Prove that $ 3\mid \sigma_{1}(3k+2) $ for each positive integer $ k $.
Here's my proof:
Let $3k+2=p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}.$ Then $3\equiv 0\pmod {3}$ and $3k+2\equiv 2\pmod {3}.$ Thus $p_{i}^{k_{i}}\not\equiv 0\pmod {3} $ for $ i=1, 2, ..., s.$ Suppose all $p_{i}^{k_{i}}\equiv 1\pmod {3}.$ Then $ p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}\equiv 1\pmod {3} $. Since $ p_{1}^{k_{1}}p_{2}^{k_{2}}\dotsb p_{s}^{k_{s}}\equiv 2\pmod {3} $, it follows that there exists one $p_{i}$ satisfying $ p_{i}^{k_{i}}\equiv 2\pmod {3} $. This means $ p_{i}\equiv 2\pmod {3} $. Observe that $ p_{i}^{2}\equiv 4\equiv 1\pmod {3} $ and $ p_{i}^{3}\equiv 8\equiv 2\pmod {3} $. If $ p_{i}^{r}\equiv 2\pmod {3} $, then $r$ must be odd. This implies $ p_{i}^{k_{i}}\equiv 2\pmod {3} $ where $k_{i}$ is odd. Now we have $ \sigma_{1}(p_{i}^{k_{i}})=p_{i}^{k_{i}}+p_{i}^{k_{i}-1}+\dotsb +p_{i}+1 $ $\equiv [2+1+\dotsb +2+1]\pmod {3}$ $\equiv 0\pmod {3}$, because $ p_{i}^{r}\equiv 2\pmod {3} $ if $r$ is odd and $ p_{i}^{r}\equiv 1\pmod {3} $ if $r$ is even. Thus \begin{align*} \ 3\mid \sigma_{1}(p_{i}^{k_{i}})\implies 3\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*} where $\sigma_{1}$ is multiplicative. Therefore, $ 3\mid \sigma_{1}(3k+2) $ for each positive integer $k$.
Above is my proof for this question/problem. Can anyone please take a look at this and check/verify to see if it's correct?
Comment: For particular cases we may say :
1-If $3k+2$ is prime then $\sigma=3k+2+1=3(k+1)\Rightarrow 3|\sigma( 3k+2)$
2- We rewrite $3k+2$ as:
$3k+2=3k+3-1=3m-1$
Suppose we can write:
$3m-1=(3t-1)^{2n+1} $
The sum of divisors will be:
$\sigma=\frac{[(3t-1)^{2n+2}=3S+1]-1}{(3t-1)-1}=3 x\Rightarrow 3|\sigma(3k+2)$