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ThePerfectHacker 2. Since $\displaystyle n$ is prime it means $\displaystyle \sigma (n) = n+1$. Thus, $\displaystyle n+2\sigma (n) = n+2(n+1)$. Show this is not divisible by three by contradiction.
If $\displaystyle n=p^2$ then $\displaystyle \sigma (p^2)=1+p+p^2$. And thus, $\displaystyle n+2\sigma (n) = 3p^2+2p+2 \equiv 2p+2 \equiv p+1\equiv 0 (\bmod 3)$.
Try to do the last one by substituting $\displaystyle \sigma (pq) = 1+p+q+pq$ into the expression and seeing what you get.