我懂了,完整作法如下: 因 P 必為奇數 → 2 ^ P = 2 ( mod 3 ) 設 P = 3N + K (K=1,2) ( 3N + 1 ) ^ 2 = 9N^2 + 6N + 1 = 1 ( mod 3 ) ( 3N + 2 ) ^ 2 = 9N^2 + 12N + 1 = 1 ( mod 3 ) 1+2=0 ( mod 3 )
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2003-04-02 18:06
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