{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261014-advance-20261111-r06-quadratic-residue-0053", "pilot_number": "R0045-23353", "number": 23353, "family_id": "qf-mod42-quadratic-paired-roots", "title": { "zh-Hant": "第 23353 題：二次剩餘", "en": "Question 23353: A Nonzero Residue Modulo an Odd Prime Has Two Distinct Opposite Roots" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261014-advance-20261111-r06-quadratic-residue-0053", "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-paired-roots；資料={\"p\":7,\"a\":2}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "answer": { "kind": "expression", "value": "(7, 2, [3, 4], 4, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)", "explanation_zh": "Wave6第五層核對=(7, 2, [3, 4], 4, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；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": 1, "skills": [ "二次剩餘" ] }, "variant": { "family_id": "qf-mod42-quadratic-paired-roots", "seed": 202611116053, "parameters": { "p": 7, "a": 2, "domain": "Odd_prime_nonzero_quadratic_residue", "operation": "quadratic-paired-roots", "semantic_template": "qf-mod42-quadratic-paired-roots/{\"a\":2,\"p\":7}" } }, "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-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." } }, "givens": [ { "id": "G001", "text": { "zh-Hant": "odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={\"p\":7,\"a\":2}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。", "en": "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)." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "source operation=quadratic-paired-roots;original snapshot and mathematics retained", "after": "explicit ordered source output fields=[p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate]", "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 modulus:ORIGINAL integer=7;complete possible trial divisors=[2]", "after": "ALL [divisor,remainder] rows=[[2, 1]];prime=是;a composite>1 has a factor at most its square root", "relation": "implies", "rule_id": "finite_prime_ALL_trial_divisors_through_square_root", "conditions": [], "explanation": { "zh-Hant": "逐完整平方根範圍核整除；prime2的範圍為空而通過，0/1不通過。不能由少數取樣、平方表或Euler判準代替此資格。", "en": "Check the entire square-root divisor range.The empty range qualifies prime2,but not0 or1.Samples,square tables or an Euler criterion do not replace primality qualification." }, "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": "Quadratic ORIGINAL nonzero root1 square := mul(1, 1)", "after": "Quadratic ORIGINAL nonzero root1 square = 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 nonzero root1 square，不以小數近似替代。", "en": "Recompute Quadratic ORIGINAL nonzero root1 square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "1", "1" ], "output": "1", "label": "Quadratic ORIGINAL nonzero root1 square" } }, { "id": "S004", "uses": [ "G001", "S003" ], "before": "ORIGINAL nonzero root x=1 in1..6;actual square=1", "after": "Euclidean square=0*7+1;nonzero square class=1;complete root-index order retained", "relation": "implies", "rule_id": "quadratic_ALL_nonzero_square_rows_not_sample", "conditions": [], "explanation": { "zh-Hant": "逐每個非零root，列真平方與mod分解；prime下非零x不會平方成0。不是只抽幾個residues或預用Euler／Legendre。", "en": "Enumerate every nonzero root with its exact square and modular decomposition.A nonzero x cannot square to0 modulo a prime.This is neither a residue sample nor an Euler/Legendre shortcut." }, "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": "Quadratic ORIGINAL nonzero root2 square := mul(2, 2)", "after": "Quadratic ORIGINAL nonzero root2 square = 4", "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 nonzero root2 square，不以小數近似替代。", "en": "Recompute Quadratic ORIGINAL nonzero root2 square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "2", "2" ], "output": "4", "label": "Quadratic ORIGINAL nonzero root2 square" } }, { "id": "S006", "uses": [ "G001", "S005" ], "before": "ORIGINAL nonzero root x=2 in1..6;actual square=4", "after": "Euclidean square=0*7+4;nonzero square class=4;complete root-index order retained", "relation": "implies", "rule_id": "quadratic_ALL_nonzero_square_rows_not_sample", "conditions": [], "explanation": { "zh-Hant": "逐每個非零root，列真平方與mod分解；prime下非零x不會平方成0。