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This will be my second question here in math.stackexchange (so far).

This time, I am trying to consider the cases which give $\sigma(n) \equiv 0 \pmod 4$ whenever $n$ is odd. I get the following "lemma":

$\mathbf{Lemma:}$ If $n = \displaystyle\prod_{i = 1}^{r}{{p_i}^{\alpha_i}}$ is odd, then $\sigma(n) \equiv 0 \pmod 4$ when:

(i) there exists an $i$ such that the prime $p_i \equiv 3 \pmod 4$ has corresponding $\mathbf{odd}$ exponent $\alpha_i$ with ${p_i}^{\alpha_i}||n$; or

(ii) there exists an $i$ such that the prime $p_i \equiv 1 \pmod 4$ has corresponding exponent $\alpha_i \equiv 3 \pmod 4$ with ${p_i}^{\alpha_i}||n$; or

(iii) there exist $i$ and $j$ (with $i \neq j$) such that the primes $p_i$ and $p_j$ satisfy $p_i \equiv p_j \equiv 1 \pmod 4$ and have corresponding exponents $\alpha_i$ and $\alpha_j$ satisfying $\alpha_i \equiv \alpha_j \equiv 1 \pmod 4$ with ${p_i}^{\alpha_i}||n$ and ${p_j}^{\alpha_j}||n$.

My question now is: Is this (already) an exhaustive list of conditions for $\sigma(n) \equiv 0 \pmod 4$, whenever $n$ is odd?

Any ideas/comments/suggestions are most welcome.

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up vote 3 down vote accepted

$\sigma$ is a multiplicative function, i.e. if $m$ and $n$ are coprime, $\sigma(mn) = \sigma(m) \sigma(n)$. If prime $p \equiv 1 \mod 4$, $\sigma(p^k) = 1 + p + \ldots + p^k \equiv k+1 \mod 4$, while if prime $p \equiv 3 \mod 4$, $\sigma(p^k) \equiv 0 \mod 4$ if $k$ is odd, $\equiv 1 \mod 4$ if $k$ is even. Also note that $\sigma(2^k) \equiv 3 \mod 4$ for all $k \ge 1$.

The only way a product of integers can be divisible by $4$ is that one of the factors is divisible by $4$ or at least two are divisible by $2$. So, regardless of whether $n$ is odd or even, if $n = \prod_{i=1}^r p_i^{\alpha_i}$, $\sigma(n) \equiv 0 \mod 4$ iff at least one of the following holds:

  • there is at least one $p_i \equiv 3 \mod 4$ for which $\alpha_i$ is odd

  • there is at least one $p_i \equiv 1 \mod 4$ for which $\alpha_i \equiv 3 \mod 4$

  • there are at least two $p_i \equiv 1 \mod 4$ for which $\alpha_i \equiv 1 \mod 4$.

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Thank you very much for the detailed explanation Robert! So, can we say now that my list of conditions above is indeed exhaustive? :) –  Jose Arnaldo Dris Jun 20 '12 at 14:53
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Yes, it is exhaustive. Of course you should clarify that $i \ne j$. –  Robert Israel Jun 20 '12 at 16:35
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