第 23355 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":13,"a":12}。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":13,"a":12}。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=13;complete possible trial divisors=[2, 3]

ALL [divisor,remainder] rows=[[2, 1], [3, 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..12;actual square=1

Euclidean square=0*13+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..12;actual square=4

Euclidean square=0*13+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..12;actual square=9

Euclidean square=0*13+9;nonzero square class=9;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..12;actual square=16

Euclidean square=1*13+3;nonzero square class=3;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..12;actual square=25

Euclidean square=1*13+12;nonzero square class=12;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..12;actual square=36

Euclidean square=2*13+10;nonzero square class=10;complete root-index order retained

Quadratic ORIGINAL nonzero root7 square := mul(7, 7)

Quadratic ORIGINAL nonzero root7 square = 49

ORIGINAL nonzero root x=7 in1..12;actual square=49

Euclidean square=3*13+10;nonzero square class=10;complete root-index order retained

Quadratic ORIGINAL nonzero root8 square := mul(8, 8)

Quadratic ORIGINAL nonzero root8 square = 64

ORIGINAL nonzero root x=8 in1..12;actual square=64

Euclidean square=4*13+12;nonzero square class=12;complete root-index order retained

Quadratic ORIGINAL nonzero root9 square := mul(9, 9)

Quadratic ORIGINAL nonzero root9 square = 81

ORIGINAL nonzero root x=9 in1..12;actual square=81

Euclidean square=6*13+3;nonzero square class=3;complete root-index order retained

Quadratic ORIGINAL nonzero root10 square := mul(10, 10)

Quadratic ORIGINAL nonzero root10 square = 100

ORIGINAL nonzero root x=10 in1..12;actual square=100

Euclidean square=7*13+9;nonzero square class=9;complete root-index order retained

Quadratic ORIGINAL nonzero root11 square := mul(11, 11)

Quadratic ORIGINAL nonzero root11 square = 121

ORIGINAL nonzero root x=11 in1..12;actual square=121

Euclidean square=9*13+4;nonzero square class=4;complete root-index order retained

Quadratic ORIGINAL nonzero root12 square := mul(12, 12)

Quadratic ORIGINAL nonzero root12 square = 144

ORIGINAL nonzero root x=12 in1..12;actual square=144

Euclidean square=11*13+1;nonzero square class=1;complete root-index order retained

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

Quadratic distinct nonzero residue count = 6

Quadratic ORIGINAL modulus minus1 := sub(13, 1)

Quadratic ORIGINAL modulus minus1 = 12

Quadratic half of all nonzero roots := div(12, 2)

Quadratic half of all nonzero roots = 6

ORIGINAL ODD PRIME p=13;complete nonzero square table=[[1, 1], [2, 4], [3, 9], [4, 3], [5, 12], [6, 10], [7, 10], [8, 12], [9, 3], [10, 9], [11, 4], [12, 1]];distinct classes=[1, 3, 4, 9, 10, 12]

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=6=(13-1)/2=6

Quadratic ORIGINAL candidate class Euclidean quotient times modulus := mul(0, 13)

Quadratic ORIGINAL candidate class Euclidean quotient times modulus = 0

Quadratic ORIGINAL candidate class ORIGINAL signed value minus multiple := sub(12, 0)

Quadratic ORIGINAL candidate class ORIGINAL signed value minus multiple = 12

Quadratic ORIGINAL candidate class:ORIGINAL signed value=12;positive modulus=13;actual quotient=0

canonical residue=12;value=quotient*modulus+residue and0<=residue<modulus

Quadratic negative of FIRST source root Euclidean quotient times modulus := mul(-1, 13)

Quadratic negative of FIRST source root Euclidean quotient times modulus = -13

Quadratic negative of FIRST source root ORIGINAL signed value minus multiple := sub(-5, -13)

Quadratic negative of FIRST source root ORIGINAL signed value minus multiple = 8

Quadratic negative of FIRST source root:ORIGINAL signed value=-5;positive modulus=13;actual quotient=-1

canonical residue=8;value=quotient*modulus+residue and0<=residue<modulus

逐步反向核對

• S036 · 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。
• S035 · 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。
• S034 · 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。
• S033 · 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。
• S032 · 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。
• S031 · 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。
• S030 · 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。
• S029 · 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。
• S028 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL modulus minus1，不以小數近似替代。
• S027 · 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。
• S026 · 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。
• S025 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL nonzero root12 square，不以小數近似替代。
• S024 · 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。
• S023 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL nonzero root11 square，不以小數近似替代。
• S022 · 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。
• S021 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL nonzero root10 square，不以小數近似替代。
• S020 · 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。
• S019 · premise_and_result_audit — 以精確有理數重算 Quadratic ORIGINAL nonzero root9 square，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 ORIGINAL nonzero root8 square，不以小數近似替代。
• 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 ORIGINAL nonzero root7 square，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

(13, 12, [5, 8], 8, 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。