- #1

Math100

- 793

- 220

- 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 ##.

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 ##.