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Prove if p is a prime number that does not divide a, then $a^p^2$ congruent to $a^pmod p^2$

How to prove that if $aequiv b pmodkn$ then $a^kequiv b^k pmodk^2n$Prove that $x^2 equiv -1 pmod p$ has no solutions if prime $p equiv 3 pmod 4$.Show that Fermat's Little Theorem doesn't hold if p is not prime …Let $p$ be a prime and $q$ a prime divisor of $2^p -1$. Use Fermat's Little Theorem to prove that $qequiv 1 (mod space p)$How to select the right modulus to prove that there do not exist integers $a$ and $b$ such that $a^2+b^2=1234567$?Prove that if a $in mathbbZ$ then $a^3 equiv a(mod 3)$A prime number generator algorithm based on $x^2+(x-1)^2$ that generates only primesProof that there are infinitely many prime numbers $p$ such that $p-2$ is not prime.Prove that for integers $a$, $b$, and $n$, if $a$ and $b$ are each relatively prime to $n$, then the product $ab$ is also relatively prime to $n$.Prove that $4^n+ 1$ is not divisible by $3$How does one prove that a prime $p$ does not generate the multiplicative group of integers mod $4p-1$?

0

1

$begingroup$

I tried to do proof by contradiction, so I started to do use real arbitrary numbers. I decided I use p for prime still, but a being a multiple of p and by doing this, the statement is still true. I tried it with p=3, a =6 or p=2 and a =6. Anyways to solve it by contradiction or should I use another method

modular-arithmetic

share|cite|improve this question

edited Mar 13 at 18:32

Wuestenfux

5,1721513

asked Mar 13 at 18:29

Kevin WangKevin Wang

241

New contributor

Kevin Wang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment | 

0

1

$begingroup$

I tried to do proof by contradiction, so I started to do use real arbitrary numbers. I decided I use p for prime still, but a being a multiple of p and by doing this, the statement is still true. I tried it with p=3, a =6 or p=2 and a =6. Anyways to solve it by contradiction or should I use another method

modular-arithmetic

share|cite|improve this question

edited Mar 13 at 18:32

Wuestenfux

5,1721513

asked Mar 13 at 18:29

Kevin WangKevin Wang

241

New contributor

Kevin Wang 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 | 

$begingroup$

I tried to do proof by contradiction, so I started to do use real arbitrary numbers. I decided I use p for prime still, but a being a multiple of p and by doing this, the statement is still true. I tried it with p=3, a =6 or p=2 and a =6. Anyways to solve it by contradiction or should I use another method

modular-arithmetic

share|cite|improve this question

edited Mar 13 at 18:32

Wuestenfux

5,1721513

asked Mar 13 at 18:29

Kevin WangKevin Wang

241

New contributor

Kevin Wang 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 question

edited Mar 13 at 18:32

Wuestenfux

5,1721513

asked Mar 13 at 18:29

Kevin WangKevin Wang

241

New contributor

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

share|cite|improve this question

edited Mar 13 at 18:32

Wuestenfux

5,1721513

asked Mar 13 at 18:29

Kevin WangKevin Wang

241

New contributor

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

edited Mar 13 at 18:32

Wuestenfux

5,1721513

edited Mar 13 at 18:32

Wuestenfux

5,1721513

asked Mar 13 at 18:29

Kevin WangKevin Wang

241

New contributor

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

asked Mar 13 at 18:29

Kevin WangKevin Wang

241

4 Answers
4

active

oldest

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3

$begingroup$

You have

$$a^p^2 - a^p =a^p(a^p^2-p -1) =$$ $$ a^p(a^phi(p^2)-1)equiv a^p(1-1) pmodp^2,$$

where Euler's theorem is used in the last step.

share|cite|improve this answer

answered Mar 13 at 18:36

B. GoddardB. Goddard

19.7k21442

$endgroup$

add a comment | 

1

$begingroup$

Recall $ bequiv a pmod!m,Rightarrow,b^large mequiv a^large m pmod!m^large 2.,$ Put $,b=a^large p, m = p$

share|cite|improve this answer

answered Mar 13 at 22:07

Bill DubuqueBill Dubuque

212k29195654

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's a special case of the polynomial double root test - as I explain in the linked answer.
  $endgroup$
  – Bill Dubuque
  Mar 13 at 22:13
  
  

  
  
  
  

add a comment | 

0

$begingroup$

As by Fermat's little theorem $a^pequiv apmod p$ for any integer $a$

$(a^p)^pequiv( a)^p$

So, we don't need $pnmid a$

share|cite|improve this answer

answered Mar 13 at 18:32

lab bhattacharjeelab bhattacharjee

227k15158275

$endgroup$

add a comment | 

-1

$begingroup$

Simple, $phi(p^2)=(p-1)p$ and $p^2-(p-1)p=p$

Therefore $a^pequiv a^p bmod p^2$ which is true.

