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 :)



Giuga and Carmichael numbers
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. 


