# Has $\sigma\left(\sigma_0(n)^4\right)=n$ infinitely many solutions?

In this post, for integers $n\geq 1$ we consider the sum of divisors function $\sum_{d\mid n}d$ denoted as $\sigma(n)$ and the divisor-counting function $\sum_{d\mid n}1$ as $\sigma_0(n)$. Then I wondered about the solutions of $$\sigma\left((\sigma_0(n))^4\right)=n.\tag{1}$$

Here is the Wikipedia's article Divisor function dedicated to the sum of the positive divisors of an integer $n\geq 1$, that is $\sigma(n)$ and the arithmetic function $\sigma_0(n)$ (also denoted in the literature as $\tau(n)$) that counts the number of those positive divisors.

I was inspired in [1].

Question. Can you prove or refute that there exist infinitely many solutions $n\geq 1$ of $$\sigma\left((\sigma_0(n))^4\right)=n\,?$$ Many thanks.

Computational evidence. Using a Pari/GP program I know that the first few solutions of previous equation $\sigma\left(\sigma_0(n)^4\right)=n,$ are $$1 ,31 ,121 ,511, 3751$$ and $61831.$ Thus I've no enough computational evidence showing that maybe should be infinitely many solutions of $(1)$. Finally seems that previous sequence isn't in the OEIS.

## References:

[1] Problem 3565, Crux Mathematicorum, Volume 37, Number 6 (2011).

• Since $\sigma$ is multiplicative function, the product of two relatively prime solutions is also solution. For example, $31 \times 121=3751$. However, I feel like there are only finite number of solutions... – didgogns Jun 4 '18 at 13:49
• I think that also there are only a finite number of solutions ( and I know that my words contradict my conjecture, but my computations don't offer me more), I tried to combine the size of the divisor-counting function and inequalities from the literature about the maximal size of the sum of divisors function @didgogns , but my attempt was failed. – user243301 Jun 4 '18 at 13:52
• By the way, what was the original Problem 3565, Crux Mathematicorum, Volume 37, Number 6? – didgogns Jun 6 '18 at 5:52
• Many thanks @didgogns again for your new calculations, your work was a professional work $100$% and thanks to the user editing the post. On the other hand you can see it from the web of Crux Mathematicorum, choose the Volume 37 (that is the year 2011)> Oct > Solutions... Feel free to study equations that can be stated as generalization, I am an amateur and I cann't do it. – user243301 Jun 6 '18 at 9:03

Let $\sigma_0(n)=m$, and we need to solve the following equation:$$\sigma_0(\sigma(m^4))=m$$ with known solutions $m=1, 2, 3, 4, 6, 12$. We have a good estimation on the upper bound of both $\sigma_0$ and $\sigma$ functions. For $\sigma_0$, $\sigma_0(x)\le x^{ \left( \frac{1.07}{\log \log x} \right) }$ and for $\sigma$, $\sigma(x)<e^\gamma x\ln \ln x+\frac{0.6483x}{\ln \ln x}$ ($\gamma$ is Euler-Mascheroni constant). For sufficiently large $x$ (at least $e^{e^{5}}$), $\sigma(x)<x^{1.1}$ and $\sigma_0(x)\le x^{0.214}$, therefore$$\sigma_0(\sigma(x^4))<\sigma_0(x^{4.4})\le x^{0.9416}<x$$ so there is only finite number of solution.

Edit June 6

I'll retry using a better upper bound, $\sigma_0(n) \leq n^{ \left( \frac{\log 2}{\log \log n} \right) \left( 1 + \frac{1.935}{\log \log n} \right)}$. Indeed, if $x>e^{e^{4.4}/4.4}$, then again$$\sigma_0(\sigma(x^4))<\sigma_0(x^{4.4})\le x^{0.9979}<x$$so we need to check only for $x<e^{e^{4.4}/4.4}\approx1.1\times10^8$. For such $x$, $\ln(\ln(x^4))<4.3047$ and so $\sigma(x^4)<4.3047e^\gamma x^4+\frac{0.6483x^4}{4.3047}<7.8176x^4<1.1245\times10^{33}$. Now I hope I could get a better upper bound for $\sigma_0(\sigma(x^4))$ using the fact that $\sigma(x^4)$ is odd number, but there is no researches available and the numerical result (https://oeis.org/A053624) is also limited to like $30$ digits.

Therefore, I bruted forced to find that odd number under $1.1245\times10^{33}$ can have at most $11796480$ divisors. It again gives maximum $\sigma(x^4)$ value of about $1.4706\times10^{29}$, which gives maximum divisor number of $2654208$, which gives maximum $\sigma(x^4)$ value of $3.6857\times10^{26}$, which gives maximum divisor number of $983040$, which gives maximum $\sigma(x^4)$ value of $6.8223\times10^{24}$, which gives maximum divisor number of $497664$, which gives maximum $\sigma(x^4)$ value of $4.4271\times10^{23}$, which gives maximum divisor number of $307200$, which gives maximum $\sigma(x^4)$ value of $6.3697\times10^{22}$, which gives maximum divisor number of $221184$, which gives maximum $\sigma(x^4)$ value of $1.7009\times10^{22}$, which gives maximum divisor number of $184320$, which gives maximum $\sigma(x^4)$ value of $8.1725\times10^{21}$, which gives maximum divisor number of $153600$, which gives maximum $\sigma(x^4)$ value of $3.9265\times10^{21}$, which gives maximum divisor number of $138240$, which gives maximum $\sigma(x^4)$ value of $2.5706\times10^{21}$, which gives maximum divisor number of $122880$. That is, $m \le 122880$.

Also, $2.5706\times10^{21}<3^{45}$ so $m$ cannot have prime factor greater than $45$. Computer-checking all integers under $122880$ confirms that they are indeed the only solutions.

• You have magic in your soccer boots! Many thanks I am going to study it, and the linked MSE and Wikipedia. On the other hand $4.4\cdot 0.214$ was close to $1$. If you want to add some observation/calculation in next future feel free to do it. – user243301 Jun 4 '18 at 16:24
• Can you elaborate on why $n=\sigma(m^4)$ is equivalent to the statement $\sigma_0(\sigma(m^4))=m$ ? I would think you'd need $\sigma_0$ to be injective for that to be true :) – N8tron Jun 4 '18 at 17:05
• From the original equation, applying $\sigma_0$ to both side gives $\sigma_0(\sigma\left((\sigma_0(n))^4\right))=\sigma_0(n)$ so $\sigma_0(\sigma(m^4))=m$ is derived. From $\sigma_0(\sigma(m^4))=m$, $4$-power both sides and applying $\sigma$ gives $\sigma(\sigma_0(\sigma(m^4))^4)=\sigma(m^4)$ so original equation is derived. – didgogns Jun 5 '18 at 0:08