不是只抽幾個residues或預用Euler／Legendre。", "en": "Enumerate every nonzero root with its exact square and modular decomposition.A nonzero x cannot square to0 modulo a prime.This is neither a residue sample nor an Euler/Legendre shortcut." }, "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": "Quadratic ORIGINAL nonzero root3 square := mul(3, 3)", "after": "Quadratic ORIGINAL nonzero root3 square = 9", "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 nonzero root3 square，不以小數近似替代。", "en": "Recompute Quadratic ORIGINAL nonzero root3 square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "3", "3" ], "output": "9", "label": "Quadratic ORIGINAL nonzero root3 square" } }, { "id": "S008", "uses": [ "G001", "S007" ], "before": "ORIGINAL nonzero root x=3 in1..6;actual square=9", "after": "Euclidean square=1*7+2;nonzero square class=2;complete root-index order retained", "relation": "implies", "rule_id": "quadratic_ALL_nonzero_square_rows_not_sample", "conditions": [], "explanation": { "zh-Hant": "逐每個非零root，列真平方與mod分解；prime下非零x不會平方成0。不是只抽幾個residues或預用Euler／Legendre。", "en": "Enumerate every nonzero root with its exact square and modular decomposition.A nonzero x cannot square to0 modulo a prime.This is neither a residue sample nor an Euler/Legendre shortcut." }, "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": "Quadratic ORIGINAL nonzero root4 square := mul(4, 4)", "after": "Quadratic ORIGINAL nonzero root4 square = 16", "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 nonzero root4 square，不以小數近似替代。", "en": "Recompute Quadratic ORIGINAL nonzero root4 square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "4", "4" ], "output": "16", "label": "Quadratic ORIGINAL nonzero root4 square" } }, { "id": "S010", "uses": [ "G001", "S009" ], "before": "ORIGINAL nonzero root x=4 in1..6;actual square=16", "after": "Euclidean square=2*7+2;nonzero square class=2;complete root-index order retained", "relation": "implies", "rule_id": "quadratic_ALL_nonzero_square_rows_not_sample", "conditions": [], "explanation": { "zh-Hant": "逐每個非零root，列真平方與mod分解；prime下非零x不會平方成0。不是只抽幾個residues或預用Euler／Legendre。", "en": "Enumerate every nonzero root with its exact square and modular decomposition.A nonzero x cannot square to0 modulo a prime.This is neither a residue sample nor an Euler/Legendre shortcut." }, "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": "Quadratic ORIGINAL nonzero root5 square := mul(5, 5)", "after": "Quadratic ORIGINAL nonzero root5 square = 25", "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 nonzero root5 square，不以小數近似替代。", "en": "Recompute Quadratic ORIGINAL nonzero root5 square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "5", "5" ], "output": "25", "label": "Quadratic ORIGINAL nonzero root5 square" } }, { "id": "S012", "uses": [ "G001", "S011" ], "before": "ORIGINAL nonzero root x=5 in1..6;actual square=25", "after": "Euclidean square=3*7+4;nonzero square class=4;complete root-index order retained", "relation": "implies", "rule_id": "quadratic_ALL_nonzero_square_rows_not_sample", "conditions": [], "explanation": { "zh-Hant": "逐每個非零root，列真平方與mod分解；prime下非零x不會平方成0。不是只抽幾個residues或預用Euler／Legendre。", "en": "Enumerate every nonzero root with its exact square and modular decomposition.A nonzero x cannot square to0 modulo a prime.This is neither a residue sample nor an Euler/Legendre shortcut." }, "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": "Quadratic ORIGINAL nonzero root6 square := mul(6, 6)", "after": "Quadratic ORIGINAL nonzero root6 square = 36", "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 nonzero root6 square，不以小數近似替代。", "en": "Recompute Quadratic ORIGINAL nonzero root6 square with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "6", "6" ], "output": "36", "label": "Quadratic ORIGINAL nonzero root6 square" } }, { "id": "S014", "uses": [ "G001", "S013" ], "before": "ORIGINAL nonzero root x=6 in1..6;actual square=36", "after": "Euclidean square=5*7+1;nonzero square class=1;complete root-index order retained", "relation": "implies", "rule_id": "quadratic_ALL_nonzero_square_rows_not_sample", "conditions": [], "explanation": { "zh-Hant": "逐每個非零root，列真平方與mod分解；prime下非零x不會平方成0。