Mixing
Proof of Bertrand's postulate
$begingroup$
The following proof is from the 19th page of
Everest, Graham; Ward, Thomas, An introduction to number theory, Graduate Texts in Mathematics 232. London: Springer (ISBN 1-85233-917-9/hbk). x, 294 p. (2005). ZBL1089.11001.
In fact, I think this proof is not finished. For the red line, only $k(p)ge 2$ is disproved. For the case $k(p)=1$, there is not any talk. How to understand it ? Thanks.
$textbf{Theorem 1.9.} [text{B}{scriptstyle{text{ERTRAND'S}}} text{ P}scriptstyle{text{OSTULATE}}] $ If $ngeqslant1$, then there is at least one prime $p$ with the property that $$n<pleqslant2n.tag{1.13}$$ $text{P}scriptstyle{text{ROOF}}$. For any real number $x$, let $lfloor xrfloor$ denote the integer part of $x$. Thus $lfloor xrfloor$ is the greatest integer less than or equal to $x$. Let $p$ be any prime. Then $$leftlfloordfrac nprightrfloor+leftlfloordfrac n{p^2}rightrfloor+leftlfloordfrac n{p^3}rightrfloor+cdots$$ is the largest power of $p$ dividing $n!$ (see Exercise $8.7(a)$ on p. $162$). Fix $ngeqslant 1$ and let $$N=prod_{pleqslant2n}p^{k(p)}$$ be the prime decomposition of $N=(2n)!/(n!)^2$. The number of times that a given prime $p$ divides $N$ is the difference between the number of times it divides $(2n)!$ and $(n!)^2$, so $$k(p)=sum_{m=1}^inftyleft(leftlfloordfrac{2n}{p^m}rightrfloor-2leftlfloordfrac n{p^m}rightrfloorright),tag{1.14}$$ and each of the terms in the sum is either $0$ or $1$, depending on whether $leftlfloorfrac{2n}{p^m}rightrfloor$ is odd or even. If $p^m>2n$ the term is certainly $0$, so $$k(p)leqslantleftlfloordfrac{log2n}{log p}rightrfloor.tag{1.15}$$
number-theory
edited Jan 27 at 9:58
Citation Needed
31
asked Jan 9 at 9:58
lanse7ptylanse7pty
1,7811823
$endgroup$
add a comment |
$begingroup$
The following proof is from the 19th page of
Everest, Graham; Ward, Thomas, An introduction to number theory, Graduate Texts in Mathematics 232. London: Springer (ISBN 1-85233-917-9/hbk). x, 294 p. (2005). ZBL1089.11001.
In fact, I think this proof is not finished. For the red line, only $k(p)ge 2$ is disproved. For the case $k(p)=1$, there is not any talk. How to understand it ? Thanks.
$textbf{Theorem 1.9.} [text{B}{scriptstyle{text{ERTRAND'S}}} text{ P}scriptstyle{text{OSTULATE}}] $ If $ngeqslant1$, then there is at least one prime $p$ with the property that $$n<pleqslant2n.tag{1.13}$$ $text{P}scriptstyle{text{ROOF}}$. For any real number $x$, let $lfloor xrfloor$ denote the integer part of $x$. Thus $lfloor xrfloor$ is the greatest integer less than or equal to $x$. Let $p$ be any prime. Then $$leftlfloordfrac nprightrfloor+leftlfloordfrac n{p^2}rightrfloor+leftlfloordfrac n{p^3}rightrfloor+cdots$$ is the largest power of $p$ dividing $n!$ (see Exercise $8.7(a)$ on p. $162$). Fix $ngeqslant 1$ and let $$N=prod_{pleqslant2n}p^{k(p)}$$ be the prime decomposition of $N=(2n)!/(n!)^2$. The number of times that a given prime $p$ divides $N$ is the difference between the number of times it divides $(2n)!$ and $(n!)^2$, so $$k(p)=sum_{m=1}^inftyleft(leftlfloordfrac{2n}{p^m}rightrfloor-2leftlfloordfrac n{p^m}rightrfloorright),tag{1.14}$$ and each of the terms in the sum is either $0$ or $1$, depending on whether $leftlfloorfrac{2n}{p^m}rightrfloor$ is odd or even. If $p^m>2n$ the term is certainly $0$, so $$k(p)leqslantleftlfloordfrac{log2n}{log p}rightrfloor.tag{1.15}$$
number-theory
edited Jan 27 at 9:58
Citation Needed
31
asked Jan 9 at 9:58
lanse7ptylanse7pty
1,7811823
$endgroup$
$begingroup$
Already the definition of $k(p)$ is confusing.
