# On prime-perfect numbers and the equation $\frac{\varphi(n)}{n}=\frac{\varphi(\operatorname{rad}(n))}{\operatorname{rad}(\sigma(n))}$

While I was exploring equations involving multiple compositions of number theoretic functions that satisfy the sequence of even perfect numbers, I wondered next question (below in the Appendix I add a different equation if you want to explore it).

For integers $n\geq 1$ in this post we denote the Euler's totient function as $\varphi(n)$, the sum of divisors function $\sum_{d\mid n}d$ as $\sigma(n)$ and 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 last arithmetic function has title Radical of an integer).

As tell us the article dedicated to the sequence A027598 from The On-Line Encyclopedia of Integer Sequences, these are known by the mathematicians as prime-perfect numbers.

Question. Prove or refute that an integer $n\geq 1$ satisfies $$\frac{\varphi(n)}{n}=\frac{\varphi(\operatorname{rad}(n))}{\operatorname{rad}(\sigma(n))}$$ if and only if $n$ is an element of the sequence $A027598$. Many thanks.

I would like to know what work can be done, thus if there are some answers but isn't possible to deduce the full conjecture if and only if, I even should accept an answer.

## Appendix:

The other equation that I wrote and I add here just if some user want to know it, is

$$2\sigma(n)=\operatorname{rad}(\sigma(n))\cdot \varphi(\sigma(\operatorname{rad}(n))).$$

Also it is easy to prove that each even perfect number satisfies it. There are more solutions of previous equation (the equation in this Appendix) that aren't perfect numbers, but I believe that are abundant numbers. I believe that the resulting sequences isn't in the OEIS.

• I add here next reference R. K. Guy, Unsolved problems in number theory, third ed., Problem Books in Mathematics, Springer-Verlag, New York, 2004. – user243301 Jun 1 '18 at 20:54

It's easily proven that $$\frac{\varphi(n)}{n}=\prod_{p\mid n}\frac{p-1}{p}.$$ So if $rad(n)=rad(m)$, then $\frac{\varphi(n)}{n}=\frac{\varphi(m)}{m}$.
Now, suppose that $$\frac{\varphi(n)}{n}=\frac{\varphi(rad(n))}{rad(\sigma(n))}.$$ Since $rad(n)=rad(rad(n))$, it follows that $rad(n)=rad(rad(\sigma(n)))=rad(\sigma(n))$, as desired.