Problems 11.3, No. 14 (a) and (b) from page 237 of David M. Burton's "Elementary Number Theory" (7th Edition) 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)?

 A: 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.$\quad\blacksquare$
