(Note: This was cross-posted from MO, because it was not well-received there. Will delete the MO post in a few.)
Observe that an even perfect number $M = (2^p - 1)\cdot{2^{p - 1}}$ and an odd perfect number $N = q^k n^2$ have similar multiplicative forms. (Indeed, it is conjectured that $k=1$, and this prediction goes back to Descartes ($1638$).)
Here is my initial question:
INITIAL QUESTION
Is there any other known relationship between even perfect numbers and odd perfect numbers, apart from their multiplicative forms?
I ask because of a related query here.
Indeed, if we could show that $$\left|2^r - t\right| = 1,$$ for $r$ and $t$ satisfying $n^2 - q^k = 2^r t$ and $\gcd(2, t)=1$, then we would have a proof for $n < q^k$, which together with Brown's estimate $q < n$ ($2016$) would yield a refutation of Descartes's Conjecture that $k = 1$.
Note that, for even perfect numbers, we actually have $$(2^p - 1) - 2^{p-1} = 2^{p-1} - 1 = \bigg(2^{\dfrac{p-1}{2}} + 1\bigg)\bigg(2^{\dfrac{p-1}{2}} - 1\bigg) = ab,$$ where the factorization works whenever $M \neq 6$. (Note that $\gcd(a, b) = 1$.)
We compute that $$\left|a - b\right| = 2.$$
DISCUSSION
In the hyperlinked MO question, the following (summarized) cases were considered for odd perfect numbers, which we now examine for even perfect numbers:
Case 1: $$2^{\dfrac{p-1}{2}} < \min(a,b) = 2^{\dfrac{p-1}{2}} - 1$$
Notice that Case 1 clearly does not hold.
Case 2: $$\min(a,b) = 2^{\dfrac{p-1}{2}} - 1 < 2^{\dfrac{p-1}{2}} < \max(a,b) = 2^{\dfrac{p-1}{2}} + 1$$
Notice that Case 2 clearly holds.
It follows (from mimicking the resulting inequality $q^k < n\cdot{\left|2^r - t\right|}$ for odd perfect numbers) that $$2^p - 1 < {2^{\dfrac{p-1}{2}}}\cdot{\left|a - b\right|} = {2^{\dfrac{p-1}{2}}}\cdot{2} = {2^{\dfrac{p+1}{2}}}.$$
This last inequality implies that $$p < -\dfrac{2\bigg(\log(2) - \log(\sqrt{2} + \sqrt{6})\bigg)}{\log(2)} \approx 1.89997,$$ which is a contradiction to $p \geq 2$.
Case 3: $$\max(a,b) = 2^{\dfrac{p-1}{2}} + 1 < 2^{\dfrac{p-1}{2}}$$
Notice that Case 3 clearly does not hold.
Note that for even perfect numbers, we do have $$2^{\dfrac{p-1}{2}} < 2^p - 1$$ which "mimics" the conjecture $n < q^k$ for odd perfect numbers above.
Here is my final question for this post:
FINAL QUESTION
Does the exhaustion of all possible cases in the DISCUSSION section essentially disprove the existence of odd perfect numbers?
Edit (Added on July 25, 2022 - 7:15 PM Manila time)
On second thought, I think the argument above only rules out $\left|a - b\right| = 2$. Since $ab = 2^{p-1} - 1$ is always odd, and moreover, that $ab \equiv 3 \pmod 4$ always holds (for $p > 2$), then we can write $$\left|a - b\right| = 2^y z$$ for some $y \geq 1$ and $z$ satisfying $\gcd(2,z)=1$. (Notice that $\left|a - b\right| = 2^y z$ is not a square.)
THIS WILL THEN ENTAIL A CAREFUL CONSIDERATION OF VARIOUS CONSEQUENT CASES.
