For integers $n\geq 1$ in this post we denote the square-free kernel as $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p,$$ that is the product of distinct primes dividing an integer $n>1$ with the definition $\operatorname{rad}(1)=1$ (the Wikipedia's article dedicated to this multiplicative function is Radical of an integer).

And I denote the Euler's totient function as $\varphi(n)$. We consider the solutions over integers $n\geq 1$ of the equation $$\sigma(\varphi(n))=\sigma(\operatorname{rad}(n)).\tag{1}$$

The first few solutions are $n=1,4,18,87,260,362$ and $732$. We denote the set of all solutions of $(1)$ as $\mathcal{A}$ and for a real number $x>1$ let $\mathcal{A}(x)$ defined as $\mathcal{A}(x)=\mathcal{A}\cap[1,x]$ with cardinality denoted as $\#\mathcal{A}(x)$.

After I did a table using a program with my computer and since my equation is a variant of an equation from the literature [1], I wrote next conjecture (any case I believe that it can be wrong thus I am asking my Question).

Conjecture. The estimate $$\#\mathcal{A}(x)=O\left(\frac{x}{(\log x)^3}\right)\tag{2}$$ is true for enough large $x>1$.

Question. Can you prove or refute previous Conjecture? Many thanks.

I'm curious to know if we can refute previous conjeture, what methods/reasonings can one use to refute it?


[1] Jean-Marie De Koninck and Florian Luca, Positive Integers $n$ Such That $\sigma(\phi(n))=\sigma(n)$, Journal of Integer Sequences, Vol. 11 (2008), Article 08.1.5.

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    $\begingroup$ Upto $10^8$ , $2\ 942$ solutions exist $\endgroup$ – Peter Jun 7 '18 at 11:17
  • $\begingroup$ Many thanks for your calculation @Peter $\endgroup$ – user243301 Jun 7 '18 at 14:28
  • $\begingroup$ My suggestion is that you walk through the method in the paper you cited, and try on your equation. $\endgroup$ – Sungjin Kim Jun 24 '18 at 13:34
  • $\begingroup$ Thanks for your suggestion @i707107 $\endgroup$ – user243301 Jun 24 '18 at 18:09

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