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\sum_{j=1}^p j^m is divisible by p

Source: 1986 Polish MO Finals p3

January 21, 2020
number theorySumdivisibleprimeprime divisor

Problem Statement

pp is a prime and mm is a non-negative integer <p−1< p-1. Show that ∑j=1pjm \sum_{j=1}^p j^m is divisible by pp.
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