# Equations involving arithmetic functions, totatives and even perfect numbers

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.

• 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. – Peter Apr 20 '18 at 11:42
• 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 – user243301 Apr 20 '18 at 12:24
• 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. – Konstantinos Gaitanas Apr 20 '18 at 18:32
• Many thanks for your attention and remarks @KonstantinosGaitanas – user243301 Apr 20 '18 at 20:19
• @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)? – user243301 May 5 '18 at 17:23

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.