I've deduced simple relationships that satisfy each even perfecf number (even numbers $n$ for which $\sum_{d\mid n}d=2n$) and now I wondered about related conjectures.
For each integer $m\geq 1$ we denote the sum of divisors function $\sum_{d\mid m}d$ as $\sigma(m)$, and the Euler's totient function as $\varphi(m)$.
Claim 1. It's easy to prove that each even perfect number $n$ satisfies $$\sigma(n)=\frac{1}{2}\left(1+8\varphi(n)+\sqrt{1+8n}\right).\tag{1}$$
Claim 2. Thus each even perfect number satisfies also $$4n=1+8\varphi(n)+\sqrt{1+8n}.\tag{2}$$
Question. I would like to know what work can be done about the following conjectures (prove it or provide us what calculations/reasonings can be done, or refute these finding a counterexample):
C1) If an integer $m\geq 1$ satifies $(1)$, then $m$ is an even perfect number.
C2) Similarly to previous conjecture, each integer $l\geq 1$ satisfying $(2)$ is an even perfect number.
Many thanks.
I've tested that seems there are no counterexamples in my experiments, say us $\leq 10^5$.
I don't know if these equations are in the literature, thus answer as a reference request this question adding the articles where there are information about these equations and I try to search and read those from the literature.
This is not a complete answer, just some comments that are too long to fit in the appropriate section.
From the paper On the Ratio of the Sum of Divisors and Euler's Totient Function I, we have the following inequality: $$\sum_{p \mid n}{\log\bigg(1+\frac{1}{p}\bigg)} \leq \log\bigg(\frac{\sigma(n)}{\phi(n)}\bigg) = \sum_{p \mid n}{\log\bigg(\frac{p^2 - p^{1 - \nu_p(n)}}{(p-1)^2}\bigg)} \leq 2\sum_{p \mid n}{\log\bigg(1 + \frac{1}{p-1}\bigg)}.$$
From the paper On the Ratio of the Sum of Divisors and Euler's Totient Function II, we have that if $\sigma(N) = {a}\cdot{\phi(N)}$ and $a = 4$, then $\omega(N) = 2$ and $N$ is of the form ${2^{q-2}}{M_q}$ where $M_q = 2^q - 1$ is a prime.
I am guessing that you can use these results for showing that if $n$ satisfies $$\sigma(n) = \frac{1}{2}\cdot\bigg(1+8\phi(n)+\sqrt{1+8n}\bigg),$$ then $n$ must be an even perfect number.