I've seen this posted but I haven't seen this in depth as i need it. I turned this in as homework but only got 1 out of 3 on it, so any clarification would be wonderful.
Show that if the sum of all divisors of n is a prime, say p, then the number of divisors of n is also a prime.
So we know n=$p_1^{a_1}p_2^{a_2}...p_k^{a_k}$ and we have our divisor function $\sigma$(n) = $\sigma(p_1^{a_1}p_2^{a_2}...p_k^{a_k})$ = $\sigma(p_1^{a_1})\sigma(p_2^{a_2})...\sigma(p_k^{a_k})$, since $\sigma$ is a multiplicative function.
But, $\sigma(n)$ is a prime, say p. So at most, $\sigma(n)$ = $\sigma(p^a)$ = q, where q is also a prime, since $\sigma(p_1^{a_1})\sigma(p_2^{a_2})...\sigma(p_k^{a_k})$ is a composite number.
Therefore, $\sigma(p^a)$ = $\frac{p^{a+1}-1}{p-1}$ = $\frac{(p^{a}+p^{a-1}+...+p+1)(p-1)}{p-1}$ = $p^{a}+p^{a-1}+...+p+1$ = q.
let $f(p)$=$p^{a}+p^{a-1}+...+p+1$.
Cases: If $deg(f(p))$ is odd then our cyclotomic polynomial $f(p)$ has an even number of terms and can be factored which implies it is composite. Thus it is a necessary codition for $a$ to be even.
If it is such that $a+1$ is a composite number, we can still factor $f(p)$. (i think this is where my error is because I have to validate this statement and I can't think of how to show this is the case...).
Thus, the only way for $f(p)$ to be irreducible is for $a+1$ to be prime. (again, what is my proof here?)
This implies $d(n) = d(p^a) = a+1$, which is a prime and where $d(n)$ is the divisor function.