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能. 證明:若質數n+1不整除a,即a與0關於模n+1不同餘.於是,根據費馬小定理,有a^n與1關於模(n+1)同餘.同理有b^n與1關於模(n+1)同餘.於是必有: a^n-b^n與0關於模(n+1)同餘.即(n+1)整除a^n-b^n
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