(Preamble: This question is an offshoot of this answer by mathlove to an earlier post.)
Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of the divisors of the positive integer $x$.
If $\sigma(m)=2m$ and $m$ is odd, then $m$ is called an odd perfect number. It is currently unknown whether there are any odd perfect numbers, although it is widely believed that there is none.
Euler proved that an odd perfect number $m$, if one exists, must have the so-called Eulerian form
$$m = q^k n^2$$
where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since the divisor sum function $\sigma$ is a multiplicative function and $m = q^k n^2$ is perfect, we obtain
$$2 q^k n^2 = 2m=\sigma(m)=\sigma(q^k n^2)=\sigma(q^k)\sigma(n^2).$$
From the hyperlinked answer by mathlove, we have:
We can have
$$\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\Bigg)$$
$$=\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}{n^2}\Bigg)-\frac{q\Bigg(q^{\dfrac{k-1}{2}}-1\Bigg)\Bigg(q^{\dfrac{k+1}{2}}-1\Bigg)}{(q-1)^2}\sigma(n^2)$$
since we have
$$\begin{align}&\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}{n^2}\Bigg)-\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\Bigg)
\\\\&=\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}\Bigg)\sigma({n^2})-\sigma(q^k)\sigma(n^2)
\\\\&=\Bigg(\frac{q^{\dfrac{k+3}{2}}-1}{q-1}\cdot\frac{q^{\dfrac{k+1}{2}}-1}{q-1}-\frac{q^{k+1}-1}{q-1}\Bigg)\sigma(n^2)
\\\\&=\frac{\Bigg(q^{\dfrac{k+3}{2}}-1\Bigg)\Bigg(q^{\dfrac{k+1}{2}}-1\Bigg)-(q^{k+1}-1)(q-1)}{(q-1)^2}\sigma(n^2)
\\\\&=\frac{q\Bigg(q^{\dfrac{k-1}{2}}-1\Bigg)\Bigg(q^{\dfrac{k+1}{2}}-1\Bigg)}{(q-1)^2}\sigma(n^2)\end{align}$$
Note that the last equation is equivalent to
$$\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}\Bigg)\sigma(n^2)-\sigma(q^k)\sigma(n^2)={q\sigma\Bigg({q^{\dfrac{k-3}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}\Bigg)}{\sigma(n^2)} \text{ (*)},$$
and that $\gcd(q,\sigma(q^j))=1$ for all positive integers $j$.
Now, assume that $k \neq 1$. Since $k \equiv 1 \pmod 4$, then $k \geq 5$.
But we know that $q \mid \sigma(n^2)$ holds because
$$\dfrac{\sigma(n^2)}{q^k}=\dfrac{2n^2}{\sigma(q^k)}=\gcd(n^2,\sigma(n^2))$$
(which follows from $\gcd(q^k,\sigma(q^k))=1$) implies that
$$\dfrac{\sigma(n^2)}{q}=q^{k-1} \cdot{\dfrac{2n^2}{\sigma(q^k)}}=q^{k-1} \cdot{\gcd(n^2,\sigma(n^2))}.$$
We obtain
$${q} \mid \text{LHS of Equation (*)},$$
which means that
$$q^2 \mid \text{RHS of Equation (*)}.$$
We therefore get
$$q^3 \mid \text{LHS of Equation (*)},$$
which implies that
$$q^4 \mid \text{RHS of Equation (*)}.$$
$$\ldots$$
$$\ldots$$
$$\ldots$$
This results in too many $q$'s, dividing $\sigma(n^2)$, on each side of the equation, resulting to a contradiction. (???) (Notice that we know that the maximal power of $q$ that can divide $\sigma(n^2)$ is $q^k$.)
We therefore conclude that $k=1$.
Here is my inquiry:
Have I indeed stumbled upon a contradiction? If my proof is not sound, how can it be mended so as to produce a valid argument?
Best Answer
Let $T:=\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}\Bigg)-\sigma(q^k)$. Dividing the both sides of $\text{(*)}$ by $\sigma(n^2)$, one has $T={q\sigma\Bigg({q^{\dfrac{k-3}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}\Bigg)}$. Since $q\ ||\ {q\sigma\Bigg({q^{\dfrac{k-3}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}\Bigg)}$, one has $q\ ||\ T$. This implies that $q^{k+1}\ ||\ \text{LHS of (*)}$ and $q^{k+1}\ ||\ \text{RHS of (*)}$.
In your argument, I don't get how you got $q^3\ \mid\ \text{LHS of (*)}$ from $q^2\ \mid\ \text{RHS of (*)}$.