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Application of Unique Factorisation Theorem in Proof

1

0

$begingroup$

CONTEXT: Proof question made up by uni math lecturer

Suppose you have $x+y=2z$ (where $x$ and $y$ are consecutive odd primes) for some integer $z>1$, and that you need to prove that $x+y$ has at least three prime divisors (that don't have to be distinct).

Is it sufficient to say that, according to the unique factorisation theorem for integers, since $z$ can be expressed as a product of primes (and we already have the factor of $2$ which is prime), we know that $x+y$ has at least three prime divisors?

Or, would you need to do further working to show that $z=ab$ for some primes $a$ and $b$?

discrete-mathematics proof-verification proof-writing prime-numbers prime-factorization

share|cite|improve this question

edited Apr 8 at 12:55

Ruby Pa

asked Apr 8 at 12:49

Ruby PaRuby Pa

467

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is certainly not the case that $z$ is always $ab$ for some primes $a,b$.
  $endgroup$
  – Gerry Myerson
  Apr 8 at 13:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Anyway, you don't need the uniqueness part of the Unique Factorization Theorem. The existence part will do.
  $endgroup$
  – Gerry Myerson
  Apr 8 at 13:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No. Knowing that $2z$ is even only tells you that $z$ is… a natural. A correct proof would be to show that $z=ab$, where $a,b$ are at least $2$ (but don't need to be primes).
  $endgroup$
  – Yves Daoust
  Apr 8 at 13:29
  

  
  
  
  

add a comment | 

1

0

$begingroup$

CONTEXT: Proof question made up by uni math lecturer

Suppose you have $x+y=2z$ (where $x$ and $y$ are consecutive odd primes) for some integer $z>1$, and that you need to prove that $x+y$ has at least three prime divisors (that don't have to be distinct).

Is it sufficient to say that, according to the unique factorisation theorem for integers, since $z$ can be expressed as a product of primes (and we already have the factor of $2$ which is prime), we know that $x+y$ has at least three prime divisors?

Or, would you need to do further working to show that $z=ab$ for some primes $a$ and $b$?

discrete-mathematics proof-verification proof-writing prime-numbers prime-factorization

share|cite|improve this question

edited Apr 8 at 12:55

Ruby Pa

asked Apr 8 at 12:49

Ruby PaRuby Pa

467

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It is certainly not the case that $z$ is always $ab$ for some primes $a,b$.
  $endgroup$
  – Gerry Myerson
  Apr 8 at 13:14
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Anyway, you don't need the uniqueness part of the Unique Factorization Theorem. The existence part will do.
  $endgroup$
  – Gerry Myerson
  Apr 8 at 13:16
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No. Knowing that $2z$ is even only tells you that $z$ is… a natural. A correct proof would be to show that $z=ab$, where $a,b$ are at least $2$ (but don't need to be primes).
  $endgroup$
  – Yves Daoust
  Apr 8 at 13:29
  

  
  
  
  

add a comment | 

$begingroup$

CONTEXT: Proof question made up by uni math lecturer

Suppose you have $x+y=2z$ (where $x$ and $y$ are consecutive odd primes) for some integer $z>1$, and that you need to prove that $x+y$ has at least three prime divisors (that don't have to be distinct).

Is it sufficient to say that, according to the unique factorisation theorem for integers, since $z$ can be expressed as a product of primes (and we already have the factor of $2$ which is prime), we know that $x+y$ has at least three prime divisors?

Or, would you need to do further working to show that $z=ab$ for some primes $a$ and $b$?

discrete-mathematics proof-verification proof-writing prime-numbers prime-factorization

share|cite|improve this question

edited Apr 8 at 12:55

Ruby Pa

asked Apr 8 at 12:49

Ruby PaRuby Pa

467

$endgroup$

share|cite|improve this question

edited Apr 8 at 12:55

Ruby Pa

asked Apr 8 at 12:49

Ruby PaRuby Pa

467

share|cite|improve this question

edited Apr 8 at 12:55

Ruby Pa

asked Apr 8 at 12:49

Ruby PaRuby Pa

467

asked Apr 8 at 12:49

Ruby PaRuby Pa

467

asked Apr 8 at 12:49

Ruby PaRuby Pa

467

2 Answers
2

active

oldest

votes

4

$begingroup$

$z$ is strictly between $x$ and $y$, hence, not a prime.

share|cite|improve this answer

answered Apr 8 at 13:18

Gerry MyersonGerry Myerson

148k8152306

$endgroup$

add a comment | 

-1

$begingroup$

Is it sufficient to say that, according to the unique factorisation
theorem for integers, since $z$ can be expressed as a product of
primes (and we already have the factor of $2$ which is prime), we know
that $x+y$ has at least three prime divisors?

