Answer by Ivan Loh for $m \in \{2,6,42,1806,...\} $ - a problem of...
Quick answer: No for all three questions.Explanation: Let us suppose $r(m)=1$. Let $p$ be a prime factor of $m$, then we must have $$1 \equiv f(m)=\sum_{k=0}^{m-1}{k^m} \equiv...
View Article$m \in \{2,6,42,1806,...\} $ - a problem of sum-of-$m$'th powers modulo $m$
(continuing the work for an answer for a question here in MSE and also in MO)I'm (re-)viewing the function $$ f(m) = \sum_{k=0}^{m-1} k^m $$ considering its residue modulo $m$:$$ r(m) \equiv f(m) \pmod...
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