Edit (Added on July 26, 2022 - 10:25 PM Manila time)
Since $ab \equiv 3 \pmod 4$ holds, then either $$\begin{cases} {a \equiv 1 \pmod 4 \\ b \equiv 3 \pmod 4 } \end{cases} $$ or $$\begin{cases} {a \equiv 3 \pmod 4 \\ b \equiv 1 \pmod 4 } \end{cases} $$ Either way, we obtain $$2^y z = \left|a - b\right| \equiv 2 \pmod 4$$ so that we know that $y = 1$.
By the considerations in the DISCUSSION above, we finally obtain that $$z \geq 3 > 2 = 2^y.$$
Therefore, we know that $$\min(2^y, z) = 2^y = 2$$ while $$\max(2^y, z) = z.$$
(This is a partial answer.)
Let $(2^p - 1)\cdot{2^{p-1}}$ be an even perfect number distinct from $6$.
Recall that $\left|a - b\right| = 2z$ for some integer $z$ satisfying $\gcd(2,z)=1$, and that $z \geq 3$. Note that it is given that $$ab = 2^{p-1} - 1 = \bigg(2^{\dfrac{p-1}{2}} + 1\bigg)\bigg(2^{\dfrac{p-1}{2}} - 1\bigg).$$
Notice that $$\gcd\Bigg(2^{\dfrac{p-1}{2}} + 1, 2^{\dfrac{p-1}{2}} - 1\Bigg) = 1.$$
We consider the following subcases.
Case 1: $2^{\dfrac{p-1}{2}} > z > 2$
We obtain from Case 1 that $$\bigg(2^{\dfrac{p-1}{2}} - z\bigg)\bigg(2^{\dfrac{p-1}{2}} + 2\bigg) > 0$$ which can be re-expressed as $$2^{p-1} - 2z > {2^{\dfrac{p-1}{2}}}\cdot\left|2 - z\right|$$ from which it follows that $$ab + 1 - \left|a - b\right| > \sqrt{ab + 1}\cdot\left|2 - \dfrac{\left|a - b\right|}{2}\right|. \tag{1}$$
Note: I tried asking WolframAlpha and it does seem to yield solutions to Inequality (1) where all the variables are positive-valued. I am therefore led to conjecture that this the correct case to consider (see Case 2 below). Indeed, we know that $2^{\dfrac{p-1}{2}} < 2^p - 1$ trivially holds.
Case 2: $2^{\dfrac{p-1}{2}} \geq 2 > z$
We obtain from Case 2 that $$2^{\dfrac{p-1}{2}} \geq 2 > z \geq 3$$ which is a contradiction.
Case 3: $z > 2^{\dfrac{p-1}{2}} \geq 2$
We obtain from Case 3 that $$\bigg(2^{\dfrac{p-1}{2}} + 2\bigg)\bigg(2^{\dfrac{p-1}{2}} - z\bigg) < 0$$ which can be re-expressed as $$2^{p-1} - 2z < {2^{\dfrac{p-1}{2}}}\cdot{\left|2 - z\right|}$$ from which it follows that $$ab + 1 - \left|a - b\right| < \sqrt{ab + 1}\cdot\left|2 - \dfrac{\left|a - b\right|}{2}\right|. \tag{2}$$
Note: I tried asking WolframAlpha and it does seem to yield solutions to Inequality (2) where all the variables are positive-valued.
Case 4: $2 \geq 2^{\dfrac{p-1}{2}} > z$
We obtain from Case 4 that $$2 \geq 2^{\dfrac{p-1}{2}} > z \geq 3,$$ which is a contradiction.
Case 5: $z > 2 \geq 2^{\dfrac{p-1}{2}}$
We obtain from Case 5 that $$2z = \left|a - b\right| > 2^{p-1} = ab + 1$$ which yields the contradiction $$\bigg(a \geq 1\bigg) \land \bigg(b < \dfrac{a-1}{a+1}\bigg).$$
Case 6: $2 > z > 2^{\dfrac{p-1}{2}}$
We obtain from Case 6 that $$2 > z > 2^{\dfrac{p-1}{2}}$$ which implies that $$1> \dfrac{p-1}{2}.$$ We obtain finally that $$p < 3$$ which contradicts our requirement that $p > 2$.