第 20002 題：Euler定理

原題與已知條件

n大於等於2、exact gcd、phi與安全modular exponentiation。操作=eulerthm-congruence；資料={"n":6,"a":5}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — n大於等於2、exact gcd、phi與安全modular exponentiation。操作=eulerthm-congruence；資料={"n":6,"a":5}。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-euler-theorem-number-theory

逐步推導

source operation=eulerthm-congruence;original snapshot and mathematics retained

explicit ordered source output fields=[n, normalized_base, totient, residue, congruent_to_one]

Euler ORIGINAL signed base normalization exact quotient := div(5, 6)

Euler ORIGINAL signed base normalization exact quotient = 5/6

Euler ORIGINAL signed base normalization floor quotient multiple := mul(0, 6)

Euler ORIGINAL signed base normalization floor quotient multiple = 0

Euler ORIGINAL signed base normalization normalized remainder := sub(5, 0)

Euler ORIGINAL signed base normalization normalized remainder = 5

Euler ORIGINAL signed base normalization:ORIGINAL signed integer=5;modulus=6;exact ratio=5/6;floor quotient=0

value=q*n+remainder;normalized residue=5 in[0,6);same congruence class,NOT absolute-value replacement

Euler ORIGINAL base applicability gcd:ORIGINAL gcd inputs=[5, 6];nonnegative initial pair=[5, 6]

ALL exact Euclidean rows[left,right,q,r]=[[5, 6, 0, 5], [6, 5, 1, 1], [5, 1, 5, 0]];terminal gcd=1

Euler residue representative0:ORIGINAL gcd inputs=[0, 6];nonnegative initial pair=[0, 6]

ALL exact Euclidean rows[left,right,q,r]=[[0, 6, 0, 0]];terminal gcd=6

Euler residue representative1:ORIGINAL gcd inputs=[1, 6];nonnegative initial pair=[1, 6]

ALL exact Euclidean rows[left,right,q,r]=[[1, 6, 0, 1], [6, 1, 6, 0]];terminal gcd=1

Euler residue representative2:ORIGINAL gcd inputs=[2, 6];nonnegative initial pair=[2, 6]

ALL exact Euclidean rows[left,right,q,r]=[[2, 6, 0, 2], [6, 2, 3, 0]];terminal gcd=2

Euler residue representative3:ORIGINAL gcd inputs=[3, 6];nonnegative initial pair=[3, 6]

ALL exact Euclidean rows[left,right,q,r]=[[3, 6, 0, 3], [6, 3, 2, 0]];terminal gcd=3

Euler residue representative4:ORIGINAL gcd inputs=[4, 6];nonnegative initial pair=[4, 6]

ALL exact Euclidean rows[left,right,q,r]=[[4, 6, 0, 4], [6, 4, 1, 2], [4, 2, 2, 0]];terminal gcd=2

Euler residue representative5:ORIGINAL gcd inputs=[5, 6];nonnegative initial pair=[5, 6]

ALL exact Euclidean rows[left,right,q,r]=[[5, 6, 0, 5], [6, 5, 1, 1], [5, 1, 5, 0]];terminal gcd=1

Euler complete reduced residue count phi := sum(1, 1)

Euler complete reduced residue count phi = 2

COMPLETE original residue carrier0..5;ALL [representative,gcd,unit] rows=[[0, 6, 否], [1, 1, 是], [2, 2, 否], [3, 3, 否], [4, 2, 否], [5, 1, 是]]

increasing COMPLETE U(n)=[1, 5];phi(n)=2;source count over1..n agrees because0 and n are BOTH nonunits for n>=2

ORIGINAL normalized base=5;COMPLETE totient exponent=2;modulus=6

initialize accumulator1,current_base=base,remaining=phi;invariant accumulator*current_base^remaining congruent to the original totient power

Euler power bit0 base square := mul(5, 5)

Euler power bit0 base square = 25

Euler power bit0 squared-base reduction exact quotient := div(25, 6)

Euler power bit0 squared-base reduction exact quotient = 25/6

Euler power bit0 squared-base reduction floor quotient multiple := mul(4, 6)

Euler power bit0 squared-base reduction floor quotient multiple = 24

Euler power bit0 squared-base reduction normalized remainder := sub(25, 24)

Euler power bit0 squared-base reduction normalized remainder = 1

Euler power bit0 squared-base reduction:ORIGINAL signed integer=25;modulus=6;exact ratio=25/6;floor quotient=4

value=q*n+remainder;normalized residue=1 in[0,6);same congruence class,NOT absolute-value replacement

step=0;previous[accumulator,base,exponent]=[1, 5, 2];low bit=0

next[accumulator,base,exponent]=[1, 1, 1];split exponent=2*q+bit preserves original-power invariant

Euler power bit1 accumulator product := mul(1, 1)

Euler power bit1 accumulator product = 1

Euler power bit1 accumulator reduction exact quotient := div(1, 6)

Euler power bit1 accumulator reduction exact quotient = 1/6

Euler power bit1 accumulator reduction floor quotient multiple := mul(0, 6)

Euler power bit1 accumulator reduction floor quotient multiple = 0

Euler power bit1 accumulator reduction normalized remainder := sub(1, 0)

