$a equiv b$ (mod $p$) implies $a^{p^n} equiv b^{p^n}$ (mod $p^n$)?

1

$begingroup$

Let $p$ be a prime number. If $a equiv b$ (mod $p$), does that imply $a^{p^n} equiv b^{p^n}$ (mod $p^n$)?

I think the answer will be yes, and I suspect that the way of proving it will involve writing $a^{p^n}-b^{p^n}$ as a multiple of $(a-b)^n$.

I also noticed that $n<p^n$, and I'm wondering if this will make the proof easier or not.

prime-numbers modular-arithmetic cyclotomic-polynomials

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asked Nov 8 '18 at 18:19

Pascal's WagerPascal's Wager

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1

$begingroup$

Let $p$ be a prime number. If $a equiv b$ (mod $p$), does that imply $a^{p^n} equiv b^{p^n}$ (mod $p^n$)?

I think the answer will be yes, and I suspect that the way of proving it will involve writing $a^{p^n}-b^{p^n}$ as a multiple of $(a-b)^n$.

I also noticed that $n<p^n$, and I'm wondering if this will make the proof easier or not.

prime-numbers modular-arithmetic cyclotomic-polynomials

share|cite|improve this question

asked Nov 8 '18 at 18:19

Pascal's WagerPascal's Wager

357315

$endgroup$

add a comment | 

1

1

1

$begingroup$

Let $p$ be a prime number. If $a equiv b$ (mod $p$), does that imply $a^{p^n} equiv b^{p^n}$ (mod $p^n$)?

I think the answer will be yes, and I suspect that the way of proving it will involve writing $a^{p^n}-b^{p^n}$ as a multiple of $(a-b)^n$.

I also noticed that $n<p^n$, and I'm wondering if this will make the proof easier or not.

prime-numbers modular-arithmetic cyclotomic-polynomials

share|cite|improve this question

asked Nov 8 '18 at 18:19

Pascal's WagerPascal's Wager

357315

$endgroup$

Let $p$ be a prime number. If $a equiv b$ (mod $p$), does that imply $a^{p^n} equiv b^{p^n}$ (mod $p^n$)?

I think the answer will be yes, and I suspect that the way of proving it will involve writing $a^{p^n}-b^{p^n}$ as a multiple of $(a-b)^n$.

I also noticed that $n<p^n$, and I'm wondering if this will make the proof easier or not.

prime-numbers modular-arithmetic cyclotomic-polynomials

prime-numbers modular-arithmetic cyclotomic-polynomials

share|cite|improve this question

asked Nov 8 '18 at 18:19

Pascal's WagerPascal's Wager

357315

share|cite|improve this question

asked Nov 8 '18 at 18:19

Pascal's WagerPascal's Wager

357315

share|cite|improve this question

share|cite|improve this question

asked Nov 8 '18 at 18:19

Pascal's WagerPascal's Wager

357315

asked Nov 8 '18 at 18:19

Pascal's WagerPascal's Wager

357315

asked Nov 8 '18 at 18:19

Pascal's WagerPascal's Wager

357315

357315

add a comment | 

add a comment | 

2 Answers
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4

$begingroup$

$a=b+kp$ where $k$ is an integer

$(b+kp)^{p^n}=b^{p^n}+binom{p^n}1b^{p^n-1}kp+binom{p^n}2b^{p^n-2}(kp)^{2}+cdots+(kp)^{p^n}$

$equiv b^{p^n}pmod{p^{n+1}}$

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answered Nov 8 '18 at 18:28

lab bhattacharjeelab bhattacharjee

225k15157275

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So simple yet so useful! Thank you!
  $endgroup$
  – Pascal's Wager
  Nov 8 '18 at 23:45
  

  
  
  
  

add a comment | 

0

$begingroup$

There is following theorem (Lifting The Exponent Lemma).
We will use notation $p^{alpha}||n$ for positive integer $n$, nonnegative integer $alpha$ and prime $p$, which equivalent to $p^{alpha}|n$ and $p^{alpha+1}nmid n$.

