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Prove that if $p$ is prime and $a^7-b^3=p^2$ then $textgcd(a,b)=1$

1

$begingroup$

Prove that if $p$ is prime and $a^7-b^3=p^2$ then $textgcd(a,b)=1$.

Any help is appreciated.

elementary-number-theory proof-explanation

share|cite|improve this question

edited Apr 2 at 17:08

Maria Mazur

50.5k1361126

asked Apr 1 at 18:59

Mathstudent123Mathstudent123

93

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Highest common factor, he means that $a$ and $b$ are coprime.
  $endgroup$
  – stuart stevenson
  Apr 1 at 19:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
  $endgroup$
  – fleablood
  Apr 1 at 19:13
  

  
  
  
  

add a comment | 

1

$begingroup$

Prove that if $p$ is prime and $a^7-b^3=p^2$ then $textgcd(a,b)=1$.

Any help is appreciated.

elementary-number-theory proof-explanation

share|cite|improve this question

edited Apr 2 at 17:08

Maria Mazur

50.5k1361126

asked Apr 1 at 18:59

Mathstudent123Mathstudent123

93

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Highest common factor, he means that $a$ and $b$ are coprime.
  $endgroup$
  – stuart stevenson
  Apr 1 at 19:01
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.
  $endgroup$
  – fleablood
  Apr 1 at 19:13
  

  
  
  
  

add a comment | 

$begingroup$

Prove that if $p$ is prime and $a^7-b^3=p^2$ then $textgcd(a,b)=1$.

Any help is appreciated.

elementary-number-theory proof-explanation

share|cite|improve this question

edited Apr 2 at 17:08

Maria Mazur

50.5k1361126

asked Apr 1 at 18:59

Mathstudent123Mathstudent123

93

$endgroup$

share|cite|improve this question

edited Apr 2 at 17:08

Maria Mazur

50.5k1361126

asked Apr 1 at 18:59

Mathstudent123Mathstudent123

93

share|cite|improve this question

edited Apr 2 at 17:08

Maria Mazur

50.5k1361126

asked Apr 1 at 18:59

Mathstudent123Mathstudent123

93

edited Apr 2 at 17:08

Maria Mazur

50.5k1361126

edited Apr 2 at 17:08

Maria Mazur

50.5k1361126

asked Apr 1 at 18:59

Mathstudent123Mathstudent123

93

asked Apr 1 at 18:59

Mathstudent123Mathstudent123

93

2 Answers
2

active

oldest

votes

2

$begingroup$

If $q$ is a prime which divides both $a$ and $b$ then $q^3$ divides $p^2$. A contradiction.

So there is no prime that divides $a$ and $b$ and thus $gcd (a,b)=1$.

share|cite|improve this answer

edited Apr 1 at 19:08

answered Apr 1 at 19:04

Maria MazurMaria Mazur

50.5k1361126

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this an empty conclusion, i.e., are there perhaps no integers $a,b$ with $a^7-b^3=p^2$?
  $endgroup$
  – Dietrich Burde
  Apr 1 at 19:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DietrichBurde All integral solutions to $a^7 - b^3 = c^2$ are known; the only possible prime values of $c$ are $3$ ($1^7 - (-2)^3 = 3^2$) and $71$ ($2^7 - (-17)^3 = 71^2$).
  $endgroup$
  – FredH
  Apr 1 at 22:36
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Well the $hcf(a,b)$ will divide both $a^7$ and $b^3$ so it will divide $a^7 - b^3$ so it will divide $p^2$. But $p$ is prime so....

What are the only divisors of $p^2$? $hcf(a,b)$ will have to be one of these.

For each of these choices what do you get when you look at $frac a^7hcf(a,b) - frac b^3hcf(a,b) = frac p^2hcf(a,b)$?

Which of those choices will allow this to be true?

===

Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.