$endgroup$
– Peter
Jan 9 at 10:06
$begingroup$
@Peter I think the definition of $k(p)$ is very ok
$endgroup$
– lanse7pty
Jan 9 at 10:11
$begingroup$
But hard to actually understand. We should find out what $k(p)=1$ means. Perhaps this case is trivial for the desired proof.
$endgroup$
– Peter
Jan 9 at 10:13
$begingroup$
$k(p)$ is the exponent of the prime $p$ in the prime decomposition of $N$. $k(p)$ is well-defined, unique and greater or equal to $0$.
$endgroup$
– Snake707
Jan 9 at 10:30
$begingroup$
@Snake707 I did not claim that $k(p)$ is not well defined.
$endgroup$
– Peter
Jan 9 at 10:57
add a comment |
$begingroup$
The following proof is from the 19th page of
Everest, Graham; Ward, Thomas, An introduction to number theory, Graduate Texts in Mathematics 232. London: Springer (ISBN 1-85233-917-9/hbk). x, 294 p. (2005). ZBL1089.11001.
In fact, I think this proof is not finished. For the red line, only $k(p)ge 2$ is disproved. For the case $k(p)=1$, there is not any talk. How to understand it ? Thanks.
$textbf{Theorem 1.9.} [text{B}{scriptstyle{text{ERTRAND'S}}} text{ P}scriptstyle{text{OSTULATE}}] $ If $ngeqslant1$, then there is at least one prime $p$ with the property that $$n<pleqslant2n.tag{1.13}$$ $text{P}scriptstyle{text{ROOF}}$. For any real number $x$, let $lfloor xrfloor$ denote the integer part of $x$. Thus $lfloor xrfloor$ is the greatest integer less than or equal to $x$. Let $p$ be any prime. Then $$leftlfloordfrac nprightrfloor+leftlfloordfrac n{p^2}rightrfloor+leftlfloordfrac n{p^3}rightrfloor+cdots$$ is the largest power of $p$ dividing $n!$ (see Exercise $8.7(a)$ on p. $162$). Fix $ngeqslant 1$ and let $$N=prod_{pleqslant2n}p^{k(p)}$$ be the prime decomposition of $N=(2n)!/(n!)^2$. The number of times that a given prime $p$ divides $N$ is the difference between the number of times it divides $(2n)!$ and $(n!)^2$, so $$k(p)=sum_{m=1}^inftyleft(leftlfloordfrac{2n}{p^m}rightrfloor-2leftlfloordfrac n{p^m}rightrfloorright),tag{1.14}$$ and each of the terms in the sum is either $0$ or $1$, depending on whether $leftlfloorfrac{2n}{p^m}rightrfloor$ is odd or even. If $p^m>2n$ the term is certainly $0$, so $$k(p)leqslantleftlfloordfrac{log2n}{log p}rightrfloor.tag{1.15}$$
number-theory
edited Jan 27 at 9:58
Citation Needed
31
asked Jan 9 at 9:58
lanse7ptylanse7pty
1,7811823
$endgroup$
The following proof is from the 19th page of
Everest, Graham; Ward, Thomas, An introduction to number theory, Graduate Texts in Mathematics 232. London: Springer (ISBN 1-85233-917-9/hbk). x, 294 p. (2005). ZBL1089.11001.
In fact, I think this proof is not finished. For the red line, only $k(p)ge 2$ is disproved. For the case $k(p)=1$, there is not any talk. How to understand it ? Thanks.
