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If I am right it is easy to prove the following facts involging the Euler's totient function $\varphi(m)$ and the sum of divisors function $\sigma(m)$.

Fact 1. If there exists an odd perfect number $n$ such that $\gcd(105,n)=1$, then our odd perfect number satisfies $$\varphi\left(103\frac{\sigma(n)}{2}+107n\right)=48\varphi(n).\tag{1}$$

Fact 2. If $n>6$ is an even perfect number then $$\varphi\left(103\frac{\sigma(n)}{2}+107n\right)=48\varphi(2n)\tag{2}$$ holds.

Question. I would like to know what work can be done to know the veracity about the following conjectures:

C1) If $n\geq 1$ is an integer satisfying $2\mid \sigma(n)$ and $\gcd(n,105)=1$ such that $(1)$ holds, then $n$ is an odd perfect number.

C2) If $n\geq 1$ is an integer satisfying $2\mid \sigma(n)$ such that $(2)$ holds, then $n$ is an even perfect number.

Many thanks.

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    $\begingroup$ Feel free to add what computational effort can be proposed to motivate or refute the conjectures. $\endgroup$ – user686930 Jul 7 at 19:15
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    $\begingroup$ Can anything be said about the tuned-down versions of these conjectures, with $(105,103,107,48)$ replaced with $(15,13,17,8)$? Or even $(3,1,5,2)$? $\endgroup$ – Hagen von Eitzen Jul 7 at 19:31
  • $\begingroup$ I should see/study it more slowly, if I understand well your words maybe you want to say that the mathematical content of this post isn't the best. Many thanks for your words @HagenvonEitzen $\endgroup$ – user686930 Jul 7 at 19:38
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The facts seem correct. By definition $\sigma(n)=2n$ for any perfect number $n$. So if $n$ is perfect and odd, the first fact reduces to:

$$\phi(210n)=48n$$

So by $(105,n)=1$ and $n$ is odd:

$$\phi(210n)=\phi(210)\phi(n)=48\phi(n)$$.

A similar argument follows for the second fact.

It is well known that if $n$ is an odd perfect number, then 105 does not divide it (however it can maybe still have some factors 3,5,7, just not all of them). So your restriction of $(105,n)$ focuses on a narrower set of odd perfect numbers (if they even exist in the first place).

Your question though seems about as difficult as figuring out whether or not there are odd perfect numbers. I would try at least showing the relations hold for $n\leq 10^6$.

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  • $\begingroup$ Maybe some user can to find a counterexample for this new statement: If $n\geq 1$ is an integer satisfying $2\mid \sigma(n)$ and the identity $(1)$ holds, then $n$ is an odd perfect number. $\endgroup$ – user686930 Jul 7 at 22:34

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