# 二次剩餘基礎 · 脈衝 R06

本批已通過 Schema、去重、來源快照與獨立答案重算；仍未標記為人工審定。

本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留；AI 覆核、有限計算與 Lean 形式證明分開標示。

batch-20261014-advance-20261111-r06-quadratic-residue-foundations · 2026-10-14 · 100 題

欄序為來源綁定的呈現補充；原始 JSONL 與數學答案不變。

[欄序與來源綁定資料](https://ocme.evemisslab.com/data/i18n/question-output-protocols.json)



## 1. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":3}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(3, [[1, 1], [2, 1]], [1], 1, 1, 是)

Wave6第五層核對=(3, [[1, 1], [2, 1]], [1], 1, 1, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0001/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 2. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":5}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(5, [[1, 1], [2, 4], [3, 4], [4, 1]], [1, 4], 2, 2, 是)

Wave6第五層核對=(5, [[1, 1], [2, 4], [3, 4], [4, 1]], [1, 4], 2, 2, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0002/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 3. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":7}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(7, [[1, 1], [2, 4], [3, 2], [4, 2], [5, 4], [6, 1]], [1, 2, 4], 3, 3, 是)

Wave6第五層核對=(7, [[1, 1], [2, 4], [3, 2], [4, 2], [5, 4], [6, 1]], [1, 2, 4], 3, 3, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0003/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 4. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":11}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(11, [[1, 1], [2, 4], [3, 9], [4, 5], [5, 3], [6, 3], [7, 5], [8, 9], [9, 4], [10, 1]], [1, 3, 4, 5, 9], 5, 5, 是)

Wave6第五層核對=(11, [[1, 1], [2, 4], [3, 9], [4, 5], [5, 3], [6, 3], [7, 5], [8, 9], [9, 4], [10, 1]], [1, 3, 4, 5, 9], 5, 5, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0004/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 5. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":13}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(13, [[1, 1], [2, 4], [3, 9], [4, 3], [5, 12], [6, 10], [7, 10], [8, 12], [9, 3], [10, 9], [11, 4], [12, 1]], [1, 3, 4, 9, 10, 12], 6, 6, 是)

Wave6第五層核對=(13, [[1, 1], [2, 4], [3, 9], [4, 3], [5, 12], [6, 10], [7, 10], [8, 12], [9, 3], [10, 9], [11, 4], [12, 1]], [1, 3, 4, 9, 10, 12], 6, 6, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0005/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 6. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":17}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(17, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 8], [6, 2], [7, 15], [8, 13], [9, 13], [10, 15], [11, 2], [12, 8], [13, 16], [14, 9], [15, 4], [16, 1]], [1, 2, 4, 8, 9, 13, 15, 16], 8, 8, 是)

Wave6第五層核對=(17, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 8], [6, 2], [7, 15], [8, 13], [9, 13], [10, 15], [11, 2], [12, 8], [13, 16], [14, 9], [15, 4], [16, 1]], [1, 2, 4, 8, 9, 13, 15, 16], 8, 8, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0006/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 7. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":19}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(19, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 6], [6, 17], [7, 11], [8, 7], [9, 5], [10, 5], [11, 7], [12, 11], [13, 17], [14, 6], [15, 16], [16, 9], [17, 4], [18, 1]], [1, 4, 5, 6, 7, 9, 11, 16, 17], 9, 9, 是)

Wave6第五層核對=(19, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 6], [6, 17], [7, 11], [8, 7], [9, 5], [10, 5], [11, 7], [12, 11], [13, 17], [14, 6], [15, 16], [16, 9], [17, 4], [18, 1]], [1, 4, 5, 6, 7, 9, 11, 16, 17], 9, 9, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0007/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 8. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":23}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(23, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 2], [6, 13], [7, 3], [8, 18], [9, 12], [10, 8], [11, 6], [12, 6], [13, 8], [14, 12], [15, 18], [16, 3], [17, 13], [18, 2], [19, 16], [20, 9], [21, 4], [22, 1]], [1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18], 11, 11, 是)

Wave6第五層核對=(23, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 2], [6, 13], [7, 3], [8, 18], [9, 12], [10, 8], [11, 6], [12, 6], [13, 8], [14, 12], [15, 18], [16, 3], [17, 13], [18, 2], [19, 16], [20, 9], [21, 4], [22, 1]], [1, 2, 3, 4, 6, 8, 9, 12, 13, 16, 18], 11, 11, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0008/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 9. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":29}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(29, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 7], [7, 20], [8, 6], [9, 23], [10, 13], [11, 5], [12, 28], [13, 24], [14, 22], [15, 22], [16, 24], [17, 28], [18, 5], [19, 13], [20, 23], [21, 6], [22, 20], [23, 7], [24, 25], [25, 16], [26, 9], [27, 4], [28, 1]], [1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28], 14, 14, 是)

Wave6第五層核對=(29, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 7], [7, 20], [8, 6], [9, 23], [10, 13], [11, 5], [12, 28], [13, 24], [14, 22], [15, 22], [16, 24], [17, 28], [18, 5], [19, 13], [20, 23], [21, 6], [22, 20], [23, 7], [24, 25], [25, 16], [26, 9], [27, 4], [28, 1]], [1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28], 14, 14, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0009/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 10. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":31}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(31, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 5], [7, 18], [8, 2], [9, 19], [10, 7], [11, 28], [12, 20], [13, 14], [14, 10], [15, 8], [16, 8], [17, 10], [18, 14], [19, 20], [20, 28], [21, 7], [22, 19], [23, 2], [24, 18], [25, 5], [26, 25], [27, 16], [28, 9], [29, 4], [30, 1]], [1, 2, 4, 5, 7, 8, 9, 10, 14, 16, 18, 19, 20, 25, 28], 15, 15, 是)

Wave6第五層核對=(31, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 5], [7, 18], [8, 2], [9, 19], [10, 7], [11, 28], [12, 20], [13, 14], [14, 10], [15, 8], [16, 8], [17, 10], [18, 14], [19, 20], [20, 28], [21, 7], [22, 19], [23, 2], [24, 18], [25, 5], [26, 25], [27, 16], [28, 9], [29, 4], [30, 1]], [1, 2, 4, 5, 7, 8, 9, 10, 14, 16, 18, 19, 20, 25, 28], 15, 15, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0010/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 11. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":37}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(37, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 12], [8, 27], [9, 7], [10, 26], [11, 10], [12, 33], [13, 21], [14, 11], [15, 3], [16, 34], [17, 30], [18, 28], [19, 28], [20, 30], [21, 34], [22, 3], [23, 11], [24, 21], [25, 33], [26, 10], [27, 26], [28, 7], [29, 27], [30, 12], [31, 36], [32, 25], [33, 16], [34, 9], [35, 4], [36, 1]], [1, 3, 4, 7, 9, 10, 11, 12, 16, 21, 25, 26, 27, 28, 30, 33, 34, 36], 18, 18, 是)

