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Can we always find an integer x such that … [closed]

Prove that $x$ exists such that the sequence is divisible by $n$proving there exists an integer $ k$ such that $ak^2+bk+cequiv 0 pmod2^n$ by inductionProve that $(a,m) = 1$ iff there exists an integer $n$ such that $na equiv 1 pmod m$Given $a, b, c, d, m inmathbbZ$such that $5mid (am^3 + bm^2 + cm + d)$, prove that there exists integer $n$ such that…For every prime $p$ there exists integer numbers $x,y,z$ such that $(x^2+1)(y^2+1)(z^2+1)equiv1pmod p$Why there exist integer $b,c$ such $a=b+c$A positive integer $k$ such that the last $2015$ digits of $53^k$ are all $3$There exists an integer $n$ such that $n$ and $n+2$ are quadratic residue modulo $p$if and only if there exists an integer $x$ such that $x^2016 + x^2015 + cdots + 1 equiv p^2016 pmodp^2017$.Showing that there exists a positive integer $t$ such that $5^tequiv -3pmod 2^n+4$Prove that there exists an integer $ x$ such that $ fracx^2-2p$ is the square of an integer

3

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Can we prove that for every integer $n>0$, there exists an integer $x$ such that for every integer $a$ :

$$aequiv -1pmod nimplies x^aequiv -1pmod n$$

number-theory modular-arithmetic

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edited Mar 29 at 21:00

zwim

12.7k832

asked Mar 29 at 20:55

Abdallah krichenAbdallah krichen

182

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closed as off-topic by Sil, John Omielan, Cesareo, Jyrki Lahtonen, Eevee Trainer Mar 30 at 8:51

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, Jyrki Lahtonen, Eevee Trainer

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course not. $x^6 not equiv -1pmod 7$ ever.
  $endgroup$
  – fleablood
  Mar 29 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related to OP's next question.
  $endgroup$
  – Bill Dubuque
  Mar 29 at 22:15
  

  
  
  
  

add a comment | 

3

$begingroup$

Can we prove that for every integer $n>0$, there exists an integer $x$ such that for every integer $a$ :

$$aequiv -1pmod nimplies x^aequiv -1pmod n$$

number-theory modular-arithmetic

share|cite|improve this question

edited Mar 29 at 21:00

zwim

12.7k832

asked Mar 29 at 20:55

Abdallah krichenAbdallah krichen

182

$endgroup$

closed as off-topic by Sil, John Omielan, Cesareo, Jyrki Lahtonen, Eevee Trainer Mar 30 at 8:51

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, Jyrki Lahtonen, Eevee Trainer

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

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Of course not. $x^6 not equiv -1pmod 7$ ever.
  $endgroup$
  – fleablood
  Mar 29 at 21:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related to OP's next question.
  $endgroup$
  – Bill Dubuque
  Mar 29 at 22:15
  

  
  
  
  

add a comment | 

$begingroup$

Can we prove that for every integer $n>0$, there exists an integer $x$ such that for every integer $a$ :

$$aequiv -1pmod nimplies x^aequiv -1pmod n$$

number-theory modular-arithmetic

share|cite|improve this question

edited Mar 29 at 21:00

zwim

12.7k832

asked Mar 29 at 20:55

Abdallah krichenAbdallah krichen

182

$endgroup$

share|cite|improve this question

edited Mar 29 at 21:00

zwim

12.7k832

asked Mar 29 at 20:55

Abdallah krichenAbdallah krichen

182

share|cite|improve this question

edited Mar 29 at 21:00

zwim

12.7k832

asked Mar 29 at 20:55

Abdallah krichenAbdallah krichen

182

edited Mar 29 at 21:00

zwim

12.7k832

edited Mar 29 at 21:00

zwim

12.7k832

asked Mar 29 at 20:55

Abdallah krichenAbdallah krichen

182

asked Mar 29 at 20:55

Abdallah krichenAbdallah krichen

182

3 Answers
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No, it is not possible. If $n=3$, we can choose $a = 2$, but there is no solution to $x^2 equiv -1 pmod 3$.

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answered Mar 29 at 21:04

FredHFredH

3,7201023

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1

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True iff $n$ is even. If so $,aequiv -1pmod!n,Rightarrow,a,$ odd $,Rightarrow,(-1)^aequiv -1pmod!n $ so $,x = -1,$ works.