$textbf{Theorem 1.9.} [text{B}{scriptstyle{text{ERTRAND'S}}} text{ P}scriptstyle{text{OSTULATE}}] $ If $ngeqslant1$, then there is at least one prime $p$ with the property that $$n<pleqslant2n.tag{1.13}$$ $text{P}scriptstyle{text{ROOF}}$. For any real number $x$, let $lfloor xrfloor$ denote the integer part of $x$. Thus $lfloor xrfloor$ is the greatest integer less than or equal to $x$. Let $p$ be any prime. Then $$leftlfloordfrac nprightrfloor+leftlfloordfrac n{p^2}rightrfloor+leftlfloordfrac n{p^3}rightrfloor+cdots$$ is the largest power of $p$ dividing $n!$ (see Exercise $8.7(a)$ on p. $162$). Fix $ngeqslant 1$ and let $$N=prod_{pleqslant2n}p^{k(p)}$$ be the prime decomposition of $N=(2n)!/(n!)^2$. The number of times that a given prime $p$ divides $N$ is the difference between the number of times it divides $(2n)!$ and $(n!)^2$, so $$k(p)=sum_{m=1}^inftyleft(leftlfloordfrac{2n}{p^m}rightrfloor-2leftlfloordfrac n{p^m}rightrfloorright),tag{1.14}$$ and each of the terms in the sum is either $0$ or $1$, depending on whether $leftlfloorfrac{2n}{p^m}rightrfloor$ is odd or even. If $p^m>2n$ the term is certainly $0$, so $$k(p)leqslantleftlfloordfrac{log2n}{log p}rightrfloor.tag{1.15}$$
number-theory
number-theory
edited Jan 27 at 9:58
Citation Needed
31
asked Jan 9 at 9:58
lanse7ptylanse7pty
1,7811823
edited Jan 27 at 9:58
Citation Needed
31
asked Jan 9 at 9:58
lanse7ptylanse7pty
1,7811823
edited Jan 27 at 9:58
Citation Needed
31
edited Jan 27 at 9:58
Citation Needed
31
edited Jan 27 at 9:58
Citation Needed
31
31
asked Jan 9 at 9:58
lanse7ptylanse7pty
1,7811823
asked Jan 9 at 9:58
lanse7ptylanse7pty
1,7811823
asked Jan 9 at 9:58
lanse7ptylanse7pty
1,7811823
1,7811823
$begingroup$
Already the definition of $k(p)$ is confusing.
$endgroup$
– Peter
Jan 9 at 10:06
$begingroup$
@Peter I think the definition of $k(p)$ is very ok
$endgroup$
– lanse7pty
Jan 9 at 10:11
$begingroup$
But hard to actually understand. We should find out what $k(p)=1$ means. Perhaps this case is trivial for the desired proof.
$endgroup$
– Peter
Jan 9 at 10:13
$begingroup$
$k(p)$ is the exponent of the prime $p$ in the prime decomposition of $N$. $k(p)$ is well-defined, unique and greater or equal to $0$.
$endgroup$
– Snake707
Jan 9 at 10:30
$begingroup$
@Snake707 I did not claim that $k(p)$ is not well defined.
$endgroup$
– Peter
Jan 9 at 10:57
add a comment |
$begingroup$
Already the definition of $k(p)$ is confusing.
$endgroup$
– Peter
Jan 9 at 10:06
$begingroup$
@Peter I think the definition of $k(p)$ is very ok
$endgroup$
– lanse7pty
Jan 9 at 10:11
$begingroup$
But hard to actually understand. We should find out what $k(p)=1$ means. Perhaps this case is trivial for the desired proof.
$endgroup$
– Peter
Jan 9 at 10:13
$begingroup$
$k(p)$ is the exponent of the prime $p$ in the prime decomposition of $N$. $k(p)$ is well-defined, unique and greater or equal to $0$.
$endgroup$
– Snake707
Jan 9 at 10:30
$begingroup$
@Snake707 I did not claim that $k(p)$ is not well defined.
$endgroup$
– Peter
Jan 9 at 10:57
$begingroup$
Already the definition of $k(p)$ is confusing.
$endgroup$
– Peter
Jan 9 at 10:06
$begingroup$
Already the definition of $k(p)$ is confusing.
$endgroup$
– Peter
Jan 9 at 10:06
$begingroup$
@Peter I think the definition of $k(p)$ is very ok
$endgroup$
– lanse7pty
Jan 9 at 10:11
$begingroup$
@Peter I think the definition of $k(p)$ is very ok
$endgroup$
– lanse7pty
Jan 9 at 10:11
$begingroup$
But hard to actually understand. We should find out what $k(p)=1$ means. Perhaps this case is trivial for the desired proof.
$endgroup$
– Peter
Jan 9 at 10:13
$begingroup$
But hard to actually understand. We should find out what $k(p)=1$ means. Perhaps this case is trivial for the desired proof.
$endgroup$
– Peter
Jan 9 at 10:13
$begingroup$
$k(p)$ is the exponent of the prime $p$ in the prime decomposition of $N$. $k(p)$ is well-defined, unique and greater or equal to $0$.
$endgroup$
– Snake707
Jan 9 at 10:30
$begingroup$
$k(p)$ is the exponent of the prime $p$ in the prime decomposition of $N$. $k(p)$ is well-defined, unique and greater or equal to $0$.
$endgroup$
– Snake707
Jan 9 at 10:30
$begingroup$
@Snake707 I did not claim that $k(p)$ is not well defined.
