Problem Statement
Prove that
(a) Any odd perfect number $n$ can be represented in the form $n = pa^2$, where $p$ is a prime.
(b) If $n = pa^2$ is an odd perfect number, then $n \equiv p \pmod 8$.
My Attempt
It was Euler who proved that an odd perfect number, if one exists, must necessarily have the form $$n = p^k m^2$$ where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Proof for (a):
If $k=1$, then we can just take $a=m$, and let $p$ be the special prime. So, no problem.
If $k>1$, we proceed as follows. We write $n$ as $$n = p\bigg(p^{\frac{k-1}{2}} m\bigg)^2$$ and then take $$a = p^{\frac{k-1}{2}} m.$$ Note that $a$ is still an integer here as $k \equiv 1 \pmod 4$.
Thus, in both cases we see that the odd perfect number $n$ can be represented in the form $n = pa^2$, where $p$ is the special/Euler prime.
This finishes the proof for (a).
Proof for (b):
Let $n = pa^2$ be an odd perfect number. We wish to show that $n \equiv p \pmod 8$. This is equivalent to showing that $n - p \equiv 0 \pmod 8$. But we have $$n - p = pa^2 - p = p(a^2 - 1)$$ and we know that $$a^2 \equiv 1 \pmod 8$$ since $a$ is odd, being a divisor of the odd perfect number $n$.
This ends the proof for (b).
Inquiry
Are my proofs for (a) and (b) correct (i.e. logically sound and valid)?
Your proofs look correct to me.
Using the fact proved by Euler should be an overkill. I would prove the fact to prove (a) as follows :
Since $n$ is an odd perfect number, $n$ can be written as $n=\prod_{j=1}^{k}p_j^{m_j}$ satisfying $$2n=\prod_{j=1}^{k}(1+p_j+p_j^2+\cdots +p_j^{m_j})$$ where $p_1,p_2,\cdots, p_k$ are distinct odd primes and $m_j$ are positive integers.
Here, note that, in mod $4$,
If $p_j\equiv 1$ and $m_j$ is even, then $1+p_j+p_j^2+\cdots +p_j^{m_j}$ is odd.
If $p_j\equiv 1$ and $m_j\equiv 1$, then $1+p_j+p_j^2+\cdots +p_j^{m_j}\equiv 2$
If $p_j\equiv 1$ and $m_j\equiv 3$, then $1+p_j+p_j^2+\cdots +p_j^{m_j}\equiv 0$
If $p_j\equiv -1$ and $m_j$ is even, then $1+p_j+p_j^2+\cdots +p_j^{m_j}$ is odd.
If $p_j\equiv -1$ and $m_j$ is odd, then $1+p_j+p_j^2+\cdots +p_j^{m_j}\equiv 0$
Since $2n$ is divisible by $2$ and not divisible by $4$, we see that there is only one $j$ such that $p_j\equiv m_j\equiv 1\pmod 4$.
We also see that $m_j$ is even for each of all the other $j$s.
It follows from these that $n$ is of the form $$n=p^{4m+1}b^2=p(p^{2m}b)^2$$ where $p$ is a prime.$\quad\blacksquare$