Wave6第五層核對=(37, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 12], [8, 27], [9, 7], [10, 26], [11, 10], [12, 33], [13, 21], [14, 11], [15, 3], [16, 34], [17, 30], [18, 28], [19, 28], [20, 30], [21, 34], [22, 3], [23, 11], [24, 21], [25, 33], [26, 10], [27, 26], [28, 7], [29, 27], [30, 12], [31, 36], [32, 25], [33, 16], [34, 9], [35, 4], [36, 1]], [1, 3, 4, 7, 9, 10, 11, 12, 16, 21, 25, 26, 27, 28, 30, 33, 34, 36], 18, 18, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0011/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 12. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":41}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(41, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 8], [8, 23], [9, 40], [10, 18], [11, 39], [12, 21], [13, 5], [14, 32], [15, 20], [16, 10], [17, 2], [18, 37], [19, 33], [20, 31], [21, 31], [22, 33], [23, 37], [24, 2], [25, 10], [26, 20], [27, 32], [28, 5], [29, 21], [30, 39], [31, 18], [32, 40], [33, 23], [34, 8], [35, 36], [36, 25], [37, 16], [38, 9], [39, 4], [40, 1]], [1, 2, 4, 5, 8, 9, 10, 16, 18, 20, 21, 23, 25, 31, 32, 33, 36, 37, 39, 40], 20, 20, 是)

Wave6第五層核對=(41, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 8], [8, 23], [9, 40], [10, 18], [11, 39], [12, 21], [13, 5], [14, 32], [15, 20], [16, 10], [17, 2], [18, 37], [19, 33], [20, 31], [21, 31], [22, 33], [23, 37], [24, 2], [25, 10], [26, 20], [27, 32], [28, 5], [29, 21], [30, 39], [31, 18], [32, 40], [33, 23], [34, 8], [35, 36], [36, 25], [37, 16], [38, 9], [39, 4], [40, 1]], [1, 2, 4, 5, 8, 9, 10, 16, 18, 20, 21, 23, 25, 31, 32, 33, 36, 37, 39, 40], 20, 20, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0012/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 13. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":43}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(43, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 6], [8, 21], [9, 38], [10, 14], [11, 35], [12, 15], [13, 40], [14, 24], [15, 10], [16, 41], [17, 31], [18, 23], [19, 17], [20, 13], [21, 11], [22, 11], [23, 13], [24, 17], [25, 23], [26, 31], [27, 41], [28, 10], [29, 24], [30, 40], [31, 15], [32, 35], [33, 14], [34, 38], [35, 21], [36, 6], [37, 36], [38, 25], [39, 16], [40, 9], [41, 4], [42, 1]], [1, 4, 6, 9, 10, 11, 13, 14, 15, 16, 17, 21, 23, 24, 25, 31, 35, 36, 38, 40, 41], 21, 21, 是)

Wave6第五層核對=(43, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 6], [8, 21], [9, 38], [10, 14], [11, 35], [12, 15], [13, 40], [14, 24], [15, 10], [16, 41], [17, 31], [18, 23], [19, 17], [20, 13], [21, 11], [22, 11], [23, 13], [24, 17], [25, 23], [26, 31], [27, 41], [28, 10], [29, 24], [30, 40], [31, 15], [32, 35], [33, 14], [34, 38], [35, 21], [36, 6], [37, 36], [38, 25], [39, 16], [40, 9], [41, 4], [42, 1]], [1, 4, 6, 9, 10, 11, 13, 14, 15, 16, 17, 21, 23, 24, 25, 31, 35, 36, 38, 40, 41], 21, 21, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0013/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 14. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":47}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(47, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 2], [8, 17], [9, 34], [10, 6], [11, 27], [12, 3], [13, 28], [14, 8], [15, 37], [16, 21], [17, 7], [18, 42], [19, 32], [20, 24], [21, 18], [22, 14], [23, 12], [24, 12], [25, 14], [26, 18], [27, 24], [28, 32], [29, 42], [30, 7], [31, 21], [32, 37], [33, 8], [34, 28], [35, 3], [36, 27], [37, 6], [38, 34], [39, 17], [40, 2], [41, 36], [42, 25], [43, 16], [44, 9], [45, 4], [46, 1]], [1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 17, 18, 21, 24, 25, 27, 28, 32, 34, 36, 37, 42], 23, 23, 是)

Wave6第五層核對=(47, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 2], [8, 17], [9, 34], [10, 6], [11, 27], [12, 3], [13, 28], [14, 8], [15, 37], [16, 21], [17, 7], [18, 42], [19, 32], [20, 24], [21, 18], [22, 14], [23, 12], [24, 12], [25, 14], [26, 18], [27, 24], [28, 32], [29, 42], [30, 7], [31, 21], [32, 37], [33, 8], [34, 28], [35, 3], [36, 27], [37, 6], [38, 34], [39, 17], [40, 2], [41, 36], [42, 25], [43, 16], [44, 9], [45, 4], [46, 1]], [1, 2, 3, 4, 6, 7, 8, 9, 12, 14, 16, 17, 18, 21, 24, 25, 27, 28, 32, 34, 36, 37, 42], 23, 23, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0014/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 15. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":53}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(53, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 11], [9, 28], [10, 47], [11, 15], [12, 38], [13, 10], [14, 37], [15, 13], [16, 44], [17, 24], [18, 6], [19, 43], [20, 29], [21, 17], [22, 7], [23, 52], [24, 46], [25, 42], [26, 40], [27, 40], [28, 42], [29, 46], [30, 52], [31, 7], [32, 17], [33, 29], [34, 43], [35, 6], [36, 24], [37, 44], [38, 13], [39, 37], [40, 10], [41, 38], [42, 15], [43, 47], [44, 28], [45, 11], [46, 49], [47, 36], [48, 25], [49, 16], [50, 9], [51, 4], [52, 1]], [1, 4, 6, 7, 9, 10, 11, 13, 15, 16, 17, 24, 25, 28, 29, 36, 37, 38, 40, 42, 43, 44, 46, 47, 49, 52], 26, 26, 是)