Theorem.
Let $a,b,n$ be a positive integers and $p^{alpha}||a-b$, $p^{beta}||n$. Then:

• if $p>2$ and $alphageq 1$ then $p^{alpha+beta}||a^n-b^n$;


• if $p=2$ and $alphageq 2$ then $2^{alpha+beta}||a^n-b^n$.

Note. If $p=2$ and $alpha=1$ then $2^{beta+1}||a^n-b^n$.

Your statement is a consequence of this theorem.

Some links about LTE lemma:

What can I do with the lifting the exponent lemma?

https://brilliant.org/wiki/lifting-the-exponent/

share|cite|improve this answer

answered Jan 12 at 15:00

richrowrichrow

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2 Answers
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active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

4

$begingroup$

$a=b+kp$ where $k$ is an integer

$(b+kp)^{p^n}=b^{p^n}+binom{p^n}1b^{p^n-1}kp+binom{p^n}2b^{p^n-2}(kp)^{2}+cdots+(kp)^{p^n}$

$equiv b^{p^n}pmod{p^{n+1}}$

share|cite|improve this answer

answered Nov 8 '18 at 18:28

lab bhattacharjeelab bhattacharjee

225k15157275

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So simple yet so useful! Thank you!
  $endgroup$
  – Pascal's Wager
  Nov 8 '18 at 23:45
  

  
  
  
  

add a comment | 

4

$begingroup$

$a=b+kp$ where $k$ is an integer

$(b+kp)^{p^n}=b^{p^n}+binom{p^n}1b^{p^n-1}kp+binom{p^n}2b^{p^n-2}(kp)^{2}+cdots+(kp)^{p^n}$

$equiv b^{p^n}pmod{p^{n+1}}$

share|cite|improve this answer

answered Nov 8 '18 at 18:28

lab bhattacharjeelab bhattacharjee

225k15157275

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So simple yet so useful! Thank you!
  $endgroup$
  – Pascal's Wager
  Nov 8 '18 at 23:45
  

  
  
  
  

add a comment | 

4

4

4

$begingroup$

$a=b+kp$ where $k$ is an integer

$(b+kp)^{p^n}=b^{p^n}+binom{p^n}1b^{p^n-1}kp+binom{p^n}2b^{p^n-2}(kp)^{2}+cdots+(kp)^{p^n}$

$equiv b^{p^n}pmod{p^{n+1}}$

share|cite|improve this answer

answered Nov 8 '18 at 18:28

lab bhattacharjeelab bhattacharjee

225k15157275

$endgroup$

$a=b+kp$ where $k$ is an integer

$(b+kp)^{p^n}=b^{p^n}+binom{p^n}1b^{p^n-1}kp+binom{p^n}2b^{p^n-2}(kp)^{2}+cdots+(kp)^{p^n}$

$equiv b^{p^n}pmod{p^{n+1}}$

share|cite|improve this answer

answered Nov 8 '18 at 18:28

lab bhattacharjeelab bhattacharjee

225k15157275

share|cite|improve this answer

share|cite|improve this answer

answered Nov 8 '18 at 18:28

lab bhattacharjeelab bhattacharjee

225k15157275

answered Nov 8 '18 at 18:28

lab bhattacharjeelab bhattacharjee

225k15157275

answered Nov 8 '18 at 18:28

lab bhattacharjeelab bhattacharjee

225k15157275

225k15157275

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So simple yet so useful! Thank you!
  $endgroup$
  – Pascal's Wager
  Nov 8 '18 at 23:45
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So simple yet so useful! Thank you!
  $endgroup$
  – Pascal's Wager
  Nov 8 '18 at 23:45
  

  
  
  
  

$begingroup$
So simple yet so useful! Thank you!
$endgroup$
– Pascal's Wager
Nov 8 '18 at 23:45

$begingroup$
So simple yet so useful! Thank you!
$endgroup$
– Pascal's Wager
Nov 8 '18 at 23:45

add a comment | 

0

$begingroup$

There is following theorem (Lifting The Exponent Lemma).
We will use notation $p^{alpha}||n$ for positive integer $n$, nonnegative integer $alpha$ and prime $p$, which equivalent to $p^{alpha}|n$ and $p^{alpha+1}nmid n$.

