The number and amount of dividers - a power of two For a positive integer $n$ is known that the sum of all divisors of that number is a power of $2$. Prove that the number of these divisors is also a power of $2$.
My work so far:
Several of these numbers I found
Let $\tau(n) -$ the number of divisors of $n$
1) $n=3; 1+3=4=2^2$ and $\tau (3)=2$
2) $n=7; 1+7=8=2^3$ and $\tau (7)=2$
From $1$ to $20$ such numbers do not have. 
3) $n=21=3 \cdot 7; \tau(21)=4 (1,3,7,21)$ and $1+3+7+21=32=2^5$
4) $n=31; 1+31=32=2^5$ and $\tau(31)=2$

The first two divisor functions of a number with prime factorization $n=\prod_ip_i^{a_i}$ are
$$
\sigma_0(n)=\prod_i(a_i+1)
$$
and
$$
\sigma_1(n)=\prod_i\left(1+p_i+\cdots+p_i^{a_i}\right)\;.
$$
Since $\sigma_0$ is clearly a power of $2$ if and only if the $a_i+1$ are, we need to show that $a_i+1$ is a power of $2$ if $1+p_i+\cdots+p_i^{a_i}$ is a power of $2$. Now if $p_i=2$, then $1+p_i+\cdots+p_i^{a_i}=2^{a_i+1}-1$, which is not a power of $2$, so we can focus on odd primes. For odd $p_i$, if $1+p_i+\cdots+p_i^{a_i}$ is a power (and hence a multiple) of $2$, then $a_i+1$ is even, and $1+p_i+\cdots+p_i^{a_i}=(1+p_i)(1+p_i^2+\cdots+p_i^{a_i-1})$. For this to be a power of $2$, both factors must be. But then we can apply the same reasoning to the factor $1+p_i^2+\cdots+p_i^{a_i-1}$ and factor out $1+p_i^2$. We an continue to factorize the entire sum like this, and it follows that $a_i+1$, the number of summands, is a power of $2$, as required.
