Fdtxjr

How to add theme from github with composer

I am the person who abides by rules but breaks the rules . Who am I

Why does this boat have a landing pad? (SpaceX's GO Searcher) Any plans for propulsive capsule landings?

Why does a car's steering wheel get lighter with increasing speed

Paper published similar to PhD thesis

If nine coins are tossed, what is the probability that the number of heads is even?

Is "cogitate" used appropriately in "I cogitate that success relies on hard work"?

How to educate team mate to take screenshots for bugs with out unwanted stuff

Is it a Cyclops number? "Nobody" knows!

Why is my explanation wrong?

How can I portion out frozen cookie dough?

What is Tony Stark injecting into himself in Iron Man 3?

EXM headers adding bounce@spe.sitecoremail.com as the sender

An Undercover Army

How do you make a gun that shoots melee weapons and/or swords?

Can I challenge the interviewer to give me a proper technical feedback?

Why would /etc/passwd be used every time someone executes `ls -l` command?

How to recover against Snake as a heavyweight character?

Too soon for a plot twist?

Generating a list with duplicate entries

What is better: yes / no radio, or simple checkbox?

Geological Explanation for an Unusually Temperate Northern Mountain Valley

Having the player face themselves after the mid-game

Are small insurances worth it?

How to prove that $p! (mod p^2) = p(p-1) $

question about congruencesHow can I find $x$ such that $ax equiv 1 pmodbx+c$, given $a,b,c$?prove a=b (mod n) then ra=rb(mod n)Demostrating that a number x is the smallest such that 24x mod (59) = 2 mod (59)Intuition behind modular arithmetic $21z equiv 3 mod 16 iff 5z equiv 3 mod 16$If $a^p-1 equiv 1 pmod p$ and then $a^fracp-12 equiv 1$ or $p-1 pmod p$?Is this possible to solve through algebra?How does modular arithmetic work - Fermat's last theorem near misses?Prove that $(t$ mod $N) $ mod $ p = (t$ mod $p)$About Congruences of $fracabmod n$

0

$begingroup$

While doing my Math homework about Modular arithmetic. I accidentally found this

$p!equiv p(p-1)pmodp^2$

It's help me save time to find $ 21! (mod 361) $ a lot.

The question is how can I prove an equation above . Thank you in advance.

number-theory prime-numbers modular-arithmetic

share|cite|improve this question

edited 13 hours ago

Fabio Lucchini

8,66311426

asked 13 hours ago

ABCDEFG user157844ABCDEFG user157844

454310

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Wilson's theorem will get you most of the way
  $endgroup$
  – Henry
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Also note that your formula can be simplified: $p(p-1)=p^2-pequiv -p pmodp^2$. But pay attention: the equation holds if and only if $p$ is a prime number!
  $endgroup$
  – AlessioDV
  13 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

While doing my Math homework about Modular arithmetic. I accidentally found this

$p!equiv p(p-1)pmodp^2$

It's help me save time to find $ 21! (mod 361) $ a lot.

The question is how can I prove an equation above . Thank you in advance.

number-theory prime-numbers modular-arithmetic

share|cite|improve this question

edited 13 hours ago

Fabio Lucchini

8,66311426

asked 13 hours ago

ABCDEFG user157844ABCDEFG user157844

454310

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Wilson's theorem will get you most of the way
  $endgroup$
  – Henry
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Also note that your formula can be simplified: $p(p-1)=p^2-pequiv -p pmodp^2$. But pay attention: the equation holds if and only if $p$ is a prime number!
  $endgroup$
  – AlessioDV
  13 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

While doing my Math homework about Modular arithmetic. I accidentally found this

$p!equiv p(p-1)pmodp^2$

It's help me save time to find $ 21! (mod 361) $ a lot.

