推 Desperato : 推推推推推想超久qw q 10/05 12:26
推 rfvbgtsport : 感謝大大們 10/05 15:35
※ 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.
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