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Is this congruence equation has solutions?

0

0

$begingroup$

For any two positive integers $n ,N$ consider the congruence $n!X equiv p^n (text mod p^N)$ ($n < N,$ $p$ is a prime).

Does this congruence equation have solutions? If it does, how to prove it?

I tried to prove $v_p(n!) le n$. But it seems not true.

Where $v_p(n)$ is maximum power $k$ of $p$ where $p^k$ divides $n.$

Please help!

number-theory

share|cite|improve this question

edited Apr 1 at 7:21

Dbchatto67

3,205625

asked Apr 1 at 7:05

ogadaogada

31

$endgroup$

add a comment | 

0

0

$begingroup$

For any two positive integers $n ,N$ consider the congruence $n!X equiv p^n (text mod p^N)$ ($n < N,$ $p$ is a prime).

Does this congruence equation have solutions? If it does, how to prove it?

I tried to prove $v_p(n!) le n$. But it seems not true.

Where $v_p(n)$ is maximum power $k$ of $p$ where $p^k$ divides $n.$

Please help!

number-theory

share|cite|improve this question

edited Apr 1 at 7:21

Dbchatto67

3,205625

asked Apr 1 at 7:05

ogadaogada

31

$endgroup$

add a comment | 

$begingroup$

For any two positive integers $n ,N$ consider the congruence $n!X equiv p^n (text mod p^N)$ ($n < N,$ $p$ is a prime).

Does this congruence equation have solutions? If it does, how to prove it?

I tried to prove $v_p(n!) le n$. But it seems not true.

Where $v_p(n)$ is maximum power $k$ of $p$ where $p^k$ divides $n.$

Please help!

number-theory

share|cite|improve this question

edited Apr 1 at 7:21

Dbchatto67

3,205625

asked Apr 1 at 7:05

ogadaogada

31

$endgroup$

share|cite|improve this question

edited Apr 1 at 7:21

Dbchatto67

3,205625

asked Apr 1 at 7:05

ogadaogada

31

share|cite|improve this question

edited Apr 1 at 7:21

Dbchatto67

3,205625

asked Apr 1 at 7:05

ogadaogada

31

edited Apr 1 at 7:21

Dbchatto67

3,205625

edited Apr 1 at 7:21

Dbchatto67

3,205625

asked Apr 1 at 7:05

ogadaogada

31

asked Apr 1 at 7:05

ogadaogada

31

1 Answer
1

active

oldest

votes

1

$begingroup$

Hint $:$ First observe that the highest power $v_p (n!)$ of $p$ dividing $n!$ is

$$v_p (n!) = sumlimits_k=1^infty left lfloor frac n p^k right rfloor = sumlimits_k=1^m left lfloor frac n p^k right rfloor < sumlimits_k=1^infty frac n p^k = frac n p-1 leq n$$ since $p geq 2,$ where $m=left lfloor frac log_e n log_e p right rfloor.$

Now if $text gcd (n!,p^N)=1$ you are through since then $n!$ is a unit in $Bbb Z/ p^N Bbb Z.$ If $text gcd (n!,p^N)=p.$ Since $N geq 2$ it follows that one is the highest power of $p$ that can divide $n!.$ But then $text gcd left (frac n! p,p^N right ) = 1.$ So the congruence $frac n! p X equiv p^n-1 (text mod p^N)$ has a unique solution. Hence the given congruence relation admits a solution. This process can be similarly extended to the cases where $text gcd(n!,p^N) = p^k,$ where $1 leq k leq n.$

share|cite|improve this answer

edited Apr 1 at 9:56

answered Apr 1 at 7:37

Dbchatto67Dbchatto67

3,205625

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See my edited answer above @ogada.
  $endgroup$
  – Dbchatto67
  Apr 1 at 8:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you very much!
  $endgroup$
  – ogada
  Apr 1 at 9:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you understood all my reasonings clearly @ogada? If not please let me inform without any hesitation.
  $endgroup$
  – Dbchatto67
  Apr 1 at 9:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. I just struggled with proving first observation. and your explanation is very clear!
  $endgroup$
  – ogada
  Apr 1 at 9:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really very glad to help you.
  $endgroup$
  – Dbchatto67
  Apr 1 at 9:54
  

  
  
  
  

add a comment | 

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1

$begingroup$

Hint $:$ First observe that the highest power $v_p (n!)$ of $p$ dividing $n!$ is

$$v_p (n!) = sumlimits_k=1^infty left lfloor frac n p^k right rfloor = sumlimits_k=1^m left lfloor frac n p^k right rfloor < sumlimits_k=1^infty frac n p^k = frac n p-1 leq n$$ since $p geq 2,$ where $m=left lfloor frac log_e n log_e p right rfloor.$

