# 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 $$i$$?

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 $$N$$.

• Unfortunately, this does not answer my question, as it will not allow me to compare $p$ with all $q_i$ such that $\alpha_i = 2$. – Jose Arnaldo Bebita-Dris Nov 27 '18 at 3:52
• 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. – Keith Backman Nov 27 '18 at 17:04
• Thank you for the added details, @KeithBackman. Will accept your answer in a few, if there are no others. – Jose Arnaldo Bebita-Dris Nov 28 '18 at 4:58