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Prove that $x$ exists such that the sequence is divisible by $n$ [closed]

Can we always find an integer x such that …Proving that there exists $a_i in a_1,dots,a_k$ so that for any positive integer $n$ $F(n)$ is divisible by $a_i$Prove that for all odd $n$, there is an $m$ such that $2^m - 1$ is divisible by $n$Prove that for each integer $n ge 2$ there exists a prime number $p$ dividing $a_n$Sum of $k$ consecutive term divisible by $k+1$Prove that the sequence is purely periodicIndex $i$ such that $prod_k=1^p a_k+a_i$ is divisible by $p^2$Prove that for any given positive integer $n$,there exists a number having digits $0,1$ which is divisible by $n$.Prove that there exists a positive integer $a$ such that $n|a^2-a$Minimum value of $n$ such that $nq+k$ is divisible by $p$How do I prove that such a sequence exists?

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Prove that for every positive integer $n$ there exists a positive integer $x$ such that each of the terms of the infinite sequence : $$x+1,x^x+1,x^x^x+1,ldots$$ is divisible by $n$.

number-theory divisibility

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edited Mar 29 at 21:31

FredH

3,7201023

asked Mar 29 at 21:23

Abdallah krichenAbdallah krichen

182

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closed as off-topic by Sil, John Omielan, Cesareo, YiFan, Shailesh Mar 30 at 2:36

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." – Sil, John Omielan, Cesareo, YiFan, Shailesh

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Where does this question come from?
  $endgroup$
  – coffeemath
  Mar 29 at 21:27
  

  
  
  
  


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  What has been tried ? I know it wasn't learning MathJax, fixed it for you ...
  $endgroup$
  – Roddy MacPhee
  Mar 29 at 21:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You have already asked couple questions here, all of them are lacking any context or effort of your own... You might want to read How to ask a good question.
  $endgroup$
  – Sil
  Mar 29 at 21:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related to OP's prior question.
  $endgroup$
  – Bill Dubuque
  Mar 29 at 22:17
  

  
  
  
  

add a comment | 

-2

1

$begingroup$

Prove that for every positive integer $n$ there exists a positive integer $x$ such that each of the terms of the infinite sequence : $$x+1,x^x+1,x^x^x+1,ldots$$ is divisible by $n$.

number-theory divisibility

share|cite|improve this question

edited Mar 29 at 21:31

FredH

3,7201023

asked Mar 29 at 21:23

Abdallah krichenAbdallah krichen

182

$endgroup$

closed as off-topic by Sil, John Omielan, Cesareo, YiFan, Shailesh Mar 30 at 2:36

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." – Sil, John Omielan, Cesareo, YiFan, Shailesh

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

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Where does this question come from?
  $endgroup$
  – coffeemath
  Mar 29 at 21:27
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What has been tried ? I know it wasn't learning MathJax, fixed it for you ...
  $endgroup$
  – Roddy MacPhee
  Mar 29 at 21:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You have already asked couple questions here, all of them are lacking any context or effort of your own... You might want to read How to ask a good question.
  $endgroup$
  – Sil
  Mar 29 at 21:40
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related to OP's prior question.
  $endgroup$
  – Bill Dubuque
  Mar 29 at 22:17
  

  
  
  
  

add a comment | 

$begingroup$

Prove that for every positive integer $n$ there exists a positive integer $x$ such that each of the terms of the infinite sequence : $$x+1,x^x+1,x^x^x+1,ldots$$ is divisible by $n$.

number-theory divisibility

share|cite|improve this question

edited Mar 29 at 21:31

FredH

3,7201023

asked Mar 29 at 21:23

Abdallah krichenAbdallah krichen

182

$endgroup$

share|cite|improve this question

edited Mar 29 at 21:31

FredH

3,7201023

asked Mar 29 at 21:23

Abdallah krichenAbdallah krichen

182

share|cite|improve this question

edited Mar 29 at 21:31

FredH

3,7201023

asked Mar 29 at 21:23

Abdallah krichenAbdallah krichen

182

edited Mar 29 at 21:31

FredH

3,7201023

edited Mar 29 at 21:31

FredH

3,7201023

asked Mar 29 at 21:23

Abdallah krichenAbdallah krichen

182

asked Mar 29 at 21:23

Abdallah krichenAbdallah krichen

182

2 Answers
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If $n$ is even, $x=n-1$ will work.

If $n$ is odd, choose $x=2n-1$.

