On miscellaneous questions about perfect numbers II Let $\varphi(m)$ the Euler's totient function and $\sigma(m)$ the sum of divisors function. We also denote the product of primes dividing an integer $m>1$ as $\operatorname{rad}(m)$, that is the radical of an integer, see this Wikipedia, with the definition  $\operatorname{rad}(1)=1$.
For each integer $m>1$ one has $$\varphi(m)=m\prod_{p\mid m}\left(1-\frac{1}{p}\right).$$
If $n$ is an even perfect number it is well known that $2\varphi(n)=\varphi(\sigma(n))$ (sequence A067704 in The On-Line Encyclopedia of Integer Sequences) and one writes that $n$ then that also satisfies 
$$\frac{\sigma(n)}{\operatorname{rad}(n)}=\frac{\varphi(\sigma(n))}{\varphi(\operatorname{rad}(n))}.\tag{1}$$
I think that it also is well-known.

Question. I would like to know if we can prove or discard that the sequence of integers $m$ satisfying $(1)$ is the OEIS sequence A027598. That is if you can to find a counterexample or justify these are the same sequence. Many thanks.

 A: It might be helpful to rewrite your $(1)$ as
$$\frac{\varphi(\operatorname{rad}(n))}{\operatorname{rad}(n)}=\frac{\varphi(\sigma(n))}{\sigma(n)}.\tag{1}$$
OEIS sequence A027598 is the sequence of numbers satisfying
$$\operatorname{rad}(n)=\operatorname{rad}(\sigma(n)).\tag{2}$$
If a number $m$ is in this sequence we have
$$\frac{\varphi(\operatorname{rad}(m))}{\operatorname{rad}(m)}=\frac{\varphi(\operatorname{rad}(\sigma(m)))}{\operatorname{rad}(\sigma(m))}.$$
We can use the equation
$\frac{\varphi(\operatorname{rad}(n))}{\operatorname{rad}(n)}=\frac{\varphi(n)}{n}$
to obtain
$$\frac{\varphi(\operatorname{rad}(m))}{\operatorname{rad}(m)}=\frac{\varphi(\sigma(m))}{\sigma(m)}$$
which means $m$ satisfies $(1)$. Thus any number in the sequence satisfies $(1)$.
Suppose $m$ is a (different) number satisfying $(1)$. Then we have
$$\frac{\varphi(\operatorname{rad}(m))}{\operatorname{rad}(m)}=\frac{\varphi(\operatorname{rad}(\sigma(m)))}{\operatorname{rad}(\sigma(m))}.\tag{3}$$
Suppose also $(2)$ does not hold for $m$ so that
$$\frac{\operatorname{rad}(m)}{p_1 p_2 p_3...p_r}=\frac{\operatorname{rad}(\sigma(m))}{q_1 q_2 q_3...q_r}\tag{4}$$
Where the $q_i$ are prime factors of $\sigma(m)$ but not of $m$ and the $p_i$ are factors of $m$ but not of $\sigma(m)$.
We can rewrite $(3)$ as 
$$\frac{\operatorname{rad}(\sigma(m))}{\operatorname{rad}(m)}=\frac{\varphi(\operatorname{rad}(\sigma(m)))}{\varphi(\operatorname{rad}(m))}$$
and using $(4)$ we can write it as
$$\frac{q_1 q_2 q_3...q_r}{p_1p_2p_3...p_r}=\frac{\varphi(\operatorname{rad}(\sigma(m)))}{\varphi(\operatorname{rad}(m))}$$
where the fraction on the left is in lowest terms. But from $(4)$ we also have
$$\frac{\varphi(\operatorname{rad}(\sigma(m)))}{\varphi(\operatorname{rad}(m))}=\frac{(q_1-1)(q_2-1)(q_3-1)...(q_r-1)}{(p_1-1)(p_2-1)(p_3-1)...(p_r-1)}.$$
Combining these two we obtain
$$\frac{q_1 q_2 q_3...q_r}{p_1p_2p_3...p_r}=\frac{(q_1-1)(q_2-1)(q_3-1)...(q_r-1)}{(p_1-1)(p_2-1)(p_3-1)...(p_r-1)}.$$
This cannot be true because the fraction on the left is in lowest terms, and the fraction on the right has a smaller numerator and a smaller denominator. Thus our assumption was false and no such $p_i$ and $q_i$ exist. Therefore $(2)$ holds.
We have now shown that if $(1)$ holds for a number $m$ then $(2)$ holds i.e. $m$ is in A027598. We have also shown previously that if $m$ is in A027598 then $(1)$ holds. Thus the sequence of integers satisfying $(1)$ is the same sequence as A027598.
