Sum and number of divisors

Question

If the sum of divisors is prime, how can I show that the number of divisors is also prime? I've tried to use definitons, but I'm not getting anything.

Answer 1

The sum of the divisiors of the number $$N=p_1^{a_1}\cdots p_n^{a_n}$$ is $$(1+p_1+\cdots +p_1^{a_1})\cdots (1+p_n+\cdots +p_n^{a_n})$$ All the factors are greater than $1$, so if $N$ has more than one prime factor, the sum of the divisors cannot be prime. Hence we can concentrate on the prime powers. For a prime power $p^n$ , the sum of divisors is $$\frac{p^{n+1}-1}{p-1}$$ and the number of divisors is $n+1$. If $n+1$ is composite, let us say $n+1=ab$ with $a,b>1$ , then $p^a-1\mid p^{n+1}-1$ giving the non-trivial factor $\frac{p^a-1}{p-1}$. This completes the proof of the statement.