Is it true that if $p\neq5$ is a prime number then $1^p+2^{p-1}+\cdots+(p-1)^2+p^1\not\equiv0\pmod p$?

If $p=5$ then $1^5+2^4+3^3+4^2+5^1\equiv0\pmod 5,$ but there is no such prime $p\leq40000$ any more, can you prove it or give a counterexample?

PS: If $p>5$ could be a composite number, then $p\in \{16,208,688,784,2864,9555\cdots\}$ also works.


$p=81239$ is a counterexample.

  • 2
    $\begingroup$ How did you come up with this? $\endgroup$ – Kunnysan Jul 30 '13 at 7:42
  • $\begingroup$ @Kunnysan Just a brute-force search, nothing sophisticated. $\endgroup$ – Zander Jul 30 '13 at 11:18

This sounds like a $\sum\frac 1p$ distribution, similar to "If $p \mid 10^x-1$ then so does $p^2$". This has solutions for $3$, $487$, and $56598313$, and no other values under $120^4$.

One just has to go out and have a look.


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