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For \(p>1\), integer \(p\) can only be prime iff\[\sum_{i=1}^{p-1}i^{p-1}\equiv-1\pmod p\]Prove?
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Whoops, ahaha, fixed the TeX.
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lol..
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Use fermat's little theorem. a^p = a which is equivalent to a^(p-1) = 1 Now proceed with the summation.
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