Let $n=\prod_{i=1}^{k}{p_i^{a_i}}$ be the canonical form of $n$. Since $n$ is odd, each $p_i$ is odd. We've $2n=\prod_{i=1}^{k}{(1+p_i+p_i^2+...+p_i^{a_i})}$. Note that $\nu_2 (2n)=1$. Since $1+p_i+p_i^2+...+p_i^{a_i} \equiv a_i+1\pmod{2}$, there exists at least one $j\in \{ 1,2,...,k\}$ that $a_j+1$ is even. If there exists two distinct such $j$, we'll get $4\mid 2n$, contradiction with $\nu_2 (2n)=1$. Hence, there exists exactly one $j\in \{ 1,2,...,k\}$ that $a_j$ is odd. So, $a_i+1$ is odd for all $i\in \{ 1,2,...,k\}$ that $i\neq j$, this gives $n=p_j^{a_j}m^2$ for some $m\in \mathbb{Z}^+$. If $p_j\equiv 3\pmod{4}$, we get $1+p_j+p_j^2+...+p_j^{a_j}\equiv \frac{3^{a_j+1}-1}{2}\pmod{4}$. We've $2n\equiv 2 \pmod{4} \implies (1+p_j+p_j^2+...+p_j^{a_j})m^2\equiv 2\pmod{4}\implies m^2\equiv 1\pmod{4}$. Not hard to see that there's no even number $u$ that $\frac{3^u-1}{2}\equiv 2\pmod{4}$, done. Hence, $p_j\equiv 1\pmod{4}$, this implies $$n\equiv 1\pmod{4}\implies (1+p_j+p_j^2+...+p_j^{a_j})m^2\equiv 2\pmod{4}\implies 1+a_j\equiv 2\pmod{4}.$$Hence, $a_j\equiv 1\pmod{4}$, and this complete part a).

First, note that any even perfect number must be of the from $2^{n-1}(2^n-1)$ where $2^n-1$ is a prime number. We've two possible cases depend on parity of $n$. 1. If $n$ is even, $n-1$ is odd, by a), we get that $n=p^am^2+1$ for some prime number $p$ and positive integers $a,m$ with $p,a\equiv 1\pmod{4}$. Note that $n\equiv 2\pmod{4}\implies \frac{n(n+1)}{2}\equiv 1\pmod{2}$. So $\frac{n(n+1)}{2}=q^b\ell^2$ for some prime number $q$ and positive integers $b,\ell$ with $q,b\equiv 1\pmod{4}$. Since $\gcd (\frac{n}{2} ,n+1)=1$, we get that one of $\frac{n}{2},n+1$ must be of the form $q^bk^2$ for some positive integer $k$. We've $q^bk^2\equiv k^2\pmod{4}$ and it can't be even number, so $q^bk^2\equiv 1\pmod{4}$. If $n+1=q^bk^2\equiv 1\pmod{4}$, we get $n\equiv 0\pmod{4}$, impossible. So $\frac{n}{2}=q^bk^2\implies n+1=v^2=p^am^2+2$ for some positive integer $v$. Consider the equation in modulo $4$ gives us the contradiction. 2. If $n$ is odd, $n-1$ is even, we get that $n-1=2^p(2^{p+1}-1)$ for some positive integer $p$ that $2^{p+1}-1$ is a prime number. If $p\geq 2$, we've $n\equiv 1\pmod{4}\implies \frac{n(n+1)}{2}\equiv 1\pmod{2}$. So $\frac{n(n+1)}{2}=q^b\ell^2$ for some prime number $q$ and positive integers $b,\ell$ with $q,b\equiv 1\pmod{4}$. We've $(2^p(2^{p+1}-1)+1)(2^{p-1}(2^{p+1}-1)+1)=q^b\ell^2$. So there exists $\epsilon \in \{ 0,1\}$ that $2^{p-\epsilon }(2^{p+1}-1)+1=v^2$ for some positive integer $v$. So, $2^{p-\epsilon }(2^{p+1}-1)=(v-1)(v+1)$. Recall that $2^{p+1}-1$ is prime number. We get that $2(2^{p+1}-1)$ divide one of $v-1,v+1$. So, $v\geq 2(2^{p+1}-1)+1$, we easily arrive at the contradiction from $2^p(2^{p+1}-1)+1\geq v^2\geq (2(2^{p+1}-1)+1)^2$ when $p\geq 2$. Hence, $p=1$, this gives $n=7$, easy to see that this works and so is the only solution to our problem.