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?
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.
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.