The question is how can I prove an equation above . Thank you in advance.

number-theory prime-numbers modular-arithmetic

share|cite|improve this question

edited 13 hours ago

Fabio Lucchini

8,66311426

asked 13 hours ago

ABCDEFG user157844ABCDEFG user157844

454310

$endgroup$

share|cite|improve this question

edited 13 hours ago

Fabio Lucchini

8,66311426

asked 13 hours ago

ABCDEFG user157844ABCDEFG user157844

454310

share|cite|improve this question

edited 13 hours ago

Fabio Lucchini

8,66311426

asked 13 hours ago

ABCDEFG user157844ABCDEFG user157844

454310

edited 13 hours ago

Fabio Lucchini

8,66311426

edited 13 hours ago

Fabio Lucchini

8,66311426

asked 13 hours ago

ABCDEFG user157844ABCDEFG user157844

454310

asked 13 hours ago

ABCDEFG user157844ABCDEFG user157844

454310

3 Answers
3

active

oldest

votes

3

$begingroup$

Warning ! You statement is true only when $n$ is prime. For example for $n=4$, you don't have $n! = n(n-1) quad (mathrmmod text n^2)$. So I guess you can't use it with $21$ ("by hand", you can see that $21!=323 neq 21times 20 quad (mathrmmod text 361)$).

Proof when $n$ is prime :

By Wilson's theorem,
$$(p-1)! = -1 quad (mathrmmod text p)$$

Multiplying by $p$, this implies
$$p! = -p quad (mathrmmod text p^2)$$

i.e.
$$p! = p^2-p = p(p-1) quad (mathrmmod text p^2)$$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

TheSilverDoeTheSilverDoe

3,220112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I never knew that we can multiply p throughout the whole equation including (mod p) , that's my truly open-eyed indeed. Wonderful! Thank you.
  $endgroup$
  – ABCDEFG user157844
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, $a=b quad (mathrmmod text p)$ means that there exists $n$ such that $a-b = pn$. So multiplying by any number $q$, you get $aq-bq=pqn$, so $aq=bq quad (mathrmmod text pq)$.
  $endgroup$
  – TheSilverDoe
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can we avoid to using Chinese remainder theorem by this principle?
  $endgroup$
  – ABCDEFG user157844
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No, I don't think so, Chinese remainder theorem helps to solve a system of congruence equations. It does not consist in just algebraic modifications of one equality.
  $endgroup$
  – TheSilverDoe
  12 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

Wilson's Theorem states that:

$(p-1)! = -1 (mod p^2)$ ......(a)

and

$p!=p(p-1)!$, $-1=p-1(mod p^2)$ .......(b)

applying (b) to (a), we have

$p!=p(p-1)!=p*-1 (mod p^2)$

which implies that

$p!=p(p-1) (mod p^2)$

Note that the $p$ above stands for prime numbers and $21$ is not a prime number since $21=3*7$. Therefore the above statements will not apply, but take for example, 5 is a prime number and

$5!=5*4*3*2*1=120=5(5-1)=20 (mod 25)$

share|cite|improve this answer

answered 13 hours ago

Goodness ThankyouGoodness Thankyou

11

New contributor

Goodness Thankyou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

$$p! equiv p(p-1) pmodp^2 Leftrightarrow p^2 mid p!-p(p-1) Leftrightarrow p mid (p-1)!-(p-1) Leftrightarrow (p-1)! equiv -1 pmodp$$
If $p$ is a prime, then except for the two solutions of $x^2 equiv 1 pmodp$ (which are equal $pmodp$ for $p=2$), we can pair the numbers in $2, ... (p-2)$ in pairs $(x,y)$ with $x notequiv y pmodp$ such that $xy equiv 1 pmodp$, so $(p-2)! equiv 1 pmodp$ and so $(p-1)! equiv (p-1)(p-2)! equiv (p-1) equiv -1 pmodp$.

share|cite|improve this answer

answered 12 hours ago

Wolfgang KaisWolfgang Kais

6715

$endgroup$

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
);



);

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%2f3139955%2fhow-to-prove-that-p-mod-p2-pp-1%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

3

$begingroup$

Warning ! You statement is true only when $n$ is prime. For example for $n=4$, you don't have $n! = n(n-1) quad (mathrmmod text n^2)$. So I guess you can't use it with $21$ ("by hand", you can see that $21!=323 neq 21times 20 quad (mathrmmod text 361)$).

