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 $4 \nmid \left(\sigma(n) - 2\right)$.