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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.

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    $\begingroup$ We can show that $n$ must be a triangular number, hence is of the form $$n=\frac{t(t+1)}{2}$$ With $s:=\sigma(\frac{t(t+1)}{2})$ and $p:=\phi(\frac{t(t+1)}{2})$ , the equations turn into $$s=4p+t+1$$ and $$2t^2-2=8p$$ and we have to show that they can only hold if $t$ is a Mersenne prime, which seems to be the case, but I have no proof yet. $\endgroup$ – Peter Apr 20 '18 at 11:42
  • $\begingroup$ Many thanks for your calculations and attention, since I am asking about what work can be done about the conjectures feel free to share your work with this community as soon as you want. Many thanks one more time @Peter $\endgroup$ – user243301 Apr 20 '18 at 12:24
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    $\begingroup$ I would guess that the right hand side is in general much smaller than the left hand side. Both of the two conjectures should be true. $\endgroup$ – Konstantinos Gaitanas Apr 20 '18 at 18:32
  • $\begingroup$ Many thanks for your attention and remarks @KonstantinosGaitanas $\endgroup$ – user243301 Apr 20 '18 at 20:19
  • $\begingroup$ @Peter I am computing the first few solutions $(n,m)$ of the equation $8\varphi(n)+\sqrt{1+8n}=2m^2+2m-1$, here $1\leq n,m$ are integers. I know $(n,m)=(1,2),(6,3),(28,7),(496,31)$ and $(n,m)=(2145,62)$. I know that if $N$ is an even perfect number and $M$ its Mersenne prime, then $(N,M)$ is a solution. That I would like to know more solutions (one or two more) with $n$ not an even perfect number, likes $(2145,62)$. I would like to know the binary representation of such $n'$s in the pair $(n,m)$ (here $2145=100001100001_2$). Can you calculate the next solution of this kind ($n$ isn't EPN)? $\endgroup$ – user243301 May 5 '18 at 17:23
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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.

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    $\begingroup$ Many thanks for your attention and help. I am going o study your reasonings. $\endgroup$ – user243301 May 1 '18 at 11:02

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