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Math Help - sum of divisors

  1. #1
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    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).
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  2. #2
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    Quote Originally Posted by Cairo View Post
    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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