Let $\sigma(n)$ denote the sum of divisors of $n$.
Here is my question:
If $\sigma(n) \equiv 2 \pmod 4$ and $n$ is odd, does this imply that $n = p^k m^2$ where $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$?
I think the proof must closely parallel the proof for the form of odd perfect numbers (as initially proved by Euler), but I don't know how to begin the proof.
Any hints will be appreciated.
Write $n = p_1^{e_1}p_2^{e_2} \cdots p_t^{e_t}$, with $p_1, p_2, \cdots, p_t$ different odd primes.
Note that $\sigma(n) = \sigma(p_1^{e_1}p_2^{e_2} \cdots p_t^{e_t}) = \sigma(p_1^{e_1})\sigma(p_2^{e_2})\cdots \sigma(p_t^{e_t})$.
If $\sigma(n) \equiv 2 \mod 4$, exactly one of these factors has exactly one factor two and the other factors are odd. Note that $\sigma(p_l^{e_l}) = p_l^{e_l} + p_l^{e_l-1} + \cdots + p_l +1$, all terms in this sum are odd so the sum is odd iff there is an odd number of terms, so then $e_l$ is even. Without loss of generality $p_1$ is the prime such that $e_1$ is not even, the rest must be even so $p_2^{e_2} \cdots p_t^{e_t} = m^2$ for some $m$.
$\sigma(p_1^{e_1})$ must have one factor two. Therefore, $e_1$ can't be even and must be odd. Write $\sigma(p_1^{e_1}) = p_1^{e_1} + p_l^{e_1-1} + \cdots + p_1 +1 = (p_1+1)(p_1^{e_1-1}+p_1^{e_1-3}+\cdots + 1)$.
This is possible because $e_1$ is odd. If $p_1 \equiv 3 \mod 4$, $(p_1+1)$ is divisible by $4$, which isn't possible. So $p_1 \equiv 1 \mod 4$. If $e_1 \equiv 3 \mod 4$, then $(p_1^{e_1-1}+p_1^{e_1-3}+\cdots + 1)$ is the sum of an even number of odd numbers and hence even. But $(p_1+1)$ is also even, so then $\sigma(p_1^{e_1})$ is divisible by $4$, which isn't possible, so also $e_1 \equiv 1 \mod 4$. This completes the proof.