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 $p > {10}^{60}$.


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 $i$?


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 $N$.

  • $\begingroup$ Unfortunately, this does not answer my question, as it will not allow me to compare $p$ with all $q_i$ such that $\alpha_i = 2$. $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 27 '18 at 3:52
  • $\begingroup$ I've added detail to what I meant by "generalizing". If this does not answer the question, then I do not understand what you are asking with the words "What happens...?" I thought you were asking whether or not $p$ could be shown to be the largest prime factor of $N$ in that case. $\endgroup$ – Keith Backman Nov 27 '18 at 17:04
  • $\begingroup$ Thank you for the added details, @KeithBackman. Will accept your answer in a few, if there are no others. $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 28 '18 at 4:58
  • $\begingroup$ Gladly accepting your answer now, @KeithBackman. $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 28 '18 at 10:53

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