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Congruence modulo p^2

Source: Russian TST 2019, Day 6 P2

March 22, 2023
number theorycongruence

Problem Statement

Prove that for every odd prime number pp{}, the following congruence holds ∑n=1p−1np−1≡(p−1)!+p(modp2).\sum_{n=1}^{p-1}n^{p-1}\equiv (p-1)!+p\pmod{p^2}.
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