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Prove that if $(n-1)! + 1$ is divisible by $n$, then $n$ is prime. [closed]

Prove $left(frac{q(q+1)}{p}right) =left(frac{1+q^{-1}}{p}right )$ for $pgt2$ a prime, and any $q in mathbb{Z^+} $.Let $n=p^rm$, where $p$ is a prime not dividing an integer $mge 1$. Prove that $p$ is a prime not dividing $C_n^{p^r}$Erdős Prime Sieve Conjectureprove that if $a^n+1$ is prime then $a$ is even and $n=2^k$Twin prime conjecture hypothesisHow do I prove that “If prime p does not divide natural number m, then gcd(p,m) = 1”Prove that $forall k = m^2 + 1. space m in mathbb{Z}^+$, if $k$ is divisible by any prime then that prime is congruent to $1, 2 pmod 4$.If $2p+1$ is a prime number, then $(p!)^2+(-1)^p$ is divisible by $2p+1$.Can we prove Prime number theorem by prime factorization?Convert $50!$ to product of the powers of respective prime numbers ? (without direct calculation )

-2

$begingroup$

For prime $n$, I took $n = 3mpm1$ or $6mpm1$. But I was struck in the factorial part.

Please help

number-theory

share|cite|improve this question

edited Mar 13 at 6:47

Yadati Kiran

2,1121622

asked Mar 13 at 6:17

Krushna GoelKrushna Goel

133

$endgroup$

closed as off-topic by Saad, Shailesh, José Carlos Santos, Gibbs, Vinyl_cape_jawa Mar 13 at 9:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Shailesh, José Carlos Santos, Gibbs, Vinyl_cape_jawa

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that Wilson's Theorem states that $(n-1)! + 1$ is a multiple of $n$ if and only if $n$ is prime.
  $endgroup$
  – John Omielan
  Mar 13 at 7:55
  

  
  
  
  

add a comment | 

-2

$begingroup$

For prime $n$, I took $n = 3mpm1$ or $6mpm1$. But I was struck in the factorial part.

Please help

number-theory

share|cite|improve this question

edited Mar 13 at 6:47

Yadati Kiran

2,1121622

asked Mar 13 at 6:17

Krushna GoelKrushna Goel

133

$endgroup$

closed as off-topic by Saad, Shailesh, José Carlos Santos, Gibbs, Vinyl_cape_jawa Mar 13 at 9:15

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Shailesh, José Carlos Santos, Gibbs, Vinyl_cape_jawa

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Note that Wilson's Theorem states that $(n-1)! + 1$ is a multiple of $n$ if and only if $n$ is prime.
  $endgroup$
  – John Omielan
  Mar 13 at 7:55
  

  
  
  
  

add a comment | 

$begingroup$

For prime $n$, I took $n = 3mpm1$ or $6mpm1$. But I was struck in the factorial part.

Please help

number-theory

share|cite|improve this question

edited Mar 13 at 6:47

Yadati Kiran

2,1121622

asked Mar 13 at 6:17

Krushna GoelKrushna Goel

133

$endgroup$

share|cite|improve this question

edited Mar 13 at 6:47

Yadati Kiran

2,1121622

asked Mar 13 at 6:17

Krushna GoelKrushna Goel

133

share|cite|improve this question

edited Mar 13 at 6:47

Yadati Kiran

2,1121622

asked Mar 13 at 6:17

Krushna GoelKrushna Goel

133

edited Mar 13 at 6:47

Yadati Kiran

2,1121622

edited Mar 13 at 6:47

Yadati Kiran

2,1121622

asked Mar 13 at 6:17

Krushna GoelKrushna Goel

133

asked Mar 13 at 6:17

Krushna GoelKrushna Goel

133

1 Answer
1

active

oldest

votes

0

$begingroup$

If $n$ is not a prime it has prime divisor which is necessarily $<n$. It must divide $(n-1)!+1$ but it also divides $(n-1)!$. What is the conclusion?

share|cite|improve this answer

answered Mar 13 at 6:22

Kavi Rama MurthyKavi Rama Murthy

68.9k53169

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not saying that $n$ divides $(n-1)!$. In your case $2$ is a prime divisor of $4$ and it does divide $(4-1)!$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 13 at 6:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.G: suppose $1<kmid n$ then since $nmid (n-1)!+1$ also $kmid (n-1)!+1$ but as $kmid n$ and $k<n$ it is obvious that $kmid (n-1)!$. But clearly a natural number (different from $1$) cannot divide two consecutive numbers!
  $endgroup$
  – b00n heT
  Mar 13 at 6:31
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

If $n$ is not a prime it has prime divisor which is necessarily $<n$. It must divide $(n-1)!+1$ but it also divides $(n-1)!$. What is the conclusion?

share|cite|improve this answer

answered Mar 13 at 6:22

Kavi Rama MurthyKavi Rama Murthy

68.9k53169

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not saying that $n$ divides $(n-1)!$. In your case $2$ is a prime divisor of $4$ and it does divide $(4-1)!$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 13 at 6:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.G: suppose $1<kmid n$ then since $nmid (n-1)!+1$ also $kmid (n-1)!+1$ but as $kmid n$ and $k<n$ it is obvious that $kmid (n-1)!$. But clearly a natural number (different from $1$) cannot divide two consecutive numbers!
  $endgroup$
  – b00n heT
  Mar 13 at 6:31
  
  
  

  
  
  
  

add a comment | 

0

$begingroup$

If $n$ is not a prime it has prime divisor which is necessarily $<n$. It must divide $(n-1)!+1$ but it also divides $(n-1)!$. What is the conclusion?

share|cite|improve this answer

answered Mar 13 at 6:22

Kavi Rama MurthyKavi Rama Murthy

68.9k53169

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not saying that $n$ divides $(n-1)!$. In your case $2$ is a prime divisor of $4$ and it does divide $(4-1)!$.
  $endgroup$
  – Kavi Rama Murthy
  Mar 13 at 6:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @J.G: suppose $1<kmid n$ then since $nmid (n-1)!+1$ also $kmid (n-1)!+1$ but as $kmid n$ and $k<n$ it is obvious that $kmid (n-1)!$. But clearly a natural number (different from $1$) cannot divide two consecutive numbers!
  $endgroup$
  – b00n heT
  Mar 13 at 6:31
  
  
  

  
  
  
  

add a comment | 

$begingroup$

If $n$ is not a prime it has prime divisor which is necessarily $<n$. It must divide $(n-1)!+1$ but it also divides $(n-1)!$. What is the conclusion?

share|cite|improve this answer

answered Mar 13 at 6:22

Kavi Rama MurthyKavi Rama Murthy

68.9k53169

$endgroup$

share|cite|improve this answer

answered Mar 13 at 6:22

Kavi Rama MurthyKavi Rama Murthy

68.9k53169

answered Mar 13 at 6:22

Kavi Rama MurthyKavi Rama Murthy

68.9k53169

answered Mar 13 at 6:22

Kavi Rama MurthyKavi Rama Murthy

68.9k53169