": "64", "label": "Legendre ORIGINAL nonzero root8 exact square" } }, { "id": "S018", "uses": [ "G001", "S017" ], "before": "ORIGINAL root x=8 in1..30;exact square=64", "after": "square=2*31+2;canonical square class=2;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S019", "uses": [ "G001", "S018" ], "before": "Legendre ORIGINAL nonzero root9 exact square := mul(9, 9)", "after": "Legendre ORIGINAL nonzero root9 exact square = 81", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root9 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root9 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "9", "9" ], "output": "81", "label": "Legendre ORIGINAL nonzero root9 exact square" } }, { "id": "S020", "uses": [ "G001", "S019" ], "before": "ORIGINAL root x=9 in1..30;exact square=81", "after": "square=2*31+19;canonical square class=19;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S021", "uses": [ "G001", "S020" ], "before": "Legendre ORIGINAL nonzero root10 exact square := mul(10, 10)", "after": "Legendre ORIGINAL nonzero root10 exact square = 100", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root10 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root10 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "10", "10" ], "output": "100", "label": "Legendre ORIGINAL nonzero root10 exact square" } }, { "id": "S022", "uses": [ "G001", "S021" ], "before": "ORIGINAL root x=10 in1..30;exact square=100", "after": "square=3*31+7;canonical square class=7;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S023", "uses": [ "G001", "S022" ], "before": "Legendre ORIGINAL nonzero root11 exact square := mul(11, 11)", "after": "Legendre ORIGINAL nonzero root11 exact square = 121", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root11 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root11 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "11", "11" ], "output": "121", "label": "Legendre ORIGINAL nonzero root11 exact square" } }, { "id": "S024", "uses": [ "G001", "S023" ], "before": "ORIGINAL root x=11 in1..30;exact square=121", "after": "square=3*31+28;canonical square class=28;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S025", "uses": [ "G001", "S024" ], "before": "Legendre ORIGINAL nonzero root12 exact square := mul(12, 12)", "after": "Legendre ORIGINAL nonzero root12 exact square = 144", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root12 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root12 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "12", "12" ], "output": "144", "label": "Legendre ORIGINAL nonzero root12 exact square" } }, { "id": "S026", "uses": [ "G001", "S025" ], "before": "ORIGINAL root x=12 in1..30;exact square=144", "after": "square=4*31+20;canonical square class=20;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S027", "uses": [ "G001", "S026" ], "before": "Legendre ORIGINAL nonzero root13 exact square := mul(13, 13)", "after": "Legendre ORIGINAL nonzero root13 exact square = 169", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root13 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root13 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "13", "13" ], "output": "169", "label": "Legendre ORIGINAL nonzero root13 exact square" } }, { "id": "S028", "uses": [ "G001", "S027" ], "before": "ORIGINAL root x=13 in1..30;exact square=169", "after": "square=5*31+14;canonical square class=14;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S029", "uses": [ "G001", "S028" ], "before": "Legendre ORIGINAL nonzero root14 exact square := mul(14, 14)", "after": "Legendre ORIGINAL nonzero root14 exact square = 196", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root14 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root14 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "14", "14" ], "output": "196", "label": "Legendre ORIGINAL nonzero root14 exact square" } }, { "id": "S030", "uses": [ "G001", "S029" ], "before": "ORIGINAL root x=14 in1..30;exact square=196", "after": "square=6*31+10;canonical square class=10;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S031", "uses": [ "G001", "S030" ], "before": "Legendre ORIGINAL nonzero root15 exact square := mul(15, 15)", "after": "Legendre ORIGINAL nonzero root15 exact square = 225", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root15 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root15 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "15", "15" ], "output": "225", "label": "Legendre ORIGINAL nonzero root15 exact square" } }, { "id": "S032", "uses": [ "G001", "S031" ], "before": "ORIGINAL root x=15 in1..30;exact square=225", "after": "square=7*31+8;canonical square class=8;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S033", "uses": [ "G001", "S032" ], "before": "Legendre ORIGINAL nonzero root16 exact square := mul(16, 16)", "after": "Legendre ORIGINAL nonzero root16 exact square = 256", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root16 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root16 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "16", "16" ], "output": "256", "label": "Legendre ORIGINAL nonzero root16 exact square" } }, { "id": "S034", "uses": [ "G001", "S033" ], "before": "ORIGINAL root x=16 in1..30;exact square=256", "after": "square=8*31+8;canonical