{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261014-advance-20261111-r06-quadratic-residue-0086", "pilot_number": "R0045-23311", "number": 23311, "family_id": "qf-mod42-quadratic-domain-boundary", "title": { "zh-Hant": "第 23311 題：二次剩餘", "en": "Question 23311: Zero, Modulus Two and Composite Moduli Need Separate Conventions" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261014-advance-20261111-r06-quadratic-residue-0086", "batch_id": "batch-20261014-advance-20261111-r06-quadratic-residue-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-quadratic-residue" ], "title_zh": "二次剩餘", "stem_zh": "odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={\"mode\":\"even_prime\",\"modulus\":2,\"a\":11}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "answer": { "kind": "expression", "value": "(even_prime, 2, 1, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)", "explanation_zh": "Wave6第五層核對=(even_prime, 2, 1, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。" }, "difficulty": { "band": "undergraduate_core", "level": 2, "skills": [ "二次剩餘" ] }, "variant": { "family_id": "qf-mod42-quadratic-domain-boundary", "seed": 202611116086, "parameters": { "mode": "even_prime", "modulus": 2, "a": 11, "domain": "Odd_prime_nonzero_quadratic_residue", "operation": "quadratic-domain-boundary", "semantic_template": "qf-mod42-quadratic-domain-boundary/{\"a\":11,\"mode\":\"even_prime\",\"modulus\":2}" } }, "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、Lagrange converse existence、higher-order analyticity、sampled Hausdorff、CDF density、unchecked Catalan division、Euler criterion、bounded NFA rejection、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-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." } }, "givens": [ { "id": "G001", "text": { "zh-Hant": "odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={\"mode\":\"even_prime\",\"modulus\":2,\"a\":11}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "en": "Use mode=even_prime, modulus=2, original a=11. Give (mode, modulus, normalized_class, declared_model_qualified, boundary)." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "source operation=quadratic-domain-boundary;original snapshot and mathematics retained", "after": "explicit ordered source output fields=[mode, modulus, normalized_class, declared_model_qualified, boundary]", "relation": "implies", "rule_id": "source_ordered_interface_explicit_without_javascript", "conditions": [], "explanation": { "zh-Hant": "原題快照保留歷史介面；此處直接列來源欄序，不依賴JavaScript。tag與No保聲稱範圍，固定來源排序不冒充數學證明。", "en": "Retain the historical interface marker while listing source fields without JavaScript.Tags and No retain their claim scope;source-protocol ordering is not a mathematical proof." }, "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": "Quadratic ORIGINAL boundary class Euclidean quotient times modulus := mul(5, 2)", "after": "Quadratic ORIGINAL boundary class Euclidean quotient times modulus = 10", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依本題原始角色、次序與已核中間量精確代值，不用浮點近似或另一題的資料。", "en": "Substitute this question's original roles,order and established intermediates exactly,not floats or another question's data." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Quadratic ORIGINAL boundary class Euclidean quotient times modulus，不以小數近似替代。", "en": "Recompute Quadratic ORIGINAL boundary class Euclidean quotient times modulus with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "5", "2" ], "output": "10", "label": "Quadratic ORIGINAL boundary class Euclidean quotient times modulus" } }, { "id": "S003", "uses": [ "G001", "S002" ], "before": "Quadratic ORIGINAL boundary class ORIGINAL signed value minus multiple := sub(11, 10)", "after": "Quadratic ORIGINAL boundary class ORIGINAL signed value minus multiple = 1", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "依本題原始角色、次序與已核中間量精確代值，不用浮點近似或另一題的資料。", "en": "Substitute this question's original roles,order and established intermediates exactly,not floats or another question's data." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 Quadratic ORIGINAL boundary class ORIGINAL signed value minus multiple，不以小數近似替代。", "en": "Recompute Quadratic ORIGINAL boundary class ORIGINAL signed value minus multiple with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sub", "inputs": [ "11", "10" ], "output": "1", "label": "Quadratic ORIGINAL boundary class ORIGINAL signed value minus multiple" } }, { "id": "S004", "uses": [ "G001", "S003" ], "before": "Quadratic ORIGINAL boundary class:ORIGINAL signed value=11;positive modulus=2;actual quotient=5", "after": "canonical residue=1;value=quotient*modulus+residue and0<=residue