Wave6第五層核對=(53, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 11], [9, 28], [10, 47], [11, 15], [12, 38], [13, 10], [14, 37], [15, 13], [16, 44], [17, 24], [18, 6], [19, 43], [20, 29], [21, 17], [22, 7], [23, 52], [24, 46], [25, 42], [26, 40], [27, 40], [28, 42], [29, 46], [30, 52], [31, 7], [32, 17], [33, 29], [34, 43], [35, 6], [36, 24], [37, 44], [38, 13], [39, 37], [40, 10], [41, 38], [42, 15], [43, 47], [44, 28], [45, 11], [46, 49], [47, 36], [48, 25], [49, 16], [50, 9], [51, 4], [52, 1]], [1, 4, 6, 7, 9, 10, 11, 13, 15, 16, 17, 24, 25, 28, 29, 36, 37, 38, 40, 42, 43, 44, 46, 47, 49, 52], 26, 26, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0015/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 16. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":59}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(59, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 5], [9, 22], [10, 41], [11, 3], [12, 26], [13, 51], [14, 19], [15, 48], [16, 20], [17, 53], [18, 29], [19, 7], [20, 46], [21, 28], [22, 12], [23, 57], [24, 45], [25, 35], [26, 27], [27, 21], [28, 17], [29, 15], [30, 15], [31, 17], [32, 21], [33, 27], [34, 35], [35, 45], [36, 57], [37, 12], [38, 28], [39, 46], [40, 7], [41, 29], [42, 53], [43, 20], [44, 48], [45, 19], [46, 51], [47, 26], [48, 3], [49, 41], [50, 22], [51, 5], [52, 49], [53, 36], [54, 25], [55, 16], [56, 9], [57, 4], [58, 1]], [1, 3, 4, 5, 7, 9, 12, 15, 16, 17, 19, 20, 21, 22, 25, 26, 27, 28, 29, 35, 36, 41, 45, 46, 48, 49, 51, 53, 57], 29, 29, 是)

Wave6第五層核對=(59, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 5], [9, 22], [10, 41], [11, 3], [12, 26], [13, 51], [14, 19], [15, 48], [16, 20], [17, 53], [18, 29], [19, 7], [20, 46], [21, 28], [22, 12], [23, 57], [24, 45], [25, 35], [26, 27], [27, 21], [28, 17], [29, 15], [30, 15], [31, 17], [32, 21], [33, 27], [34, 35], [35, 45], [36, 57], [37, 12], [38, 28], [39, 46], [40, 7], [41, 29], [42, 53], [43, 20], [44, 48], [45, 19], [46, 51], [47, 26], [48, 3], [49, 41], [50, 22], [51, 5], [52, 49], [53, 36], [54, 25], [55, 16], [56, 9], [57, 4], [58, 1]], [1, 3, 4, 5, 7, 9, 12, 15, 16, 17, 19, 20, 21, 22, 25, 26, 27, 28, 29, 35, 36, 41, 45, 46, 48, 49, 51, 53, 57], 29, 29, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0016/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 17. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":61}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(61, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 3], [9, 20], [10, 39], [11, 60], [12, 22], [13, 47], [14, 13], [15, 42], [16, 12], [17, 45], [18, 19], [19, 56], [20, 34], [21, 14], [22, 57], [23, 41], [24, 27], [25, 15], [26, 5], [27, 58], [28, 52], [29, 48], [30, 46], [31, 46], [32, 48], [33, 52], [34, 58], [35, 5], [36, 15], [37, 27], [38, 41], [39, 57], [40, 14], [41, 34], [42, 56], [43, 19], [44, 45], [45, 12], [46, 42], [47, 13], [48, 47], [49, 22], [50, 60], [51, 39], [52, 20], [53, 3], [54, 49], [55, 36], [56, 25], [57, 16], [58, 9], [59, 4], [60, 1]], [1, 3, 4, 5, 9, 12, 13, 14, 15, 16, 19, 20, 22, 25, 27, 34, 36, 39, 41, 42, 45, 46, 47, 48, 49, 52, 56, 57, 58, 60], 30, 30, 是)

Wave6第五層核對=(61, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 3], [9, 20], [10, 39], [11, 60], [12, 22], [13, 47], [14, 13], [15, 42], [16, 12], [17, 45], [18, 19], [19, 56], [20, 34], [21, 14], [22, 57], [23, 41], [24, 27], [25, 15], [26, 5], [27, 58], [28, 52], [29, 48], [30, 46], [31, 46], [32, 48], [33, 52], [34, 58], [35, 5], [36, 15], [37, 27], [38, 41], [39, 57], [40, 14], [41, 34], [42, 56], [43, 19], [44, 45], [45, 12], [46, 42], [47, 13], [48, 47], [49, 22], [50, 60], [51, 39], [52, 20], [53, 3], [54, 49], [55, 36], [56, 25], [57, 16], [58, 9], [59, 4], [60, 1]], [1, 3, 4, 5, 9, 12, 13, 14, 15, 16, 19, 20, 22, 25, 27, 34, 36, 39, 41, 42, 45, 46, 47, 48, 49, 52, 56, 57, 58, 60], 30, 30, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0017/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 18. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":67}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(67, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 14], [10, 33], [11, 54], [12, 10], [13, 35], [14, 62], [15, 24], [16, 55], [17, 21], [18, 56], [19, 26], [20, 65], [21, 39], [22, 15], [23, 60], [24, 40], [25, 22], [26, 6], [27, 59], [28, 47], [29, 37], [30, 29], [31, 23], [32, 19], [33, 17], [34, 17], [35, 19], [36, 23], [37, 29], [38, 37], [39, 47], [40, 59], [41, 6], [42, 22], [43, 40], [44, 60], [45, 15], [46, 39], [47, 65], [48, 26], [49, 56], [50, 21], [51, 55], [52, 24], [53, 62], [54, 35], [55, 10], [56, 54], [57, 33], [58, 14], [59, 64], [60, 49], [61, 36], [62, 25], [63, 16], [64, 9], [65, 4], [66, 1]], [1, 4, 6, 9, 10, 14, 15, 16, 17, 19, 21, 22, 23, 24, 25, 26, 29, 33, 35, 36, 37, 39, 40, 47, 49, 54, 55, 56, 59, 60, 62, 64, 65], 33, 33, 是)