Euler power bit1 accumulator reduction normalized remainder = 1

Euler power bit1 accumulator reduction:ORIGINAL signed integer=1;modulus=6;exact ratio=1/6;floor quotient=0

value=q*n+remainder;normalized residue=1 in[0,6);same congruence class,NOT absolute-value replacement

Euler power bit1 base square := mul(1, 1)

Euler power bit1 base square = 1

Euler power bit1 squared-base reduction exact quotient := div(1, 6)

Euler power bit1 squared-base reduction exact quotient = 1/6

Euler power bit1 squared-base reduction floor quotient multiple := mul(0, 6)

Euler power bit1 squared-base reduction floor quotient multiple = 0

Euler power bit1 squared-base reduction normalized remainder := sub(1, 0)

Euler power bit1 squared-base reduction normalized remainder = 1

Euler power bit1 squared-base reduction:ORIGINAL signed integer=1;modulus=6;exact ratio=1/6;floor quotient=0

value=q*n+remainder;normalized residue=1 in[0,6);same congruence class,NOT absolute-value replacement

step=1;previous[accumulator,base,exponent]=[1, 1, 1];low bit=1

next[accumulator,base,exponent]=[1, 1, 0];split exponent=2*q+bit preserves original-power invariant

remaining exponent0;final normalized accumulator=1

EXACT original normalized base^2 modulo6=1

Euler ACTUAL base-unit inverse product := mul(5, 5)

Euler ACTUAL base-unit inverse product = 25

Euler ACTUAL inverse check exact quotient := div(25, 6)

Euler ACTUAL inverse check exact quotient = 25/6

Euler ACTUAL inverse check floor quotient multiple := mul(4, 6)

Euler ACTUAL inverse check floor quotient multiple = 24

Euler ACTUAL inverse check normalized remainder := sub(25, 24)

Euler ACTUAL inverse check normalized remainder = 1

Euler ACTUAL inverse check:ORIGINAL signed integer=25;modulus=6;exact ratio=25/6;floor quotient=4

value=q*n+remainder;normalized residue=1 in[0,6);same congruence class,NOT absolute-value replacement

ACTUAL modulus n=6;gcd(original a=5,n)=1;normalized base=5;ACTUAL inverse=5;COMPLETE U(n)=[1, 5]

for ANY r in U(n),a*r remains a unit;actual inverse makes multiplication injective;finite self-injection on U(n) is surjective,so a*U(n)=U(n)

SAME complete U(n)=[1, 5] of size phi=2;multiplication by original a=5 permutes it

multiply ALL unit images:a^(2)*product(U) congruent product(U);EVERY U factor is a UNIT,so product(U) has reverse-product inverse;cancel ONLY this unit to get a^phi congruent1 modulo6

逐步反向核對

• S040 · 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。
• S039 · 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。
• S038 · 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。
• S037 · premise_and_result_audit — 以精確有理數重算 Euler ACTUAL inverse check normalized remainder，不以小數近似替代。
• S036 · premise_and_result_audit — 以精確有理數重算 Euler ACTUAL inverse check floor quotient multiple，不以小數近似替代。
• S035 · premise_and_result_audit — 以精確有理數重算 Euler ACTUAL inverse check exact quotient，不以小數近似替代。
• S034 · premise_and_result_audit — 以精確有理數重算 Euler ACTUAL base-unit inverse product，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 — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S031 · 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。
• S030 · premise_and_result_audit — 以精確有理數重算 Euler power bit1 squared-base reduction normalized remainder，不以小數近似替代。
• S029 · premise_and_result_audit — 以精確有理數重算 Euler power bit1 squared-base reduction floor quotient multiple，不以小數近似替代。
• S028 · premise_and_result_audit — 以精確有理數重算 Euler power bit1 squared-base reduction exact quotient，不以小數近似替代。
• S027 · premise_and_result_audit — 以精確有理數重算 Euler power bit1 base square，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 Euler power bit1 accumulator reduction normalized remainder，不以小數近似替代。
• S024 · premise_and_result_audit — 以精確有理數重算 Euler power bit1 accumulator reduction floor quotient multiple，不以小數近似替代。
• S023 · premise_and_result_audit — 以精確有理數重算 Euler power bit1 accumulator reduction exact quotient，不以小數近似替代。
• S022 · premise_and_result_audit — 以精確有理數重算 Euler power bit1 accumulator product，不以小數近似替代。
• 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 — 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 Euler power bit0 squared-base reduction normalized remainder，不以小數近似替代。
• S018 · premise_and_result_audit — 以精確有理數重算 Euler power bit0 squared-base reduction floor quotient multiple，不以小數近似替代。
• S017 · premise_and_result_audit — 以精確有理數重算 Euler power bit0 squared-base reduction exact quotient，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 Euler power bit0 base square，不以小數近似替代。
• S015 · 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。
• 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 Euler complete reduced residue count phi，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 — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• 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 — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• 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 — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• 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 — 回到原始資料與本步條件核對，不把單向推論誤稱等價。
• S004 · premise_and_result_audit — 以精確有理數重算 Euler ORIGINAL signed base normalization normalized remainder，不以小數近似替代。
• S003 · premise_and_result_audit — 以精確有理數重算 Euler ORIGINAL signed base normalization floor quotient multiple，不以小數近似替代。
• S002 · premise_and_result_audit — 以精確有理數重算 Euler ORIGINAL signed base normalization exact quotient，不以小數近似替代。
• 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

(6, 5, 2, 1, 是)

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。