This is my third question here.
For the past two days, I have asked two questions here, both of which have a bearing on my current question. I have been working on deriving an equation relating $\sigma(n^2)$ to $\sigma(n)$, if $n$ is odd.
Now, to begin with, we know that $\sigma(n^2)$ is odd, but we can't really be sure about the residue of $\sigma(n^2)$ modulo $4$, even if I know (by my current considerations) that $\sigma(n)$ is incongruent to $2$ modulo $4$. [Edit: As of today [June 22 2012], I am no longer sure about $\sigma(n) \not\equiv 2 \pmod 4$! :-(] I tried tackling the various cases that need to be considered to cover this problem in its full generality, but I think this mammoth task is too complex for me. Hence, I decided to ask this third question:
Is there a specific equation relating $\sigma(n^2)$ to $\sigma(n)$, if $n$ is odd?
Since $n \mid n^2$ and $n < n^2$, all the divisors of $n$ also divide $n^2$. Thus, intuitively, I know that $\sigma(n)$ forms part of the sum that is $\sigma(n^2)$. In other words, I have:
$\sigma(n^2) = \sigma(n) + \sum_{\substack{d \mid n^2 \\ n < d \le n^2}}{d}.$
(I am not so sure about that second addend, though.) Furthermore, since $I(n) < I(n^2) < (I(n))^2$, where $I(x) = \frac{\sigma(x)}{x}$ is the abundancy index of the positive integer $x$, then $n\sigma(n) < \sigma(n^2) < (\sigma(n))^2$. So then, I have:
$\sigma(n^2) = n\sigma(n) + A, \hspace{0.10in} (**)$
and
$\sigma(n^2) = (\sigma(n))^2 - B. \hspace{0.10in} (***)$
My obvious question at this point is: Is it possible to express $A$ or $B$ in terms of $n$, without using $(**)$ and $(***)$?
Appreciate any feedback/replies on this. Thanks!