$endgroup$
– Peter
Jan 9 at 10:57
$begingroup$
@Snake707 I did not claim that $k(p)$ is not well defined.
$endgroup$
– Peter
Jan 9 at 10:57
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The idea is to notice that
$$log(N) leq sum_{p|N}{log(p)} + sum_{k(p) geq 2}{k(p)log(p)}.$$
The first term is dealt with by $(1.16)$, the second one by the last estimate of the image before last.
answered Jan 9 at 10:30
MindlackMindlack
3,49217
$endgroup$
1
$begingroup$
Could you talk it detail ? I can't understand you. Thanks.
$endgroup$
– lanse7pty
Jan 9 at 11:44
$begingroup$
Have you read your page $21$?
$endgroup$
– Mindlack
Jan 9 at 11:45
$begingroup$
Yes, I have read it, but seemly, there is not anything about the case $k(p)=1$.
$endgroup$
– lanse7pty
Jan 9 at 11:58
$begingroup$
In the derivation of $(1.17)$, you count every prime once, and then you count all primes with $k(p) geq 2$ $k(p)$ times. So primes with $k(p)=1$ are counted once and primes with $k(p) > 1$ are counted $k(p)+1$ times.
$endgroup$
– Mindlack
Jan 9 at 12:08
$begingroup$
I understand you, thanks.
$endgroup$
– lanse7pty
Jan 9 at 12:46
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%2f3067273%2fproof-of-bertrands-postulate%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
$begingroup$
The idea is to notice that
$$log(N) leq sum_{p|N}{log(p)} + sum_{k(p) geq 2}{k(p)log(p)}.$$
The first term is dealt with by $(1.16)$, the second one by the last estimate of the image before last.
answered Jan 9 at 10:30
MindlackMindlack
3,49217
$endgroup$
1
$begingroup$
Could you talk it detail ? I can't understand you. Thanks.
$endgroup$
– lanse7pty
Jan 9 at 11:44
$begingroup$
Have you read your page $21$?
$endgroup$
– Mindlack
Jan 9 at 11:45
$begingroup$
Yes, I have read it, but seemly, there is not anything about the case $k(p)=1$.
$endgroup$
– lanse7pty
Jan 9 at 11:58
$begingroup$
In the derivation of $(1.17)$, you count every prime once, and then you count all primes with $k(p) geq 2$ $k(p)$ times. So primes with $k(p)=1$ are counted once and primes with $k(p) > 1$ are counted $k(p)+1$ times.
$endgroup$
– Mindlack
Jan 9 at 12:08
$begingroup$
I understand you, thanks.
$endgroup$
– lanse7pty
Jan 9 at 12:46
add a comment |
$begingroup$
The idea is to notice that
$$log(N) leq sum_{p|N}{log(p)} + sum_{k(p) geq 2}{k(p)log(p)}.$$
The first term is dealt with by $(1.16)$, the second one by the last estimate of the image before last.
answered Jan 9 at 10:30
MindlackMindlack
3,49217
$endgroup$
1
$begingroup$
Could you talk it detail ? I can't understand you. Thanks.
$endgroup$
– lanse7pty
Jan 9 at 11:44
$begingroup$
Have you read your page $21$?
$endgroup$
– Mindlack
Jan 9 at 11:45
$begingroup$
Yes, I have read it, but seemly, there is not anything about the case $k(p)=1$.
$endgroup$
– lanse7pty
Jan 9 at 11:58
$begingroup$
In the derivation of $(1.17)$, you count every prime once, and then you count all primes with $k(p) geq 2$ $k(p)$ times. So primes with $k(p)=1$ are counted once and primes with $k(p) > 1$ are counted $k(p)+1$ times.
$endgroup$
– Mindlack
Jan 9 at 12:08
$begingroup$
I understand you, thanks.
$endgroup$
– lanse7pty
Jan 9 at 12:46
add a comment |
$begingroup$
The idea is to notice that
$$log(N) leq sum_{p|N}{log(p)} + sum_{k(p) geq 2}{k(p)log(p)}.$$
The first term is dealt with by $(1.16)$, the second one by the last estimate of the image before last.
answered Jan 9 at 10:30
MindlackMindlack
3,49217
$endgroup$
The idea is to notice that
$$log(N) leq sum_{p|N}{log(p)} + sum_{k(p) geq 2}{k(p)log(p)}.$$
The first term is dealt with by $(1.16)$, the second one by the last estimate of the image before last.
answered Jan 9 at 10:30
MindlackMindlack
3,49217
answered Jan 9 at 10:30
MindlackMindlack
3,49217
answered Jan 9 at 10:30
MindlackMindlack
3,49217
answered Jan 9 at 10:30
MindlackMindlack
3,49217
3,49217
1
$begingroup$
Could you talk it detail ? I can't understand you. Thanks.
