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.
