- #1
Math100
- 770
- 219
- Homework Statement
- Is the following true, for each positive integer ## k ##?
## 5\mid \sigma_{1} (5k+4) ##
Briefly justify your answer.
- Relevant Equations
- None.
Let ## 5k+4=p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}} ##.
Then ## 5\equiv 0\pmod {5} ## and ## 5k+4\equiv 4\pmod {5} ##.
Thus ## p_{i}^{k_{i}}\not \equiv 0\pmod {5} ## for ## i=1, 2,..., s ##.
Suppose all ## p_{i}^{k_{i}}\equiv 1\pmod {5} ##.
Then ## p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}\equiv 1\pmod {5} ##.
Since ## p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}\equiv 4\pmod {5} ##,
it follows that ## \exists ## one ## p_{i} ## satisfying ## p_{i}^{k_{i}}\equiv 4\pmod {5} ##.
This means ## p_{i}\equiv 4\pmod {5} ##.
Observe that ## p_{i}^{2}\equiv 16\equiv 1\pmod {5} ## and ## p_{i}^{3}\equiv 4\pmod {5} ##.
If ## p_{i}^{r}\equiv 4\pmod {5} ##, then ## r ## must be odd.
This implies ## p_{i}^{k_{i}}\equiv 4\pmod {5} ## 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 (4+1+\dotsb +4+1)\pmod {5}\\
&\equiv 0\pmod {5},\\
\end{align*}
because ## p_{i}^{r}\equiv 4\pmod {5} ## if ## r ## is odd and ## p_{i}^{r}\equiv 1\pmod {5} ## if ## r ## is even.
Thus
\begin{align*}
&5\mid \sigma_{1} (p_{i}^{k_{i}})\implies 5\mid [\sigma_{1} (p_{1}^{k_{1}})\dotsb \sigma_{1} (p_{i}^{k_{i}})\dotsb \sigma_{1} (p_{s}^{k_{s}})]\\
&\implies \sigma_{1} (p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}).\\
\end{align*}
Therefore, ## 5\mid \sigma_{1} (5k+4) ## for each positive integer ## k ##. And the following is true.
Then ## 5\equiv 0\pmod {5} ## and ## 5k+4\equiv 4\pmod {5} ##.
Thus ## p_{i}^{k_{i}}\not \equiv 0\pmod {5} ## for ## i=1, 2,..., s ##.
Suppose all ## p_{i}^{k_{i}}\equiv 1\pmod {5} ##.
Then ## p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}\equiv 1\pmod {5} ##.
Since ## p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}\equiv 4\pmod {5} ##,
it follows that ## \exists ## one ## p_{i} ## satisfying ## p_{i}^{k_{i}}\equiv 4\pmod {5} ##.
This means ## p_{i}\equiv 4\pmod {5} ##.
Observe that ## p_{i}^{2}\equiv 16\equiv 1\pmod {5} ## and ## p_{i}^{3}\equiv 4\pmod {5} ##.
If ## p_{i}^{r}\equiv 4\pmod {5} ##, then ## r ## must be odd.
This implies ## p_{i}^{k_{i}}\equiv 4\pmod {5} ## 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 (4+1+\dotsb +4+1)\pmod {5}\\
&\equiv 0\pmod {5},\\
\end{align*}
because ## p_{i}^{r}\equiv 4\pmod {5} ## if ## r ## is odd and ## p_{i}^{r}\equiv 1\pmod {5} ## if ## r ## is even.
Thus
\begin{align*}
&5\mid \sigma_{1} (p_{i}^{k_{i}})\implies 5\mid [\sigma_{1} (p_{1}^{k_{1}})\dotsb \sigma_{1} (p_{i}^{k_{i}})\dotsb \sigma_{1} (p_{s}^{k_{s}})]\\
&\implies \sigma_{1} (p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}).\\
\end{align*}
Therefore, ## 5\mid \sigma_{1} (5k+4) ## for each positive integer ## k ##. And the following is true.