Let $\sigma(x)=\sigma_1(x)$ denote the classical sum of divisors of the positive integer $x$.
If $\sigma(M)=2M$ and $M$ is odd, then $M$ is called an odd perfect number (hereinafter abbreviated as OPN). It is currently unknown if there are any OPNs. Euler proved that a hypothetical OPN, if one exists, must have the so-called Eulerian form
$$M = p^k m^2$$
where $p$ is the special/Euler prime satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p,m)=1$.
Dris conjectured in his M. Sc. thesis that $p^k < m$ always holds for an OPN $M$ given in Eulerian form.
Consider the following "proof argument" for $p^k < m$:
Let $M = p^k m^2$ be an OPN given in Eulerian form. Suppose that we have $m^2 < 5p^k$.
Then we have
$$\Bigg(m^2 < 5p^k\Bigg) \iff \Bigg(\frac{m^2}{p^k} < 5 \leq p^k\Bigg) \implies \Bigg(m^2 < p^{2k}\Bigg) \iff \Bigg(m < p^k\Bigg).$$
However, in the paper titled Improving the Chen and Chen result for odd perfect numbers, Broughan et al. show that $\sigma(m^2)/p^k \geq 315$. This implies that
$$\frac{p^k}{m^2} < \frac{2}{315},$$
which contradicts
$$\frac{1}{5} < \frac{p^k}{m^2}.$$
Here is my question:
Does this "proof" indeed show that $p^k < m$ holds? If it does not, can it be mended so as to produce a valid argument?
Best Answer
I think that it does not.
You know that supposing that $m^2\lt 5p^k$ gives a contradiction. So, you have $m^2\geqslant 5p^k$ and "$m^2\lt 5p^k\implies m\lt p^k$" from which you cannot conclude $p^k\lt m$. (If you can show that "$m\lt p^k\implies m^2\lt 5p^k$", then you can say that $m\gt p^k$.)