On the congruence $$1^m + 2^m + \cdots + m^m\equiv n \pmod {m}$$ 1 m + 2 m + ⋯ + m m ≡ n ( mod m ) with $$n\mid m$$ n ∣ m

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ژورنال

عنوان ژورنال: Monatshefte für Mathematik

سال: 2014

ISSN: 0026-9255,1436-5081

DOI: 10.1007/s00605-014-0660-0