A problem about $\sigma (n)$ 
If $\sigma(n)$ is odd, then $n$ is a square or the double of a square.

I don't know how to begin with it... Any suggestions?
 A: Use the multiplicativity of $\sigma$. Let $p$ be a prime divisor of $n$. Write $n=p^e m$, with $p$ not dividing $m$. Then $\sigma(n) = \sigma(p^e) \sigma(m)$. If $\sigma(n)$ is odd, what can you say about $\sigma(p^e)$ and so about $p$ and $e$? Consider the case $p=2$ and $p$ odd separately, as usual.
Or just use the multiplicativity of $\sigma$ and solve the problem for prime powers.
A: Let $n=2^k \cdot m$, with $m$ odd. We need to prove that $m$ is a perfect square.
Let $d_1,.., d_k$ be all the divisors of $m$. Then, all the divisors of $n$, excepting $d_1,..,d_k$ are even.
This shows that if $\sigma(n)$ is odd then $d_1+..+d_k$ is odd.
Now, assume by contradiction that $m$ is not a perfect square. Then for any $d$ divisor of $m$, $\frac{m}{d}$ is also a divisor and $d \neq \frac{m}{d}$. This allows you to pair all the divisors in disjoint pairs $(d, \frac{m}{d})$, which means there is an even number of divisors. 
Now we have our contradiction, we have an even number of divisors, each is odd, and their sum is even. 
Thus our assumption is wrong, which proves $m$ is a perfect square.
A: Let $n=p_1^{a_1}\ldots p_k^{a_k+1}$ be the prime factorization of $n$. Thus $\sigma(n)=(a_1+1)\ldots(a_k+1)$. If this is odd, then every $a_i$ is even.