不是只抽幾個residues或預用Euler／Legendre。", "en": "Enumerate every nonzero root with its exact square and modular decomposition.A nonzero x cannot square to0 modulo a prime.This is neither a residue sample nor an Euler/Legendre shortcut." }, "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": "Quadratic distinct nonzero residue count := sum(1, 1, 1)", "after": "Quadratic distinct nonzero residue count = 3", "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 distinct nonzero residue count，不以小數近似替代。", "en": "Recompute Quadratic distinct nonzero residue count with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sum", "inputs": [ "1", "1", "1" ], "output": "3", "label": "Quadratic distinct nonzero residue count" } }, { "id": "S016", "uses": [ "G001", "S015" ], "before": "Quadratic ORIGINAL modulus minus1 := sub(7, 1)", "after": "Quadratic ORIGINAL modulus minus1 = 6", "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 modulus minus1，不以小數近似替代。", "en": "Recompute Quadratic ORIGINAL modulus minus1 with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sub", "inputs": [ "7", "1" ], "output": "6", "label": "Quadratic ORIGINAL modulus minus1" } }, { "id": "S017", "uses": [ "G001", "S016" ], "before": "Quadratic half of all nonzero roots := div(6, 2)", "after": "Quadratic half of all nonzero roots = 3", "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 half of all nonzero roots，不以小數近似替代。", "en": "Recompute Quadratic half of all nonzero roots with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "div", "inputs": [ "6", "2" ], "output": "3", "label": "Quadratic half of all nonzero roots" } }, { "id": "S018", "uses": [ "G001", "S017" ], "before": "ORIGINAL ODD PRIME p=7;complete nonzero square table=[[1, 1], [2, 4], [3, 2], [4, 2], [5, 4], [6, 1]];distinct classes=[1, 2, 4]", "after": "x^2=y^2 implies p divides(x-y)(x+y),so x=y or x=-y modp;odd p and nonzero x make opposite roots DISTINCT;each image has2 roots,so residue count=3=(7-1)/2=3", "relation": "implies", "rule_id": "quadratic_prime_field_opposite_root_pairs_exactly_two", "conditions": [], "explanation": { "zh-Hant": "prime給無零因子的乘積消去，odd保2x不等0；±配對互不重疊覆蓋全部p−1非零root。這證計數與兩根性，不外推到zero、p2或composite。", "en": "The prime-field no-zero-divisor property gives x=±y;oddness makes2x nonzero.The opposite pairs partition all p−1 nonzero roots.This proves both counting and two-root statements,not their extension to zero,2 or composite moduli." }, "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": "Quadratic ORIGINAL candidate class Euclidean quotient times modulus := mul(0, 7)", "after": "Quadratic ORIGINAL candidate class Euclidean quotient times modulus = 0", "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 candidate class Euclidean quotient times modulus，不以小數近似替代。", "en": "Recompute Quadratic ORIGINAL candidate class Euclidean quotient times modulus with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "0", "7" ], "output": "0", "label": "Quadratic ORIGINAL candidate class Euclidean quotient times modulus" } }, { "id": "S020", "uses": [ "G001", "S019" ], "before": "Quadratic ORIGINAL candidate class ORIGINAL signed value minus multiple := sub(2, 0)", "after": "Quadratic ORIGINAL candidate class ORIGINAL signed value minus multiple = 2", "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 candidate class ORIGINAL signed value minus multiple，不以小數近似替代。", "en": "Recompute Quadratic ORIGINAL candidate class ORIGINAL signed value minus multiple with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sub", "inputs": [ "2", "0" ], "output": "2", "label": "Quadratic ORIGINAL candidate class ORIGINAL signed value minus multiple" } }, { "id": "S021", "uses": [ "G001", "S020" ], "before": "Quadratic ORIGINAL candidate class:ORIGINAL signed value=2;positive modulus=7;actual quotient=0", "after": "canonical residue=2;value=quotient*modulus+residue and0<=residue