Let $N = p^k m^2$ be a hypothetical odd perfect number with special prime $p$ satisfying $p \equiv k \equiv 1 \pmod 4$ and $\gcd(p, m) = 1$. Note that, since $\gcd(p, m) = 1$ and $p$ is (the special) prime, then $p^k \neq m$. (This also follows from the fact that prime powers are deficient, contradicting $N$ is perfect.) By trichotomy, either $(p^k < m) \oplus (m < p^k)$ is true, where $\oplus$ denotes exclusive-OR. (That is, $A \oplus B$ is true if and only if exactly one of $A$ or $B$ holds.)

We will denote the *classical sum of divisors* of the positive integer $z$ by $\sigma(z)=\sigma_1(z)$, and the *abundancy index* of $z$ by $I(z)=\sigma(z)/z$. Furthermore, we will denote the *deficiency* of $z$ by $D(z)=2z-\sigma(z)$, and the *aliquot sum* of $z$ by $s(z)=\sigma(z)-z$.

Here is a summary of some new results on odd perfect numbers, which were realized by the author on May 5, 2023:

- If $p < m$, then the quantity $m^2 - p^k$ is not a square. (Kindly note the contrapositive.)
- If $m < p$, then the following statements hold:
- Descartes's conjecture holds (i.e. $k = 1$).
- Dris conjecture (i.e. $p^k < m$) is false.
- The quantity $m^2 - p^k$ is a square.
- The square root of the non-Euler part $m^2$ is almost perfect.
- Define the following GCDs:
- $G=\gcd\left(\sigma(p^k),\sigma(m^2)\right)$
- $H=\gcd\left(m,\sigma(m^2)\right)$
- $I=\gcd\left(m^2,\sigma(m^2)\right)$

- Dris proved in February 10, 2022 that the chain of divisibility conditions
- $G \mid H \mid I$

- $G=\gcd\left(\sigma(p^k)/2,\sigma(m^2)/p^k\right)=\sigma(p^k)/2=\gcd(G,I)$
- $H=\gcd\left(m,\sigma(m^2)/p^k\right)=m=\gcd(H,I)$.
- In particular, the divisibility constraint $\sigma(p^k)/2 \mid m$ holds.
- The divisibility condition $\sigma(p^k) \mid 2m$ is equivalent to $m \mid \sigma(m^2)$.
- ( To be continued$\ldots$ )