第 23351 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-paired-roots；資料={"p":3,"a":1}。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-paired-roots；資料={"p":3,"a":1}。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-paired-roots;original snapshot and mathematics retained

explicit ordered source output fields=[p, normalized_class, roots, opposite_first_root, opposite_present, distinct_roots, certificate]

Quadratic ORIGINAL modulus:ORIGINAL integer=3;complete possible trial divisors=[]

ALL [divisor,remainder] rows=[];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..2;actual square=1

Euclidean square=0*3+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..2;actual square=4

Euclidean square=1*3+1;nonzero square class=1;complete root-index order retained

Quadratic distinct nonzero residue count := sum(1)

Quadratic distinct nonzero residue count = 1

Quadratic ORIGINAL modulus minus1 := sub(3, 1)

Quadratic ORIGINAL modulus minus1 = 2

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

Quadratic half of all nonzero roots = 1

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

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=1=(3-1)/2=1

Quadratic ORIGINAL candidate class Euclidean quotient times modulus := mul(0, 3)

Quadratic ORIGINAL candidate class Euclidean quotient times modulus = 0

Quadratic ORIGINAL candidate class ORIGINAL signed value minus multiple := sub(1, 0)

Quadratic ORIGINAL candidate class ORIGINAL signed value minus multiple = 1

Quadratic ORIGINAL candidate class:ORIGINAL signed value=1;positive modulus=3;actual quotient=0

canonical residue=1;value=quotient*modulus+residue and0<=residue<modulus

Quadratic negative of FIRST source root Euclidean quotient times modulus := mul(-1, 3)

Quadratic negative of FIRST source root Euclidean quotient times modulus = -3

Quadratic negative of FIRST source root ORIGINAL signed value minus multiple := sub(-1, -3)

Quadratic negative of FIRST source root ORIGINAL signed value minus multiple = 2

Quadratic negative of FIRST source root:ORIGINAL signed value=-1;positive modulus=3;actual quotient=-1

canonical residue=2;value=quotient*modulus+residue and0<=residue<modulus

逐步反向核對

• S016 · 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。
• S015 · premise_and_result_audit — 以精確有理數重算 Quadratic negative of FIRST source root ORIGINAL signed value minus multiple，不以小數近似替代。
• S014 · premise_and_result_audit — 以精確有理數重算 Quadratic negative of FIRST source root Euclidean quotient times modulus，不以小數近似替代。
• S013 · 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。
• S012 · 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。
• S011 · 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。
• 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 half of all nonzero roots，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL modulus minus1，不以小數近似替代。
• S007 · 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。
• 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

(3, 1, [1, 2], 2, 是, 是, odd_prime_nonzero_residue_has_two_distinct_opposite_roots)

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。