$endgroup$
– lanse7pty
Jan 9 at 11:44
$begingroup$
Have you read your page $21$?
$endgroup$
– Mindlack
Jan 9 at 11:45
$begingroup$
Yes, I have read it, but seemly, there is not anything about the case $k(p)=1$.
$endgroup$
– lanse7pty
Jan 9 at 11:58
$begingroup$
In the derivation of $(1.17)$, you count every prime once, and then you count all primes with $k(p) geq 2$ $k(p)$ times. So primes with $k(p)=1$ are counted once and primes with $k(p) > 1$ are counted $k(p)+1$ times.
$endgroup$
– Mindlack
Jan 9 at 12:08
$begingroup$
I understand you, thanks.
$endgroup$
– lanse7pty
Jan 9 at 12:46
add a comment |
1
$begingroup$
Could you talk it detail ? I can't understand you. Thanks.
$endgroup$
– lanse7pty
Jan 9 at 11:44
$begingroup$
Have you read your page $21$?
$endgroup$
– Mindlack
Jan 9 at 11:45
$begingroup$
Yes, I have read it, but seemly, there is not anything about the case $k(p)=1$.
$endgroup$
– lanse7pty
Jan 9 at 11:58
$begingroup$
In the derivation of $(1.17)$, you count every prime once, and then you count all primes with $k(p) geq 2$ $k(p)$ times. So primes with $k(p)=1$ are counted once and primes with $k(p) > 1$ are counted $k(p)+1$ times.
$endgroup$
– Mindlack
Jan 9 at 12:08
$begingroup$
I understand you, thanks.
$endgroup$
– lanse7pty
Jan 9 at 12:46
1
1
$begingroup$
Could you talk it detail ? I can't understand you. Thanks.
$endgroup$
– lanse7pty
Jan 9 at 11:44
$begingroup$
Could you talk it detail ? I can't understand you. Thanks.
$endgroup$
– lanse7pty
Jan 9 at 11:44
$begingroup$
Have you read your page $21$?
$endgroup$
– Mindlack
Jan 9 at 11:45
$begingroup$
Have you read your page $21$?
$endgroup$
– Mindlack
Jan 9 at 11:45
$begingroup$
Yes, I have read it, but seemly, there is not anything about the case $k(p)=1$.
$endgroup$
– lanse7pty
Jan 9 at 11:58
$begingroup$
Yes, I have read it, but seemly, there is not anything about the case $k(p)=1$.
$endgroup$
– lanse7pty
Jan 9 at 11:58
$begingroup$
In the derivation of $(1.17)$, you count every prime once, and then you count all primes with $k(p) geq 2$ $k(p)$ times. So primes with $k(p)=1$ are counted once and primes with $k(p) > 1$ are counted $k(p)+1$ times.
$endgroup$
– Mindlack
Jan 9 at 12:08
$begingroup$
In the derivation of $(1.17)$, you count every prime once, and then you count all primes with $k(p) geq 2$ $k(p)$ times. So primes with $k(p)=1$ are counted once and primes with $k(p) > 1$ are counted $k(p)+1$ times.
$endgroup$
– Mindlack
Jan 9 at 12:08
$begingroup$
I understand you, thanks.
$endgroup$
– lanse7pty
Jan 9 at 12:46
$begingroup$
I understand you, thanks.
$endgroup$
– lanse7pty
Jan 9 at 12:46
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.
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%2f3067273%2fproof-of-bertrands-postulate%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
$begingroup$
Already the definition of $k(p)$ is confusing.
$endgroup$
– Peter
Jan 9 at 10:06
$begingroup$
@Peter I think the definition of $k(p)$ is very ok
$endgroup$
– lanse7pty
Jan 9 at 10:11
$begingroup$
But hard to actually understand. We should find out what $k(p)=1$ means. Perhaps this case is trivial for the desired proof.
$endgroup$
– Peter
Jan 9 at 10:13
$begingroup$
$k(p)$ is the exponent of the prime $p$ in the prime decomposition of $N$. $k(p)$ is well-defined, unique and greater or equal to $0$.
$endgroup$
– Snake707
Jan 9 at 10:30
$begingroup$
@Snake707 I did not claim that $k(p)$ is not well defined.
$endgroup$
– Peter
Jan 9 at 10:57