If $p$ is both Giuga and Carmichael number

then its known that

$1^{p-1}+2^{p-1}+3^{p-1}+\cdots+(p-1)^{p-1} \equiv -1\pmod{p}$

is it true that

if $p$ is both Giuga and Carmichael number then

$1^{p-1}+2^{p-1}+3^{p-1}+\cdots+(r-1)^{p-1} \equiv (r-1)\pmod{p}$ where $2\le r\le p-2$

Thanks in advance :)


closed as off-topic by Marco Golla, Joonas Ilmavirta, Will Sawin, Boris Bukh, Yemon Choi Aug 15 '15 at 23:34

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  • $\begingroup$ So your sum should run from 1 to $k-1$, not to $k$, right? $\endgroup$ – Gerry Myerson Feb 4 '14 at 12:21
  • $\begingroup$ i don't know the range . i have doubt whether it runs from 0 to $k-1$ or 0 to $p-1$. $\endgroup$ – hanugm Feb 4 '14 at 12:24
  • $\begingroup$ Crossposted on MSE math.stackexchange.com/questions/663086/… $\endgroup$ – Tobias Kildetoft Feb 4 '14 at 12:24
  • $\begingroup$ yeah i had not get solution there, thats why i asked here. $\endgroup$ – hanugm Feb 4 '14 at 12:26
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    $\begingroup$ You asked here only 3 hours after asking there, and without giving any sort of indication of this. Please do not do this, as it can cause duplication of effort. $\endgroup$ – Tobias Kildetoft Feb 4 '14 at 12:27

If $n$ is Carmichael, then $a^{n-1}\equiv1\pmod n$ for all $a$ with $\gcd(a,n)=1$. If $\gcd(a,n)\ne1$, then it is clearly impossible to have $a^{n-1}\equiv1\pmod n$. So, let $n$ be Carmichael, let $r$ be the smallest divisor of $n$ (other than 1); then it is impossible to have $1^{n-1}+2^{n-1}+\cdots+r^{n-1}\equiv r\pmod n$, since the first $r-1$ terms are 1 (mod $n$) and the last term isn't.

  • $\begingroup$ Its also giuga number , any difference ? $\endgroup$ – hanugm Feb 5 '14 at 5:55
  • $\begingroup$ @hanu If it always fails for Carmichael numbers, then it does not help to restrict to a subset of the Carmichael numbers. $\endgroup$ – Tobias Kildetoft Feb 5 '14 at 8:28
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    $\begingroup$ The only difference it makes is that if there aren't any numbers that are both Carmichael and Giuga (and that's what Giuga's conjecture asserts, isn't it?), then the premise is vacuous, and the conclusion follows vacuously. $\endgroup$ – Gerry Myerson Feb 5 '14 at 10:41

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