Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
In response to Marvis, I am hereby copying my comment to his answer here: I am trying to determine if either $\sigma(n^2) \equiv 1 \pmod 4$ XOR $\sigma(n^2) \equiv 3 \pmod 4$, given that I also know that $4 \nmid \left(\sigma(n) - 2\right)$. –  Jose Arnaldo Dris Jun 19 '12 at 23:49
    
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. –  Jose Arnaldo Dris Jun 19 '12 at 23:57
add comment

1 Answer

up vote 1 down vote accepted

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?

share|improve this answer
    
I am trying to determine if either $\sigma(n^2) \equiv 1 \pmod 4$ XOR $\sigma(n^2) \equiv 3 \pmod 4$, given that I also know that $4 \nmid \left(\sigma(n) - 2\right)$. –  Jose Arnaldo Dris Jun 19 '12 at 23:46
1  
@ArnieDris $\alpha = 1 \implies (-1)^{\alpha} = -1$ and where did I say that $(p+1) \vert \sigma(p^{2 \alpha})$? –  user17762 Jun 19 '12 at 23:56
    
Oops sorry, got confused. Yes you're right. How do I delete these comments? :( –  Jose Arnaldo Dris Jun 20 '12 at 0:04
1  
@ArnieDris No problem. I just wanted to point it out. –  user17762 Jun 20 '12 at 0:06
1  
@ArnieDris Yes $\pmod{4}$. It was a typo. –  user17762 Jun 20 '12 at 0:36
show 5 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.