number theory proof Does this proof work?
Prove or disprove that if $\sigma(n)$ is a prime number, n must be a power of a prime.
Since $\sigma(n)$ is prime, $n$ can not be prime unless it is the only even prime, $2$, since $\sigma(n)$ for prime $n=p+1$ which will always be even and therefore not a power of a prime.
Now, assume that n is a power of a prime.
Then $\sigma(n)=\sigma(p^a)$ for prime number $p$ 
so since $\sigma$ is a multiplicative function,
$\sigma(n)=\sigma(p^a)=\sigma(p)\sigma(p)...\sigma(p)$
which is not prime since it has multiple factors.
$n$ must therefore not be a power of a prime
 A: Suppose that $n\gt 1$ is not a prime power. We will show that $\sigma(n)$ cannot be prime.
Let $p$ be a prime divisor of $n$, and let $p^k$ be the largest power of $p$ that divides $n$. Let $n=p^k m$. Since $n$ is not a prime power, we have $m\gt 1$. 
Note that $p^k$ and $m$ are relatively prime. Thus $\sigma(n)=\sigma(p^k)\sigma(m)$. Each of $\sigma(p^k)$ and $\sigma(m)$ is greater than $1$, so $\sigma(n)$ cannot be prime. 
Remark: Note that for prime powers $n$, $\sigma(n)$ can certainly be prime. For example, $\sigma(4)=7$, $\sigma(9)=13$, $\sigma(25)=31$.  It is true that $\sigma(n)$ cannot be prime if $n$ is prime, except, as you pointed out, in the special case $n=2$. 
We cannot write $\sigma(p^k)=(\sigma(p))^k$. The multiplicativeness of $\sigma$ says that $\sigma(ab)=\sigma(a)\sigma(b)$ when $a$ and $b$ are relatively prime. It is for example never the case that $\sigma(p^2)=(\sigma(p))^2$ if $p$ is prime.  
A: Not quite. $\sigma$ is a multiplicative function, as you said, which means that if $\textrm{gcd}(a,b)=1$, then $\sigma(a)\sigma(b)=\sigma(ab)$. But with $a=b=p$, we clearly don't have $\textrm{gcd}(a,b)=1$. By looking at the contrapositive of that statement, you can see that you're being asked to show is that if $n$ is not a prime power - or in other words, that $n$ has multiple prime factors - then $\sigma(n)$ is not prime. Can you show this using your multiplicativity hypothesis?
(As a side note, if a function $\phi: \mathbb{Z}^+ \rightarrow \mathbb{Z}^+$ has $\phi(ab)=\phi(a)\phi(b)$ for all $a,b \in \mathbb{Z}^+$, $\phi$ is called "completely multiplicative".)
A: But $\sigma(3^2)=1+3+9=13$ is prime, so it’s certainly possible that $n$ is a power of a prime when $\sigma(n)$ is prime. You have the right ingredients, but you’ve not put them together quite right.
Suppose that $\sigma(n)$ is a prime. Let the prime factorization of $n$ be
$$n=p_1^{r_1}p_2^{r_2}\ldots p_m^{r_m}\;;$$
then
$$\sigma(n)=\prod_{k=1}^m\sigma(p_k^{r_k})=\prod_{k=1}^m\frac{p_k^{r_k+1}-1}{p_k-1}\;.\tag{1}$$
Now prove that each factor in the product in $(2)$ is greater than $1$ and conclude that there can be only one factor. It immediately follows that $n=p_1^{r_1}$ is a power of a prime.
