Under what conditions is the sum of all the proper divisors an odd number?

Question

Under what conditions is σ(n) odd? σ(n) = the sum of all proper divisors of n

My Answer: I found out that σ(n) is odd iff n is a square number

Is that the only case, or are there more?

Answer 1

Recall that $\sigma$ is a multiplicative function. Thus, let us make use of the prime factorization of $n$ and write $$n=p_1^{m_1}\cdot p_2^{m_2}\cdot...\cdot p_k^{m_k}$$ with distinct primes and nonzero multiplicities. Then we have that $$\sigma(n)=\sigma(p_1^{m_1})\cdot \sigma(p_2^{m_2})\cdot ...\cdot \sigma(p_k^{m^k})$$ and so $\sigma(n)$ is odd if and only if each $\sigma(p_i^{m_i})$ is odd. Note that $$\sigma(p_i^{m_i})=1+p_i+p_i^2+...+p_i^{m_i}$$ If $p_i=2$, then this is always odd, so the multiplicity of $2$ in $n$ can be anything. If $p_i\ne 2$, then $p_i$ is odd, and so $\sigma(p_i^{m_i})$ is odd if and only if $m_i$ is even (which can be seen from its expression as a sum written above). Thus, the multiplicity of each prime factor of $n$ must be even with the exception of $2$, meaning that $n$ is in the form $n=k^2$ or $n=2k^2$ with $k\in\mathbb N$.