The point is that $xequiv -1pmod n$ implies $x^aequiv -1pmod n$ if $a$ is odd.

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answered Mar 29 at 21:36

BerciBerci

61.9k23776

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add a comment | 

0

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Note that for any positive integer $n$, there exists an odd positive integer $x$ such that $n|x+1$. Such an $x$ can be constructed as follows$$x=2nk+1quad,quad kinBbb Z$$Furthermore, if $x,k$ are odd positive integers, then so is $x^k$. By a simple induction we can show that $x,x^x,x^x^x,cdots$ are all odd positive integers and hence we can write $$x+1|x^x^x^x^dots+1$$which is obtained from $a+b|a^n+b^n$ for odd $ninBbb N$. This completes our proof.

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answered Mar 29 at 21:37

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

0

$begingroup$

If $n$ is even, $x=n-1$ will work.

If $n$ is odd, choose $x=2n-1$.

The point is that $xequiv -1pmod n$ implies $x^aequiv -1pmod n$ if $a$ is odd.

share|cite|improve this answer

answered Mar 29 at 21:36

BerciBerci

61.9k23776

$endgroup$

add a comment | 

0

$begingroup$

If $n$ is even, $x=n-1$ will work.

If $n$ is odd, choose $x=2n-1$.

The point is that $xequiv -1pmod n$ implies $x^aequiv -1pmod n$ if $a$ is odd.

share|cite|improve this answer

answered Mar 29 at 21:36

BerciBerci

61.9k23776

$endgroup$

add a comment | 

$begingroup$

If $n$ is even, $x=n-1$ will work.

If $n$ is odd, choose $x=2n-1$.

The point is that $xequiv -1pmod n$ implies $x^aequiv -1pmod n$ if $a$ is odd.

share|cite|improve this answer

answered Mar 29 at 21:36

BerciBerci

61.9k23776

$endgroup$

share|cite|improve this answer

answered Mar 29 at 21:36

BerciBerci

61.9k23776

answered Mar 29 at 21:36

BerciBerci

61.9k23776

answered Mar 29 at 21:36

BerciBerci

61.9k23776

0

$begingroup$

Note that for any positive integer $n$, there exists an odd positive integer $x$ such that $n|x+1$. Such an $x$ can be constructed as follows$$x=2nk+1quad,quad kinBbb Z$$Furthermore, if $x,k$ are odd positive integers, then so is $x^k$. By a simple induction we can show that $x,x^x,x^x^x,cdots$ are all odd positive integers and hence we can write $$x+1|x^x^x^x^dots+1$$which is obtained from $a+b|a^n+b^n$ for odd $ninBbb N$. This completes our proof.

share|cite|improve this answer

answered Mar 29 at 21:37

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

0

$begingroup$

Note that for any positive integer $n$, there exists an odd positive integer $x$ such that $n|x+1$. Such an $x$ can be constructed as follows$$x=2nk+1quad,quad kinBbb Z$$Furthermore, if $x,k$ are odd positive integers, then so is $x^k$. By a simple induction we can show that $x,x^x,x^x^x,cdots$ are all odd positive integers and hence we can write $$x+1|x^x^x^x^dots+1$$which is obtained from $a+b|a^n+b^n$ for odd $ninBbb N$. This completes our proof.

share|cite|improve this answer

answered Mar 29 at 21:37

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

add a comment | 

$begingroup$

Note that for any positive integer $n$, there exists an odd positive integer $x$ such that $n|x+1$. Such an $x$ can be constructed as follows$$x=2nk+1quad,quad kinBbb Z$$Furthermore, if $x,k$ are odd positive integers, then so is $x^k$. By a simple induction we can show that $x,x^x,x^x^x,cdots$ are all odd positive integers and hence we can write $$x+1|x^x^x^x^dots+1$$which is obtained from $a+b|a^n+b^n$ for odd $ninBbb N$. This completes our proof.

share|cite|improve this answer

answered Mar 29 at 21:37

Mostafa AyazMostafa Ayaz

18.2k31040

$endgroup$

share|cite|improve this answer

answered Mar 29 at 21:37

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 29 at 21:37

Mostafa AyazMostafa Ayaz

18.2k31040

answered Mar 29 at 21:37

Mostafa AyazMostafa Ayaz

18.2k31040