# Let $f(n)$ be the sum of positive divisors of n. Prove that $f(3n+2)$ is divisible by $3$ and $f(4n+3)$ is divisible by $4$ for any positive $n$.

I have a question about in Euler-phi function chapter in the book 'Elementary Number Theory' by David M. Burton(6th Edition) p.145 exercise number 9.

The problem is :

Let $f(n)$ be the sum of positive divisors of n. Prove that $f(3n+2)$ is divisible by $3$ and $f(4n+3)$ is divisible by $4$ for any positive $n$.

Let $d_1, \ldots, d_m$ be the distinct positive divisors of $k$. If $k \equiv 2 \mod 3$ then $k$ is not a square, so therefore for each $i$ there is a $j \neq i$ such that $k = d_id_j$. Then $d_i$ and $d_j$ must be distinct mod $3$ (because their product is 2) and they are not zero mod 3 (else $k$ would be divisible by $3$) so their sum is $0$ mod 3.
Use a similar trick if $k \equiv 3 \mod 4$.