sum of divisors

**Cairo** · Feb 11th 2009, 11:59 PM

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).

**NonCommAlg** · Feb 12th 2009, 01:10 AM

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

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

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