Mfcttrf

Is it possible that Curiosity measured its own methane or failed doing the spectrometry?

what is the meaning of "stock" dilution on the Massive Dev Chart Website?

Pandas merge and fillna

How can I know (without going to the station) if RATP is offering the Anti Pollution tickets?

How long had Bertha Mason been in the attic at the point of the events in Jane Eyre

Why would a propellor have blades of different lengths?

Why did my leaking pool light trip the circuit breaker, but not the GFCI?

What is the right way to query an I2C device from an interrupt service routine?

What is meaning of 4 letter acronyms in Roman names like Titus Flavius T. f. T. n. Sabinus?

Blood-based alcohol for vampires?

Are the plates of a battery really charged?

Was Wolfgang Unzicker the last Amateur GM?

What is the difference between a historical drama and a period drama?

Olive oil in Japanese cooking

How might boat designs change in order to allow them to be pulled by dragons?

Which high-degree derivatives play an essential role?

Phrase origin: "You ain't got to go home but you got to get out of here."

gzip compress a local folder and extract it to remote server

What is a "tittering order"?

How did sloshing prevent the Apollo Service Module from moving safely away from the Command Module and how was this fixed?

What is the difference between figures illustration and images?

Are there advantages in writing by hand over typing out a story?

How frequently do Russian people still refer to others by their patronymic (отчество)?

What instances can be solved today by modern solvers (pure LP)?

Prime number raised to a power

question on prime numberPrime number questionFind the prime-power decomposition of 999999999999If all pairs of addends that sum up to $N$ are coprime, then $N$ is prime.Is the square of a prime necessarily larger than the next prime?There are infinitely many odd numbers not expressible as the sum of a prime number and a power of $2$Is there any deep reason that 23456789 is prime?Is this number prime?Advise on determining prime numbers such that the product of their geometric series is equivalent to another prime number raised to a power.Is there a name for a multiplicative function $psi(p^n) = (p^n-1)(p^n-1-1)cdots(p-1)$?

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

2

1

$begingroup$

Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.

I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.

number-theory elementary-number-theory

share|cite|improve this question

asked 8 hours ago

Math GuyMath Guy

18077 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
  $endgroup$
  – Piquito
  8 hours ago
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.

I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.

number-theory elementary-number-theory

share|cite|improve this question

asked 8 hours ago

Math GuyMath Guy

18077 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $p^n|aiff a=p^nmiff dfrac ap=p^n-1minmathbb N$
  $endgroup$
  – Piquito
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Let $p$ be a prime number. If $n, a$ are two positive integers such that $p^n | a$, then $p|a$.

I've recently come across this result in a problem. They didn't give any proof for it, they just asserted that it is true. I am not too good at number theory, so please tell me whether this is some well-known result.

number-theory elementary-number-theory

share|cite|improve this question

asked 8 hours ago

Math GuyMath Guy

18077 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

Math GuyMath Guy

18077 bronze badges

share|cite|improve this question

asked 8 hours ago

Math GuyMath Guy

18077 bronze badges

asked 8 hours ago

Math GuyMath Guy

18077 bronze badges

asked 8 hours ago

Math GuyMath Guy

18077 bronze badges

5 Answers
5

active

oldest

votes

4

$begingroup$

We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.

share|cite|improve this answer

answered 8 hours ago

Azif00Azif00

94799 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you! Which part uses specifically that $p$ is a prime number?
  $endgroup$
  – Math Guy
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This does not depend on $p$ prime. It works for any $n$.
  $endgroup$
  – Antonios-Alexandros Robotis
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Nowhere, we do not need prime numbers here.
  $endgroup$
  – Azif00
  8 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.

Otherwise, let me prove it for you here (at least for you to compare):

Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.

share|cite|improve this answer

answered 8 hours ago

ThorWittichThorWittich

1,80611 silver badge1212 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

If $n = 1$, we have $p mid a$ and we are done.

For $n ge 2$:

$p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$

This result, being so simple, is very well-known.

We note that $p$ needn't be prime.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Robert LewisRobert Lewis

50.8k22 gold badges3333 silver badges6969 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Maybe this is an easier way to look at this:
For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.

share|cite|improve this answer

answered 8 hours ago

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

If $p^n | a$ then $a = p^nk$ for some integer $k$.

As $p | p^kk$, so we have $p|a$.

share|cite|improve this answer

answered 7 hours ago

mlchristiansmlchristians

4441414 bronze badges

$endgroup$

add a comment | 

Your Answer

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
);



);

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3280142%2fprime-number-raised-to-a-power%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

4

$begingroup$

We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.

share|cite|improve this answer

answered 8 hours ago

Azif00Azif00

94799 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you! Which part uses specifically that $p$ is a prime number?
  $endgroup$
  – Math Guy
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This does not depend on $p$ prime. It works for any $n$.
  $endgroup$
  – Antonios-Alexandros Robotis
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Nowhere, we do not need prime numbers here.
  $endgroup$
  – Azif00
  8 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.

share|cite|improve this answer

answered 8 hours ago

Azif00Azif00

94799 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you! Which part uses specifically that $p$ is a prime number?
  $endgroup$
  – Math Guy
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  This does not depend on $p$ prime. It works for any $n$.
  $endgroup$
  – Antonios-Alexandros Robotis
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Nowhere, we do not need prime numbers here.
  $endgroup$
  – Azif00
  8 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

We have $p^nmid a$ and obviously $pmid p^n$. Then $pmid a$ by transitivity.

share|cite|improve this answer

answered 8 hours ago

Azif00Azif00

94799 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Azif00Azif00

94799 bronze badges

answered 8 hours ago

Azif00Azif00

94799 bronze badges

answered 8 hours ago

Azif00Azif00

94799 bronze badges

4

$begingroup$

We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.

