MQ70 --- Number Theory
回答 (2)
2013-01-19 4:46 am
✔ 最佳答案
Let x = a + d where a >= 1, 0 < d < 1[x]^2 = x {x}
a^2 = d(a + d) = ad + d^2
a(a - d) = d^2
As 0 < d^2 < 1=> a < 2 => a = 1
So, we can set x = 1 + d
1 = d + d^2
d^2 + d - 1 = 0
d = (√5 - 1)/2
x = (√5 + 1)/2 ~ 1.618033989
2013-01-19 9:47 am
Good solution - missed the x = 0 case though.
收錄日期: 2021-04-13 19:15:32
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https://hk.answers.yahoo.com/question/index?qid=20130118000051KK00227