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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 :)

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  • $\begingroup$ So your sum should run from 1 to $k-1$, not to $k$, right? $\endgroup$ Feb 4, 2014 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, 2014 at 12:24
  • $\begingroup$ Crossposted on MSE math.stackexchange.com/questions/663086/… $\endgroup$ Feb 4, 2014 at 12:24
  • $\begingroup$ yeah i had not get solution there, thats why i asked here. $\endgroup$
    – hanugm
    Feb 4, 2014 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. $\endgroup$ Feb 4, 2014 at 12:27

1 Answer 1

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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.

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  • $\begingroup$ Its also giuga number , any difference ? $\endgroup$
    – hanugm
    Feb 5, 2014 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. $\endgroup$ Feb 5, 2014 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. $\endgroup$ Feb 5, 2014 at 10:41

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