{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0001","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=1、b=1。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(1, 1, 2)","explanation_zh":"股平方與斜邊平方=(1, 1, 2)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131001,"parameters":{"a":1,"b":1,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"1\",\"1\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0002","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=3、b=4。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(9, 16, 25)","explanation_zh":"股平方與斜邊平方=(9, 16, 25)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131002,"parameters":{"a":3,"b":4,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"9\",\"16\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0003","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=4、b=4。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(16, 16, 32)","explanation_zh":"股平方與斜邊平方=(16, 16, 32)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131003,"parameters":{"a":4,"b":4,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"16\",\"16\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0004","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=5、b=5。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(25, 25, 50)","explanation_zh":"股平方與斜邊平方=(25, 25, 50)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131004,"parameters":{"a":5,"b":5,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"25\",\"25\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0005","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=6、b=6。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(36, 36, 72)","explanation_zh":"股平方與斜邊平方=(36, 36, 72)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131005,"parameters":{"a":6,"b":6,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"36\",\"36\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0006","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=7、b=7。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(49, 49, 98)","explanation_zh":"股平方與斜邊平方=(49, 49, 98)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131006,"parameters":{"a":7,"b":7,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"49\",\"49\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0007","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=8、b=8。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(64, 64, 128)","explanation_zh":"股平方與斜邊平方=(64, 64, 128)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131007,"parameters":{"a":8,"b":8,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"64\",\"64\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0008","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=9、b=2。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(81, 4, 85)","explanation_zh":"股平方與斜邊平方=(81, 4, 85)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131008,"parameters":{"a":9,"b":2,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"81\",\"4\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0009","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=10、b=3。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(100, 9, 109)","explanation_zh":"股平方與斜邊平方=(100, 9, 109)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131009,"parameters":{"a":10,"b":3,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"100\",\"9\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0010","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=11、b=4。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(121, 16, 137)","explanation_zh":"股平方與斜邊平方=(121, 16, 137)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131010,"parameters":{"a":11,"b":4,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"121\",\"16\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0011","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=12、b=5。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(144, 25, 169)","explanation_zh":"股平方與斜邊平方=(144, 25, 169)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131011,"parameters":{"a":12,"b":5,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"144\",\"25\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0012","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=13、b=6。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(169, 36, 205)","explanation_zh":"股平方與斜邊平方=(169, 36, 205)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131012,"parameters":{"a":13,"b":6,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"169\",\"36\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0013","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=14、b=7。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(196, 49, 245)","explanation_zh":"股平方與斜邊平方=(196, 49, 245)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131013,"parameters":{"a":14,"b":7,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"196\",\"49\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0014","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=15、b=8。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(225, 64, 289)","explanation_zh":"股平方與斜邊平方=(225, 64, 289)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131014,"parameters":{"a":15,"b":8,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"225\",\"64\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0015","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=16、b=2。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(256, 4, 260)","explanation_zh":"股平方與斜邊平方=(256, 4, 260)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131015,"parameters":{"a":16,"b":2,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"256\",\"4\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0016","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=17、b=3。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(289, 9, 298)","explanation_zh":"股平方與斜邊平方=(289, 9, 298)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131016,"parameters":{"a":17,"b":3,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"289\",\"9\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0017","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=18、b=4。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(324, 16, 340)","explanation_zh":"股平方與斜邊平方=(324, 16, 340)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131017,"parameters":{"a":18,"b":4,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"324\",\"16\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0018","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=19、b=5。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(361, 25, 386)","explanation_zh":"股平方與斜邊平方=(361, 25, 386)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131018,"parameters":{"a":19,"b":5,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"361\",\"25\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0019","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=20、b=6。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(400, 36, 436)","explanation_zh":"股平方與斜邊平方=(400, 36, 436)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131019,"parameters":{"a":20,"b":6,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"400\",\"36\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0020","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=21、b=7。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(441, 49, 490)","explanation_zh":"股平方與斜邊平方=(441, 49, 490)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131020,"parameters":{"a":21,"b":7,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"441\",\"49\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0021","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=22、b=8。