square class=8;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S035", "uses": [ "G001", "S034" ], "before": "Legendre ORIGINAL nonzero root17 exact square := mul(17, 17)", "after": "Legendre ORIGINAL nonzero root17 exact square = 289", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root17 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root17 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "17", "17" ], "output": "289", "label": "Legendre ORIGINAL nonzero root17 exact square" } }, { "id": "S036", "uses": [ "G001", "S035" ], "before": "ORIGINAL root x=17 in1..30;exact square=289", "after": "square=9*31+10;canonical square class=10;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S037", "uses": [ "G001", "S036" ], "before": "Legendre ORIGINAL nonzero root18 exact square := mul(18, 18)", "after": "Legendre ORIGINAL nonzero root18 exact square = 324", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root18 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root18 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "18", "18" ], "output": "324", "label": "Legendre ORIGINAL nonzero root18 exact square" } }, { "id": "S038", "uses": [ "G001", "S037" ], "before": "ORIGINAL root x=18 in1..30;exact square=324", "after": "square=10*31+14;canonical square class=14;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S039", "uses": [ "G001", "S038" ], "before": "Legendre ORIGINAL nonzero root19 exact square := mul(19, 19)", "after": "Legendre ORIGINAL nonzero root19 exact square = 361", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root19 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root19 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "19", "19" ], "output": "361", "label": "Legendre ORIGINAL nonzero root19 exact square" } }, { "id": "S040", "uses": [ "G001", "S039" ], "before": "ORIGINAL root x=19 in1..30;exact square=361", "after": "square=11*31+20;canonical square class=20;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S041", "uses": [ "G001", "S040" ], "before": "Legendre ORIGINAL nonzero root20 exact square := mul(20, 20)", "after": "Legendre ORIGINAL nonzero root20 exact square = 400", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root20 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root20 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "20", "20" ], "output": "400", "label": "Legendre ORIGINAL nonzero root20 exact square" } }, { "id": "S042", "uses": [ "G001", "S041" ], "before": "ORIGINAL root x=20 in1..30;exact square=400", "after": "square=12*31+28;canonical square class=28;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S043", "uses": [ "G001", "S042" ], "before": "Legendre ORIGINAL nonzero root21 exact square := mul(21, 21)", "after": "Legendre ORIGINAL nonzero root21 exact square = 441", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root21 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root21 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "21", "21" ], "output": "441", "label": "Legendre ORIGINAL nonzero root21 exact square" } }, { "id": "S044", "uses": [ "G001", "S043" ], "before": "ORIGINAL root x=21 in1..30;exact square=441", "after": "square=14*31+7;canonical square class=7;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S045", "uses": [ "G001", "S044" ], "before": "Legendre ORIGINAL nonzero root22 exact square := mul(22, 22)", "after": "Legendre ORIGINAL nonzero root22 exact square = 484", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root22 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root22 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "22", "22" ], "output": "484", "label": "Legendre ORIGINAL nonzero root22 exact square" } }, { "id": "S046", "uses": [ "G001", "S045" ], "before": "ORIGINAL root x=22 in1..30;exact square=484", "after": "square=15*31+19;canonical square class=19;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S047", "uses": [ "G001", "S046" ], "before": "Legendre ORIGINAL nonzero root23 exact square := mul(23, 23)", "after": "Legendre ORIGINAL nonzero root23 exact square = 529", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root23 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root23 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "23", "23" ], "output": "529", "label": "Legendre ORIGINAL nonzero root23 exact square" } }, { "id": "S048", "uses": [ "G001", "S047" ], "before": "ORIGINAL root x=23 in1..30;exact square=529", "after": "square=17*31+2;canonical square class=2;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S049", "uses": [ "G001", "S048" ], "before": "Legendre ORIGINAL nonzero root24 exact square := mul(24, 24)", "after": "Legendre ORIGINAL nonzero root24 exact square = 576", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root24 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root24 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "24", "24" ], "output": "576", "label": "Legendre ORIGINAL nonzero root24 exact square" } }, { "id": "S050", "uses": [ "G001", "S049" ], "before": "ORIGINAL root x=24 in1..30;exact square=576", "after": "square=18*31+18;canonical square class=18;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S051", "uses": [ "G001", "S050" ], "before": "Legendre ORIGINAL nonzero root25 