# Giuga and Carmichael numbers [closed]

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 $2\le r\le p-2$

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

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• So your sum should run from 1 to $k-1$, not to $k$, right? – Gerry Myerson Feb 4 '14 at 12:21
• i don't know the range . i have doubt whether it runs from 0 to $k-1$ or 0 to $p-1$. – hanugm Feb 4 '14 at 12:24
• Crossposted on MSE math.stackexchange.com/questions/663086/… – Tobias Kildetoft Feb 4 '14 at 12:24
• yeah i had not get solution there, thats why i asked here. – hanugm Feb 4 '14 at 12:26
• 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. – 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.