※ 引述《billiechick (比利小鸡)》之铭言： : 利用数学归纳法 : 对任意正整数n,证明n^5-n为30的倍数 : 拜托各位了! 将证对所有整数, 命题成立 Let g_j(x)= Π(x-i) and f(x)=x^5-x. i=0~4 i!=j 1. f(-2),f(-1),f(0),f(1),f(2) 皆为30的倍数. 2. 假设命题对於 -2,-1,..,n-1 (>=2) 皆成立 By Lagrange, f(n)= sum f(j)g_j(n)/g_j(0) j=0~4 note that g_j(n)/g_j(0) is an integer and 30|f(j) by assumption, therefore, 30|f(n). 3. since f(x) is odd, the case of negative is obvious. --

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