exact square := mul(25, 25)", "after": "Legendre ORIGINAL nonzero root25 exact square = 625", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root25 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root25 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "25", "25" ], "output": "625", "label": "Legendre ORIGINAL nonzero root25 exact square" } }, { "id": "S052", "uses": [ "G001", "S051" ], "before": "ORIGINAL root x=25 in1..30;exact square=625", "after": "square=20*31+5;canonical square class=5;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S053", "uses": [ "G001", "S052" ], "before": "Legendre ORIGINAL nonzero root26 exact square := mul(26, 26)", "after": "Legendre ORIGINAL nonzero root26 exact square = 676", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root26 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root26 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "26", "26" ], "output": "676", "label": "Legendre ORIGINAL nonzero root26 exact square" } }, { "id": "S054", "uses": [ "G001", "S053" ], "before": "ORIGINAL root x=26 in1..30;exact square=676", "after": "square=21*31+25;canonical square class=25;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S055", "uses": [ "G001", "S054" ], "before": "Legendre ORIGINAL nonzero root27 exact square := mul(27, 27)", "after": "Legendre ORIGINAL nonzero root27 exact square = 729", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root27 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root27 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "27", "27" ], "output": "729", "label": "Legendre ORIGINAL nonzero root27 exact square" } }, { "id": "S056", "uses": [ "G001", "S055" ], "before": "ORIGINAL root x=27 in1..30;exact square=729", "after": "square=23*31+16;canonical square class=16;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S057", "uses": [ "G001", "S056" ], "before": "Legendre ORIGINAL nonzero root28 exact square := mul(28, 28)", "after": "Legendre ORIGINAL nonzero root28 exact square = 784", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root28 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root28 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "28", "28" ], "output": "784", "label": "Legendre ORIGINAL nonzero root28 exact square" } }, { "id": "S058", "uses": [ "G001", "S057" ], "before": "ORIGINAL root x=28 in1..30;exact square=784", "after": "square=25*31+9;canonical square class=9;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S059", "uses": [ "G001", "S058" ], "before": "Legendre ORIGINAL nonzero root29 exact square := mul(29, 29)", "after": "Legendre ORIGINAL nonzero root29 exact square = 841", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root29 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root29 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "29", "29" ], "output": "841", "label": "Legendre ORIGINAL nonzero root29 exact square" } }, { "id": "S060", "uses": [ "G001", "S059" ], "before": "ORIGINAL root x=29 in1..30;exact square=841", "after": "square=27*31+4;canonical square class=4;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S061", "uses": [ "G001", "S060" ], "before": "Legendre ORIGINAL nonzero root30 exact square := mul(30, 30)", "after": "Legendre ORIGINAL nonzero root30 exact square = 900", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL nonzero root30 exact square，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL nonzero root30 exact square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "30", "30" ], "output": "900", "label": "Legendre ORIGINAL nonzero root30 exact square" } }, { "id": "S062", "uses": [ "G001", "S061" ], "before": "ORIGINAL root x=30 in1..30;exact square=900", "after": "square=29*31+1;canonical square class=1;ALL root indices retained", "relation": "implies", "rule_id": "legendre_ALL_nonzero_square_rows_before_symbol_definition", "conditions": [], "explanation": { "zh-Hant": "每個非零root均列入，不能由少量平方樣本宣稱另一類無root；不預用Euler criterion。", "en": "Include every nonzero root.A few squares do not prove another class rootless;no Euler criterion is pre-used." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S063", "uses": [ "G001", "S062" ], "before": "ORIGINAL prime=31;COMPLETE nonzero square table=[[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]];nonzero square classes=[1, 2, 4, 5, 7, 8, 9, 10, 14, 16, 18, 19, 20, 25, 28]", "after": "for EACH normalized class r,first r0 ->symbol0;else square-table member ->symbol1;otherwise ->symbol-1;these cases partition ALL0..p-1", "relation": "implies", "rule_id": "legendre_ZERO_then_residue_nonresidue_three_way_definition", "conditions": [], "explanation": { "zh-Hant": "先分離zero類，不能因0也是平方便給+1；非零類再依完整square table分residue／nonresidue。這是定義與完整見證，不靠後續乘法性或Euler公式。", "en": "Separate zero first;the fact that0 is a square does not give symbol+1.Nonzero classes are classified by the complete square table.This is definition plus exhaustive evidence,not a later multiplicativity or Euler formula." