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\sum{i^{-p}} = 0 mod p^3

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September 28, 2005
modular arithmeticalgebranumber theoryDivisibility

Problem Statement

Let p>3p > 3 be a prime. Prove that if ∑i=1p−11ip=nm, \sum_{i=1 }^{p-1}{1\over i^p} = {n\over m}, with \gdc(n,m) = 1, then p3p^3 divides nn.
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