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a的平方-1一定被8整除(a為奇數) 證明:設a為2n+1,(n為整數) (2n+1)的平方-1=2n(2n+2)=4n(n+1) n,n+1為連續整數,所以其中必有一個為偶數 所以n(n+1)是2的倍數 囙此(2n+1)的平方一定整除8 即a的平方一定整除8 得證
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