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S064", "uses": [ "G001", "S063" ], "before": "Legendre ORIGINAL candidate class quotient times modulus := mul(10, 31)", "after": "Legendre ORIGINAL candidate class quotient times modulus = 310", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL candidate class quotient times modulus，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL candidate class quotient times modulus with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "10", "31" ], "output": "310", "label": "Legendre ORIGINAL candidate class quotient times modulus" } }, { "id": "S065", "uses": [ "G001", "S064" ], "before": "Legendre ORIGINAL candidate class ORIGINAL signed value minus multiple := sub(310, 310)", "after": "Legendre ORIGINAL candidate class ORIGINAL signed value minus multiple = 0", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依原始角色與已證中間量精確代值，不用浮點近似，也不把有限測試冒充全稱極限。", "en": "Substitute original roles and established intermediates exactly,without floats or treating finite tests as universal limits." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Legendre ORIGINAL candidate class ORIGINAL signed value minus multiple，不以小數近似替代。", "en": "Recompute Legendre ORIGINAL candidate class ORIGINAL signed value minus multiple with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sub", "inputs": [ "310", "310" ], "output": "0", "label": "Legendre ORIGINAL candidate class ORIGINAL signed value minus multiple" } }, { "id": "S066", "uses": [ "G001", "S065" ], "before": "Legendre ORIGINAL candidate class:ORIGINAL a=310,positive modulus=31;actual quotient=10", "after": "normalized class=0;a=quotient*modulus+class;ORIGINAL a retained in output where declared", "relation": "implies", "rule_id": "legendre_ORIGINAL_signed_class_Euclidean_normalization", "conditions": [], "explanation": { "zh-Hant": "負a亦保同一congruence class，以0到p−1表示；不把負餘數直接當標準類。", "en": "Retain the same class for negative a and normalize into0..p−1,rather than using a negative language-specific remainder." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null } ], "math_model": { "result": [ 31, 310, 0, 0, [], "p_divides_a_so_Legendre_symbol_is_zero_before_residue_classification" ] }, "conclusion": { "canonical_value": "(31, 310, 0, 0, [], p_divides_a_so_Legendre_symbol_is_zero_before_residue_classification)", "explanation": { "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。", "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged." } }, "definition_note": { "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。", "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate." }, "source": { "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db", "question_binding": { "batch_id": "batch-20261014-advance-20261112-r06-legendre-symbol-foundations", "batch_path": "public/data/questions/batches/batch-20261014-advance-20261112-r06-legendre-symbol-foundations/questions.jsonl", "batch_sha256": "ece4cc3b3fcf3c5b1dd2d6a1e92fe38d880eb9a488ab04606bc610adb75aede4", "original_line": 60, "original_line_sha256": "1fc028c9d880a60a8a7d463a005d70da10f57a34cdc21109023f8cd505329dd6", "question_compact_sha256": "a5dc4943bbbd7522e7b5350b1b9c469a8026a5cc915dc40a7261b25c84cd2a64", "english_row_compact_sha256": "cee11e8488cd7ece971aec12c64bcdd1b7f8912a250717cb17b3bc315cf1f8e2", "english_manifest_path": "public/data/i18n/editions/batch-20261014-advance-20261112-r06-legendre-symbol-foundations.en.json", "english_rows_path": "public/data/i18n/questions/batch-20261014-advance-20261112-r06-legendre-symbol-foundations.en.jsonl" }, "mko_bindings": { "mko-legendre-symbol": { "sha256": "56ce955775c56864a29a398a2adae655ad1f8d53a82d5d9970bc7d0e1a4cebde", "english_sha256": "c00926f8c5fb9b2c966956a02566cfdbfe62191fd6143c23260619d3a0da2cee", "original_path": "public/data/mko/legendre-symbol.json", "english_path": "public/data/i18n/editions/mko-legendre-symbol.en.json" } }, "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff", "selected_part_sha256": "04d5fe00084f382c9e6853d5f11c80b558d9cf857160ceb644fa20b64948d1fe", "original_candidate_sha256": "a256c57f3cdb2ef5ac0ee3223d3217e21f419ee8ff01b5b44d0e85b30fecc4d2", "original_candidate_line_sha256": "a256c57f3cdb2ef5ac0ee3223d3217e21f419ee8ff01b5b44d0e85b30fecc4d2" }, "historical_source_state": { "review_state": { "derivation": "candidate_authored_and_locally_checked", "source_question_state": "candidate", "independent_AI_or_human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN" }, "scope_note": { "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。", "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged." }, "label": "Frozen producer metadata; current per-question review is separate below." }, "revision": { "id": "review-20261026-009-p47", "review_date": "2026-10-26", "owner_authorized_reader_publication": true, "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.", "review_receipts": { "semantic_review_sha256": "a7c0d888ade6d307703c4222c4137c78c0d2448babdcba2fa3446fe6fdd343c1", "question_first_witness_sha256": "c9571f97a37d503ad6a61034cc00108543a05c37ef246d1c4962f55136a3de02" }, "status": "second_context_reviewed", "mathematical_semantics": true, "bilingual_semantics": true, "steps_reviewed": 66, "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction", "exact_question_first_check": { "model_sha256": "5601b3c611fc854b86ff0ee24361d351e9a4b3c5e21187b336dc2b8e431c31a9", "arithmetic_steps": 32 }, "uses_answer_as_premise": false, "human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN", "Canonical_truth_promotion": false, "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged." } }