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(484, 64, 548)","explanation_zh":"股平方與斜邊平方=(484, 64, 548)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131021,"parameters":{"a":22,"b":8,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"484\",\"64\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0022","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=23、b=2。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(529, 4, 533)","explanation_zh":"股平方與斜邊平方=(529, 4, 533)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131022,"parameters":{"a":23,"b":2,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"529\",\"4\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0023","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=24、b=3。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(576, 9, 585)","explanation_zh":"股平方與斜邊平方=(576, 9, 585)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131023,"parameters":{"a":24,"b":3,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"576\",\"9\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0024","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=25、b=4。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(625, 16, 641)","explanation_zh":"股平方與斜邊平方=(625, 16, 641)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131024,"parameters":{"a":25,"b":4,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"625\",\"16\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0025","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正股的有序平方與斜邊平方","stem_zh":"歐氏直角邊長模型；正股長，a=26、b=5。依序求 (a², b², a²+b²)，最後欄是模型的斜邊平方，不交換股序。","answer":{"kind":"expression","value":"(676, 25, 701)","explanation_zh":"股平方與斜邊平方=(676, 25, 701)。分別平方兩正股再加總；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正股的有序平方與斜邊平方"]},"variant":{"family_id":"qf-density13-pythagorean-leg-square-sum","seed":202610131025,"parameters":{"a":26,"b":5,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_leg_square_sum","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-leg-square-sum/[\"676\",\"25\"]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0026","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=3、b=4，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"5","explanation_zh":"正整數斜邊=5。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131026,"parameters":{"a":3,"b":4,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"25\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0027","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=6、b=8，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"10","explanation_zh":"正整數斜邊=10。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131027,"parameters":{"a":6,"b":8,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"100\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0028","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=9、b=12，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"15","explanation_zh":"正整數斜邊=15。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131028,"parameters":{"a":9,"b":12,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"225\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0029","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=12、b=16，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"20","explanation_zh":"正整數斜邊=20。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131029,"parameters":{"a":12,"b":16,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"400\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0030","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=15、b=20，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"25","explanation_zh":"正整數斜邊=25。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131030,"parameters":{"a":15,"b":20,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"625\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0031","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=18、b=24，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"30","explanation_zh":"正整數斜邊=30。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131031,"parameters":{"a":18,"b":24,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"900\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0032","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=21、b=28，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"35","explanation_zh":"正整數斜邊=35。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131032,"parameters":{"a":21,"b":28,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"1225\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0033","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=24、b=32，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"40","explanation_zh":"正整數斜邊=40。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131033,"parameters":{"a":24,"b":32,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"1600\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0034","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=27、b=36，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"45","explanation_zh":"正整數斜邊=45。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131034,"parameters":{"a":27,"b":36,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"2025\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0035","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=30、b=40，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"50","explanation_zh":"正整數斜邊=50。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131035,"parameters":{"a":30,"b":40,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"2500\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0036","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=33、b=44，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"55","explanation_zh":"正整數斜邊=55。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131036,"parameters":{"a":33,"b":44,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"3025\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0037","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=36、b=48，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"60","explanation_zh":"正整數斜邊=60。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131037,"parameters":{"a":36,"b":48,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"3600\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0038","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=39、b=52，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"65","explanation_zh":"正整數斜邊=65。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131038,"parameters":{"a":39,"b":52,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"4225\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0039","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=42、b=56，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"70","explanation_zh":"正整數斜邊=70。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131039,"parameters":{"a":42,"b":56,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"4900\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0040","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=45、b=60，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"75","explanation_zh":"正整數斜邊=75。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131040,"parameters":{"a":45,"b":60,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"5625\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0041","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=48、b=64，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"80","explanation_zh":"正整數斜邊=80。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131041,"parameters":{"a":48,"b":64,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"6400\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0042","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=51、b=68，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"85","explanation_zh":"正整數斜邊=85。