So $a^6*frac ahcf(a,b) - b^2frac bhcf(a,b) = frac p^2hfc(a,b)$. So that means $frac p^2hfc(a,b)$ is also a multiple of $hcf(a,b)$.

So $a^5 (frac ahcf(a,b) )^2 - b(frac bhcf(a,b))^2 = frac p^2hcf^2(a,b)$. So that means that $frac p^2hcf^2(a,b)$ is also a multiple of $hcf(a,b)$.

So $a^5(frac ahcf(a,b) )^3- (frac bhcf(a,b))^3 = frac p^2hcf^3(a,b)$ is an integer.

How can $frac p^2hcf^3(a,b)$ possibly be an integer?

share|cite|improve this answer

edited Apr 1 at 19:20

answered Apr 1 at 19:10

fleabloodfleablood

1

$endgroup$

add a comment | 

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2

$begingroup$

If $q$ is a prime which divides both $a$ and $b$ then $q^3$ divides $p^2$. A contradiction.

So there is no prime that divides $a$ and $b$ and thus $gcd (a,b)=1$.

share|cite|improve this answer

edited Apr 1 at 19:08

answered Apr 1 at 19:04

Maria MazurMaria Mazur

50.5k1361126

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this an empty conclusion, i.e., are there perhaps no integers $a,b$ with $a^7-b^3=p^2$?
  $endgroup$
  – Dietrich Burde
  Apr 1 at 19:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DietrichBurde All integral solutions to $a^7 - b^3 = c^2$ are known; the only possible prime values of $c$ are $3$ ($1^7 - (-2)^3 = 3^2$) and $71$ ($2^7 - (-17)^3 = 71^2$).
  $endgroup$
  – FredH
  Apr 1 at 22:36
  
  

  
  
  
  

add a comment | 

2

$begingroup$

If $q$ is a prime which divides both $a$ and $b$ then $q^3$ divides $p^2$. A contradiction.

So there is no prime that divides $a$ and $b$ and thus $gcd (a,b)=1$.

share|cite|improve this answer

edited Apr 1 at 19:08

answered Apr 1 at 19:04

Maria MazurMaria Mazur

50.5k1361126

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is this an empty conclusion, i.e., are there perhaps no integers $a,b$ with $a^7-b^3=p^2$?
  $endgroup$
  – Dietrich Burde
  Apr 1 at 19:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DietrichBurde All integral solutions to $a^7 - b^3 = c^2$ are known; the only possible prime values of $c$ are $3$ ($1^7 - (-2)^3 = 3^2$) and $71$ ($2^7 - (-17)^3 = 71^2$).
  $endgroup$
  – FredH
  Apr 1 at 22:36
  
  

  
  
  
  

add a comment | 

$begingroup$

If $q$ is a prime which divides both $a$ and $b$ then $q^3$ divides $p^2$. A contradiction.

So there is no prime that divides $a$ and $b$ and thus $gcd (a,b)=1$.

share|cite|improve this answer

edited Apr 1 at 19:08

answered Apr 1 at 19:04

Maria MazurMaria Mazur

50.5k1361126

$endgroup$

share|cite|improve this answer

edited Apr 1 at 19:08

answered Apr 1 at 19:04

Maria MazurMaria Mazur

50.5k1361126

answered Apr 1 at 19:04

Maria MazurMaria Mazur

50.5k1361126

answered Apr 1 at 19:04

Maria MazurMaria Mazur

50.5k1361126

1

$begingroup$

Well the $hcf(a,b)$ will divide both $a^7$ and $b^3$ so it will divide $a^7 - b^3$ so it will divide $p^2$. But $p$ is prime so....

What are the only divisors of $p^2$? $hcf(a,b)$ will have to be one of these.

For each of these choices what do you get when you look at $frac a^7hcf(a,b) - frac b^3hcf(a,b) = frac p^2hcf(a,b)$?

Which of those choices will allow this to be true?

===

Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.