Proof when $n$ is prime :

By Wilson's theorem,
$$(p-1)! = -1 quad (mathrmmod text p)$$

Multiplying by $p$, this implies
$$p! = -p quad (mathrmmod text p^2)$$

i.e.
$$p! = p^2-p = p(p-1) quad (mathrmmod text p^2)$$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

TheSilverDoeTheSilverDoe

3,220112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I never knew that we can multiply p throughout the whole equation including (mod p) , that's my truly open-eyed indeed. Wonderful! Thank you.
  $endgroup$
  – ABCDEFG user157844
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, $a=b quad (mathrmmod text p)$ means that there exists $n$ such that $a-b = pn$. So multiplying by any number $q$, you get $aq-bq=pqn$, so $aq=bq quad (mathrmmod text pq)$.
  $endgroup$
  – TheSilverDoe
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can we avoid to using Chinese remainder theorem by this principle?
  $endgroup$
  – ABCDEFG user157844
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No, I don't think so, Chinese remainder theorem helps to solve a system of congruence equations. It does not consist in just algebraic modifications of one equality.
  $endgroup$
  – TheSilverDoe
  12 hours ago
  

  
  
  
  

add a comment | 

3

$begingroup$

Warning ! You statement is true only when $n$ is prime. For example for $n=4$, you don't have $n! = n(n-1) quad (mathrmmod text n^2)$. So I guess you can't use it with $21$ ("by hand", you can see that $21!=323 neq 21times 20 quad (mathrmmod text 361)$).

Proof when $n$ is prime :

By Wilson's theorem,
$$(p-1)! = -1 quad (mathrmmod text p)$$

Multiplying by $p$, this implies
$$p! = -p quad (mathrmmod text p^2)$$

i.e.
$$p! = p^2-p = p(p-1) quad (mathrmmod text p^2)$$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

TheSilverDoeTheSilverDoe

3,220112

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I never knew that we can multiply p throughout the whole equation including (mod p) , that's my truly open-eyed indeed. Wonderful! Thank you.
  $endgroup$
  – ABCDEFG user157844
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Well, $a=b quad (mathrmmod text p)$ means that there exists $n$ such that $a-b = pn$. So multiplying by any number $q$, you get $aq-bq=pqn$, so $aq=bq quad (mathrmmod text pq)$.
  $endgroup$
  – TheSilverDoe
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Can we avoid to using Chinese remainder theorem by this principle?
  $endgroup$
  – ABCDEFG user157844
  12 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  No, I don't think so, Chinese remainder theorem helps to solve a system of congruence equations. It does not consist in just algebraic modifications of one equality.
  $endgroup$
  – TheSilverDoe
  12 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Warning ! You statement is true only when $n$ is prime. For example for $n=4$, you don't have $n! = n(n-1) quad (mathrmmod text n^2)$. So I guess you can't use it with $21$ ("by hand", you can see that $21!=323 neq 21times 20 quad (mathrmmod text 361)$).

Proof when $n$ is prime :

By Wilson's theorem,
$$(p-1)! = -1 quad (mathrmmod text p)$$

Multiplying by $p$, this implies
$$p! = -p quad (mathrmmod text p^2)$$

i.e.
$$p! = p^2-p = p(p-1) quad (mathrmmod text p^2)$$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

TheSilverDoeTheSilverDoe

3,220112

$endgroup$

share|cite|improve this answer

edited 13 hours ago

answered 13 hours ago

TheSilverDoeTheSilverDoe

3,220112

answered 13 hours ago

TheSilverDoeTheSilverDoe

3,220112

answered 13 hours ago

TheSilverDoeTheSilverDoe

3,220112

0

$begingroup$

Wilson's Theorem states that:

$(p-1)! = -1 (mod p^2)$ ......(a)

and

$p!=p(p-1)!$, $-1=p-1(mod p^2)$ .......(b)

applying (b) to (a), we have

$p!=p(p-1)!=p*-1 (mod p^2)$

which implies that

$p!=p(p-1) (mod p^2)$

Note that the $p$ above stands for prime numbers and $21$ is not a prime number since $21=3*7$. Therefore the above statements will not apply, but take for example, 5 is a prime number and

$5!=5*4*3*2*1=120=5(5-1)=20 (mod 25)$

share|cite|improve this answer

answered 13 hours ago

Goodness ThankyouGoodness Thankyou

11

New contributor

Goodness Thankyou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

Wilson's Theorem states that:

$(p-1)! = -1 (mod p^2)$ ......(a)

and

$p!=p(p-1)!$, $-1=p-1(mod p^2)$ .......(b)

applying (b) to (a), we have

$p!=p(p-1)!=p*-1 (mod p^2)$

which implies that

$p!=p(p-1) (mod p^2)$

Note that the $p$ above stands for prime numbers and $21$ is not a prime number since $21=3*7$. Therefore the above statements will not apply, but take for example, 5 is a prime number and