This generalizes to $a^p^nequiv a^p^n-1bmod p^n$

share|cite|improve this answer

answered Mar 14 at 18:03

Roddy MacPheeRoddy MacPhee

436117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can mark me down all you want it still holds.
  $endgroup$
  – Roddy MacPhee
  Mar 14 at 21:31
  

  
  
  
  

add a comment | 

Your Answer

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3

$begingroup$

You have

$$a^p^2 - a^p =a^p(a^p^2-p -1) =$$ $$ a^p(a^phi(p^2)-1)equiv a^p(1-1) pmodp^2,$$

where Euler's theorem is used in the last step.

share|cite|improve this answer

answered Mar 13 at 18:36

B. GoddardB. Goddard

19.7k21442

$endgroup$

add a comment | 

3

$begingroup$

You have

$$a^p^2 - a^p =a^p(a^p^2-p -1) =$$ $$ a^p(a^phi(p^2)-1)equiv a^p(1-1) pmodp^2,$$

where Euler's theorem is used in the last step.

share|cite|improve this answer

answered Mar 13 at 18:36

B. GoddardB. Goddard

19.7k21442

$endgroup$

add a comment | 

$begingroup$

You have

$$a^p^2 - a^p =a^p(a^p^2-p -1) =$$ $$ a^p(a^phi(p^2)-1)equiv a^p(1-1) pmodp^2,$$

where Euler's theorem is used in the last step.

share|cite|improve this answer

answered Mar 13 at 18:36

B. GoddardB. Goddard

19.7k21442

$endgroup$

share|cite|improve this answer

answered Mar 13 at 18:36

B. GoddardB. Goddard

19.7k21442

answered Mar 13 at 18:36

B. GoddardB. Goddard

19.7k21442

answered Mar 13 at 18:36

B. GoddardB. Goddard

19.7k21442

1

$begingroup$

Recall $ bequiv a pmod!m,Rightarrow,b^large mequiv a^large m pmod!m^large 2.,$ Put $,b=a^large p, m = p$

share|cite|improve this answer

answered Mar 13 at 22:07

Bill DubuqueBill Dubuque

212k29195654

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's a special case of the polynomial double root test - as I explain in the linked answer.
  $endgroup$
  – Bill Dubuque
  Mar 13 at 22:13
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Recall $ bequiv a pmod!m,Rightarrow,b^large mequiv a^large m pmod!m^large 2.,$ Put $,b=a^large p, m = p$

share|cite|improve this answer

answered Mar 13 at 22:07

Bill DubuqueBill Dubuque

212k29195654

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It's a special case of the polynomial double root test - as I explain in the linked answer.
  $endgroup$
  – Bill Dubuque
  Mar 13 at 22:13
  
  

  
  
  
  

add a comment | 

$begingroup$

Recall $ bequiv a pmod!m,Rightarrow,b^large mequiv a^large m pmod!m^large 2.,$ Put $,b=a^large p, m = p$

share|cite|improve this answer

answered Mar 13 at 22:07

Bill DubuqueBill Dubuque

212k29195654

$endgroup$

share|cite|improve this answer

answered Mar 13 at 22:07

Bill DubuqueBill Dubuque

212k29195654

answered Mar 13 at 22:07

Bill DubuqueBill Dubuque

212k29195654

answered Mar 13 at 22:07

Bill DubuqueBill Dubuque

212k29195654

0

$begingroup$

As by Fermat's little theorem $a^pequiv apmod p$ for any integer $a$

$(a^p)^pequiv( a)^p$

So, we don't need $pnmid a$

share|cite|improve this answer

answered Mar 13 at 18:32

lab bhattacharjeelab bhattacharjee

227k15158275

$endgroup$

add a comment | 

0

$begingroup$

As by Fermat's little theorem $a^pequiv apmod p$ for any integer $a$

$(a^p)^pequiv( a)^p$

So, we don't need $pnmid a$

share|cite|improve this answer

answered Mar 13 at 18:32

lab bhattacharjeelab bhattacharjee

227k15158275

$endgroup$

add a comment | 

$begingroup$

As by Fermat's little theorem $a^pequiv apmod p$ for any integer $a$

$(a^p)^pequiv( a)^p$

So, we don't need $pnmid a$

share|cite|improve this answer

answered Mar 13 at 18:32

lab bhattacharjeelab bhattacharjee

227k15158275

$endgroup$

share|cite|improve this answer

answered Mar 13 at 18:32

lab bhattacharjeelab bhattacharjee

227k15158275

answered Mar 13 at 18:32

lab bhattacharjeelab bhattacharjee

227k15158275

answered Mar 13 at 18:32

lab bhattacharjeelab bhattacharjee

227k15158275

-1

$begingroup$

Simple, $phi(p^2)=(p-1)p$ and $p^2-(p-1)p=p$

Therefore $a^pequiv a^p bmod p^2$ which is true.

This generalizes to $a^p^nequiv a^p^n-1bmod p^n$

share|cite|improve this answer

answered Mar 14 at 18:03

Roddy MacPheeRoddy MacPhee

436117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can mark me down all you want it still holds.
  $endgroup$
  – Roddy MacPhee
  Mar 14 at 21:31
  

  
  
  
  

add a comment | 

-1

$begingroup$

Simple, $phi(p^2)=(p-1)p$ and $p^2-(p-1)p=p$

Therefore $a^pequiv a^p bmod p^2$ which is true.

This generalizes to $a^p^nequiv a^p^n-1bmod p^n$

share|cite|improve this answer

answered Mar 14 at 18:03

Roddy MacPheeRoddy MacPhee

436117

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can mark me down all you want it still holds.
  $endgroup$
  – Roddy MacPhee
  Mar 14 at 21:31
  

  
  
  
  

add a comment | 

$begingroup$

Simple, $phi(p^2)=(p-1)p$ and $p^2-(p-1)p=p$

Therefore $a^pequiv a^p bmod p^2$ which is true.

This generalizes to $a^p^nequiv a^p^n-1bmod p^n$

share|cite|improve this answer

answered Mar 14 at 18:03

Roddy MacPheeRoddy MacPhee

436117

$endgroup$

share|cite|improve this answer

answered Mar 14 at 18:03

Roddy MacPheeRoddy MacPhee

436117

answered Mar 14 at 18:03

Roddy MacPheeRoddy MacPhee

436117

answered Mar 14 at 18:03

Roddy MacPheeRoddy MacPhee

436117