Wave6第五層核對=(67, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 14], [10, 33], [11, 54], [12, 10], [13, 35], [14, 62], [15, 24], [16, 55], [17, 21], [18, 56], [19, 26], [20, 65], [21, 39], [22, 15], [23, 60], [24, 40], [25, 22], [26, 6], [27, 59], [28, 47], [29, 37], [30, 29], [31, 23], [32, 19], [33, 17], [34, 17], [35, 19], [36, 23], [37, 29], [38, 37], [39, 47], [40, 59], [41, 6], [42, 22], [43, 40], [44, 60], [45, 15], [46, 39], [47, 65], [48, 26], [49, 56], [50, 21], [51, 55], [52, 24], [53, 62], [54, 35], [55, 10], [56, 54], [57, 33], [58, 14], [59, 64], [60, 49], [61, 36], [62, 25], [63, 16], [64, 9], [65, 4], [66, 1]], [1, 4, 6, 9, 10, 14, 15, 16, 17, 19, 21, 22, 23, 24, 25, 26, 29, 33, 35, 36, 37, 39, 40, 47, 49, 54, 55, 56, 59, 60, 62, 64, 65], 33, 33, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0018/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 19. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":71}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(71, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 10], [10, 29], [11, 50], [12, 2], [13, 27], [14, 54], [15, 12], [16, 43], [17, 5], [18, 40], [19, 6], [20, 45], [21, 15], [22, 58], [23, 32], [24, 8], [25, 57], [26, 37], [27, 19], [28, 3], [29, 60], [30, 48], [31, 38], [32, 30], [33, 24], [34, 20], [35, 18], [36, 18], [37, 20], [38, 24], [39, 30], [40, 38], [41, 48], [42, 60], [43, 3], [44, 19], [45, 37], [46, 57], [47, 8], [48, 32], [49, 58], [50, 15], [51, 45], [52, 6], [53, 40], [54, 5], [55, 43], [56, 12], [57, 54], [58, 27], [59, 2], [60, 50], [61, 29], [62, 10], [63, 64], [64, 49], [65, 36], [66, 25], [67, 16], [68, 9], [69, 4], [70, 1]], [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 19, 20, 24, 25, 27, 29, 30, 32, 36, 37, 38, 40, 43, 45, 48, 49, 50, 54, 57, 58, 60, 64], 35, 35, 是)

Wave6第五層核對=(71, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 10], [10, 29], [11, 50], [12, 2], [13, 27], [14, 54], [15, 12], [16, 43], [17, 5], [18, 40], [19, 6], [20, 45], [21, 15], [22, 58], [23, 32], [24, 8], [25, 57], [26, 37], [27, 19], [28, 3], [29, 60], [30, 48], [31, 38], [32, 30], [33, 24], [34, 20], [35, 18], [36, 18], [37, 20], [38, 24], [39, 30], [40, 38], [41, 48], [42, 60], [43, 3], [44, 19], [45, 37], [46, 57], [47, 8], [48, 32], [49, 58], [50, 15], [51, 45], [52, 6], [53, 40], [54, 5], [55, 43], [56, 12], [57, 54], [58, 27], [59, 2], [60, 50], [61, 29], [62, 10], [63, 64], [64, 49], [65, 36], [66, 25], [67, 16], [68, 9], [69, 4], [70, 1]], [1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 19, 20, 24, 25, 27, 29, 30, 32, 36, 37, 38, 40, 43, 45, 48, 49, 50, 54, 57, 58, 60, 64], 35, 35, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0019/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 20. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":73}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(73, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 8], [10, 27], [11, 48], [12, 71], [13, 23], [14, 50], [15, 6], [16, 37], [17, 70], [18, 32], [19, 69], [20, 35], [21, 3], [22, 46], [23, 18], [24, 65], [25, 41], [26, 19], [27, 72], [28, 54], [29, 38], [30, 24], [31, 12], [32, 2], [33, 67], [34, 61], [35, 57], [36, 55], [37, 55], [38, 57], [39, 61], [40, 67], [41, 2], [42, 12], [43, 24], [44, 38], [45, 54], [46, 72], [47, 19], [48, 41], [49, 65], [50, 18], [51, 46], [52, 3], [53, 35], [54, 69], [55, 32], [56, 70], [57, 37], [58, 6], [59, 50], [60, 23], [61, 71], [62, 48], [63, 27], [64, 8], [65, 64], [66, 49], [67, 36], [68, 25], [69, 16], [70, 9], [71, 4], [72, 1]], [1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 19, 23, 24, 25, 27, 32, 35, 36, 37, 38, 41, 46, 48, 49, 50, 54, 55, 57, 61, 64, 65, 67, 69, 70, 71, 72], 36, 36, 是)

Wave6第五層核對=(73, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 8], [10, 27], [11, 48], [12, 71], [13, 23], [14, 50], [15, 6], [16, 37], [17, 70], [18, 32], [19, 69], [20, 35], [21, 3], [22, 46], [23, 18], [24, 65], [25, 41], [26, 19], [27, 72], [28, 54], [29, 38], [30, 24], [31, 12], [32, 2], [33, 67], [34, 61], [35, 57], [36, 55], [37, 55], [38, 57], [39, 61], [40, 67], [41, 2], [42, 12], [43, 24], [44, 38], [45, 54], [46, 72], [47, 19], [48, 41], [49, 65], [50, 18], [51, 46], [52, 3], [53, 35], [54, 69], [55, 32], [56, 70], [57, 37], [58, 6], [59, 50], [60, 23], [61, 71], [62, 48], [63, 27], [64, 8], [65, 64], [66, 49], [67, 36], [68, 25], [69, 16], [70, 9], [71, 4], [72, 1]], [1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 19, 23, 24, 25, 27, 32, 35, 36, 37, 38, 41, 46, 48, 49, 50, 54, 55, 57, 61, 64, 65, 67, 69, 70, 71, 72], 36, 36, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0020/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 21. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":79}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(79, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 2], [10, 21], [11, 42], [12, 65], [13, 11], [14, 38], [15, 67], [16, 19], [17, 52], [18, 8], [19, 45], [20, 5], [21, 46], [22, 10], [23, 55], [24, 23], [25, 72], [26, 44], [27, 18], [28, 73], [29, 51], [30, 31], [31, 13], [32, 76], [33, 62], [34, 50], [35, 40], [36, 32], [37, 26], [38, 22], [39, 20], [40, 20], [41, 22], [42, 26], [43, 32], [44, 40], [45, 50], [46, 62], [47, 76], [48, 13], [49, 31], [50, 51], [51, 73], [52, 18], [53, 44], [54, 72], [55, 23], [56, 55], [57, 10], [58, 46], [59, 5], [60, 45], [61, 8], [62, 52], [63, 19], [64, 67], [65, 38], [66, 11], [67, 65], [68, 42], [69, 21], [70, 2], [71, 64], [72, 49], [73, 36], [74, 25], [75, 16], [76, 9], [77, 4], [78, 1]], [1, 2, 4, 5, 8, 9, 10, 11, 13, 16, 18, 19, 20, 21, 22, 23, 25, 26, 31, 32, 36, 38, 40, 42, 44, 45, 46, 49, 50, 51, 52, 55, 62, 64, 65, 67, 72, 73, 76], 39, 39, 是)

