标题:
求教数学寂静219
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作者:
nancy5130
时间:
2013-1-24 20:40
标题:
求教数学寂静219
219. 已知x,y都是整数,问y是否是odd?
(1)2y-x=x2-y2
(2)x为偶数
【解释】
(1)2y-x=(x-y)(x+y)
若x=even,y=even =》(x-y)(x+y)=even 2y-x=even
若x=even,y=odd =》(x-y)(x+y)=odd 2y-x=even 舍去
若x=odd ,y=even =》(x-y)(x+y)=odd 2y-x=odd
若x=odd ,y=odd =》(x-y)(x+y)=even 2y-x=odd 舍去
满足条件的情况下 y=even S
2)NS
【确定】A
看了半天觉得不太对劲。有没有哪位大虾可以解释一下!拜谢!
作者:
winnielxx
时间:
2013-1-25 06:57
since: 2y-x=(x-y)(x+y)
so 1) x-y)(x+y)=odd 2y-x=even 舍去
2) (x-y)(x+y)=even 2y-x=odd 舍去
自相矛盾,舍去。
剩下的, y=even, bingo!
作者:
nancy5130
时间:
2013-1-25 21:48
你这么一说突然间就明白了
谢谢!
作者:
louisehu
时间:
2013-1-27 20:35
(1) 2y-x = x^2-y^2 => y^2+2y= x^2+x => y(y+2)= x(x+1)
since x(x+1) must be even given x is integer, therefore two possibilities: a) between y and y+2, one is odd and one is even; or b) both y and y+2 are even. a) can not be true given y is integer, so y must be even.
作者:
ozymen
时间:
2013-1-29 06:43
额, 2y-x=x2-y2 = > x(x+1) = y(y+2)
x与x+1必定一个even, 一个odd. 所以x(x+1)必定是even, 那么y(y+2)也必定even,所以y必须even. 因为如果y是odd的话,y+2也是odd, odd*odd = odd
作者:
Cecile8chen
时间:
2013-1-30 06:38
看完你的答案突然就。。豁然开朗。。
JJ上的步骤太复杂啦
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