第 19695 題：線性Diophantine方程

原題與已知條件

exact integer coefficients、normalized gcd與extended Euclidean證書。操作=dioph-solvability；資料={"a":-126,"b":189,"c":127}。請依 exact_ordered_tuple_declared_by_interface 欄序回答。

• G001 — exact integer coefficients、normalized gcd與extended Euclidean證書。操作=dioph-solvability；資料={"a":-126,"b":189,"c":127}。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-linear-diophantine-equation

逐步推導

source operation=dioph-solvability;original snapshot and mathematics retained

explicit ordered source output fields=[a, b, c, gcd, both_coefficients_zero, solvable, boundary]

ORIGINAL signed coefficients a=-126,b=189

start exact Euclidean chain on absolute values=[126, 189];nonnegative gcd;source deterministic coefficient recurrence

Dioph Euclid step0 exact ratio := div(126, 189)

Dioph Euclid step0 exact ratio = 2/3

Dioph Euclid step0 divisor multiple := mul(0, 189)

Dioph Euclid step0 divisor multiple = 0

Dioph Euclid step0 remainder := sub(126, 0)

Dioph Euclid step0 remainder = 126

Dioph Euclid step0 first coefficient multiple := mul(0, 0)

Dioph Euclid step0 first coefficient multiple = 0

Dioph Euclid step0 first coefficient difference := sub(1, 0)

Dioph Euclid step0 first coefficient difference = 1

Dioph Euclid step0 second coefficient multiple := mul(0, 1)

Dioph Euclid step0 second coefficient multiple = 0

Dioph Euclid step0 second coefficient difference := sub(0, 0)

Dioph Euclid step0 second coefficient difference = 0

step=0;positive divisor=189;old pair=[126, 189];exact ratio=2/3;floor quotient=0;old coefficient pairs=[[1, 0], [0, 1]]

next remainder=126 in[0,189);next signed absolute-input coefficients=[1, 0];common divisor set unchanged

Dioph Euclid step1 exact ratio := div(189, 126)

Dioph Euclid step1 exact ratio = 3/2

Dioph Euclid step1 divisor multiple := mul(1, 126)

Dioph Euclid step1 divisor multiple = 126

Dioph Euclid step1 remainder := sub(189, 126)

Dioph Euclid step1 remainder = 63

Dioph Euclid step1 first coefficient multiple := mul(1, 1)

Dioph Euclid step1 first coefficient multiple = 1

Dioph Euclid step1 first coefficient difference := sub(0, 1)

Dioph Euclid step1 first coefficient difference = -1

Dioph Euclid step1 second coefficient multiple := mul(1, 0)

Dioph Euclid step1 second coefficient multiple = 0

Dioph Euclid step1 second coefficient difference := sub(1, 0)

Dioph Euclid step1 second coefficient difference = 1

step=1;positive divisor=126;old pair=[189, 126];exact ratio=3/2;floor quotient=1;old coefficient pairs=[[0, 1], [1, 0]]

next remainder=63 in[0,126);next signed absolute-input coefficients=[-1, 1];common divisor set unchanged

Dioph Euclid step2 exact ratio := div(126, 63)

Dioph Euclid step2 exact ratio = 2

Dioph Euclid step2 divisor multiple := mul(2, 63)

Dioph Euclid step2 divisor multiple = 126

Dioph Euclid step2 remainder := sub(126, 126)

Dioph Euclid step2 remainder = 0

Dioph Euclid step2 first coefficient multiple := mul(2, -1)

Dioph Euclid step2 first coefficient multiple = -2

Dioph Euclid step2 first coefficient difference := sub(1, -2)

Dioph Euclid step2 first coefficient difference = 3

Dioph Euclid step2 second coefficient multiple := mul(2, 1)

Dioph Euclid step2 second coefficient multiple = 2

Dioph Euclid step2 second coefficient difference := sub(0, 2)

Dioph Euclid step2 second coefficient difference = -2

step=2;positive divisor=63;old pair=[126, 63];exact ratio=2;floor quotient=2;old coefficient pairs=[[1, 0], [-1, 1]]

next remainder=0 in[0,63);next signed absolute-input coefficients=[3, -2];common divisor set unchanged

Dioph restore original a sign := mul(-1, -1)

Dioph restore original a sign = 1

Dioph restore original b sign := mul(1, 1)

Dioph restore original b sign = 1

Dioph ORIGINAL a times Bezout x := mul(-126, 1)

Dioph ORIGINAL a times Bezout x = -126

Dioph ORIGINAL b times Bezout y := mul(189, 1)

Dioph ORIGINAL b times Bezout y = 189

Dioph original signed Bezout identity := sum(-126, 189)

Dioph original signed Bezout identity = 63

complete exact chain=[[126, 189, 0, 126], [189, 126, 1, 63], [126, 63, 2, 0]];terminal pair=[63, 0];sign-restored x=1,y=1

a*x+b*y=63=nonnegative gcd d=63;gcd from invariance and terminal gcd(d,0)=d

Dioph target divided by positive gcd := div(127, 63)

Dioph target divided by positive gcd = 127/63

ACTUAL original equation-126*x+189*y=127;positive gcd=63;exact c/d=127/63;Bezout coefficients=[1, 1]

d divides both coefficients,so ANY integer solution requires d divides c;conversely integer c/d scales Bezout to a solution;solvable=否

逐步反向核對

• S034 · 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。
• S033 · premise_and_result_audit — 以精確有理數重算 Dioph target divided by positive gcd，不以小數近似替代。
• 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 — 以精確有理數重算 Dioph original signed Bezout identity，不以小數近似替代。
• S030 · premise_and_result_audit — 以精確有理數重算 Dioph ORIGINAL b times Bezout y，不以小數近似替代。
• S029 · premise_and_result_audit — 以精確有理數重算 Dioph ORIGINAL a times Bezout x，不以小數近似替代。
• S028 · premise_and_result_audit — 以精確有理數重算 Dioph restore original b sign，不以小數近似替代。
• S027 · premise_and_result_audit — 以精確有理數重算 Dioph restore original a sign，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 Dioph Euclid step2 second coefficient difference，不以小數近似替代。
• S024 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step2 second coefficient multiple，不以小數近似替代。
• S023 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step2 first coefficient difference，不以小數近似替代。
• S022 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step2 first coefficient multiple，不以小數近似替代。
• S021 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step2 remainder，不以小數近似替代。
• S020 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step2 divisor multiple，不以小數近似替代。
• S019 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step2 exact ratio，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 Dioph Euclid step1 second coefficient difference，不以小數近似替代。
• S016 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step1 second coefficient multiple，不以小數近似替代。
• S015 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step1 first coefficient difference，不以小數近似替代。
• S014 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step1 first coefficient multiple，不以小數近似替代。
• S013 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step1 remainder，不以小數近似替代。
• S012 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step1 divisor multiple，不以小數近似替代。
• S011 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step1 exact ratio，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 Dioph Euclid step0 second coefficient difference，不以小數近似替代。
• S008 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step0 second coefficient multiple，不以小數近似替代。
• S007 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step0 first coefficient difference，不以小數近似替代。
• S006 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step0 first coefficient multiple，不以小數近似替代。
• S005 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step0 remainder，不以小數近似替代。
• S004 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step0 divisor multiple，不以小數近似替代。
• S003 · premise_and_result_audit — 以精確有理數重算 Dioph Euclid step0 exact ratio，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

(-126, 189, 127, 63, 否, 否, gcd_divides_c_iff_integer_solution)

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。