$5!=5*4*3*2*1=120=5(5-1)=20 (mod 25)$

share|cite|improve this answer

answered 13 hours ago

Goodness ThankyouGoodness Thankyou

11

New contributor

Goodness Thankyou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Wilson's Theorem states that:

$(p-1)! = -1 (mod p^2)$ ......(a)

and

$p!=p(p-1)!$, $-1=p-1(mod p^2)$ .......(b)

applying (b) to (a), we have

$p!=p(p-1)!=p*-1 (mod p^2)$

which implies that

$p!=p(p-1) (mod p^2)$

Note that the $p$ above stands for prime numbers and $21$ is not a prime number since $21=3*7$. Therefore the above statements will not apply, but take for example, 5 is a prime number and

$5!=5*4*3*2*1=120=5(5-1)=20 (mod 25)$

share|cite|improve this answer

answered 13 hours ago

Goodness ThankyouGoodness Thankyou

11

New contributor

Goodness Thankyou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered 13 hours ago

Goodness ThankyouGoodness Thankyou

11

New contributor

Goodness Thankyou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 13 hours ago

Goodness ThankyouGoodness Thankyou

11

New contributor

Goodness Thankyou is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 13 hours ago

Goodness ThankyouGoodness Thankyou

11

0

$begingroup$

$$p! equiv p(p-1) pmodp^2 Leftrightarrow p^2 mid p!-p(p-1) Leftrightarrow p mid (p-1)!-(p-1) Leftrightarrow (p-1)! equiv -1 pmodp$$
If $p$ is a prime, then except for the two solutions of $x^2 equiv 1 pmodp$ (which are equal $pmodp$ for $p=2$), we can pair the numbers in $2, ... (p-2)$ in pairs $(x,y)$ with $x notequiv y pmodp$ such that $xy equiv 1 pmodp$, so $(p-2)! equiv 1 pmodp$ and so $(p-1)! equiv (p-1)(p-2)! equiv (p-1) equiv -1 pmodp$.

share|cite|improve this answer

answered 12 hours ago

Wolfgang KaisWolfgang Kais

6715

$endgroup$

add a comment | 

0

$begingroup$

$$p! equiv p(p-1) pmodp^2 Leftrightarrow p^2 mid p!-p(p-1) Leftrightarrow p mid (p-1)!-(p-1) Leftrightarrow (p-1)! equiv -1 pmodp$$
If $p$ is a prime, then except for the two solutions of $x^2 equiv 1 pmodp$ (which are equal $pmodp$ for $p=2$), we can pair the numbers in $2, ... (p-2)$ in pairs $(x,y)$ with $x notequiv y pmodp$ such that $xy equiv 1 pmodp$, so $(p-2)! equiv 1 pmodp$ and so $(p-1)! equiv (p-1)(p-2)! equiv (p-1) equiv -1 pmodp$.

share|cite|improve this answer

answered 12 hours ago

Wolfgang KaisWolfgang Kais

6715

$endgroup$

add a comment | 

$begingroup$

$$p! equiv p(p-1) pmodp^2 Leftrightarrow p^2 mid p!-p(p-1) Leftrightarrow p mid (p-1)!-(p-1) Leftrightarrow (p-1)! equiv -1 pmodp$$
If $p$ is a prime, then except for the two solutions of $x^2 equiv 1 pmodp$ (which are equal $pmodp$ for $p=2$), we can pair the numbers in $2, ... (p-2)$ in pairs $(x,y)$ with $x notequiv y pmodp$ such that $xy equiv 1 pmodp$, so $(p-2)! equiv 1 pmodp$ and so $(p-1)! equiv (p-1)(p-2)! equiv (p-1) equiv -1 pmodp$.

share|cite|improve this answer

answered 12 hours ago

Wolfgang KaisWolfgang Kais

6715

$endgroup$

share|cite|improve this answer

answered 12 hours ago

Wolfgang KaisWolfgang Kais

6715

answered 12 hours ago

Wolfgang KaisWolfgang Kais

6715

answered 12 hours ago

Wolfgang KaisWolfgang Kais

6715