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Let an integer $m\geq 1$, and $\sigma(m)$ is the sum of positive divisors function, and $\phi(m)$ is Euler's totient function, counting the number of integers $1\leq k\leq m$ such that $gcd(k,m)=1$ (see [1], too for the definition of perfect numbers).

My question is

Question. Prove, or refute with a counterexample, the following statement: If $n\geq 1$ is an integer such that $$\left( -3+\sqrt{1+8n}\right)\sigma(n)=4\left(-1+\sqrt{1+8n}\right)\phi(n)$$ then is an even perfect number. Thanks in advance.

I excuse this question here because perhaps someone can solve o reference this exercise. It is possible assuming that $n=2^{a}m$, where $a\geq 0$ and $m$ an integer that we assume as odd, find the proof of previous? When in a past post I've use this trick, the computations was tedious. Can you try it and show some useful? I excuse this question, because in the following appendix I give details, perhaps useful for an student with easy computations, and perhaps in a future my questions are useful for someone. If you want say something about the appendix you are welcome. Currently, I haven't in my hand analytic tools to study previous conjecture.

I attempt integrate another factor 2 in the definition of even perfect numbers Appendix:

I believe that I haven't mistakes in my proof of following exercise, caution is the reciprocal of my question (I don't know if it is in the literature)

Proposition. If $n$ is an even perfect number then $$\left(\frac{-3+\sqrt{1+8n}}{2}\right)\sigma(n)=2\left(-1+\sqrt{1+8n}\right)\phi(n).$$

Skecth of my Proof. Really I build previous identity by comparison, when I've used that an even perfect number is a triangle number $\frac{q(q+1)}{2}$, where I was assuming that $q=2^p-1$ is the associted Mersenne prime with $n$, thus I've used Euler's statement for even perfect numbers (thus $n$ has the form $2^{p-1}(2^p-1)$, and $2^p-1$ is prime)

$$\frac{\sigma\left(\frac{q(q+1)}{2}\right)}{\phi\left(\frac{q(q+1)}{2}\right)}=\frac{4q}{q-1}.$$

And I've found a form for $q$ solving $q^2+q=2n$, thus $$q=\frac{-1+\sqrt{1-4\cdot(-2n)}}{2},$$

and this is the end of the proof.

After I write in my algorith for to run a program written in Pascal, I say find the integers $N\geq 1$ such that ... (perhaps the following condition remove illegally so many integers but I tell you what I've made below)

My algorithm to work. Find integers $N\geq 1$ such that $$\left( -3+\sqrt{1+8N}\right)\sigma(N)=4\left(-1+\sqrt{1+8N}\right)\phi(N).$$

Then

Computational fact. My program up 10000 only finds even perfect numbers (I tried up until $10^5$ but my program does not finish, I suspect that my procedures are bad).

References:

[1] This Mathematics Stack Exchange or Wikipedia pages for Perfect numbers, and cited arithmetic functions sum of divisor function and Euler's totient function.

[2] A related answer/question in This Mathematics Stack Exchange, B. Dris I attempt integrate another factor 2 in the definition of even perfect numbers

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  • $\begingroup$ Perhaps use some of the known inequalities relating $\sigma(n)$ and $\phi(n)$ from, e.g. Handbook of Number Theory I and Handbook of Number Theory II? $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 7 '15 at 10:13
  • $\begingroup$ Sorry I don't understand previous comment. Feel free to refeence if it is a known statement @ArnieDris Now I have no idea of this books, I will try a search in my books. Very thanks much, in true Feel free if you want use these ideas in your future post. Thanks in true $\endgroup$ – user243301 Nov 7 '15 at 10:39
  • $\begingroup$ For the audience: perhaps with analysis we can not capture the divisibility question involving the unsolved problem, that there are o no infinitely many even perfect numbers; perhaps even it is proving that my previous question is an equivalence with the definition to be an even perfect number and known propierties of previous cited arithmetical functions, the question is only a non useful equivalence. I don't know it. $\endgroup$ – user243301 Nov 7 '15 at 14:56
  • $\begingroup$ Juan, please refer to my (partial) answer to your other question here. $\endgroup$ – Jose Arnaldo Bebita-Dris Dec 30 '15 at 2:41
  • $\begingroup$ @ArnieDris hecho, que seas feliz estos días. $\endgroup$ – user243301 Dec 30 '15 at 8:00
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This is too long to fit as a comment, so I have written it out as an answer.

You misunderstood my comment Juan.

To illustrate my point, you can try to apply some of the inequalities referenced in the following papers from the Journal of Integer Sequences:

  1. On the Ratio of the Sum of Divisors and Euler’s Totient Function I by Kevin A. Broughan and Daniel Delbourgo; and

  2. On the Ratio of the Sum of Divisors and Euler’s Totient Function II by Kevin A. Broughan and Qizhi Zhou.

One such inequality is:

$$\sum_{p \mid n}{\log\left(1 + \frac{1}{p}\right)} \leq \log\left(\frac{\sigma(n)}{\phi(n)}\right) \leq 2\sum_{p \mid n}{\log\left(1 + \frac{1}{p - 1}\right)}.$$

Perhaps using one of these inequalities, together with the assumption $n = {2^a}{m}$ (where $a \geq 0$) and $$\left( {\sqrt{1 + 8n} - 3} \right)\sigma(n) = 4\left( {\sqrt{1 + 8n} - 1} \right)\phi(n),$$ would enable you to show that $n$ is an even perfect number.

Of course, a central issue in your original question would be to find integers $N$ for which $8N + 1$ is a perfect square.

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  • $\begingroup$ Very thanks much @ArnieDris I take your notes, in this week I study this, now I dont choose your contribution as a full answer, but I vote up and claim that other uses made it. Very thanks much $\endgroup$ – user243301 Nov 8 '15 at 13:43
  • $\begingroup$ You're welcome, Juan! =) $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 8 '15 at 13:45
  • $\begingroup$ Thanks for your help, this is a nice topic @ArnieDris $\endgroup$ – user243301 Nov 8 '15 at 13:55

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