Claim1: f(x1,x2,...,xn)=xi (i is not fixed) As If f(x1,x2,...,xn)=k=f(k,k,...,k),www.ddhw.com then from f(x1,x2,...,xn)=f(k,k,...,k), there at least is i, such that xi=k by condition 2 (otherwise f(x1,x2,...,xn)> So f(x1,x2,...,xn)=xi （转自：顶顶华闻 www.TopChineseNews.com ）We have to prove the i is independent from (x1,x2,...,xn). For convenience, wlog, we may assume there are distinct integers y1,y2,...,yn, such that f(y1,y2,...,yn)=y1 , then we must prove f(x1,...,xn)=x1 for any (x1,...,xn).www.ddhw.com let Z1={z1 in Z: f(z1,k,k,...k)=z1, for some k<>z1} let Z2={z2 in Z: f(k,z2,z2,...,z2)=z2, for some k<>z2} Clearly, the intersection of Z1 and Z2 is empty, otherwise , if z is a common integer f(z,k,k,...,k)=z=f(l,z,z,...,z), for some k<>z, l<>z, contradicts with condition2. Also we have: Z1UZ2=Z. (By Claim 1)www.ddhw.com （转自：顶顶华闻 www.TopChineseNews.com ）Let A={(x,y) in Z^2: f(x,y,y,...,y)=x} Let B={(x,y) in Z^2: f(x,y,y,...,y)=y} Clearly the intersection of A and B is {(x,x):xin Z} AUB=Z^2 （转自：顶顶华闻 www.TopChineseNews.com ）let z1 in Z1, z2 in Z2 If (z2,z1) is in A, f(z2,z1,z1,...,z1)=z2, then z2 is in Z1, contradiction. If (z2,z1) is in B, f(z2,z1,z1,...,z1)=z1, then z1 is in Z2, contradictionwww.ddhw.com So one of Z1, z2 is empty. （转自：顶顶华闻 www.TopChineseNews.com ）If Z1 is empty: f(z,k,k,...,k)=k for any z, k then f(m,y1,y1,...,y1)=y1=f(y1,y2,...,yn) for any m<>y1, as y1,...,yn are distinct, contradiction with condition2. So Z2 must be empty. f(z,k,...,k)=z for any z,k. If f(x1,x2,...,xn)=xi<>x1,www.ddhw.com then f(xi, m,m,...,m)=xi=f(x1,x2,...,xn) for any m beyond {x1,x2,...,xn}, contradiction with condition2. Therefore f(x1,x2,...,xn)=x1// www.ddhw.com （转自：顶顶华闻 www.TopChineseNews.com ）（转自：顶顶华闻 www.TopChineseNews.com ）（转自：顶顶华闻 www.TopChineseNews.com ）（转自：顶顶华闻 www.TopChineseNews.com ）（转自：顶顶华闻 www.TopChineseNews（转自：顶顶华闻 www.TopChineseNews.com ）（转自：顶We may prove it by induction on n n=1 trivial Define g(x2,x3...,xn)=f(x2,x2,...,xn) then g(k,k,...,k)=f(k,k,...,k)=kwww.ddhw.com if g(x2,x3,...,xn)=g(y2,y3,...,yn), then f(x2,x2,x3,...,xn)=f(y2,y2,y3,...,yn), thus at lease one pair of xi=yi, i>1 So g satisfy the two conditions, by assumption of induction, g(x2,...,xn)=xi for some fixed i>1. That is, f(x2,x2,...,xn)=xi, where i>1 is independent from x2,x3,...,xn. 顶华闻 www.TopChineseNews.com ）（转自：顶顶华闻 www.TopChineseNews.com ）www.ddhw.com |
哈哈哈哈 O(∩_∩)O哈哈~ 好久不来了，不小心看了一眼自己主页，发现自动关联了自己以往发过的帖，点进这个看了一眼，笑个半死。。。敢情我当初还发过这么搞笑的东西啊，我自己完全不记得了 
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发表于 2005-5-6 22:06:45
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发表于 2006-1-1 18:36:03
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