Now if $text gcd (n!,p^N)=1$ you are through since then $n!$ is a unit in $Bbb Z/ p^N Bbb Z.$ If $text gcd (n!,p^N)=p.$ Since $N geq 2$ it follows that one is the highest power of $p$ that can divide $n!.$ But then $text gcd left (frac n! p,p^N right ) = 1.$ So the congruence $frac n! p X equiv p^n-1 (text mod p^N)$ has a unique solution. Hence the given congruence relation admits a solution. This process can be similarly extended to the cases where $text gcd(n!,p^N) = p^k,$ where $1 leq k leq n.$

share|cite|improve this answer

edited Apr 1 at 9:56

answered Apr 1 at 7:37

Dbchatto67Dbchatto67

3,205625

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See my edited answer above @ogada.
  $endgroup$
  – Dbchatto67
  Apr 1 at 8:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you very much!
  $endgroup$
  – ogada
  Apr 1 at 9:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you understood all my reasonings clearly @ogada? If not please let me inform without any hesitation.
  $endgroup$
  – Dbchatto67
  Apr 1 at 9:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. I just struggled with proving first observation. and your explanation is very clear!
  $endgroup$
  – ogada
  Apr 1 at 9:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really very glad to help you.
  $endgroup$
  – Dbchatto67
  Apr 1 at 9:54
  

  
  
  
  

add a comment | 

1

$begingroup$

Hint $:$ First observe that the highest power $v_p (n!)$ of $p$ dividing $n!$ is

$$v_p (n!) = sumlimits_k=1^infty left lfloor frac n p^k right rfloor = sumlimits_k=1^m left lfloor frac n p^k right rfloor < sumlimits_k=1^infty frac n p^k = frac n p-1 leq n$$ since $p geq 2,$ where $m=left lfloor frac log_e n log_e p right rfloor.$

Now if $text gcd (n!,p^N)=1$ you are through since then $n!$ is a unit in $Bbb Z/ p^N Bbb Z.$ If $text gcd (n!,p^N)=p.$ Since $N geq 2$ it follows that one is the highest power of $p$ that can divide $n!.$ But then $text gcd left (frac n! p,p^N right ) = 1.$ So the congruence $frac n! p X equiv p^n-1 (text mod p^N)$ has a unique solution. Hence the given congruence relation admits a solution. This process can be similarly extended to the cases where $text gcd(n!,p^N) = p^k,$ where $1 leq k leq n.$

share|cite|improve this answer

edited Apr 1 at 9:56

answered Apr 1 at 7:37

Dbchatto67Dbchatto67

3,205625

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  See my edited answer above @ogada.
  $endgroup$
  – Dbchatto67
  Apr 1 at 8:22
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  thank you very much!
  $endgroup$
  – ogada
  Apr 1 at 9:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Have you understood all my reasonings clearly @ogada? If not please let me inform without any hesitation.
  $endgroup$
  – Dbchatto67
  Apr 1 at 9:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. I just struggled with proving first observation. and your explanation is very clear!
  $endgroup$
  – ogada
  Apr 1 at 9:52
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Really very glad to help you.
  $endgroup$
  – Dbchatto67
  Apr 1 at 9:54
  

  
  
  
  

add a comment | 

$begingroup$

Hint $:$ First observe that the highest power $v_p (n!)$ of $p$ dividing $n!$ is

$$v_p (n!) = sumlimits_k=1^infty left lfloor frac n p^k right rfloor = sumlimits_k=1^m left lfloor frac n p^k right rfloor < sumlimits_k=1^infty frac n p^k = frac n p-1 leq n$$ since $p geq 2,$ where $m=left lfloor frac log_e n log_e p right rfloor.$

Now if $text gcd (n!,p^N)=1$ you are through since then $n!$ is a unit in $Bbb Z/ p^N Bbb Z.$ If $text gcd (n!,p^N)=p.$ Since $N geq 2$ it follows that one is the highest power of $p$ that can divide $n!.$ But then $text gcd left (frac n! p,p^N right ) = 1.$ So the congruence $frac n! p X equiv p^n-1 (text mod p^N)$ has a unique solution. Hence the given congruence relation admits a solution. This process can be similarly extended to the cases where $text gcd(n!,p^N) = p^k,$ where $1 leq k leq n.$

share|cite|improve this answer

edited Apr 1 at 9:56

answered Apr 1 at 7:37

Dbchatto67Dbchatto67

3,205625

$endgroup$

share|cite|improve this answer

edited Apr 1 at 9:56

answered Apr 1 at 7:37

Dbchatto67Dbchatto67

3,205625

answered Apr 1 at 7:37

Dbchatto67Dbchatto67

3,205625

answered Apr 1 at 7:37

Dbchatto67Dbchatto67

3,205625