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131042,"parameters":{"a":51,"b":68,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"7225\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0043","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=54、b=72，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"90","explanation_zh":"正整數斜邊=90。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131043,"parameters":{"a":54,"b":72,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"8100\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0044","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=57、b=76，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"95","explanation_zh":"正整數斜邊=95。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131044,"parameters":{"a":57,"b":76,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"9025\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0045","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=60、b=80，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"100","explanation_zh":"正整數斜邊=100。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131045,"parameters":{"a":60,"b":80,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"10000\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0046","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=63、b=84，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"105","explanation_zh":"正整數斜邊=105。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131046,"parameters":{"a":63,"b":84,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"11025\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0047","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=66、b=88，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"110","explanation_zh":"正整數斜邊=110。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131047,"parameters":{"a":66,"b":88,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"12100\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0048","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=69、b=92，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"115","explanation_zh":"正整數斜邊=115。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131048,"parameters":{"a":69,"b":92,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"13225\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0049","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=72、b=96，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"120","explanation_zh":"正整數斜邊=120。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131049,"parameters":{"a":72,"b":96,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"14400\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0050","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"完全平方的正整數斜邊","stem_zh":"歐氏直角邊長模型；正股長，a=75、b=100，平方和已要求完全平方。求正整數斜邊c，不給負根。","answer":{"kind":"expression","value":"125","explanation_zh":"正整數斜邊=125。用精確整數平方根核回平方和，再取正根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["完全平方的正整數斜邊"]},"variant":{"family_id":"qf-density13-pythagorean-hypotenuse-int","seed":202610131050,"parameters":{"a":75,"b":100,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_hypotenuse_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-hypotenuse-int/\"15625\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0051","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=3、正斜邊c=5且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"4","explanation_zh":"正整數缺股=4。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131051,"parameters":{"a":3,"c":5,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"16\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0052","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=6、正斜邊c=10且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"8","explanation_zh":"正整數缺股=8。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131052,"parameters":{"a":6,"c":10,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"64\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0053","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=9、正斜邊c=15且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"12","explanation_zh":"正整數缺股=12。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131053,"parameters":{"a":9,"c":15,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"144\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0054","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=12、正斜邊c=20且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"16","explanation_zh":"正整數缺股=16。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131054,"parameters":{"a":12,"c":20,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"256\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0055","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=15、正斜邊c=25且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"20","explanation_zh":"正整數缺股=20。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131055,"parameters":{"a":15,"c":25,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"400\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0056","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=18、正斜邊c=30且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"24","explanation_zh":"正整數缺股=24。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131056,"parameters":{"a":18,"c":30,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"576\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0057","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=21、正斜邊c=35且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"28","explanation_zh":"正整數缺股=28。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131057,"parameters":{"a":21,"c":35,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"784\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0058","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=24、正斜邊c=40且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"32","explanation_zh":"正整數缺股=32。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131058,"parameters":{"a":24,"c":40,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"1024\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0059","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=27、正斜邊c=45且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"36","explanation_zh":"正整數缺股=36。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131059,"parameters":{"a":27,"c":45,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"1296\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0060","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=30、正斜邊c=50且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"40","explanation_zh":"正整數缺股=40。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131060,"parameters":{"a":30,"c":50,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"1600\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0061","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=33、正斜邊c=55且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"44","explanation_zh":"正整數缺股=44。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131061,"parameters":{"a":33,"c":55,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"1936\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0062","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=36、正斜邊c=60且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"48","explanation_zh":"正整數缺股=48。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131062,"parameters":{"a":36,"c":60,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"2304\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0063","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=39、正斜邊c=65且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"52","explanation_zh":"正整數缺股=52。