Conversely $,x^n-1equiv -1,Rightarrow, x^2n-2 equiv 1,Rightarrow, x^2n-1equiv x. $ But $,x^2n-1equiv -1,$ by $,2n!-!1equiv -1pmod! n$

hence $, xequiv -1,$ so $,-1equiv x^n-1equiv (-1)^n-1,Rightarrow n,$ even

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answered Mar 29 at 21:56

Bill DubuqueBill Dubuque

214k29196655

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also true in the degenerate case $n = 1$ but normally $bmod 1$ (trivial ring) is excluded in problems like this.
  $endgroup$
  – Bill Dubuque
  Mar 29 at 21:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We could also have used $,x^-1equiv -1,Rightarrow, xequiv -1,,$ but the above is a bit more general.
  $endgroup$
  – Bill Dubuque
  Mar 29 at 22:10
  
  

  
  
  
  

add a comment | 

0

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Prime number $p$ is an integer greater than $0$, so it is an example of $n$.

$p-1$ is an integer congruent to $-1 mod p$, so it is an example of $a$.

If $x$ is chosen such that $pmid x$, then $x^p-1 equiv 0 mod p$.

If $x$ is chosen such that $pnot mid x$, then $x^p-1 equiv 1 mod p$ (by Fermat's Little Theorem).

$n=p, a=p-1$ are numbers that don't satisfy your criterion of yielding a congruence to $-1$ for any chosen $x$, so the condition 'every integer' (twice repeated) cannot be true.

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edited Mar 29 at 21:41

answered Mar 29 at 21:32

Keith BackmanKeith Backman

1,5461812

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2

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No, it is not possible. If $n=3$, we can choose $a = 2$, but there is no solution to $x^2 equiv -1 pmod 3$.

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answered Mar 29 at 21:04

FredHFredH

3,7201023

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add a comment | 

2

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No, it is not possible. If $n=3$, we can choose $a = 2$, but there is no solution to $x^2 equiv -1 pmod 3$.

share|cite|improve this answer

answered Mar 29 at 21:04

FredHFredH

3,7201023

$endgroup$

add a comment | 

$begingroup$

No, it is not possible. If $n=3$, we can choose $a = 2$, but there is no solution to $x^2 equiv -1 pmod 3$.

share|cite|improve this answer

answered Mar 29 at 21:04

FredHFredH

3,7201023

$endgroup$

share|cite|improve this answer

answered Mar 29 at 21:04

FredHFredH

3,7201023

answered Mar 29 at 21:04

FredHFredH

3,7201023

answered Mar 29 at 21:04

FredHFredH

3,7201023

1

$begingroup$

True iff $n$ is even. If so $,aequiv -1pmod!n,Rightarrow,a,$ odd $,Rightarrow,(-1)^aequiv -1pmod!n $ so $,x = -1,$ works.

Conversely $,x^n-1equiv -1,Rightarrow, x^2n-2 equiv 1,Rightarrow, x^2n-1equiv x. $ But $,x^2n-1equiv -1,$ by $,2n!-!1equiv -1pmod! n$

hence $, xequiv -1,$ so $,-1equiv x^n-1equiv (-1)^n-1,Rightarrow n,$ even

share|cite|improve this answer

answered Mar 29 at 21:56

Bill DubuqueBill Dubuque

214k29196655

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also true in the degenerate case $n = 1$ but normally $bmod 1$ (trivial ring) is excluded in problems like this.
  $endgroup$
  – Bill Dubuque
  Mar 29 at 21:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We could also have used $,x^-1equiv -1,Rightarrow, xequiv -1,,$ but the above is a bit more general.
  $endgroup$
  – Bill Dubuque
  Mar 29 at 22:10
  
  

  
  
  
  

add a comment | 

1

$begingroup$

True iff $n$ is even. If so $,aequiv -1pmod!n,Rightarrow,a,$ odd $,Rightarrow,(-1)^aequiv -1pmod!n $ so $,x = -1,$ works.

