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
Crossref DOI link: https://doi.org/10.1007/s00605-014-0660-0
Published Online: 2014-06-27
Published Print: 2015-07
Update policy: https://doi.org/10.1007/springer_crossmark_policy
Grau, José María
Oller-Marcén, Antonio M.
Sondow, Jonathan
Text and Data Mining valid from 2014-06-27