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 $i$?
elementary-number-theory conjectures divisor-sum arithmetic-functions perfect-numbers
asked Nov 26 at 15:02
Jose Arnaldo Bebita Dris
5,31741944
add a comment |
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 $i$?
elementary-number-theory conjectures divisor-sum arithmetic-functions perfect-numbers
asked Nov 26 at 15:02
Jose Arnaldo Bebita Dris
5,31741944
add a comment |
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 $i$?
elementary-number-theory conjectures divisor-sum arithmetic-functions perfect-numbers
asked Nov 26 at 15:02
Jose Arnaldo Bebita Dris
5,31741944
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 $i$?
elementary-number-theory conjectures divisor-sum arithmetic-functions perfect-numbers
elementary-number-theory conjectures divisor-sum arithmetic-functions perfect-numbers
asked Nov 26 at 15:02
Jose Arnaldo Bebita Dris
5,31741944
asked Nov 26 at 15:02
Jose Arnaldo Bebita Dris
5,31741944
edited Nov 26 at 16:06
asked Nov 26 at 15:02
Jose Arnaldo Bebita Dris
5,31741944
asked Nov 26 at 15:02
Jose Arnaldo Bebita Dris
5,31741944
asked Nov 26 at 15:02
Jose Arnaldo Bebita Dris
5,31741944
5,31741944
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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$(N)=pq>sqrt{N}=qsqrt{p}$. 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 $m=prod q_iprod r_j^{beta_j}$ 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. $beta_jge 2$. I'm using $beta$ as the exponent to avoid any confusion with the $alpha$ of the posed question.
$N=pm^2=pprod q_i^2prod r_j^{2beta_j}$
rad$N=pprod q_iprod r_j$ and $sqrt{N}=sqrt{p}prod q_iprod r_j^{beta_j}$
For any particular $q_K$, $prod q_i=q_Kprod_{ine K}q_i=q_KQ_K$ where $Q_K=prod_{ine K}q_i$
rad$N=pq_KQ_Kprod r_j$ and $sqrt{N}=sqrt{p}q_KQ_Kprod r_j^{beta_j}$
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 $N$.
answered Nov 26 at 16:47
Keith Backman
9941612
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 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 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 at 4:58
Gladly accepting your answer now, @KeithBackman.
– Jose Arnaldo Bebita Dris
Nov 28 at 10:53
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014433%2fa-question-regarding-a-paper-of-ochem-and-rao-about-the-radical-of-an-odd-perfec%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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$(N)=pq>sqrt{N}=qsqrt{p}$. 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 $m=prod q_iprod r_j^{beta_j}$ 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. $beta_jge 2$. I'm using $beta$ as the exponent to avoid any confusion with the $alpha$ of the posed question.
$N=pm^2=pprod q_i^2prod r_j^{2beta_j}$
rad$N=pprod q_iprod r_j$ and $sqrt{N}=sqrt{p}prod q_iprod r_j^{beta_j}$
For any particular $q_K$, $prod q_i=q_Kprod_{ine K}q_i=q_KQ_K$ where $Q_K=prod_{ine K}q_i$
rad$N=pq_KQ_Kprod r_j$ and $sqrt{N}=sqrt{p}q_KQ_Kprod r_j^{beta_j}$
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 $N$.
answered Nov 26 at 16:47
Keith Backman
9941612
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 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 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 at 4:58
Gladly accepting your answer now, @KeithBackman.
– Jose Arnaldo Bebita Dris
Nov 28 at 10:53
add a comment |
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$(N)=pq>sqrt{N}=qsqrt{p}$. 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 $m=prod q_iprod r_j^{beta_j}$ 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. $beta_jge 2$. I'm using $beta$ as the exponent to avoid any confusion with the $alpha$ of the posed question.
$N=pm^2=pprod q_i^2prod r_j^{2beta_j}$
rad$N=pprod q_iprod r_j$ and $sqrt{N}=sqrt{p}prod q_iprod r_j^{beta_j}$
For any particular $q_K$, $prod q_i=q_Kprod_{ine K}q_i=q_KQ_K$ where $Q_K=prod_{ine K}q_i$
rad$N=pq_KQ_Kprod r_j$ and $sqrt{N}=sqrt{p}q_KQ_Kprod r_j^{beta_j}$
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 $N$.
answered Nov 26 at 16:47
Keith Backman
9941612
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 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 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 at 4:58
Gladly accepting your answer now, @KeithBackman.
– Jose Arnaldo Bebita Dris
Nov 28 at 10:53
add a comment |
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$(N)=pq>sqrt{N}=qsqrt{p}$. 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 $m=prod q_iprod r_j^{beta_j}$ 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. $beta_jge 2$. I'm using $beta$ as the exponent to avoid any confusion with the $alpha$ of the posed question.
$N=pm^2=pprod q_i^2prod r_j^{2beta_j}$
rad$N=pprod q_iprod r_j$ and $sqrt{N}=sqrt{p}prod q_iprod r_j^{beta_j}$
For any particular $q_K$, $prod q_i=q_Kprod_{ine K}q_i=q_KQ_K$ where $Q_K=prod_{ine K}q_i$
rad$N=pq_KQ_Kprod r_j$ and $sqrt{N}=sqrt{p}q_KQ_Kprod r_j^{beta_j}$
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 $N$.
answered Nov 26 at 16:47
Keith Backman
9941612
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$(N)=pq>sqrt{N}=qsqrt{p}$. 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 $m=prod q_iprod r_j^{beta_j}$ 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. $beta_jge 2$. I'm using $beta$ as the exponent to avoid any confusion with the $alpha$ of the posed question.
$N=pm^2=pprod q_i^2prod r_j^{2beta_j}$
rad$N=pprod q_iprod r_j$ and $sqrt{N}=sqrt{p}prod q_iprod r_j^{beta_j}$
For any particular $q_K$, $prod q_i=q_Kprod_{ine K}q_i=q_KQ_K$ where $Q_K=prod_{ine K}q_i$
rad$N=pq_KQ_Kprod r_j$ and $sqrt{N}=sqrt{p}q_KQ_Kprod r_j^{beta_j}$
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 $N$.
answered Nov 26 at 16:47
Keith Backman
9941612
edited Nov 27 at 16:59
answered Nov 26 at 16:47
Keith Backman
9941612
answered Nov 26 at 16:47
Keith Backman
9941612
answered Nov 26 at 16:47
Keith Backman
9941612
9941612
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 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 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 at 4:58
Gladly accepting your answer now, @KeithBackman.
– Jose Arnaldo Bebita Dris
Nov 28 at 10:53
add a comment |
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 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 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 at 4:58
Gladly accepting your answer now, @KeithBackman.
– Jose Arnaldo Bebita Dris
Nov 28 at 10:53
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 at 3:52
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 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 at 17:04
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 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 at 4:58
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 at 4:58
Gladly accepting your answer now, @KeithBackman.
– Jose Arnaldo Bebita Dris
Nov 28 at 10:53
Gladly accepting your answer now, @KeithBackman.
– Jose Arnaldo Bebita Dris
Nov 28 at 10:53
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3014433%2fa-question-regarding-a-paper-of-ochem-and-rao-about-the-radical-of-an-odd-perfec%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