Wave6第五層核對=(79, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 2], [10, 21], [11, 42], [12, 65], [13, 11], [14, 38], [15, 67], [16, 19], [17, 52], [18, 8], [19, 45], [20, 5], [21, 46], [22, 10], [23, 55], [24, 23], [25, 72], [26, 44], [27, 18], [28, 73], [29, 51], [30, 31], [31, 13], [32, 76], [33, 62], [34, 50], [35, 40], [36, 32], [37, 26], [38, 22], [39, 20], [40, 20], [41, 22], [42, 26], [43, 32], [44, 40], [45, 50], [46, 62], [47, 76], [48, 13], [49, 31], [50, 51], [51, 73], [52, 18], [53, 44], [54, 72], [55, 23], [56, 55], [57, 10], [58, 46], [59, 5], [60, 45], [61, 8], [62, 52], [63, 19], [64, 67], [65, 38], [66, 11], [67, 65], [68, 42], [69, 21], [70, 2], [71, 64], [72, 49], [73, 36], [74, 25], [75, 16], [76, 9], [77, 4], [78, 1]], [1, 2, 4, 5, 8, 9, 10, 11, 13, 16, 18, 19, 20, 21, 22, 23, 25, 26, 31, 32, 36, 38, 40, 42, 44, 45, 46, 49, 50, 51, 52, 55, 62, 64, 65, 67, 72, 73, 76], 39, 39, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0021/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 22. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":83}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(83, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 17], [11, 38], [12, 61], [13, 3], [14, 30], [15, 59], [16, 7], [17, 40], [18, 75], [19, 29], [20, 68], [21, 26], [22, 69], [23, 31], [24, 78], [25, 44], [26, 12], [27, 65], [28, 37], [29, 11], [30, 70], [31, 48], [32, 28], [33, 10], [34, 77], [35, 63], [36, 51], [37, 41], [38, 33], [39, 27], [40, 23], [41, 21], [42, 21], [43, 23], [44, 27], [45, 33], [46, 41], [47, 51], [48, 63], [49, 77], [50, 10], [51, 28], [52, 48], [53, 70], [54, 11], [55, 37], [56, 65], [57, 12], [58, 44], [59, 78], [60, 31], [61, 69], [62, 26], [63, 68], [64, 29], [65, 75], [66, 40], [67, 7], [68, 59], [69, 30], [70, 3], [71, 61], [72, 38], [73, 17], [74, 81], [75, 64], [76, 49], [77, 36], [78, 25], [79, 16], [80, 9], [81, 4], [82, 1]], [1, 3, 4, 7, 9, 10, 11, 12, 16, 17, 21, 23, 25, 26, 27, 28, 29, 30, 31, 33, 36, 37, 38, 40, 41, 44, 48, 49, 51, 59, 61, 63, 64, 65, 68, 69, 70, 75, 77, 78, 81], 41, 41, 是)

Wave6第五層核對=(83, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 17], [11, 38], [12, 61], [13, 3], [14, 30], [15, 59], [16, 7], [17, 40], [18, 75], [19, 29], [20, 68], [21, 26], [22, 69], [23, 31], [24, 78], [25, 44], [26, 12], [27, 65], [28, 37], [29, 11], [30, 70], [31, 48], [32, 28], [33, 10], [34, 77], [35, 63], [36, 51], [37, 41], [38, 33], [39, 27], [40, 23], [41, 21], [42, 21], [43, 23], [44, 27], [45, 33], [46, 41], [47, 51], [48, 63], [49, 77], [50, 10], [51, 28], [52, 48], [53, 70], [54, 11], [55, 37], [56, 65], [57, 12], [58, 44], [59, 78], [60, 31], [61, 69], [62, 26], [63, 68], [64, 29], [65, 75], [66, 40], [67, 7], [68, 59], [69, 30], [70, 3], [71, 61], [72, 38], [73, 17], [74, 81], [75, 64], [76, 49], [77, 36], [78, 25], [79, 16], [80, 9], [81, 4], [82, 1]], [1, 3, 4, 7, 9, 10, 11, 12, 16, 17, 21, 23, 25, 26, 27, 28, 29, 30, 31, 33, 36, 37, 38, 40, 41, 44, 48, 49, 51, 59, 61, 63, 64, 65, 68, 69, 70, 75, 77, 78, 81], 41, 41, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0022/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 23. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":89}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(89, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 11], [11, 32], [12, 55], [13, 80], [14, 18], [15, 47], [16, 78], [17, 22], [18, 57], [19, 5], [20, 44], [21, 85], [22, 39], [23, 84], [24, 42], [25, 2], [26, 53], [27, 17], [28, 72], [29, 40], [30, 10], [31, 71], [32, 45], [33, 21], [34, 88], [35, 68], [36, 50], [37, 34], [38, 20], [39, 8], [40, 87], [41, 79], [42, 73], [43, 69], [44, 67], [45, 67], [46, 69], [47, 73], [48, 79], [49, 87], [50, 8], [51, 20], [52, 34], [53, 50], [54, 68], [55, 88], [56, 21], [57, 45], [58, 71], [59, 10], [60, 40], [61, 72], [62, 17], [63, 53], [64, 2], [65, 42], [66, 84], [67, 39], [68, 85], [69, 44], [70, 5], [71, 57], [72, 22], [73, 78], [74, 47], [75, 18], [76, 80], [77, 55], [78, 32], [79, 11], [80, 81], [81, 64], [82, 49], [83, 36], [84, 25], [85, 16], [86, 9], [87, 4], [88, 1]], [1, 2, 4, 5, 8, 9, 10, 11, 16, 17, 18, 20, 21, 22, 25, 32, 34, 36, 39, 40, 42, 44, 45, 47, 49, 50, 53, 55, 57, 64, 67, 68, 69, 71, 72, 73, 78, 79, 80, 81, 84, 85, 87, 88], 44, 44, 是)

