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This will be my very first post in math.stackexchange, so please bear with me if I make any silly mistakes with my maths.

So, to proceed: I am trying to calculate $\sigma(n^2) \mod 4$, given that $n$ is odd.

If I let $n = \displaystyle\prod_{i=1}^{r}{{p_i}^{{\alpha}_i}}$, then by considering the cases $p_i \equiv 1 \pmod 4$ and $p_i \equiv 3 \pmod 4$ separately (and taking the exponents ${\alpha}_i$ into consideration as well), I am led to the final congruence relation:

$\sigma(n^2) \equiv (-1)^c \pmod 4,$

where $c = \left|\left\{i|1 \le i \le r, p_i \equiv 1 \pmod4, 2 \nmid {\alpha}_i \right\}\right|.$

My question now would be: Is this as far as we could go with this congruence? I mean, is there no further possible improvement to this congruence, as far as computing $\sigma(n^2) \pmod 4$ for odd $n$ is concerned?

Appreciate any of your replies/feedback on this.

Edit: In response to Marvis's inquiry as to what sort of improvement I am looking at - I am trying to determine whether $\sigma(n^2) \equiv 1 \pmod 4 \hspace{0.10in} XOR \hspace{0.10in} \sigma(n^2) \equiv 3 \pmod 4$, given that I also know that $4 \nmid \left(\sigma(n) - 2\right)$.

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    To clarify, so far I have: $\sigma(n^2) \equiv (-1)^c \pmod 4$, where $c$ is the number of $i$'s such that $1 \le i \le r$, for prime(s) $p_i \equiv 1 \pmod 4$ and the corresponding exponent(s) $\alpha_i$ is/are odd.2012-06-19

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If $p \equiv 1\pmod{4}$, then $\sigma (p^{2 \alpha}) = 1 + p + p^2 + \cdots + p^{2 \alpha} \equiv (2 \alpha + 1)\pmod{4} \equiv (-1)^{\alpha}\pmod{4}$, since each term is $1 \pmod{4}$.

If $p \equiv 3\pmod{4}$, then $\sigma (p^{2 \alpha}) = 1 + p + p^2 + \cdots + p^{2 \alpha} \equiv 1\pmod{4}$, since $1 + p + p^2 + \cdots + p^{2 \alpha} = 1 + p(1+p) + p^3(1+p) + p^5(1+p) + \cdots + p^{2 \alpha-1}(1+p)$ and $1+p \equiv 0 \pmod{4}$.

Hence, $\sigma(n^2) = \prod_{p_i-\text{primes of the form $4k+1$}} (-1)^{\alpha_i} \pmod{4}$

Hence, if $\displaystyle \alpha = \sum_{i} \alpha_i$ where $\alpha_i$ is the highest power of prime $p_i$ of the form $4k+1$ dividing $n$, then $\sigma(n^2) = (-1)^{\alpha} \pmod{4}$

What sort of improvement are you looking at?

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    I apologize, I didn't realize that I forgot to accept your answer to my 'first' question here at MSE. I would like you to know that I have already accepted it just now.2013-02-17