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My friend said that for any $n=p^a$, where $p$ is odd prime, $a$ is positive integer then: If $k$ is divisible by $p-1$ then $1^k+2^k+\cdots+n^k\equiv -p^{a-1}\pmod{p^a}$. I am very sure that his result is wrong. My though is simple: I use primitive root of $p^a$. But I fail to construct a counterexample for him. So my question is that: Could we construct an example of $k$ so that $p-1\mid k$ and $1^k+2^k+\cdots+n^k\not\equiv -p^{a-1}\pmod{p^a}$.

The second question is that: Does the following holds: $1^k+2^k+\cdots+n^k\not\equiv -p^{a-1}\pmod{p^a}$ if and only if $p-1\mid k$?

Hai Minh
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1 Answers1

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$n=4$, $p=2$, $a=2$, $k=3$, $1^3+2^3+3^3+4^3\equiv0\pmod4$ would seem to be a counterexample to your friend's statement.

Gerry Myerson
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