Thread: sum of divisors

Prove that 4 divides the sum of the divisors of 4k+3, where k is an integer.

Can somebody help me out here please? I think I can prove it if k is not a mulitple of 3 (i.e. a prime), bu i'm struggling with the case when k=3r (i.e. a composite number).

Originally Posted by Cairo

Prove that 4 divides the sum of the divisors of 4k+3, where k is an integer.

Can somebody help me out here please? I think I can prove it if k is not a mulitple of 3 (i.e. a prime), bu i'm struggling with the case when k=3r (i.e. a composite number).

let $\displaystyle n=4k+3=\prod_{i=1}^m p_i^{r_i}$ be the prime factorization of $\displaystyle n.$ then there exists $\displaystyle 1 \leq j \leq m$ such that $\displaystyle p_j \equiv 3 \mod 4$ and $\displaystyle r_j$ is odd. let $\displaystyle \sigma_1(n)$ be the sum of the divisors of $\displaystyle n.$ then: $\displaystyle \sum_{\ell = 0}^{r_j}p_j^{\ell} \mid \sigma_1(n).$

but we have: $\displaystyle \sum_{\ell=0}^{r_j}p_j^{\ell} \equiv \sum_{\ell=0}^{r_j}(-1)^{\ell} \equiv 0 \mod 4,$ because $\displaystyle r_j$ is odd. Q.E.D.