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Brexit - No Deal Rejection

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Assume $p, q$ are prime numbers. Decide whether it is possible for $12p + 12q$ to be equal to$ pq$ .

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1

$begingroup$

So far I have managed the following

$2(6p + 6q) = pq$

$p$ and $q$ cannot both be even as the only prime number that is even is $2$ and that does not prove this equation.

$p$ and $q$ cannot both be odd as $2$ * anything is even

So that means that one of $p$, $q$ must be odd and the other must be even.

So let $p = 2k$ and $q = 2L+1$

$2(12k + 12L + 6) = (2k) * (2L + 1) = 24k + 24L + 12 = 4kL + 2k$

Have I done this correctly and do I have sufficient proof to claim that $12p + 12q$ cannot be equal to $pq$ given that both $p$ and $q$ are both prime numbers?

elementary-number-theory

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edited Mar 11 at 9:16

dmtri

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asked Mar 11 at 8:25

AkaionAkaion

203

New contributor

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

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add a comment | 

1

$begingroup$

So far I have managed the following

$2(6p + 6q) = pq$

$p$ and $q$ cannot both be even as the only prime number that is even is $2$ and that does not prove this equation.

$p$ and $q$ cannot both be odd as $2$ * anything is even

So that means that one of $p$, $q$ must be odd and the other must be even.

So let $p = 2k$ and $q = 2L+1$

$2(12k + 12L + 6) = (2k) * (2L + 1) = 24k + 24L + 12 = 4kL + 2k$

Have I done this correctly and do I have sufficient proof to claim that $12p + 12q$ cannot be equal to $pq$ given that both $p$ and $q$ are both prime numbers?

elementary-number-theory

share|cite|improve this question

edited Mar 11 at 9:16

dmtri

1,6172521

asked Mar 11 at 8:25

AkaionAkaion

203

New contributor

Akaion 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 | 

$begingroup$

So far I have managed the following

$2(6p + 6q) = pq$

$p$ and $q$ cannot both be even as the only prime number that is even is $2$ and that does not prove this equation.

$p$ and $q$ cannot both be odd as $2$ * anything is even

So that means that one of $p$, $q$ must be odd and the other must be even.

So let $p = 2k$ and $q = 2L+1$

$2(12k + 12L + 6) = (2k) * (2L + 1) = 24k + 24L + 12 = 4kL + 2k$

Have I done this correctly and do I have sufficient proof to claim that $12p + 12q$ cannot be equal to $pq$ given that both $p$ and $q$ are both prime numbers?

elementary-number-theory

share|cite|improve this question

edited Mar 11 at 9:16

dmtri

1,6172521

asked Mar 11 at 8:25

AkaionAkaion

203

New contributor

Akaion 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 11 at 9:16

dmtri

1,6172521

asked Mar 11 at 8:25

AkaionAkaion

203

New contributor

Akaion 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 11 at 9:16

dmtri

1,6172521

asked Mar 11 at 8:25

AkaionAkaion

203

New contributor

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

edited Mar 11 at 9:16

dmtri

1,6172521

edited Mar 11 at 9:16

dmtri

1,6172521

asked Mar 11 at 8:25

AkaionAkaion

203

New contributor

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

asked Mar 11 at 8:25

AkaionAkaion

203

3 Answers
3

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3

$begingroup$

While a good answer has been given, and your own calculations are not wrong, they are just short of the goal, as it isn't apparent why your last result

$$24k+24L+12=4kL+2k$$

is any kind of contradiction. That only becomes clear once you realize that you set $p=2k$, and with $p$ being a prime, $k=1$ must follow. Then your last result becomes

$$24+24L+12=4L+2$$

and this is clearly impossible. One argument is that the left hand side is clearly bigger then the right hand side, another would be that the left hand side is divisible by 4 while the right hand side isn't.