No, that's not enough. In fact, how do you know $z$ is not prime?

To prove $z$ is not prime, consider the fact that $y=x+2$, and therefore, $2z=x+x+2=2(x+1)$. Now, since $z=x+1$, can you conclude that $z$ is not prime?

share|cite|improve this answer

answered Apr 8 at 13:01

5xum5xum

92.6k394162

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get that $y=x+2$? If $x=7$, this generates $y=9$ which is not a prime.
  $endgroup$
  – Ruby Pa
  Apr 8 at 13:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RubyPa If $x,y$ are consecutive odd primes, then either $x=y+2$ or $y=x+2$. Without loss of generality, we can assume one of these options.
  $endgroup$
  – 5xum
  Apr 8 at 14:24
  

  
  
  
  

add a comment | 

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4

$begingroup$

$z$ is strictly between $x$ and $y$, hence, not a prime.

share|cite|improve this answer

answered Apr 8 at 13:18

Gerry MyersonGerry Myerson

148k8152306

$endgroup$

add a comment | 

4

$begingroup$

$z$ is strictly between $x$ and $y$, hence, not a prime.

share|cite|improve this answer

answered Apr 8 at 13:18

Gerry MyersonGerry Myerson

148k8152306

$endgroup$

add a comment | 

$begingroup$

$z$ is strictly between $x$ and $y$, hence, not a prime.

share|cite|improve this answer

answered Apr 8 at 13:18

Gerry MyersonGerry Myerson

148k8152306

$endgroup$

share|cite|improve this answer

answered Apr 8 at 13:18

Gerry MyersonGerry Myerson

148k8152306

answered Apr 8 at 13:18

Gerry MyersonGerry Myerson

148k8152306

answered Apr 8 at 13:18

Gerry MyersonGerry Myerson

148k8152306

-1

$begingroup$

Is it sufficient to say that, according to the unique factorisation
theorem for integers, since $z$ can be expressed as a product of
primes (and we already have the factor of $2$ which is prime), we know
that $x+y$ has at least three prime divisors?

No, that's not enough. In fact, how do you know $z$ is not prime?

To prove $z$ is not prime, consider the fact that $y=x+2$, and therefore, $2z=x+x+2=2(x+1)$. Now, since $z=x+1$, can you conclude that $z$ is not prime?

share|cite|improve this answer

answered Apr 8 at 13:01

5xum5xum

92.6k394162

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get that $y=x+2$? If $x=7$, this generates $y=9$ which is not a prime.
  $endgroup$
  – Ruby Pa
  Apr 8 at 13:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RubyPa If $x,y$ are consecutive odd primes, then either $x=y+2$ or $y=x+2$. Without loss of generality, we can assume one of these options.
  $endgroup$
  – 5xum
  Apr 8 at 14:24
  

  
  
  
  

add a comment | 

-1

$begingroup$

Is it sufficient to say that, according to the unique factorisation
theorem for integers, since $z$ can be expressed as a product of
primes (and we already have the factor of $2$ which is prime), we know
that $x+y$ has at least three prime divisors?

No, that's not enough. In fact, how do you know $z$ is not prime?

To prove $z$ is not prime, consider the fact that $y=x+2$, and therefore, $2z=x+x+2=2(x+1)$. Now, since $z=x+1$, can you conclude that $z$ is not prime?

share|cite|improve this answer

answered Apr 8 at 13:01

5xum5xum

92.6k394162

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How did you get that $y=x+2$? If $x=7$, this generates $y=9$ which is not a prime.
  $endgroup$
  – Ruby Pa
  Apr 8 at 13:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @RubyPa If $x,y$ are consecutive odd primes, then either $x=y+2$ or $y=x+2$. Without loss of generality, we can assume one of these options.
  $endgroup$
  – 5xum
  Apr 8 at 14:24
  

  
  
  
  

add a comment | 

$begingroup$

Is it sufficient to say that, according to the unique factorisation
theorem for integers, since $z$ can be expressed as a product of
primes (and we already have the factor of $2$ which is prime), we know
that $x+y$ has at least three prime divisors?

No, that's not enough. In fact, how do you know $z$ is not prime?

To prove $z$ is not prime, consider the fact that $y=x+2$, and therefore, $2z=x+x+2=2(x+1)$. Now, since $z=x+1$, can you conclude that $z$ is not prime?

share|cite|improve this answer

answered Apr 8 at 13:01

5xum5xum

92.6k394162

$endgroup$

share|cite|improve this answer

answered Apr 8 at 13:01

5xum5xum

92.6k394162

answered Apr 8 at 13:01

5xum5xum

92.6k394162

answered Apr 8 at 13:01

5xum5xum

92.6k394162