A question regarding a paper of Ochem and Rao about the radical of an odd perfect number Let $\operatorname{rad}(n)$ denote the radical or square-free part of the positive integer $n$, that is, 
$$\operatorname{rad}(n) = \prod_{p \mid n}{p}$$ 
where $p$ runs over primes.
In the paper titled Another remark on the radical of an odd perfect number, part of Ochem and Rao's proof for Theorem 1.2 is as follows:
Statement of the Theorem If $N = p^e m^2$ is an odd perfect number such that $$\operatorname{rad}(N) > \sqrt{N},$$
then $p > {10}^{60}$.
Proof

Suppose that $N = p^e m^2$ is an odd perfect number such that
  $$\operatorname{rad}(N) > \sqrt{N}.$$
  This implies obviously that $e=1$. Let us write $m^2 = \Pi {q_i}^{\alpha_i}$ where the $q_i$'s are distinct primes.  We have
  $$\bigg(\operatorname{rad}(N)\bigg)^2 = \bigg(\operatorname{rad}(p(\Pi {q_i}^{\alpha_i}))\bigg)^2 = p^2 \Pi {q_i}^2 = \frac{p}{\Pi {{q_i}^{\alpha_i - 2}}}N.$$
  Hence,
  $$\operatorname{rad}(N) > \sqrt{N}$$
  implies that $p > \Pi {{q_i}^{\alpha_i - 2}}$.

Here is my question #1:

Does
  $$\operatorname{rad}(N) > \sqrt{N}$$
  also imply that $p$ is the largest prime factor of $N$?

Note that the answer is evidently YES if $\alpha_i > 2$, for all $i$.
Here then is my question #2:

What happens when $\alpha_i = 2$, for some $i$?

 A: Note that $\alpha_i$ must be even. Looking at the simplest case, if $m=q$, implying $\alpha =2$, there is no limit on the size of $q$ that results in rad$(N)=pq>\sqrt{N}=q\sqrt{p}$. Generalizing, for each $q_i$ such that $\alpha_i=2$, the same result obtains.
Added by edit: Questioner does not think the above answers his question. So a detailed look at "generalizing" is in order. Let $m=\prod q_i\prod r_j^{\beta_j}$ where $q_i$ are primes that occur as factors once in $m$ and $r_j$ are primes that occur more than once as factors in $m$; i.e. $\beta_j\ge 2$. I'm using $\beta$ as the exponent to avoid any confusion with the $\alpha$ of the posed question.
$N=pm^2=p\prod q_i^2\prod r_j^{2\beta_j}$
rad$N=p\prod q_i\prod r_j$ and $\sqrt{N}=\sqrt{p}\prod q_i\prod r_j^{\beta_j}$
For any particular $q_K$, $\prod q_i=q_K\prod_{i\ne K}q_i=q_KQ_K$ where $Q_K=\prod_{i\ne K}q_i$
rad$N=pq_KQ_K\prod r_j$ and $\sqrt{N}=\sqrt{p}q_KQ_K\prod r_j^{\beta_j}$
As in the simplest case, there is no limit on the magnitude of $q_K$ with respect to $p$ that has any influence on the relationship rad$N>\sqrt{N}$. That is to say, $p$ might be larger or smaller than any particular $q_K$. It cannot be ascertained whether $p$ is the largest prime factor of $N$. 