Otherwise, let me prove it for you here (at least for you to compare):

Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.

share|cite|improve this answer

answered 8 hours ago

ThorWittichThorWittich

1,80611 silver badge1212 bronze badges

$endgroup$

add a comment | 

4

$begingroup$

We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.

Otherwise, let me prove it for you here (at least for you to compare):

Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.

share|cite|improve this answer

answered 8 hours ago

ThorWittichThorWittich

1,80611 silver badge1212 bronze badges

$endgroup$

add a comment | 

$begingroup$

We have that $p mid p^n$ and that $p^n mid a$, such that $p mid a$. Maybe try to prove for yourself that divisibility is transitive.

Otherwise, let me prove it for you here (at least for you to compare):

Suppose that $a mid b$ and $b mid c$, that means we have $aa' = b$ and $bb' = c$ for some integers $a'$ and $b'$. We get $c = bb' = (aa')b' = a(a'b')$, such that $a mid c$.

share|cite|improve this answer

answered 8 hours ago

ThorWittichThorWittich

1,80611 silver badge1212 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

ThorWittichThorWittich

1,80611 silver badge1212 bronze badges

answered 8 hours ago

ThorWittichThorWittich

1,80611 silver badge1212 bronze badges

answered 8 hours ago

ThorWittichThorWittich

1,80611 silver badge1212 bronze badges

3

$begingroup$

If $n = 1$, we have $p mid a$ and we are done.

For $n ge 2$:

$p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$

This result, being so simple, is very well-known.

We note that $p$ needn't be prime.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Robert LewisRobert Lewis

50.8k22 gold badges3333 silver badges6969 bronze badges

$endgroup$

add a comment | 

3

$begingroup$

If $n = 1$, we have $p mid a$ and we are done.

For $n ge 2$:

$p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$

This result, being so simple, is very well-known.

We note that $p$ needn't be prime.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Robert LewisRobert Lewis

50.8k22 gold badges3333 silver badges6969 bronze badges

$endgroup$

add a comment | 

$begingroup$

If $n = 1$, we have $p mid a$ and we are done.

For $n ge 2$:

$p^n mid a Longrightarrow exists c, ; a = cp^n = (cp^n - 1)p Longrightarrow p mid a. tag 1$

This result, being so simple, is very well-known.

We note that $p$ needn't be prime.

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Robert LewisRobert Lewis

50.8k22 gold badges3333 silver badges6969 bronze badges

$endgroup$

share|cite|improve this answer

edited 8 hours ago

answered 8 hours ago

Robert LewisRobert Lewis

50.8k22 gold badges3333 silver badges6969 bronze badges

answered 8 hours ago

Robert LewisRobert Lewis

50.8k22 gold badges3333 silver badges6969 bronze badges

answered 8 hours ago

Robert LewisRobert Lewis

50.8k22 gold badges3333 silver badges6969 bronze badges

2

$begingroup$

Maybe this is an easier way to look at this:
For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.

share|cite|improve this answer

answered 8 hours ago

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

$endgroup$

add a comment | 

2

$begingroup$

Maybe this is an easier way to look at this:
For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.

share|cite|improve this answer

answered 8 hours ago

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

$endgroup$

add a comment | 

$begingroup$

Maybe this is an easier way to look at this:
For integers $a,b,c$, if $b$ divides $c$, and $a$ divides $b$, then $a$ divides $c$.

share|cite|improve this answer

answered 8 hours ago

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

answered 8 hours ago

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

answered 8 hours ago

Ruben du BurckRuben du Burck

93322 silver badges1414 bronze badges

0

$begingroup$

If $p^n | a$ then $a = p^nk$ for some integer $k$.

As $p | p^kk$, so we have $p|a$.

share|cite|improve this answer

answered 7 hours ago

mlchristiansmlchristians

4441414 bronze badges

$endgroup$

add a comment | 

0

$begingroup$

If $p^n | a$ then $a = p^nk$ for some integer $k$.

As $p | p^kk$, so we have $p|a$.

share|cite|improve this answer

answered 7 hours ago

mlchristiansmlchristians

4441414 bronze badges

$endgroup$

add a comment | 

$begingroup$

If $p^n | a$ then $a = p^nk$ for some integer $k$.

As $p | p^kk$, so we have $p|a$.

share|cite|improve this answer

answered 7 hours ago

mlchristiansmlchristians

4441414 bronze badges

$endgroup$

share|cite|improve this answer

answered 7 hours ago

mlchristiansmlchristians

4441414 bronze badges

answered 7 hours ago

mlchristiansmlchristians

4441414 bronze badges

answered 7 hours ago

mlchristiansmlchristians

4441414 bronze badges