To improve your approach for readability, I suggest just mentioning that $p$ and $q$ can't both be odd first (as you did), this means w.l.o.g $p=2$. Then the equation becomes
$24+12q=2q$, which is impossible due to the "left hand side bigger then right hand side" approach. No need to introduce $k,L$.

share|cite|improve this answer

answered Mar 11 at 8:38

IngixIngix

4,857159

$endgroup$

add a comment | 

4

$begingroup$

It is obviously false because $pq$ has only two prime factors, but $12(p+q)$ has at least $4$ prime factors.

share|cite|improve this answer

answered Mar 11 at 8:27

Peter ForemanPeter Foreman

3,7921216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You mean proper divisors or prime factors? If $p=2, q=3$, then $12(p+q)$ has only $3$ distinct prime factors, though of course it has at least $4$ proper divisors.
  $endgroup$
  – Macavity
  Mar 11 at 8:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't say distinct prime factors, I said prime factors. $6$ has two prime factors whereas $60$ has four prime factors.
  $endgroup$
  – Peter Foreman
  Mar 11 at 8:36
  

  
  
  
  

add a comment | 

0

$begingroup$

0) Let $p,q$ be prime numbers.

1) $p,q not =2$.

Then $pq$ is odd, while $12(p+q)$ is even, ruled out.

2) Assume $p=q=2$.

Then $pq =4$, ruled out (why?)

3)Assume $p=2$, $q not =2$.

Then $q= 6(p+q)$, not a prime , ruled out.

Hence?

share|cite|improve this answer

answered Mar 11 at 9:25

Peter SzilasPeter Szilas

11.5k2822

$endgroup$

add a comment | 

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3

$begingroup$

While a good answer has been given, and your own calculations are not wrong, they are just short of the goal, as it isn't apparent why your last result

$$24k+24L+12=4kL+2k$$

is any kind of contradiction. That only becomes clear once you realize that you set $p=2k$, and with $p$ being a prime, $k=1$ must follow. Then your last result becomes

$$24+24L+12=4L+2$$

and this is clearly impossible. One argument is that the left hand side is clearly bigger then the right hand side, another would be that the left hand side is divisible by 4 while the right hand side isn't.

To improve your approach for readability, I suggest just mentioning that $p$ and $q$ can't both be odd first (as you did), this means w.l.o.g $p=2$. Then the equation becomes
$24+12q=2q$, which is impossible due to the "left hand side bigger then right hand side" approach. No need to introduce $k,L$.

share|cite|improve this answer

answered Mar 11 at 8:38

IngixIngix

4,857159

$endgroup$

add a comment | 

3

$begingroup$

While a good answer has been given, and your own calculations are not wrong, they are just short of the goal, as it isn't apparent why your last result

$$24k+24L+12=4kL+2k$$

is any kind of contradiction. That only becomes clear once you realize that you set $p=2k$, and with $p$ being a prime, $k=1$ must follow. Then your last result becomes

$$24+24L+12=4L+2$$

and this is clearly impossible. One argument is that the left hand side is clearly bigger then the right hand side, another would be that the left hand side is divisible by 4 while the right hand side isn't.

To improve your approach for readability, I suggest just mentioning that $p$ and $q$ can't both be odd first (as you did), this means w.l.o.g $p=2$. Then the equation becomes
$24+12q=2q$, which is impossible due to the "left hand side bigger then right hand side" approach. No need to introduce $k,L$.

share|cite|improve this answer

answered Mar 11 at 8:38

IngixIngix

4,857159

$endgroup$

add a comment | 

$begingroup$

While a good answer has been given, and your own calculations are not wrong, they are just short of the goal, as it isn't apparent why your last result

$$24k+24L+12=4kL+2k$$

is any kind of contradiction. That only becomes clear once you realize that you set $p=2k$, and with $p$ being a prime, $k=1$ must follow. Then your last result becomes

$$24+24L+12=4L+2$$

and this is clearly impossible. One argument is that the left hand side is clearly bigger then the right hand side, another would be that the left hand side is divisible by 4 while the right hand side isn't.