Wave6第五層核對=(89, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 11], [11, 32], [12, 55], [13, 80], [14, 18], [15, 47], [16, 78], [17, 22], [18, 57], [19, 5], [20, 44], [21, 85], [22, 39], [23, 84], [24, 42], [25, 2], [26, 53], [27, 17], [28, 72], [29, 40], [30, 10], [31, 71], [32, 45], [33, 21], [34, 88], [35, 68], [36, 50], [37, 34], [38, 20], [39, 8], [40, 87], [41, 79], [42, 73], [43, 69], [44, 67], [45, 67], [46, 69], [47, 73], [48, 79], [49, 87], [50, 8], [51, 20], [52, 34], [53, 50], [54, 68], [55, 88], [56, 21], [57, 45], [58, 71], [59, 10], [60, 40], [61, 72], [62, 17], [63, 53], [64, 2], [65, 42], [66, 84], [67, 39], [68, 85], [69, 44], [70, 5], [71, 57], [72, 22], [73, 78], [74, 47], [75, 18], [76, 80], [77, 55], [78, 32], [79, 11], [80, 81], [81, 64], [82, 49], [83, 36], [84, 25], [85, 16], [86, 9], [87, 4], [88, 1]], [1, 2, 4, 5, 8, 9, 10, 11, 16, 17, 18, 20, 21, 22, 25, 32, 34, 36, 39, 40, 42, 44, 45, 47, 49, 50, 53, 55, 57, 64, 67, 68, 69, 71, 72, 73, 78, 79, 80, 81, 84, 85, 87, 88], 44, 44, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0023/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 24. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":97}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(97, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 3], [11, 24], [12, 47], [13, 72], [14, 2], [15, 31], [16, 62], [17, 95], [18, 33], [19, 70], [20, 12], [21, 53], [22, 96], [23, 44], [24, 91], [25, 43], [26, 94], [27, 50], [28, 8], [29, 65], [30, 27], [31, 88], [32, 54], [33, 22], [34, 89], [35, 61], [36, 35], [37, 11], [38, 86], [39, 66], [40, 48], [41, 32], [42, 18], [43, 6], [44, 93], [45, 85], [46, 79], [47, 75], [48, 73], [49, 73], [50, 75], [51, 79], [52, 85], [53, 93], [54, 6], [55, 18], [56, 32], [57, 48], [58, 66], [59, 86], [60, 11], [61, 35], [62, 61], [63, 89], [64, 22], [65, 54], [66, 88], [67, 27], [68, 65], [69, 8], [70, 50], [71, 94], [72, 43], [73, 91], [74, 44], [75, 96], [76, 53], [77, 12], [78, 70], [79, 33], [80, 95], [81, 62], [82, 31], [83, 2], [84, 72], [85, 47], [86, 24], [87, 3], [88, 81], [89, 64], [90, 49], [91, 36], [92, 25], [93, 16], [94, 9], [95, 4], [96, 1]], [1, 2, 3, 4, 6, 8, 9, 11, 12, 16, 18, 22, 24, 25, 27, 31, 32, 33, 35, 36, 43, 44, 47, 48, 49, 50, 53, 54, 61, 62, 64, 65, 66, 70, 72, 73, 75, 79, 81, 85, 86, 88, 89, 91, 93, 94, 95, 96], 48, 48, 是)

Wave6第五層核對=(97, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 3], [11, 24], [12, 47], [13, 72], [14, 2], [15, 31], [16, 62], [17, 95], [18, 33], [19, 70], [20, 12], [21, 53], [22, 96], [23, 44], [24, 91], [25, 43], [26, 94], [27, 50], [28, 8], [29, 65], [30, 27], [31, 88], [32, 54], [33, 22], [34, 89], [35, 61], [36, 35], [37, 11], [38, 86], [39, 66], [40, 48], [41, 32], [42, 18], [43, 6], [44, 93], [45, 85], [46, 79], [47, 75], [48, 73], [49, 73], [50, 75], [51, 79], [52, 85], [53, 93], [54, 6], [55, 18], [56, 32], [57, 48], [58, 66], [59, 86], [60, 11], [61, 35], [62, 61], [63, 89], [64, 22], [65, 54], [66, 88], [67, 27], [68, 65], [69, 8], [70, 50], [71, 94], [72, 43], [73, 91], [74, 44], [75, 96], [76, 53], [77, 12], [78, 70], [79, 33], [80, 95], [81, 62], [82, 31], [83, 2], [84, 72], [85, 47], [86, 24], [87, 3], [88, 81], [89, 64], [90, 49], [91, 36], [92, 25], [93, 16], [94, 9], [95, 4], [96, 1]], [1, 2, 3, 4, 6, 8, 9, 11, 12, 16, 18, 22, 24, 25, 27, 31, 32, 33, 35, 36, 43, 44, 47, 48, 49, 50, 53, 54, 61, 62, 64, 65, 66, 70, 72, 73, 75, 79, 81, 85, 86, 88, 89, 91, 93, 94, 95, 96], 48, 48, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0024/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 25. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-square-table；資料={"p":101}。請依 [p, nonzero_square_rows, nonzero_residues, residue_count, half_nonzero_count, equal] 欄序回答。

查看答案與解釋(101, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 100], [11, 20], [12, 43], [13, 68], [14, 95], [15, 23], [16, 54], [17, 87], [18, 21], [19, 58], [20, 97], [21, 37], [22, 80], [23, 24], [24, 71], [25, 19], [26, 70], [27, 22], [28, 77], [29, 33], [30, 92], [31, 52], [32, 14], [33, 79], [34, 45], [35, 13], [36, 84], [37, 56], [38, 30], [39, 6], [40, 85], [41, 65], [42, 47], [43, 31], [44, 17], [45, 5], [46, 96], [47, 88], [48, 82], [49, 78], [50, 76], [51, 76], [52, 78], [53, 82], [54, 88], [55, 96], [56, 5], [57, 17], [58, 31], [59, 47], [60, 65], [61, 85], [62, 6], [63, 30], [64, 56], [65, 84], [66, 13], [67, 45], [68, 79], [69, 14], [70, 52], [71, 92], [72, 33], [73, 77], [74, 22], [75, 70], [76, 19], [77, 71], [78, 24], [79, 80], [80, 37], [81, 97], [82, 58], [83, 21], [84, 87], [85, 54], [86, 23], [87, 95], [88, 68], [89, 43], [90, 20], [91, 100], [92, 81], [93, 64], [94, 49], [95, 36], [96, 25], [97, 16], [98, 9], [99, 4], [100, 1]], [1, 4, 5, 6, 9, 13, 14, 16, 17, 19, 20, 21, 22, 23, 24, 25, 30, 31, 33, 36, 37, 43, 45, 47, 49, 52, 54, 56, 58, 64, 65, 68, 70, 71, 76, 77, 78, 79, 80, 81, 82, 84, 85, 87, 88, 92, 95, 96, 97, 100], 50, 50, 是)

Wave6第五層核對=(101, [[1, 1], [2, 4], [3, 9], [4, 16], [5, 25], [6, 36], [7, 49], [8, 64], [9, 81], [10, 100], [11, 20], [12, 43], [13, 68], [14, 95], [15, 23], [16, 54], [17, 87], [18, 21], [19, 58], [20, 97], [21, 37], [22, 80], [23, 24], [24, 71], [25, 19], [26, 70], [27, 22], [28, 77], [29, 33], [30, 92], [31, 52], [32, 14], [33, 79], [34, 45], [35, 13], [36, 84], [37, 56], [38, 30], [39, 6], [40, 85], [41, 65], [42, 47], [43, 31], [44, 17], [45, 5], [46, 96], [47, 88], [48, 82], [49, 78], [50, 76], [51, 76], [52, 78], [53, 82], [54, 88], [55, 96], [56, 5], [57, 17], [58, 31], [59, 47], [60, 65], [61, 85], [62, 6], [63, 30], [64, 56], [65, 84], [66, 13], [67, 45], [68, 79], [69, 14], [70, 52], [71, 92], [72, 33], [73, 77], [74, 22], [75, 70], [76, 19], [77, 71], [78, 24], [79, 80], [80, 37], [81, 97], [82, 58], [83, 21], [84, 87], [85, 54], [86, 23], [87, 95], [88, 68], [89, 43], [90, 20], [91, 100], [92, 81], [93, 64], [94, 49], [95, 36], [96, 25], [97, 16], [98, 9], [99, 4], [100, 1]], [1, 4, 5, 6, 9, 13, 14, 16, 17, 19, 20, 21, 22, 23, 24, 25, 30, 31, 33, 36, 37, 43, 45, 47, 49, 52, 54, 56, 58, 64, 65, 68, 70, 71, 76, 77, 78, 79, 80, 81, 82, 84, 85, 87, 88, 92, 95, 96, 97, 100], 50, 50, 是)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0025/)

