Prove that ## 4\mid \sigma_{1}(4k+3) ##

[Math100]

Homework Statement

Prove that $4\mid \sigma_{1}(4k+3)$ for each positive integer $k$.

Relevant Equations

None.

Proof:

Suppose $a=4k+3$ for some $k\in\mathbb{Z^{+}}$.
Then $a=p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{r}^{k_{r}}\implies 4k+3=p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{r}^{k_{r}}$.
Observe that $\sigma_{1}(4k+3)=\sigma (p_{1}^{k_{1}})\sigma (p_{2}^{k_{2}})\dotsb \sigma (p_{r}^{k_{r}}$.
Thus
\begin{align*}
&\sigma_{1} (p_{i}^{k_{i}})=1+p_{i}+p_{i}^2+\dotsb +p_{i}^{k_{i}}\\
&\equiv (1+3+3^{2}+\dotsb +3^{k_{i}})\pmod {4}\\
&\equiv (1-1+1-1+\dotsb +1-1)\pmod {4}\\
&\equiv 0\pmod {4}.\\
\end{align*}
Therefore, $4\mid \sigma_{1}(4k+3)$ for each positive integer $k$.

 

[fresh_42]

I don't think this is correct.

We have
$$
\sigma_1(a)=\sigma_1(4k+3)=\sigma_1\left(\prod_{i=1}^r p_i^{k_i}\right)=\prod_{i=1}^r\dfrac{p_i^{k_i+1}-1}{p_i-1} =\prod_{i=1}^r \left(p_i^{k_i}+\ldots+p_i+1\right)
$$

If $[a]_4$ notes the equivalence class of an integer $a \pmod{4}$ then how do we know that $\sigma_1([a]_4)=[\sigma_1(a)]_4.$ It seems as if you would use this. It is wrong. E.g. $0=\sigma_1([8]_4)\neq [\sigma_1(8)]_4=1+2+4+8=3.$ I am not sure if you use it since the intermediate steps are missing.

Another point is: How do you know that the number of $-1$s and the number of $+1$s are equal? What if $k_i$ is even?

 

[fresh_42]

I think I know what you tried to do.

From $a=4k+3=p_1^{k_1}\cdot\ldots\cdot p_r^{k_r}$ we get
\begin{align*}
[3]_4= [p_1]_4^{k_1}\cdot\ldots\cdot [p_r]_4^{k_r}
\end{align*}
where $m:=[n]_4$ is the abbreviation of $n \equiv m \in \{0,1,2,3\}\pmod{4}.$

Now we have to examine the right-hand side.
All $[p_i]_4 \neq 0.$ Powers of $[p_i]_4=2$ result in $0$ or $2$ so that they cannot occur since we have only $[3]_4$ on the left. All $[p_i]_4=1$ are of no interest since they do not change the product. From the left-hand side we know, that at least one $[p_i]_4=3$ has to occur, say it is $p_1.$ But $[3]_4\cdot [3]_4=[1]_4$ and we need a $[3]_4.$ This means that $k_1$ has to be odd, for otherwise, we would end up with $[1]_4.$ In case $k_1$ is indeed even, then there has to be another index $j$ with $[p_j]_4=[3]_4$ and $k_j$ odd. In such a case we would choose $j$ as our first prime. Say $k_1=2l_1+1.$

Therefore, we can assume that $[p_1^{k_1}]_4=[3]_4^{2l_1+1}$. Now
\begin{align*}
\sigma_1([a]_4)&\equiv \prod_{i=1}^r \sigma_1([p_i]_4)^{k_i} \equiv \sigma_1([3^{2l_1+1}]_4) \prod_{i=2}^r \sigma_1([p_i]_4)^{k_i}\\
&\equiv \dfrac{3^{2l_1+2}-1}{3-1}\prod_{i=^2}^r \sigma_1([p_i]_4)^{k_i} \\
&\equiv \dfrac{9^m-1}{2}\prod_{i=^2}^r \sigma_1([p_i]_4)^{k_i} \\
&\equiv 0\pmod{4}
\end{align*}

with $m=l_1+1\geq 1$ and $9^m-1 \equiv 0\pmod{8},$ hence $\dfrac{9^m-1}{2}\equiv 0\pmod{4}.$

You have to explain why you may assume $p_1=3,$ which by the way is wrong. We can only assume that $[p_1]_4\equiv [3]_4.$

And make sure that you first pass the entire formula modulo four, and then apply $\sigma_1.$

 

[fresh_42]

The most important question when I read a proof, and therefore the most important question when you write a proof, is WHY. I ask this from line to line. If anybody is asking you this, you should be able to answer. In the best case, it is already written in the proof.

 

[Math100]

Okay, so I revised this proof:

Let $4k+3=p_{1}^{k_1}p_{2}^{k_2}\dotsb p_{s}^{k_s}$.
Then $4\equiv 0\pmod {4}$ and $4k+3\equiv 3\pmod {4}$.
Thus $p_{i}^{k_{i}}\not\equiv 0\pmod {4}$ for $i=1, 2,..., s$.
Suppose all $p_{i}^{k_{i}}\equiv 1\pmod {4}$.
Then $p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}\equiv 1\pmod {4}$.
Since $p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}\equiv 3\pmod {4}$,
it follows that $\exists$ one $p_{i}$ satisfying $p_{i}^{k_{i}}\equiv 3\pmod {4}$.
This means $p_{i}\equiv 3\pmod {4}$.
Observe that $p_{i}^{2}\equiv 9\equiv 1\pmod {4}$ and $p_{i}^{3}\equiv 3\pmod {4}$.
If $p_{i}^{r}\equiv 3\pmod {4}$, then $r$ must be odd.
This implies $p_{i}^{k_{i}}\equiv 3\pmod {4}$ where $k_{i}$ is odd.
Now we have
\begin{align*}
&\sigma_{1}(p_{i}^{k_{i}})=p_{i}^{k_{i}}+p_{i}^{k_{i}-1}+\dotsb +p_{i}+1\\
&\equiv (3+1+\dotsb +3+1)\pmod {4}\\
&\equiv 0\pmod {4},\\
\end{align*}
because $p_{i}^{r}\equiv 3\pmod {4}$ if $r$ is odd and $p_{i}^{r}\equiv 1\pmod {4}$ if $r$ is even.
Thus
\begin{align*}
4\mid \sigma_{1}(p_{i}^{k_{i}} ) & \implies 4\mid \sigma (p_{1}^{k_{1}} )\dotsb \sigma_{1} (p_{i}^{k_{i}})\dotsb \sigma_{1} ( p_{s}^{k_{s}} ) \\
&\implies4\mid \sigma_{1} (p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}).\\
\end{align*}
Therefore, $4\mid \sigma_{1} (4k+3)$ for each positive integer $k$.

 

[fresh_42]

Correct.

I edited the typos. If you add one # too many or set one bracket wrong, then it doesn't render properly. It's sometimes hard to find if formulas got lengthy.