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131063,"parameters":{"a":39,"c":65,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"2704\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0064","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=42、正斜邊c=70且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"56","explanation_zh":"正整數缺股=56。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131064,"parameters":{"a":42,"c":70,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"3136\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0065","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=45、正斜邊c=75且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"60","explanation_zh":"正整數缺股=60。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131065,"parameters":{"a":45,"c":75,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"3600\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0066","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=48、正斜邊c=80且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"64","explanation_zh":"正整數缺股=64。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131066,"parameters":{"a":48,"c":80,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"4096\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0067","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=51、正斜邊c=85且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"68","explanation_zh":"正整數缺股=68。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131067,"parameters":{"a":51,"c":85,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"4624\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0068","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=54、正斜邊c=90且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"72","explanation_zh":"正整數缺股=72。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131068,"parameters":{"a":54,"c":90,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"5184\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0069","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=57、正斜邊c=95且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"76","explanation_zh":"正整數缺股=76。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131069,"parameters":{"a":57,"c":95,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"5776\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0070","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=60、正斜邊c=100且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"80","explanation_zh":"正整數缺股=80。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131070,"parameters":{"a":60,"c":100,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"6400\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0071","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=63、正斜邊c=105且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"84","explanation_zh":"正整數缺股=84。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131071,"parameters":{"a":63,"c":105,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"7056\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0072","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=66、正斜邊c=110且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"88","explanation_zh":"正整數缺股=88。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131072,"parameters":{"a":66,"c":110,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"7744\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0073","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=69、正斜邊c=115且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"92","explanation_zh":"正整數缺股=92。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131073,"parameters":{"a":69,"c":115,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"8464\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0074","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=72、正斜邊c=120且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"96","explanation_zh":"正整數缺股=96。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131074,"parameters":{"a":72,"c":120,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"9216\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0075","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"正整數缺股","stem_zh":"歐氏直角邊長模型；正股長，已知a=75、正斜邊c=125且c>a，平方差為正完全平方。求另一正股b。","answer":{"kind":"expression","value":"100","explanation_zh":"正整數缺股=100。先扣已知股平方，再核正整數主根；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["正整數缺股"]},"variant":{"family_id":"qf-density13-pythagorean-missing-leg-int","seed":202610131075,"parameters":{"a":75,"c":125,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_missing_leg_int","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-missing-leg-int/\"10000\""}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0076","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=3、b=4，待核候選c=-5。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(否, 0, 否)","explanation_zh":"斜邊候選資格與殘差=(否, 0, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131076,"parameters":{"a":3,"b":4,"c":-5,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"25\",-5]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0077","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=3、b=4，待核候選c=5。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, 0, 是)","explanation_zh":"斜邊候選資格與殘差=(是, 0, 是)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131077,"parameters":{"a":3,"b":4,"c":5,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"25\",5]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0078","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=3、b=4，待核候選c=0。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(否, -25, 否)","explanation_zh":"斜邊候選資格與殘差=(否, -25, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131078,"parameters":{"a":3,"b":4,"c":0,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"25\",0]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0079","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=3、b=4，待核候選c=4。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, -9, 否)","explanation_zh":"斜邊候選資格與殘差=(是, -9, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131079,"parameters":{"a":3,"b":4,"c":4,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"25\",4]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0080","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=3、b=4，待核候選c=6。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, 11, 否)","explanation_zh":"斜邊候選資格與殘差=(是, 11, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131080,"parameters":{"a":3,"b":4,"c":6,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"25\",6]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0081","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=6、b=7，待核候選c=-8。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(否, -21, 否)","explanation_zh":"斜邊候選資格與殘差=(否, -21, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131081,"parameters":{"a":6,"b":7,"c":-8,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"85\",-8]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0082","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=7、b=8，待核候選c=9。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, -32, 否)","explanation_zh":"斜邊候選資格與殘差=(是, -32, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131082,"parameters":{"a":7,"b":8,"c":9,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"113\",9]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0083","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=8、b=9，待核候選c=-10。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(否, -45, 否)","explanation_zh":"斜邊候選資格與殘差=(否, -45, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":1,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131083,"parameters":{"a":8,"b":9,"c":-10,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"145\",-10]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0084","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=9、b=10，待核候選c=11。