<s>[INST] Find all k such that $\sigma (k)=165$ where $\sigma$ is the sum of divisors $165=(3)  (5) (11)$
$\sigma (p^a)$=$p^{a+1}-1 \over {p-1}$=$3$
$p^{a+1}-1$=$3p-3$
$p^{a+1}=3p-2$
Got stuck here. How do I proceed?
 [/INST] Since it seems that you are looking for the "hard" way:
Let $n$ be a positive integer such that $\sigma(n)=165.$ Then $\prod_{p\mid n}(p^{\alpha_p+1}-1)/(p-1)=3\cdot5\cdot11,$ where the product is taken over all primes dividing $n$ and where $\alpha_p$ is the exponent of $p$ in the prime decomposition of $n.$ Now note that if $p\mid n$ then $\alpha_p+1\geqslant2$ so $(p^{\alpha_p+1}-1)/(p-1)>1.$ Hence, since the RHS of the equality above has exactly three prime factors, $n$ has at most three prime factors. If $n=p^a$ for some prime $p$ and some integer $a>0$ then $\sigma(n)=(p^{a+1}-1)/(p-1)$ so $164=2^2\cdot41=p(165-p^a)$ so $p\in\{2,41\}.$ It is easy to check that none of these works. Thus $n$ has either two or three prime factors. If $n=p^aq^b,$ with $p$ and $q$ primes, $p\neq q$ and $a,b>0$ then we have three possibilities. If we set $\alpha:=(p^{a+1}-1)/(p-1)$ and $\beta:=(q^{b+1}-1)/(q-1),$ the three possibilities are:
$(i)$ $\alpha=15$ and $\beta=11.$ Then $14=2\cdot7=p(15-p^a)$ so $p\in\{2,7\}.$ In the same way we find that $q\in\{2,5\}.$ If $p=2$ then $a=3$ and also since $p\neq q$ and $10=2\cdot5=q(11-q^b)$ then $q=5$ so $5^b=9,$ which is impossible. If $p=7$ then $7^a=13,$ which is impossible. 
$(ii)$ $\alpha=33$ and $\beta=5.$ Then $32=2^5=p(33-p^a)$ so $p=2.$ Also $4=2^2=q(5-q^b)$ so $q=2,$ which contradicts the fact that $p\neq q.$
$(iii)$ $\alpha=3$ and $\beta=55.$ Then $2=p(3-p^a)$ so $p=2$ and $a=1.$ Also $54=2\cdot3^3=q(55-q^b)$ and since $q\neq p$ then $q=3$ and thus $3^b=37,$ which is impossible.
It remains to see what happens when $n=p^aq^br^c,$ where $p,q,r$ are pairwise distinct primes and $a,b,c$ are positive integers. Now note that there is only one possibility, which is $(p^{a+1}-1)/(p-1)=3,$ $(q^{b+1}-1)/(q-1)=5$ and $(r^{c+1}-1)/(r-1)=11.$ Then $p=q=2,$ which contradicts the fact that $p\neq q. $ Thus there is no $n$ with $\sigma(n)=165.$
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