Re: [代數] 一題數論證明
※ 引述《inwa (話太多)》之銘言：
: 請問若a、b為互質正整數
: 如何證明ab-a-b不可能寫成ax+by(其中x、y為非負整數)
: 且ab-a-b+k(k為任意正整數)可以寫成ax+by(其中x、y為非負整數)
: 感激不盡
a=1 trivial
assume a >=2
assume b >= a+1
we know if ax+by=1
then if (x,y)=(p,q) is a solution
then (x,y)=(p-bt,q+at) is also a solution
hence, we may take 1 <= y <= a-1
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ax+by=ab-a-b+1
a(x-b+1)+b(y+1)=1
we may take 1 <= y+1 <= a-1
then b <= b(y+1) <= b(a-1)
1-b(a-1) <= 1-b(y+1)
1-b(a-1) <= a(x-b+1)
1-ab+b <= ax -ab + a
ax >= 1+b-a > 0
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ax+by=k is straightforward
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a=1 trivial
a>=2
b>=a+1
if ax+by=ab-a-b
then a(x-b+1)+b(y+1)=0
x-b+1=bk , y+1=-ak
y >= 0, then k <= -1
x-b+1<=-b, x<=-1
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老梗題 old question
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◆ From: 123.194.96.239
※ 編輯: JohnMash 來自: 123.194.96.239 (03/21 16:52)
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03/21 20:59, , 1F
03/21 20:59, 1F
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