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, -60, 否)","explanation_zh":"斜邊候選資格與殘差=(是, -60, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131084,"parameters":{"a":9,"b":10,"c":11,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"181\",11]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0085","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=10、b=11，待核候選c=-12。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(否, -77, 否)","explanation_zh":"斜邊候選資格與殘差=(否, -77, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131085,"parameters":{"a":10,"b":11,"c":-12,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"221\",-12]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0086","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=11、b=12，待核候選c=13。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, -96, 否)","explanation_zh":"斜邊候選資格與殘差=(是, -96, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131086,"parameters":{"a":11,"b":12,"c":13,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"265\",13]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0087","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=12、b=13，待核候選c=-14。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(否, -117, 否)","explanation_zh":"斜邊候選資格與殘差=(否, -117, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131087,"parameters":{"a":12,"b":13,"c":-14,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"313\",-14]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0088","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=13、b=14，待核候選c=15。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, -140, 否)","explanation_zh":"斜邊候選資格與殘差=(是, -140, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131088,"parameters":{"a":13,"b":14,"c":15,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"365\",15]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0089","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=14、b=15，待核候選c=-16。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(否, -165, 否)","explanation_zh":"斜邊候選資格與殘差=(否, -165, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131089,"parameters":{"a":14,"b":15,"c":-16,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"421\",-16]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0090","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=15、b=16，待核候選c=17。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, -192, 否)","explanation_zh":"斜邊候選資格與殘差=(是, -192, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131090,"parameters":{"a":15,"b":16,"c":17,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"481\",17]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0091","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=16、b=17，待核候選c=-18。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(否, -221, 否)","explanation_zh":"斜邊候選資格與殘差=(否, -221, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131091,"parameters":{"a":16,"b":17,"c":-18,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"545\",-18]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0092","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=17、b=18，待核候選c=19。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, -252, 否)","explanation_zh":"斜邊候選資格與殘差=(是, -252, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":2,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131092,"parameters":{"a":17,"b":18,"c":19,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"613\",19]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0093","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=18、b=19，待核候選c=-20。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(否, -285, 否)","explanation_zh":"斜邊候選資格與殘差=(否, -285, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131093,"parameters":{"a":18,"b":19,"c":-20,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"685\",-20]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0094","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=19、b=20，待核候選c=21。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, -320, 否)","explanation_zh":"斜邊候選資格與殘差=(是, -320, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131094,"parameters":{"a":19,"b":20,"c":21,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"761\",21]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0095","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=20、b=21，待核候選c=-22。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(否, -357, 否)","explanation_zh":"斜邊候選資格與殘差=(否, -357, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131095,"parameters":{"a":20,"b":21,"c":-22,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"841\",-22]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0096","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=21、b=22，待核候選c=23。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, -396, 否)","explanation_zh":"斜邊候選資格與殘差=(是, -396, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131096,"parameters":{"a":21,"b":22,"c":23,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"925\",23]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0097","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=22、b=23，待核候選c=-24。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(否, -437, 否)","explanation_zh":"斜邊候選資格與殘差=(否, -437, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131097,"parameters":{"a":22,"b":23,"c":-24,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"1013\",-24]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0098","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=23、b=24，待核候選c=25。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, -480, 否)","explanation_zh":"斜邊候選資格與殘差=(是, -480, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131098,"parameters":{"a":23,"b":24,"c":25,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"1105\",25]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0099","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=24、b=25，待核候選c=-26。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(否, -525, 否)","explanation_zh":"斜邊候選資格與殘差=(否, -525, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131099,"parameters":{"a":24,"b":25,"c":-26,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"1201\",-26]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}
{"schema_version":"ocme-question-v0.1","id":"q-20261009-advance-20261013-r01-pythagorean-theorem-0100","batch_id":"batch-20261009-advance-20261013-r01-pythagorean-theorem-foundations","language":"zh-Hant","source_mko_ids":["mko-euclid-pythagorean-theorem"],"title_zh":"候選斜邊的正性與平方殘差","stem_zh":"歐氏直角邊長模型；正股長，a=25、b=26，待核候選c=27。依序答 (c是否正, c²−a²−b², 正且殘差零是否)，c只作候選。","answer":{"kind":"expression","value":"(是, -572, 否)","explanation_zh":"斜邊候選資格與殘差=(是, -572, 否)。正性與平方殘差分開核對，候選非正斜邊不合資格；有限邊長檢查非新普遍證明，未完成任意仿射橋接。"},"difficulty":{"band":"upper_secondary","level":3,"skills":["候選斜邊的正性與平方殘差"]},"variant":{"family_id":"qf-density13-pythagorean-candidate-residual","seed":202610131100,"parameters":{"a":25,"b":26,"c":27,"geometry":"euclidean_right_side_model","operation":"density13_pythagorean_candidate_residual","domain":"positive_integer_right_side_protocol","semantic_template":"qf-density13-pythagorean-candidate-residual/[\"1301\",27]"}},"verification":{"status":"candidate","checks":["producer_exact_integer_rational_arithmetic","producer_finite_enumeration","producer_source_snapshot_check","producer_substantive_template_deduplication","producer_json_schema"],"warning_zh":"僅為產生端候選及自檢；主端須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"},"provenance":{"generator_role":"question_producer","generator_model":"model_unknown","generated_at":"2026-10-09","contract_version":"ocme-question-factory-v0.1"}}