Re: [代數] 一題數論證明

[JohnMash (Paul)]

※ 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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※ 編輯: JohnMash        來自: 123.194.96.239       (03/21 16:52)
推 inwa        :thx a lot