Conversely $,x^n-1equiv -1,Rightarrow, x^2n-2 equiv 1,Rightarrow, x^2n-1equiv x. $ But $,x^2n-1equiv -1,$ by $,2n!-!1equiv -1pmod! n$

hence $, xequiv -1,$ so $,-1equiv x^n-1equiv (-1)^n-1,Rightarrow n,$ even

share|cite|improve this answer

answered Mar 29 at 21:56

Bill DubuqueBill Dubuque

214k29196655

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also true in the degenerate case $n = 1$ but normally $bmod 1$ (trivial ring) is excluded in problems like this.
  $endgroup$
  – Bill Dubuque
  Mar 29 at 21:59
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We could also have used $,x^-1equiv -1,Rightarrow, xequiv -1,,$ but the above is a bit more general.
  $endgroup$
  – Bill Dubuque
  Mar 29 at 22:10
  
  

  
  
  
  

add a comment | 

$begingroup$

True iff $n$ is even. If so $,aequiv -1pmod!n,Rightarrow,a,$ odd $,Rightarrow,(-1)^aequiv -1pmod!n $ so $,x = -1,$ works.

Conversely $,x^n-1equiv -1,Rightarrow, x^2n-2 equiv 1,Rightarrow, x^2n-1equiv x. $ But $,x^2n-1equiv -1,$ by $,2n!-!1equiv -1pmod! n$

hence $, xequiv -1,$ so $,-1equiv x^n-1equiv (-1)^n-1,Rightarrow n,$ even

share|cite|improve this answer

answered Mar 29 at 21:56

Bill DubuqueBill Dubuque

214k29196655

$endgroup$

share|cite|improve this answer

answered Mar 29 at 21:56

Bill DubuqueBill Dubuque

214k29196655

answered Mar 29 at 21:56

Bill DubuqueBill Dubuque

214k29196655

answered Mar 29 at 21:56

Bill DubuqueBill Dubuque

214k29196655

0

$begingroup$

Prime number $p$ is an integer greater than $0$, so it is an example of $n$.

$p-1$ is an integer congruent to $-1 mod p$, so it is an example of $a$.

If $x$ is chosen such that $pmid x$, then $x^p-1 equiv 0 mod p$.

If $x$ is chosen such that $pnot mid x$, then $x^p-1 equiv 1 mod p$ (by Fermat's Little Theorem).

$n=p, a=p-1$ are numbers that don't satisfy your criterion of yielding a congruence to $-1$ for any chosen $x$, so the condition 'every integer' (twice repeated) cannot be true.

share|cite|improve this answer

edited Mar 29 at 21:41

answered Mar 29 at 21:32

Keith BackmanKeith Backman

1,5461812

$endgroup$

add a comment | 

0

$begingroup$

Prime number $p$ is an integer greater than $0$, so it is an example of $n$.

$p-1$ is an integer congruent to $-1 mod p$, so it is an example of $a$.

If $x$ is chosen such that $pmid x$, then $x^p-1 equiv 0 mod p$.

If $x$ is chosen such that $pnot mid x$, then $x^p-1 equiv 1 mod p$ (by Fermat's Little Theorem).

$n=p, a=p-1$ are numbers that don't satisfy your criterion of yielding a congruence to $-1$ for any chosen $x$, so the condition 'every integer' (twice repeated) cannot be true.

share|cite|improve this answer

edited Mar 29 at 21:41

answered Mar 29 at 21:32

Keith BackmanKeith Backman

1,5461812

$endgroup$

add a comment | 

$begingroup$

Prime number $p$ is an integer greater than $0$, so it is an example of $n$.

$p-1$ is an integer congruent to $-1 mod p$, so it is an example of $a$.

If $x$ is chosen such that $pmid x$, then $x^p-1 equiv 0 mod p$.

If $x$ is chosen such that $pnot mid x$, then $x^p-1 equiv 1 mod p$ (by Fermat's Little Theorem).

$n=p, a=p-1$ are numbers that don't satisfy your criterion of yielding a congruence to $-1$ for any chosen $x$, so the condition 'every integer' (twice repeated) cannot be true.

share|cite|improve this answer

edited Mar 29 at 21:41

answered Mar 29 at 21:32

Keith BackmanKeith Backman

1,5461812

$endgroup$

share|cite|improve this answer

edited Mar 29 at 21:41

answered Mar 29 at 21:32

Keith BackmanKeith Backman

1,5461812

answered Mar 29 at 21:32

Keith BackmanKeith Backman

1,5461812

answered Mar 29 at 21:32

Keith BackmanKeith Backman

1,5461812