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

Question

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.

The naive heuristic argument says that there should be infinitely many primes with the sum congruent to 0 but this could be quite hard to prove because they're distributed very sparsely; you should only expect roughly three instances with $p\lt 10^6$! For related discussion, see http://math.stackexchange.com/questions/405944/an-infinitude-of-congruence-condition-primes and http://math.stackexchange.com/questions/337053/second-part-of-the-factorial-sum-divisibility-question/ . — Steven Stadnicki, Jul 29 '13 at 06:37

score 6 · Accepted Answer · answered Jul 29 '13 at 21:10

6

$p=81239$ is a counterexample.

answered Jul 29 '13 at 21:10

Zander

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How did you come up with this? – Kunnysan Jul 30 '13 at 07:42
@Kunnysan Just a brute-force search, nothing sophisticated. – Zander Jul 30 '13 at 11:18

wendy.krieger · Answer 2 · 2013-07-30T07:20:38.360

1

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.

edited Jul 30 '13 at 07:20

answered Jul 29 '13 at 08:11

wendy.krieger

6,856

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

2 Answers2