二次剩餘 · qf-mod42-quadratic-square-table · candidate





## 26. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":3,"a":1}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(3, 1, 1, [1, 2], 是, nonzero_square_root_witness)

Wave6第五層核對=(3, 1, 1, [1, 2], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0026/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 27. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":5,"a":2}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(5, 2, 2, [], 否, complete_nonzero_square_table_has_no_root)

Wave6第五層核對=(5, 2, 2, [], 否, complete_nonzero_square_table_has_no_root)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0027/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 28. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":7,"a":3}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(7, 3, 3, [], 否, complete_nonzero_square_table_has_no_root)

Wave6第五層核對=(7, 3, 3, [], 否, complete_nonzero_square_table_has_no_root)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0028/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 29. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":11,"a":4}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(11, 4, 4, [2, 9], 是, nonzero_square_root_witness)

Wave6第五層核對=(11, 4, 4, [2, 9], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0029/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 30. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":13,"a":5}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(13, 5, 5, [], 否, complete_nonzero_square_table_has_no_root)

Wave6第五層核對=(13, 5, 5, [], 否, complete_nonzero_square_table_has_no_root)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0030/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 31. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":17,"a":6}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(17, 6, 6, [], 否, complete_nonzero_square_table_has_no_root)

Wave6第五層核對=(17, 6, 6, [], 否, complete_nonzero_square_table_has_no_root)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0031/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 32. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":19,"a":7}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(19, 7, 7, [8, 11], 是, nonzero_square_root_witness)

Wave6第五層核對=(19, 7, 7, [8, 11], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0032/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 33. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":23,"a":8}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(23, 8, 8, [10, 13], 是, nonzero_square_root_witness)

Wave6第五層核對=(23, 8, 8, [10, 13], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0033/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 34. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":29,"a":9}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(29, 9, 9, [3, 26], 是, nonzero_square_root_witness)

Wave6第五層核對=(29, 9, 9, [3, 26], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0034/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 35. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":31,"a":10}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(31, 10, 10, [14, 17], 是, nonzero_square_root_witness)

Wave6第五層核對=(31, 10, 10, [14, 17], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0035/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 36. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":37,"a":11}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(37, 11, 11, [14, 23], 是, nonzero_square_root_witness)

Wave6第五層核對=(37, 11, 11, [14, 23], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0036/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 37. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":41,"a":12}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(41, 12, 12, [], 否, complete_nonzero_square_table_has_no_root)

Wave6第五層核對=(41, 12, 12, [], 否, complete_nonzero_square_table_has_no_root)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0037/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 38. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":43,"a":13}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(43, 13, 13, [20, 23], 是, nonzero_square_root_witness)

Wave6第五層核對=(43, 13, 13, [20, 23], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0038/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 39. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":47,"a":14}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(47, 14, 14, [22, 25], 是, nonzero_square_root_witness)

Wave6第五層核對=(47, 14, 14, [22, 25], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0039/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 40. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":53,"a":15}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(53, 15, 15, [11, 42], 是, nonzero_square_root_witness)

Wave6第五層核對=(53, 15, 15, [11, 42], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0040/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 41. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":59,"a":16}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(59, 16, 16, [4, 55], 是, nonzero_square_root_witness)

Wave6第五層核對=(59, 16, 16, [4, 55], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0041/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 42. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":61,"a":17}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(61, 17, 17, [], 否, complete_nonzero_square_table_has_no_root)

Wave6第五層核對=(61, 17, 17, [], 否, complete_nonzero_square_table_has_no_root)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0042/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 43. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":67,"a":18}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(67, 18, 18, [], 否, complete_nonzero_square_table_has_no_root)

Wave6第五層核對=(67, 18, 18, [], 否, complete_nonzero_square_table_has_no_root)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0043/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 44. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":71,"a":19}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(71, 19, 19, [27, 44], 是, nonzero_square_root_witness)

Wave6第五層核對=(71, 19, 19, [27, 44], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0044/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 45. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":73,"a":20}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(73, 20, 20, [], 否, complete_nonzero_square_table_has_no_root)

Wave6第五層核對=(73, 20, 20, [], 否, complete_nonzero_square_table_has_no_root)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0045/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 46. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":79,"a":21}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(79, 21, 21, [10, 69], 是, nonzero_square_root_witness)

Wave6第五層核對=(79, 21, 21, [10, 69], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0046/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 47. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":83,"a":22}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(83, 22, 22, [], 否, complete_nonzero_square_table_has_no_root)

Wave6第五層核對=(83, 22, 22, [], 否, complete_nonzero_square_table_has_no_root)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0047/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 48. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":89,"a":23}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(89, 23, 23, [], 否, complete_nonzero_square_table_has_no_root)

Wave6第五層核對=(89, 23, 23, [], 否, complete_nonzero_square_table_has_no_root)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0048/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 49. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":97,"a":24}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(97, 24, 24, [11, 86], 是, nonzero_square_root_witness)

Wave6第五層核對=(97, 24, 24, [11, 86], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0049/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 50. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-membership-witness；資料={"p":101,"a":25}。請依 [p, a, normalized_class, nonzero_roots, nonzero_residue, certificate] 欄序回答。

查看答案與解釋(101, 25, 25, [5, 96], 是, nonzero_square_root_witness)

Wave6第五層核對=(101, 25, 25, [5, 96], 是, nonzero_square_root_witness)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0050/)

