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.

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 $$\varphi\left(x^{\sigma(x)}\sigma(x)^x\right)=2^{x-1} x^{3x-1}\varphi(x)\tag{1}$$ holds.

Computational fact. For integers $1\leq n\leq 5000$, the only solution of $(1)$ 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 $(1)$ 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 $(1)$ 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 $(1)$ 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. Many thanks.

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 $(1)$, 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 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)))

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 who provides 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, 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, 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, question 3588192 from Mathematics Stack Exchange (Mar 20 '20).

  • $\begingroup$ Many thanks for the upvote. $\endgroup$
    – user142929
    Jul 19, 2019 at 13:40
  • $\begingroup$ As aside comment is that I believe that similar discussions are feasible in terms of the Dedekind psi function $\psi(n)$, in particular one should to have the following claim: Each odd perfect number $n$ satisfies $$\psi(n^{\sigma(n)}\sigma(n)^n )=3\cdot 2^{n-1}n^{3n-1}\psi(n).$$ And I cann't find any integer satisfying this equation over the segment $1\leq n\leq 1000$. A reformulation is using the equation $\psi(n^{\sigma(n)}\sigma(n)^n )=3\cdot 2^{n-1}n^{3n}\cdot\frac{\sigma(\text{rad}(n))}{\text{rad}(n)}$. For the definiton $\text{rad}(n)$ see the Wikipedia Radical of an integer. $\endgroup$
    – user142929
    Oct 30, 2019 at 12:56
  • $\begingroup$ I'm trying equations that satisfy odd perfect numbers $x$ (with $y=\sigma(x)$) and that harmonize with the references of MSE that I've added $$2^{\sigma(x)-x-1}x^{\sigma(x)+x-1}=\frac{1}{\varphi(y-x)}\cdot\varphi(x^y y^x)\tag{2}$$ also for integral variables (and satisfied for odd perfect numbers choosing $y=\sigma(x)=2x$ as below). The corresponding code written in Pari/GP is for(x=1, 1000, for(y=1, 1000, if(x<y&&(2^(sigma(x)-x-1))*(x^(x+y-1))*eulerphi(y-x)==eulerphi((x^y)*(y^x)),print(x," ",y)))) it seems that the code for eulerphi() is due to the Pari/GP group of Université Bordeaux I. $\endgroup$
    – user142929
    May 11, 2020 at 18:29
  • $\begingroup$ I'm sorry for the typo, and previous deleted comments. The right code corresponding to the equation $(2)$ in previous comment is for(x=1, 1000, for(y=1, 1000, if(x<y&&(2^(sigma(x)-x-1))*(x^(sigma(x)+x-1))*eulerphi(y-x)==eulerphi((x^y)*(y^x)),print(x," ",y)))) where all solutions that are showed as outputs have the form $(x,y)=(1,\text{even})$. $\endgroup$
    – user142929
    May 11, 2020 at 18:38
  • 1
    $\begingroup$ Other form to state the equation $(1)$ in our Conjecture is to rewrite it as $$y^{y-x}(y-x)^y=\frac{\sigma(x)}{\varphi(y-x)}\cdot\varphi(x^y y^x),$$ or as $$y^{y-x}(y-x)^y=\frac{\sigma(y-x)}{\varphi(y-x)}\cdot\varphi(x^y y^x)$$ for $x$ denoting an odd perfect number and $y=\sigma(x)=2x$. See if you want the related post on Mathematics Stack Exchange with title On a symmetric equation over the integer lattice that involves the Euler's totient function (asked today), with 3679325 as identificator. $\endgroup$
    – user142929
    May 17, 2020 at 17:23

2 Answers 2


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

  • $\begingroup$ Many thanks when I can I read the answer, now I'm in a call center. My intention with this kind of formulas and inequalities (see my rcent post/bounty) is try to evoke some relationship between odd perfect numbers and the growth/size of some number theoretic functions that encode Rieamnn hypothesis (while I knw that my atetmps are vague) $\endgroup$
    – user142929
    Dec 25, 2021 at 13:01

In general, problems involving the composition of multiplicative functions are very hard to analyze. I don't see any specific way to approach this problem, and I'm skeptical that this is likely to be a fruitful direction.

That said, I don't have any strong intuition of whether there will be non-odd perfect numbers satisfying this equation (aside from x=1). My guess is that there will not be, because if x is not an OPN, then $x^{\sigma(x)}$ will have primes raised to very different powers then $\sigma(x)^x$ will, and numbers where $x$ and $\sigma(x)$ have the same set of distinct prime factors are rare. Turning this idea into a proof may be difficult.

  • $\begingroup$ Many thanks for your answer. I've persuaded to myself (it is not a scientific procedure) that the problem of odd perfect numbers is related to Riemann hypothesis. My attempt thus is to create artificious equations with the intention to invoke some equivalence to the RH with the hope to get some useful information (in my imagination it is just transfer information from the odd perfect number problem to the RH problem). I think that due to the bound given by Ochem and Rao the problem of odd perfect numbers is inaccessible for the human mind, and that behind this problem there a physical meaning. $\endgroup$
    – user142929
    Jul 21, 2019 at 20:42
  • $\begingroup$ What connection are you seeing between odd perfect numbers and the Riemann Hypothesis, @user142929? Are you referring to the elementary criterion (i.e. an inequality) by Lagarias? $\endgroup$ Jul 30, 2019 at 22:41
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    $\begingroup$ I am persuaded myself that there is some relationship (and this, the persuasion, is outside the scientific field) @JoseArnaldoBebita-Dris There are in the literature more equivalences/issues related to the Riemann hypothesis similar than Lagarias theorem, because the formulations are using number theoretic functions or special sequences: highly composite numbers, Nicolas criterion, Robin's theorem. In my imagination, from my human size, my ignorance, and as a non-professional mathematicianIn, I ear that these are like a purring/murmur about the OPN. $\endgroup$
    – user142929
    Jul 31, 2019 at 8:44
  • $\begingroup$ I've updated the post adding references for similar problems that I've asked in Mathematics Stack Exchange. The purpose of those posts was try to get characterizations of certain constellations of primes. Maybe these references that I've added in the last paragraph of my post can be inspiring for some user, and we can to continue the discussion of this post of MathOverflow. Many thanks again for your excellent answer. $\endgroup$
    – user142929
    May 10, 2020 at 18:23

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