To improve your approach for readability, I suggest just mentioning that $p$ and $q$ can't both be odd first (as you did), this means w.l.o.g $p=2$. Then the equation becomes
$24+12q=2q$, which is impossible due to the "left hand side bigger then right hand side" approach. No need to introduce $k,L$.

share|cite|improve this answer

answered Mar 11 at 8:38

IngixIngix

4,857159

$endgroup$

share|cite|improve this answer

answered Mar 11 at 8:38

IngixIngix

4,857159

answered Mar 11 at 8:38

IngixIngix

4,857159

answered Mar 11 at 8:38

IngixIngix

4,857159

4

$begingroup$

It is obviously false because $pq$ has only two prime factors, but $12(p+q)$ has at least $4$ prime factors.

share|cite|improve this answer

answered Mar 11 at 8:27

Peter ForemanPeter Foreman

3,7921216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You mean proper divisors or prime factors? If $p=2, q=3$, then $12(p+q)$ has only $3$ distinct prime factors, though of course it has at least $4$ proper divisors.
  $endgroup$
  – Macavity
  Mar 11 at 8:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't say distinct prime factors, I said prime factors. $6$ has two prime factors whereas $60$ has four prime factors.
  $endgroup$
  – Peter Foreman
  Mar 11 at 8:36
  

  
  
  
  

add a comment | 

4

$begingroup$

It is obviously false because $pq$ has only two prime factors, but $12(p+q)$ has at least $4$ prime factors.

share|cite|improve this answer

answered Mar 11 at 8:27

Peter ForemanPeter Foreman

3,7921216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You mean proper divisors or prime factors? If $p=2, q=3$, then $12(p+q)$ has only $3$ distinct prime factors, though of course it has at least $4$ proper divisors.
  $endgroup$
  – Macavity
  Mar 11 at 8:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I didn't say distinct prime factors, I said prime factors. $6$ has two prime factors whereas $60$ has four prime factors.
  $endgroup$
  – Peter Foreman
  Mar 11 at 8:36
  

  
  
  
  

add a comment | 

$begingroup$

It is obviously false because $pq$ has only two prime factors, but $12(p+q)$ has at least $4$ prime factors.

share|cite|improve this answer

answered Mar 11 at 8:27

Peter ForemanPeter Foreman

3,7921216

$endgroup$

share|cite|improve this answer

answered Mar 11 at 8:27

Peter ForemanPeter Foreman

3,7921216

answered Mar 11 at 8:27

Peter ForemanPeter Foreman

3,7921216

answered Mar 11 at 8:27

Peter ForemanPeter Foreman

3,7921216

0

$begingroup$

0) Let $p,q$ be prime numbers.

1) $p,q not =2$.

Then $pq$ is odd, while $12(p+q)$ is even, ruled out.

2) Assume $p=q=2$.

Then $pq =4$, ruled out (why?)

3)Assume $p=2$, $q not =2$.

Then $q= 6(p+q)$, not a prime , ruled out.

Hence?

share|cite|improve this answer

answered Mar 11 at 9:25

Peter SzilasPeter Szilas

11.5k2822

$endgroup$

add a comment | 

0

$begingroup$

0) Let $p,q$ be prime numbers.

1) $p,q not =2$.

Then $pq$ is odd, while $12(p+q)$ is even, ruled out.

2) Assume $p=q=2$.

Then $pq =4$, ruled out (why?)

3)Assume $p=2$, $q not =2$.

Then $q= 6(p+q)$, not a prime , ruled out.

Hence?

share|cite|improve this answer

answered Mar 11 at 9:25

Peter SzilasPeter Szilas

11.5k2822

$endgroup$

add a comment | 

$begingroup$

0) Let $p,q$ be prime numbers.

1) $p,q not =2$.

Then $pq$ is odd, while $12(p+q)$ is even, ruled out.

2) Assume $p=q=2$.

Then $pq =4$, ruled out (why?)

3)Assume $p=2$, $q not =2$.

Then $q= 6(p+q)$, not a prime , ruled out.

Hence?

share|cite|improve this answer

answered Mar 11 at 9:25

Peter SzilasPeter Szilas

11.5k2822

$endgroup$

share|cite|improve this answer

answered Mar 11 at 9:25

Peter SzilasPeter Szilas

11.5k2822

answered Mar 11 at 9:25

Peter SzilasPeter Szilas

11.5k2822

answered Mar 11 at 9:25

Peter SzilasPeter Szilas

11.5k2822