Re: [代數]請前輩指點一下整數証明~急!謝謝

看板Math作者 (無聊ing ><^> .o O)時間9年前 (2016/10/05 11:36), 編輯推噓2(200)
留言2則, 2人參與, 最新討論串2/3 (看更多)
※ 引述《rfvbgtsport (uygh)》之銘言: : 若a,b,c為整數,且a/b+b/c+c/a為整數,b/a : +c/b+a/c亦為整數,証明丨a丨=丨b丨=丨c丨 : 請高手幫忙一下,謝謝 Without loss of the generality, we may assume gcd(a, b, c) = 1. Let Z1 = a/b + b/c + c/a and Z2 = b/a + c/b + a/c, so Z1 and Z2 are two integers. Note that Z1 + Z2 + 3 = (a + b + c)(ab + bc + ca) / abc is an integer. Thus a | (a + b + c)(ab + bc + ca) => a | bc(b + c). Since b(b + c)Z1 = a(b + c) + b^2(b + c)/c + bc(b + c)/a, we have c | b^3. Similarly, c | a^3. Then c | gcd(a^3, b^3, c^3) = 1 => |c| = 1. Similar argument for a and b, one has |a| = |b| = |c| = 1. -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 97.99.68.240 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1475638617.A.948.html

10/05 12:26, , 1F
推推推推推想超久qw q
10/05 12:26, 1F

10/05 15:35, , 2F
感謝大大們
10/05 15:35, 2F
文章代碼(AID): #1Nz7LPb8 (Math)
文章代碼(AID): #1Nz7LPb8 (Math)