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For integers $n\geq 1$ we denote 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 multiplicative function is Radical of an integer).

An integer $n\geq 1$ is said to be an odd perfect number if $n$ is an odd integer and satisfies $$\sigma(n)=2n,$$ where $$\sigma(m)=\sum_{d\mid m}d=\prod_{\substack{p\mid m\\p\text{ prime}}}\frac{p^{e_p+1}-1}{p-1}$$ is the sum of the positive divisors of $m=\prod_{\substack{p\mid m\\p\text{ prime}}}p^{e_p}$.

It is well-known that if there exists an odd perfect number (it is an open problem) then has a specific form, I mean mainly the Euler's theorem for odd perfect numbers but many results are known about odd perfect numbers, for example Touchard's theorem or theorems about their prime factorizations. In particular are known upper bounds for $$\operatorname{rad}(n)\leq \text{Upper bound}\tag{1}$$ where here in $(1)$ with $\text{Upper bound}=\text{Upper bound}(n)$ we denote a function of $n$. See Proposition 1 of [1], or [2] (on assumption of certain conditions that satisfy the odd perfect numbers).

Question. I would like to know if it is possible to deduce some statement of the form $$\text{Lower bound}\leq \operatorname{rad}(n),\tag{2}$$ where here in $(2)$ with $\text{Lower bound}=\text{Lower bound}(n)$ we denote a function of $n$ an odd perfect number.

Please if you know literature about it provide the references answering my question as a reference request and I try to find and read those propositions. In case that isn't in the literature, can work can be done about $(2)$ unconditionally or on assumption that our odd perfect number $n$ does satisfy any conditions? Many thanks.

References:

[1] Florian Luca and Carl Pomerance, On the radical of a perfect number, The New York Journal of Mathematics, Volume: 16 (2010), page 23-30.

[2] Ph. Ellia, A remark on the radical of a perfect numbers, The Fibonacci Quarterly VOLUME 50, NUMBER 3 (AUGUST 2012).

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    $\begingroup$ Several necessary conditions for possible odd perfect numbers have been found out, you could derive a lower bound of $rad(n)$ from them. $\endgroup$ – Peter Jun 8 '18 at 9:44
  • $\begingroup$ I don't know how get it, I know that there are many results for odd perfect numbers but I don't know what can be useful here @Peter Feel free if your want to provide an answer. $\endgroup$ – user243301 Jun 8 '18 at 9:48
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    $\begingroup$ It will be cumbersome, but I might try it. $\endgroup$ – Peter Jun 8 '18 at 9:50
  • $\begingroup$ Many thanks @Peter , I believe that isn't in the literature, thus it seems very difficult to get. As was asked in the Question you can to assume that our odd perfect number has a specific form. $\endgroup$ – user243301 Jun 8 '18 at 9:54
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    $\begingroup$ Here : en.wikipedia.org/wiki/Perfect_number#Odd_perfect_numbers are some necessary conditions $\endgroup$ – Peter Jun 8 '18 at 9:55
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Considering that the largest $3$ prime factors must exceed $10^8$, $10^4$ and $100$, and that at least $10$ distinct prime factors must exist and considering that $n$ cannot be disivisble by $105$, we get a first lower bound : $$rad(n)\ge 3\cdot 5\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23\cdot 101\cdot 10007 \cdot 100000007=1610582436734262679545>1.6\cdot 10^{21}$$

But I do not think that I considered all informations. Probably, the bound can be further improved.

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    $\begingroup$ Many thanks for your great answer, I am going to study it. I suspected that the $\text{Lower bound}$ in the answers of users it would be a constant $\text{Lower bound}=\text{constant}$. $\endgroup$ – user243301 Jun 8 '18 at 11:19
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    $\begingroup$ @user243301: Why? Were you half-expecting that there was a lower bound for $rad(n)$ in terms of $n$? $\endgroup$ – Jose Arnaldo Bebita-Dris Jun 9 '18 at 8:20
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    $\begingroup$ I don't know, I'm amatetur sometimes my thoughts harmonize with the answers that others provide me here, other times not @JoseArnaldoBebitaDris . $\endgroup$ – user243301 Jun 9 '18 at 8:28
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    $\begingroup$ @user243301, in that case, please see my answer below. $\endgroup$ – Jose Arnaldo Bebita-Dris Jun 9 '18 at 8:38
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This does not really directly answer your question, and is too long to fit in the comments. But since you asked for reference requests, I wanted to point you to other sources for the topic on the radical of an odd perfect number:

Oleksiy Klurman, Radical of perfect numbers and perfect numbers among polynomial values

Klurman proves that $rad(n) < n^{9/{14}}$, improving the previous bound of Luca and Pomerance as well as Acquaah and Konyagin.

Pascal Ochem and Michaël Rao, Another remark on the radical of an ODD perfect number

Ellia recently proved that if $n$ is an odd perfect number such that $rad(n) > \sqrt{n}$, then its special prime factor $p$ satisfies $p > 148207$ if $3 \nmid n$ and $p > 223$ otherwise. Ellia also suggested that these bounds can be improved with some computation. Ochem and Rao obtain that if $n$ is an odd perfect number such that $rad(n) > \sqrt{n}$, then $p > {10}^{60}$.

This maybe the reason why you seem to have been half-expecting some partial results on a lower bound for $rad(n)$ (in terms of $n$) when $n$ is an odd perfect number. I will keep working on it and will get back to you with an updated answer ASAP.

Added October 28 2018 I would like to revisit this answer and add another reference, this time Anirudh Prabhu's "On the Sum and Product of Distinct Prime Factors of an Odd Perfect Number":

Prabhu proves that $$rad(n) > \bigg(\frac{1}{2^{1/\omega(n)} - 1}\bigg)^{\omega(n)},$$ if $n$ is an odd perfect number with $\omega(n)$ distinct prime factors.

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  • $\begingroup$ Many thanks for your great answer. I didn't know these references. $\endgroup$ – user243301 Jun 9 '18 at 8:58
  • $\begingroup$ Note that Prabhu's lower bound for $rad(n)$, together with Nielsen's lower bound of $\omega(n) \geq 10$ (for the number of distinct prime factors of an odd perfect number $n$), gives the lower bound $$rad(n) > \bigg(\frac{1}{2^{1/10} - 1}\bigg)^{10} \approx 275651917451,$$ which is smaller compared to Peter's bound in his answer above. (Here is the WolframAlpha computation link.) $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 13 '18 at 8:04

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