第 19631 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-general-solution；資料={"a":-42,"b":63,"c":126}。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-general-solution；資料={"a":-42,"b":63,"c":126}。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-general-solution;original snapshot and mathematics retained

explicit ordered source output fields=[x0, y0, step_x, step_y, x_formula, y_formula, certificate]

ORIGINAL signed coefficients a=-42,b=63

start exact Euclidean chain on absolute values=[42, 63];nonnegative gcd;source deterministic coefficient recurrence

Dioph Euclid step0 exact ratio := div(42, 63)

Dioph Euclid step0 exact ratio = 2/3

Dioph Euclid step0 divisor multiple := mul(0, 63)

Dioph Euclid step0 divisor multiple = 0

Dioph Euclid step0 remainder := sub(42, 0)

Dioph Euclid step0 remainder = 42

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=63;old pair=[42, 63];exact ratio=2/3;floor quotient=0;old coefficient pairs=[[1, 0], [0, 1]]

next remainder=42 in[0,63);next signed absolute-input coefficients=[1, 0];common divisor set unchanged

Dioph Euclid step1 exact ratio := div(63, 42)

Dioph Euclid step1 exact ratio = 3/2

Dioph Euclid step1 divisor multiple := mul(1, 42)

Dioph Euclid step1 divisor multiple = 42

Dioph Euclid step1 remainder := sub(63, 42)

Dioph Euclid step1 remainder = 21

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=42;old pair=[63, 42];exact ratio=3/2;floor quotient=1;old coefficient pairs=[[0, 1], [1, 0]]

next remainder=21 in[0,42);next signed absolute-input coefficients=[-1, 1];common divisor set unchanged

Dioph Euclid step2 exact ratio := div(42, 21)

Dioph Euclid step2 exact ratio = 2

Dioph Euclid step2 divisor multiple := mul(2, 21)

Dioph Euclid step2 divisor multiple = 42

Dioph Euclid step2 remainder := sub(42, 42)

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=21;old pair=[42, 21];exact ratio=2;floor quotient=2;old coefficient pairs=[[1, 0], [-1, 1]]

next remainder=0 in[0,21);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(-42, 1)

Dioph ORIGINAL a times Bezout x = -42

Dioph ORIGINAL b times Bezout y := mul(63, 1)

Dioph ORIGINAL b times Bezout y = 63

Dioph original signed Bezout identity := sum(-42, 63)

Dioph original signed Bezout identity = 21

complete exact chain=[[42, 63, 0, 42], [63, 42, 1, 21], [42, 21, 2, 0]];terminal pair=[21, 0];sign-restored x=1,y=1

a*x+b*y=21=nonnegative gcd d=21;gcd from invariance and terminal gcd(d,0)=d

Dioph target divided by positive gcd := div(126, 21)

Dioph target divided by positive gcd = 6

ACTUAL original equation-42*x+63*y=126;positive gcd=21;exact c/d=6;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=是

Dioph particular x0 from scaled source Bezout := mul(1, 6)

Dioph particular x0 from scaled source Bezout = 6

Dioph particular y0 from scaled source Bezout := mul(1, 6)

Dioph particular y0 from scaled source Bezout = 6

Dioph original a times particular x0 := mul(-42, 6)

Dioph original a times particular x0 = -252

Dioph original b times particular y0 := mul(63, 6)

Dioph original b times particular y0 = 378

Dioph particular original equation check := sum(-252, 378)

Dioph particular original equation check = 126

Dioph normalized original a over gcd := div(-42, 21)

Dioph normalized original a over gcd = -2

Dioph normalized original b over gcd := div(63, 21)

Dioph normalized original b over gcd = 3

Dioph signed general y step := mul(-1, -2)

Dioph signed general y step = 2

Dioph homogeneous a times x step := mul(-42, 3)

Dioph homogeneous a times x step = -126

Dioph homogeneous b times y step := mul(63, 2)

Dioph homogeneous b times y step = 126

Dioph full homogeneous step cancellation := sum(-126, 126)

Dioph full homogeneous step cancellation = 0

particular original pair=[6, 6] solves c=126;normalized coprime A=-2,B=3;step pair=[3, 2]

for EVERY integer t,x=x0+(3)t,y=y0+(2)t gives left side=c+t*0=c;all displayed solutions valid

ANY other integer solution;differences dx=x-x0,dy=y-y0 satisfy A*dx+B*dy=0;A=-2,B=3;source Bezout gives1*A+1*B=1

dx=B*(-(1)*dy+(1)*dx),so t=dx/B is an INTEGER;B!=0 then dy=-A*t;ALL solutions have the displayed form

逐步反向核對

• S047 · 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。
• S046 · 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。
• S045 · premise_and_result_audit — 以精確有理數重算 Dioph full homogeneous step cancellation，不以小數近似替代。
• S044 · premise_and_result_audit — 以精確有理數重算 Dioph homogeneous b times y step，不以小數近似替代。
• S043 · premise_and_result_audit — 以精確有理數重算 Dioph homogeneous a times x step，不以小數近似替代。
• S042 · premise_and_result_audit — 以精確有理數重算 Dioph signed general y step，不以小數近似替代。
• S041 · premise_and_result_audit — 以精確有理數重算 Dioph normalized original b over gcd，不以小數近似替代。
• S040 · premise_and_result_audit — 以精確有理數重算 Dioph normalized original a over gcd，不以小數近似替代。
• S039 · premise_and_result_audit — 以精確有理數重算 Dioph particular original equation check，不以小數近似替代。
• S038 · premise_and_result_audit — 以精確有理數重算 Dioph original b times particular y0，不以小數近似替代。
• S037 · premise_and_result_audit — 以精確有理數重算 Dioph original a times particular x0，不以小數近似替代。
• S036 · premise_and_result_audit — 以精確有理數重算 Dioph particular y0 from scaled source Bezout，不以小數近似替代。
• S035 · premise_and_result_audit — 以精確有理數重算 Dioph particular x0 from scaled source Bezout，不以小數近似替代。
• 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

(6, 6, 3, 2, x=6+(3)t, y=6+(2)t, all_and_only_integer_solutions_for_t_in_Z)

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。