Theorem.
Let $a,b,n$ be a positive integers and $p^{alpha}||a-b$, $p^{beta}||n$. Then:

• if $p>2$ and $alphageq 1$ then $p^{alpha+beta}||a^n-b^n$;


• if $p=2$ and $alphageq 2$ then $2^{alpha+beta}||a^n-b^n$.

Note. If $p=2$ and $alpha=1$ then $2^{beta+1}||a^n-b^n$.

Your statement is a consequence of this theorem.

Some links about LTE lemma:

What can I do with the lifting the exponent lemma?

https://brilliant.org/wiki/lifting-the-exponent/

share|cite|improve this answer

answered Jan 12 at 15:00

richrowrichrow

17316

$endgroup$

add a comment | 

0

$begingroup$

There is following theorem (Lifting The Exponent Lemma).
We will use notation $p^{alpha}||n$ for positive integer $n$, nonnegative integer $alpha$ and prime $p$, which equivalent to $p^{alpha}|n$ and $p^{alpha+1}nmid n$.

Theorem.
Let $a,b,n$ be a positive integers and $p^{alpha}||a-b$, $p^{beta}||n$. Then:

• if $p>2$ and $alphageq 1$ then $p^{alpha+beta}||a^n-b^n$;


• if $p=2$ and $alphageq 2$ then $2^{alpha+beta}||a^n-b^n$.

Note. If $p=2$ and $alpha=1$ then $2^{beta+1}||a^n-b^n$.

Your statement is a consequence of this theorem.

Some links about LTE lemma:

What can I do with the lifting the exponent lemma?

https://brilliant.org/wiki/lifting-the-exponent/

share|cite|improve this answer

answered Jan 12 at 15:00

richrowrichrow

17316

$endgroup$

add a comment | 

0

0

0

$begingroup$

There is following theorem (Lifting The Exponent Lemma).
We will use notation $p^{alpha}||n$ for positive integer $n$, nonnegative integer $alpha$ and prime $p$, which equivalent to $p^{alpha}|n$ and $p^{alpha+1}nmid n$.

Theorem.
Let $a,b,n$ be a positive integers and $p^{alpha}||a-b$, $p^{beta}||n$. Then:

• if $p>2$ and $alphageq 1$ then $p^{alpha+beta}||a^n-b^n$;


• if $p=2$ and $alphageq 2$ then $2^{alpha+beta}||a^n-b^n$.

Note. If $p=2$ and $alpha=1$ then $2^{beta+1}||a^n-b^n$.

Your statement is a consequence of this theorem.

Some links about LTE lemma:

What can I do with the lifting the exponent lemma?

https://brilliant.org/wiki/lifting-the-exponent/

share|cite|improve this answer

answered Jan 12 at 15:00

richrowrichrow

17316

$endgroup$

There is following theorem (Lifting The Exponent Lemma).
We will use notation $p^{alpha}||n$ for positive integer $n$, nonnegative integer $alpha$ and prime $p$, which equivalent to $p^{alpha}|n$ and $p^{alpha+1}nmid n$.

Theorem.
Let $a,b,n$ be a positive integers and $p^{alpha}||a-b$, $p^{beta}||n$. Then:

• if $p>2$ and $alphageq 1$ then $p^{alpha+beta}||a^n-b^n$;


• if $p=2$ and $alphageq 2$ then $2^{alpha+beta}||a^n-b^n$.

Note. If $p=2$ and $alpha=1$ then $2^{beta+1}||a^n-b^n$.

Your statement is a consequence of this theorem.

Some links about LTE lemma:

What can I do with the lifting the exponent lemma?

https://brilliant.org/wiki/lifting-the-exponent/

share|cite|improve this answer

answered Jan 12 at 15:00

richrowrichrow

17316

share|cite|improve this answer

share|cite|improve this answer

answered Jan 12 at 15:00

richrowrichrow

17316

answered Jan 12 at 15:00

richrowrichrow

17316

answered Jan 12 at 15:00

richrowrichrow

17316

17316

add a comment | 

add a comment | 

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Use MathJax to format equations. MathJax reference.

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