二次剩餘 · qf-mod42-quadratic-membership-witness · candidate





## 51. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":3,"a":1}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(3, 1, [1, 2], 2, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(3, 1, [1, 2], 2, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0051/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 52. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":5,"a":4}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(5, 4, [2, 3], 3, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(5, 4, [2, 3], 3, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0052/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 53. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":7,"a":2}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(7, 2, [3, 4], 4, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(7, 2, [3, 4], 4, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0053/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 54. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":11,"a":5}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(11, 5, [4, 7], 7, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(11, 5, [4, 7], 7, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0054/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 55. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":13,"a":12}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(13, 12, [5, 8], 8, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(13, 12, [5, 8], 8, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0055/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 56. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":17,"a":2}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(17, 2, [6, 11], 11, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(17, 2, [6, 11], 11, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0056/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 57. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":19,"a":11}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(19, 11, [7, 12], 12, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(19, 11, [7, 12], 12, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0057/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 58. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":23,"a":18}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(23, 18, [8, 15], 15, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(23, 18, [8, 15], 15, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0058/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 59. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":29,"a":23}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(29, 23, [9, 20], 20, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(29, 23, [9, 20], 20, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0059/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 60. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":31,"a":7}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(31, 7, [10, 21], 21, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(31, 7, [10, 21], 21, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0060/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 61. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":37,"a":10}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(37, 10, [11, 26], 26, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(37, 10, [11, 26], 26, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0061/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 62. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":41,"a":21}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(41, 21, [12, 29], 29, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(41, 21, [12, 29], 29, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0062/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 63. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":43,"a":40}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(43, 40, [13, 30], 30, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(43, 40, [13, 30], 30, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0063/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 64. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":47,"a":8}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(47, 8, [14, 33], 33, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(47, 8, [14, 33], 33, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0064/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 65. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":53,"a":13}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(53, 13, [15, 38], 38, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(53, 13, [15, 38], 38, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0065/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 66. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":59,"a":20}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(59, 20, [16, 43], 43, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(59, 20, [16, 43], 43, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0066/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 67. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":61,"a":45}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(61, 45, [17, 44], 44, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(61, 45, [17, 44], 44, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0067/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 68. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":67,"a":56}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(67, 56, [18, 49], 49, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(67, 56, [18, 49], 49, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0068/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 69. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":71,"a":6}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(71, 6, [19, 52], 52, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(71, 6, [19, 52], 52, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0069/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 70. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":73,"a":35}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(73, 35, [20, 53], 53, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(73, 35, [20, 53], 53, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0070/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 71. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":79,"a":46}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(79, 46, [21, 58], 58, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(79, 46, [21, 58], 58, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0071/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 72. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":83,"a":69}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(83, 69, [22, 61], 61, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(83, 69, [22, 61], 61, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0072/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 73. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":89,"a":84}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(89, 84, [23, 66], 66, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(89, 84, [23, 66], 66, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0073/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 74. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":97,"a":91}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(97, 91, [24, 73], 73, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(97, 91, [24, 73], 73, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0074/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 75. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-paired-roots；資料={"p":101,"a":19}。請依 [p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate] 欄序回答。

查看答案與解釋(101, 19, [25, 76], 76, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

Wave6第五層核對=(101, 19, [25, 76], 76, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0075/)

二次剩餘 · qf-mod42-quadratic-paired-roots · candidate





## 76. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"zero_class","modulus":3,"a":0}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(zero_class, 3, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(zero_class, 3, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0076/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 77. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"even_prime","modulus":2,"a":2}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(even_prime, 2, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(even_prime, 2, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0077/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 78. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"composite_modulus","modulus":9,"a":3}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(composite_modulus, 9, 3, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(composite_modulus, 9, 3, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0078/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 79. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"zero_class","modulus":11,"a":0}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(zero_class, 11, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(zero_class, 11, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0079/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 80. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"even_prime","modulus":2,"a":5}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(even_prime, 2, 1, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(even_prime, 2, 1, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0080/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 81. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"composite_modulus","modulus":15,"a":6}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(composite_modulus, 15, 6, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(composite_modulus, 15, 6, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0081/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 82. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"zero_class","modulus":19,"a":0}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(zero_class, 19, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(zero_class, 19, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0082/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 83. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"even_prime","modulus":2,"a":8}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(even_prime, 2, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(even_prime, 2, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0083/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 84. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"composite_modulus","modulus":21,"a":9}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(composite_modulus, 21, 9, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(composite_modulus, 21, 9, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0084/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 85. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"zero_class","modulus":31,"a":0}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(zero_class, 31, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(zero_class, 31, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0085/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 86. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"even_prime","modulus":2,"a":11}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(even_prime, 2, 1, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(even_prime, 2, 1, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0086/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 87. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"composite_modulus","modulus":27,"a":12}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(composite_modulus, 27, 12, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(composite_modulus, 27, 12, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0087/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 88. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"zero_class","modulus":43,"a":0}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(zero_class, 43, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(zero_class, 43, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0088/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 89. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"even_prime","modulus":2,"a":14}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(even_prime, 2, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(even_prime, 2, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0089/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 90. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"composite_modulus","modulus":33,"a":15}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(composite_modulus, 33, 15, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(composite_modulus, 33, 15, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0090/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 91. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"zero_class","modulus":59,"a":0}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(zero_class, 59, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(zero_class, 59, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0091/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 92. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"even_prime","modulus":2,"a":17}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(even_prime, 2, 1, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(even_prime, 2, 1, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0092/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 93. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"composite_modulus","modulus":39,"a":18}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(composite_modulus, 39, 18, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(composite_modulus, 39, 18, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0093/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 94. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"zero_class","modulus":71,"a":0}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(zero_class, 71, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(zero_class, 71, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0094/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 95. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"even_prime","modulus":2,"a":20}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(even_prime, 2, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(even_prime, 2, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0095/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 96. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"composite_modulus","modulus":45,"a":21}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(composite_modulus, 45, 21, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(composite_modulus, 45, 21, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0096/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 97. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"zero_class","modulus":83,"a":0}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(zero_class, 83, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(zero_class, 83, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0097/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 98. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"even_prime","modulus":2,"a":23}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(even_prime, 2, 1, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(even_prime, 2, 1, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0098/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 99. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"composite_modulus","modulus":51,"a":24}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(composite_modulus, 51, 24, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(composite_modulus, 51, 24, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0099/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate





## 100. 二次剩餘

odd prime p上的nonzero quadratic-residue convention與complete nonzero square table。操作=quadratic-domain-boundary；資料={"mode":"zero_class","modulus":101,"a":0}。請依 [mode, modulus, normalized_class, declared_model_qualified, boundary] 欄序回答。

此否是不符合本節點的odd-prime/nonzero慣例；不宣稱a在modulus=2或composite modulus下沒有平方根。

查看答案與解釋(zero_class, 101, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)

Wave6第五層核對=(zero_class, 101, 0, 否, nonzero_quadratic_residue_convention_requires_an_odd_prime_and_nonzero_class)；依完整明示資料獨立重算；a先normalize mod p；nonzero residue有兩個distinct opposite roots；zero、p=2與composite modulus分開；不預先使用Euler criterion或Legendre symbol。



[本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261111-r06-quadratic-residue-0100/)

二次剩餘 · qf-mod42-quadratic-domain-boundary · candidate



[原始題目 JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261111-r06-quadratic-residue-foundations/questions.jsonl) · [驗收報告](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261111-r06-quadratic-residue-foundations/validation-report.json)