If $\sigma(n)$ is the sum of the divisors of $n$, show that $\sigma(6q + 5) \equiv 0 (mod 6)$ for all positive $q$. Is this an instance of a more general rule?



Sigma Congruency
The older reference I found is a review of D.H. Lehmer on a paper of Gupta:


MR0014365 (7,273f)
Gupta, Hansraj
Congruence properties of $\sigma(n)$.
Math. Student 13, (1945). 25–29.
10.0X


Let $\sigma\_a(n)$ denote, as usual, the sum of the $a$th powers of the divisors of $n$.
 Ramanathan has noted [Math. Student 11, 33--35 (1943); MR0010565 (6,37c)]
 that, for every $m>0$, $\sigma\_1(km-1)$ is divisible by $k$ whenever $k>2$ and is a divisor of 24.
 The present paper shows that this property holds for no other value of $k$. 
For the more general function $\sigma\_a(n)$, the author proves the following result noted
 by Ramanathan [see the preceding review]. In order that $\sigma\_a(km+r)$ be divisible by $k$
 for every $m>0$ it is first necessary that $r^a+1$ be divisible by $k$; for such $r$
 it is then both necessary and sufficient that $t^{2a}-1$ be divisible by $k$ 
for every $t$ prime to and less than $k$.
Reviewed by D. H. Lehmer


