第 23328 題：二次剩餘

原題與已知條件

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 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — 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 exact_ordered_tuple_declared_by_interface 欄序回答。

定義與適用範圍

沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。

• mko-quadratic-residue

逐步推導

source operation=quadratic-membership-witness;original snapshot and mathematics retained

explicit ordered source output fields=[p, a, normalized_class, nonzero_roots, nonzero_residue, certificate]

Quadratic ORIGINAL modulus:ORIGINAL integer=7;complete possible trial divisors=[2]

ALL [divisor,remainder] rows=[[2, 1]];prime=是;a composite>1 has a factor at most its square root

Quadratic ORIGINAL nonzero root1 square := mul(1, 1)

Quadratic ORIGINAL nonzero root1 square = 1

ORIGINAL nonzero root x=1 in1..6;actual square=1

Euclidean square=0*7+1;nonzero square class=1;complete root-index order retained

Quadratic ORIGINAL nonzero root2 square := mul(2, 2)

Quadratic ORIGINAL nonzero root2 square = 4

ORIGINAL nonzero root x=2 in1..6;actual square=4

Euclidean square=0*7+4;nonzero square class=4;complete root-index order retained

Quadratic ORIGINAL nonzero root3 square := mul(3, 3)

Quadratic ORIGINAL nonzero root3 square = 9

ORIGINAL nonzero root x=3 in1..6;actual square=9

Euclidean square=1*7+2;nonzero square class=2;complete root-index order retained

Quadratic ORIGINAL nonzero root4 square := mul(4, 4)

Quadratic ORIGINAL nonzero root4 square = 16

ORIGINAL nonzero root x=4 in1..6;actual square=16

Euclidean square=2*7+2;nonzero square class=2;complete root-index order retained

Quadratic ORIGINAL nonzero root5 square := mul(5, 5)

Quadratic ORIGINAL nonzero root5 square = 25

ORIGINAL nonzero root x=5 in1..6;actual square=25

Euclidean square=3*7+4;nonzero square class=4;complete root-index order retained

Quadratic ORIGINAL nonzero root6 square := mul(6, 6)

Quadratic ORIGINAL nonzero root6 square = 36

ORIGINAL nonzero root x=6 in1..6;actual square=36

Euclidean square=5*7+1;nonzero square class=1;complete root-index order retained

Quadratic distinct nonzero residue count := sum(1, 1, 1)

Quadratic distinct nonzero residue count = 3

Quadratic ORIGINAL modulus minus1 := sub(7, 1)

Quadratic ORIGINAL modulus minus1 = 6

Quadratic half of all nonzero roots := div(6, 2)

Quadratic half of all nonzero roots = 3

ORIGINAL ODD PRIME p=7;complete nonzero square table=[[1, 1], [2, 4], [3, 2], [4, 2], [5, 4], [6, 1]];distinct classes=[1, 2, 4]

x^2=y^2 implies p divides(x-y)(x+y),so x=y or x=-y modp;odd p and nonzero x make opposite roots DISTINCT;each image has2 roots,so residue count=3=(7-1)/2=3

Quadratic ORIGINAL candidate class Euclidean quotient times modulus := mul(0, 7)

Quadratic ORIGINAL candidate class Euclidean quotient times modulus = 0

Quadratic ORIGINAL candidate class ORIGINAL signed value minus multiple := sub(3, 0)

Quadratic ORIGINAL candidate class ORIGINAL signed value minus multiple = 3

Quadratic ORIGINAL candidate class:ORIGINAL signed value=3;positive modulus=7;actual quotient=0

canonical residue=3;value=quotient*modulus+residue and0<=residue<modulus

逐步反向核對

• S021 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S020 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL candidate class ORIGINAL signed value minus multiple，不以小數近似替代。
• S019 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL candidate class Euclidean quotient times modulus，不以小數近似替代。
• S018 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S017 · premise_and_result_audit — 以精確有理數重算 Quadratic half of all nonzero roots，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL modulus minus1，不以小數近似替代。
• S015 · premise_and_result_audit — 以精確有理數重算 Quadratic distinct nonzero residue count，不以小數近似替代。
• S014 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S013 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL nonzero root6 square，不以小數近似替代。
• S012 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S011 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL nonzero root5 square，不以小數近似替代。
• S010 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S009 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL nonzero root4 square，不以小數近似替代。
• S008 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S007 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL nonzero root3 square，不以小數近似替代。
• S006 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S005 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL nonzero root2 square，不以小數近似替代。
• S004 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S003 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL nonzero root1 square，不以小數近似替代。
• S002 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S001 · premise_and_result_audit — 回到原始資料與本步條件核對，不把單向推論誤稱等價。

結論

(7, 3, 3, [], 否, complete_nonzero_square_table_has_no_root)

依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。

本題覆核與來源綁定

覆核日期：2026-10-26

原題、答案、步驟表達式、數值 AST 與既有歷史保持原樣；原來的候選狀態屬於來源歷史，不覆寫為形式化證明。