So $a^6*frac ahcf(a,b) - b^2frac bhcf(a,b) = frac p^2hfc(a,b)$. So that means $frac p^2hfc(a,b)$ is also a multiple of $hcf(a,b)$.

So $a^5 (frac ahcf(a,b) )^2 - b(frac bhcf(a,b))^2 = frac p^2hcf^2(a,b)$. So that means that $frac p^2hcf^2(a,b)$ is also a multiple of $hcf(a,b)$.

So $a^5(frac ahcf(a,b) )^3- (frac bhcf(a,b))^3 = frac p^2hcf^3(a,b)$ is an integer.

How can $frac p^2hcf^3(a,b)$ possibly be an integer?

share|cite|improve this answer

edited Apr 1 at 19:20

answered Apr 1 at 19:10

fleabloodfleablood

1

$endgroup$

add a comment | 

1

$begingroup$

Well the $hcf(a,b)$ will divide both $a^7$ and $b^3$ so it will divide $a^7 - b^3$ so it will divide $p^2$. But $p$ is prime so....

What are the only divisors of $p^2$? $hcf(a,b)$ will have to be one of these.

For each of these choices what do you get when you look at $frac a^7hcf(a,b) - frac b^3hcf(a,b) = frac p^2hcf(a,b)$?

Which of those choices will allow this to be true?

===

Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.

So $a^6*frac ahcf(a,b) - b^2frac bhcf(a,b) = frac p^2hfc(a,b)$. So that means $frac p^2hfc(a,b)$ is also a multiple of $hcf(a,b)$.

So $a^5 (frac ahcf(a,b) )^2 - b(frac bhcf(a,b))^2 = frac p^2hcf^2(a,b)$. So that means that $frac p^2hcf^2(a,b)$ is also a multiple of $hcf(a,b)$.

So $a^5(frac ahcf(a,b) )^3- (frac bhcf(a,b))^3 = frac p^2hcf^3(a,b)$ is an integer.

How can $frac p^2hcf^3(a,b)$ possibly be an integer?

share|cite|improve this answer

edited Apr 1 at 19:20

answered Apr 1 at 19:10

fleabloodfleablood

1

$endgroup$

add a comment | 

$begingroup$

Well the $hcf(a,b)$ will divide both $a^7$ and $b^3$ so it will divide $a^7 - b^3$ so it will divide $p^2$. But $p$ is prime so....

What are the only divisors of $p^2$? $hcf(a,b)$ will have to be one of these.

For each of these choices what do you get when you look at $frac a^7hcf(a,b) - frac b^3hcf(a,b) = frac p^2hcf(a,b)$?

Which of those choices will allow this to be true?

===

Are you familiar with Bezouts Lemma? $ma + nb=k$ will mean that $k$ is a multiple of $hcf(a,b)$. So $a^7 - b^3= p^2$ means $p^2$ is a multiple of $hcf(a,b)$.

So $a^6*frac ahcf(a,b) - b^2frac bhcf(a,b) = frac p^2hfc(a,b)$. So that means $frac p^2hfc(a,b)$ is also a multiple of $hcf(a,b)$.

So $a^5 (frac ahcf(a,b) )^2 - b(frac bhcf(a,b))^2 = frac p^2hcf^2(a,b)$. So that means that $frac p^2hcf^2(a,b)$ is also a multiple of $hcf(a,b)$.

So $a^5(frac ahcf(a,b) )^3- (frac bhcf(a,b))^3 = frac p^2hcf^3(a,b)$ is an integer.

How can $frac p^2hcf^3(a,b)$ possibly be an integer?

share|cite|improve this answer

edited Apr 1 at 19:20

answered Apr 1 at 19:10

fleabloodfleablood

1

$endgroup$

share|cite|improve this answer

edited Apr 1 at 19:20

answered Apr 1 at 19:10

fleabloodfleablood

1

answered Apr 1 at 19:10

fleabloodfleablood

1

answered Apr 1 at 19:10

fleabloodfleablood

1