For integers $n\geq 1$ I denote the Euler's totient function as $\varphi(n)$ and the divisor function $\sum\_{1\leq d\mid n}d$ as $\sigma(n)$, that are two well-known mulitplicative functions. We assume also the theory of odd perfect numbers, see if you want the corresponding section of the Wikipedia with title *[Perfect number](https://en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers)*.


It is easy to prove the following statement, on assumption that there exists an odd perfect number $x$.



> 
> **Fact.** If $x$ is an odd perfect number then $$\varphi\left(x^{\sigma(x)}\sigma(x)^x\right)=2^{x-1} x^{3x-1}\varphi(x)\tag{1}\label{1}$$
> holds.
> 
> 
> 


**Computational fact.** For integers $1\leq n\leq 5000$, the only solution of \eqref{1} is $n=1$. To see it, after some seconds, choose *GP* as language and evaluate next code (it is just a line written in Pari/GP) in the web page Sage Cell Server


`for (x = 1, 5*10^3,if (eulerphi(x^(sigma(x))*(sigma(x))^x)==2^(x-1)*x^(3*x-1)*eulerphi(x), print(x)))`


I believe that the following conjecture holds.



> 
> **Conjecture.** *The only solution of our equation \eqref{1} is the integer $1$.*
> 
> 
> 


**Motivation for the post.** My belief is that an interesting way (but my attempts were failed) to study the unsolved problem related to odd perfect numbers (that is if there exist any of them) should be to create intrincated/artificious equations similar than \eqref{1} involving the sum of divisors functions and the Euler's totient function with the purpose to invoke inequalitites, asymptotics, heuristics or conjectures for these arithmetic functions (my belief is that the problem of odd perfect numbers is related to the distribution of prime numbers, thus maybe in the equations *similar than* \eqref{1} that previously I've evoked should be required also that arise functions as the radical of an integer $\operatorname{rad}(x)$ or even the prime-counting function $\pi(x)$, both specialized for odd perfect numbers $x$).



> 
> **Question.** What work can be done to prove of refute previous conjecture, that the only solution of $$\varphi\left(n^{\sigma(n)}\sigma(n)^n\right)=2^{n-1} n^{3n-1}\varphi(n)$$ should be $n=1$? It is welcome unconditionally statements or heuristics, but also feel free to invoke conjectures if you can get some advanced statement.
> 
> 
> 


Thus, as how it is perceived in the title of the post, previous **Question** is also an invitation to add remarkable statements about the nature of the solutions of \eqref{1}, if we are in the situation that the Question can not be solved.


**Last remarks to to emphasize my ideas.** What is saying myself thus previous **Motivation** and **Question**? That of couse I understand that the equation/characterization for odd perfect nubmers by means the equation $\sigma(x)=2x$ for odd integers $x\geq 1$ is easiest (to understand and study it) than others involving more arithmetic functions, but in my belief is that there exists a chance to get some statement for odd perfect numbers by the method to create more intrincated/artificious equations.


I think that my question is interesting, and I think that arises in a natural way when one tries to drop solutions like $2^{2^{\lambda-1}-1}$, that is the sequence *[A058891](https://oeis.org/A058891)* from the On-Line Encyclopedia of Integer Sequences, for equations like this $$\varphi(x^x\sigma(x))=x^x\varphi(x).$$
See if you want the code


`for (x = 1, 10^4,if (eulerphi((x^x)*sigma(x))==(x^x)*eulerphi(x), print(x)))`




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I would like to refer that certain characterizations of primes are feasibles from answers of next posts (the problem [2] remains as unsolved), these posts are not directly related to this my post in MathOverflow, but maybe can be inspiring for some user of MathOverflow since are similar problems. Thus I justify this last paragraph as a compilation of similar equations for constellations of primes: thanks to the excellence of the user who provides the answer of [1] we've the characterization of Sophie Germain primes and similarly for twin primes; thanks to the excellence of the user mathlove who provided the answers of problems [2] and [3] we've a characterization of Mersenne exponents, Fermat primes and near-square primes.


[1] [*From the equation* $\sigma(x^{\varphi(y)})=\frac{1}{\varphi(x)}(x^y-1)$ *involving arithmetic functions to a characterization of Sophie Germain primes*](https://math.stackexchange.com/questions/3578715/from-the-equation-sigmax-varphiy-frac1-varphixxy-1-involving), question **3578715** from Mathematics Stack Exchange (Mar 12 '20).


[2] [*From the equation* $\sigma(x^{\sigma(y)-1})=\frac{1}{\varphi(x)}(x^{y+1}-1)$ *involving arithmetic functions to a characterization of Mersenne exponents*](https://math.stackexchange.com/questions/3587159/from-the-equation-sigmax-sigmay-1-frac1-varphixxy1-1-invo), question **3587159** from Mathematics Stack Exchange (Mar 19 '20).


[3] [*On characterizations for near-square primes and Fermat primes in terms of equations involving arithmetic functions*](https://math.stackexchange.com/questions/3588192/on-characterizations-for-near-square-primes-and-fermat-primes-in-terms-of-equati), question **3588192** from Mathematics Stack Exchange (Mar 20 '20).



What work can be done to study the solutions of $\varphi\left(x^{\sigma(x)}\sigma(x)^x\right)=2^{x-1} x^{3x-1}\varphi(x)$?
Here is a proof that if an odd integer $x>1$ satisfies (1), then $x$ is a perfect number.


First, by using the property that $\varphi(nm)=n\varphi(m)$ whenever $\mathrm{rad}(n)\mid\mathrm{rad}(m)$, we can rewrite (1) as
$$\big(\frac{\sigma(x)}2\big)^{x-1}\cdot \frac{\varphi(x\sigma(x))}{\varphi(x)} = x^{3x-\sigma(x)}.$$
Since the fractions in the l.h.s. are integer, the r.h.s. is also integer and odd, and thus $\sigma(x)=2y$ for some odd $y$ such that $\mathrm{rad}(y)\mid \mathrm{rad}(x)$. Then we further simplify the above equation to
$$y^x = x^{3x-2y}.$$
In other words,
$$\big(\frac{y}x\big)^x = x^{2(x-y)}.$$


Now, if $x>y$, then lhs < 1 while rhs > 1.
Vice versa, if $x<y$, then lhs > 1 while rhs < 1.


Hence, $x=y$, meaning that $\sigma(x)=2x$. QED


