Is $n=6$ the only integer satisfies ${\sigma}_x(n) \equiv 0\bmod{n}$ for every odd integer $x > 0$ and $2 (\bmod n)$ if $x$ is even integer? After a few computations in wolfram alpha about the divisor function for some values of $n$ to look the behavior of $\sigma_x(n)\bmod n$ for $\,n=6,\,$ i got this result : $\sigma_x(6)=0 \bmod 6$ for $x$ odd and 2 mod 6 if $x$ is even 
Edit:01 :${\sigma}_x(n) =\sum_{d|n} d^x$ is the sum divisor function 
Note:01:I edited the question just to define the sum divisor function
My question here :
Is $n=6$ the only integer satisfies $\sigma_x(n)
\equiv 0\bmod n$ for every odd integer  $x > 0$  and $2 \bmod n$ if $x$ is even integer ? and if it is how do i show this ?
Note :02:I want to know more about periodicity of the divisor function 
Thank you for any help 
 A: Let $r=\gcd(k,e+1)$, and $p$ a prime. Then $\sigma_k(p^e) \equiv r\frac{p^{e+1}-1}{p^r -1} \bmod \sigma(p^e)$. Also, $r=1$ if and only if $\sigma(p^e)$ divides $\sigma_k(p^e)$.  Thus for $k$ coprime to $\tau(n)$, we have $\sigma(n)$ divides $\sigma_k(n)$.  The relation also suggests that for a given $n$ the sequence $\sigma_k(n)\bmod \sigma(n)$ is periodic in $k$ with a period dividing $L$, the least common multiple of ($1+$ each exponent) in the prime factorization of $n$.  Edit 2016.01.04: Once can show a nonreduced representation $\sigma_k(n) = a_k\sigma(n)/b_k$ where the $b_k$ are integers not necessarily coprime to the integers $a_k$ or to $a_k\sigma(n)$, with the property that the $b_k$ are bounded and periodic with period $L$.  This is not enough to show $\sigma_k(n) \bmod \sigma(n)$ is periodic with small period, unfortunately. End Edit 2016.01.04.
If now $n$ is multiperfect (so $n$ divides $\sigma(n)$) we have $n$ divides $\sigma_k(n)$ for $k$ coprime to $\tau(n)$.  In particular if $\tau(n)$ is a power of $2$, then $n$ divides $\sigma_k(n)$ for all odd $k \gt 0$.
It is still possible that $n$ can divide $\sigma_k(n)$ for $k$ not coprime
to $\tau(n)$.  However if $L$ is not prime, it seems likely that there will be more than one nonzero value of $\sigma_k(n) \bmod \sigma(n)$.  If this is so, it would be one ingredient in a proof that 6 is the unique number having the titled properties, the other ingredient being that 6 is the only nontrivial multiperfect number with $L$ a prime.
Edit 2016.01.10: I botched an earlier edit which claimed that 6 is the only known multiperfect number $n$ which satisfies $\sigma_2(n) \bmod n = 2$.  It is true, but the analysis had some flaws.  However, one expects multiperfect numbers other than 1 and 6 to be a multiple of 4; when $n$ satisfies $\sigma(n) \bmod n = 0$ and $\sigma_2(n) \bmod n = 2$, and in addition $ n \bmod 4  = 0$, then all odd prime factors of $n$ except one must occur to an even multiplicity, and the remaining odd prime factor must occur to a multiplicity of 1 mod 4 and must be a prime that is 3 mod 4.  While simple, these observations say a lot about $n$ and suggest that any numbers satisfying the title congruences are rare indeed, perhaps more so than odd multiperfect numbers.  End Edit 2016.01.10
Gerhard "Mea Culpa, Mea Maxima Culpa" Paseman, 2016.01.03
