Re: [其他] 數論

看板Math作者nonumber (空號)時間13年前 (2013/01/02 17:30)推噓2(2推 0噓 5→)

留言7則, 3人參與討論串2/2 (看更多)

※ 引述《CXTRV (C_ZERO)》之銘言： : 求所有正整數x,y (x>1)滿足 2^x+1=x^2*y : 本人完全沒想法... 2^x+1=x^2*y leftside is odd so x is odd bothside mod x 2^x≡-1 mod x imply 2^(2x)≡1 mod x by little fermat thm 2^p(x)≡1 mod x for gcd(2,x)=1 so p(x) | 2x since x is odd , rightside has only one '2' imply x must have only one prime divisor and let x=p^r there exist n>0 such that n*p(x)=2x n*(p^r-p^(r-1))=2*p^r p=n/(n-2) only one solution , n=3 and p=3 r r-1 r-1 r-1 3 3 2*3 3 2r 2 + 1 = ( 2 + 1 )(2 - 2 + 1)≡0 mod 3 r=1 done 2r r>1 no enough 3 than 3 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.113.67.128

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GaussQQ

01/03 02:27, , 1^F

01/03 02:27, 1^F

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GaussQQ

01/03 02:28, , 2^F

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GaussQQ

01/03 02:30, , 3^F

01/03 02:30, 3^F