Number of odd and even natural solutions [closed]

0

$begingroup$

I'm trying to prove or disprove the following statement:

Number of even natural solutions of $x_1+x_2+x_3+x _4=14$ is equal to the number of odd natural solutions of $x_1+x_2+x_3+x _4=14$.

I think that they should be equal because of symmetry but it does not feel true. How to approach this issue?

combinatorics

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asked Jan 11 at 19:57

kickstartkickstart

1988

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closed as off-topic by amWhy, Leucippus, Abcd, eyeballfrog, Cesareo Jan 12 at 8:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Leucippus, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Did you already try to compute the actual numbers separately?
  $endgroup$
  – Dietrich Burde
  Jan 11 at 19:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With "even natural solutions" do you mean that each x_i is even?
  $endgroup$
  – user627482
  Jan 11 at 20:03
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user627482 I guess. I just copied from the my book.
  $endgroup$
  – kickstart
  Jan 11 at 20:07
  

  
  
  
  

add a comment | 

0

$begingroup$

I'm trying to prove or disprove the following statement:

Number of even natural solutions of $x_1+x_2+x_3+x _4=14$ is equal to the number of odd natural solutions of $x_1+x_2+x_3+x _4=14$.

I think that they should be equal because of symmetry but it does not feel true. How to approach this issue?

combinatorics

share|cite|improve this question

asked Jan 11 at 19:57

kickstartkickstart

1988

$endgroup$

closed as off-topic by amWhy, Leucippus, Abcd, eyeballfrog, Cesareo Jan 12 at 8:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Leucippus, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Did you already try to compute the actual numbers separately?
  $endgroup$
  – Dietrich Burde
  Jan 11 at 19:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With "even natural solutions" do you mean that each x_i is even?
  $endgroup$
  – user627482
  Jan 11 at 20:03
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user627482 I guess. I just copied from the my book.
  $endgroup$
  – kickstart
  Jan 11 at 20:07
  

  
  
  
  

add a comment | 

0

0

0

$begingroup$

I'm trying to prove or disprove the following statement:

Number of even natural solutions of $x_1+x_2+x_3+x _4=14$ is equal to the number of odd natural solutions of $x_1+x_2+x_3+x _4=14$.

I think that they should be equal because of symmetry but it does not feel true. How to approach this issue?

combinatorics

share|cite|improve this question

asked Jan 11 at 19:57

kickstartkickstart

1988

$endgroup$

I'm trying to prove or disprove the following statement:

Number of even natural solutions of $x_1+x_2+x_3+x _4=14$ is equal to the number of odd natural solutions of $x_1+x_2+x_3+x _4=14$.

I think that they should be equal because of symmetry but it does not feel true. How to approach this issue?

combinatorics

combinatorics

share|cite|improve this question

asked Jan 11 at 19:57

kickstartkickstart

1988

share|cite|improve this question

asked Jan 11 at 19:57

kickstartkickstart

1988

share|cite|improve this question

share|cite|improve this question

asked Jan 11 at 19:57

kickstartkickstart

1988

asked Jan 11 at 19:57

kickstartkickstart

1988

asked Jan 11 at 19:57

kickstartkickstart

1988

1988

closed as off-topic by amWhy, Leucippus, Abcd, eyeballfrog, Cesareo Jan 12 at 8:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Leucippus, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by amWhy, Leucippus, Abcd, eyeballfrog, Cesareo Jan 12 at 8:04

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Leucippus, Cesareo

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Did you already try to compute the actual numbers separately?
  $endgroup$
  – Dietrich Burde
  Jan 11 at 19:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With "even natural solutions" do you mean that each x_i is even?
  $endgroup$
  – user627482
  Jan 11 at 20:03
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user627482 I guess. I just copied from the my book.
  $endgroup$
  – kickstart
  Jan 11 at 20:07
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Did you already try to compute the actual numbers separately?
  $endgroup$
  – Dietrich Burde
  Jan 11 at 19:59
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  With "even natural solutions" do you mean that each x_i is even?
  $endgroup$
  – user627482
  Jan 11 at 20:03
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user627482 I guess. I just copied from the my book.
  $endgroup$
  – kickstart
  Jan 11 at 20:07
  

  
  
  
  

1

1

$begingroup$
Did you already try to compute the actual numbers separately?
$endgroup$
– Dietrich Burde
Jan 11 at 19:59

$begingroup$
Did you already try to compute the actual numbers separately?
$endgroup$
– Dietrich Burde
Jan 11 at 19:59

$begingroup$
With "even natural solutions" do you mean that each x_i is even?
$endgroup$
– user627482
Jan 11 at 20:03

$begingroup$
With "even natural solutions" do you mean that each x_i is even?
$endgroup$
– user627482
Jan 11 at 20:03

$begingroup$
@user627482 I guess. I just copied from the my book.
$endgroup$
– kickstart
Jan 11 at 20:07

$begingroup$
@user627482 I guess. I just copied from the my book.
$endgroup$
– kickstart
Jan 11 at 20:07

add a comment | 

2 Answers
2

active

oldest

votes

7

$begingroup$

By letting $x_1'=x_1+1$, $x_2'=x_2+1$, $x_3'=x_1-1$ and $x_4'=x_4-1$, we can produce an odd solution $(x_1',x_2',x_3',x_4')$ from an even solution $(x_1,x_2,x_3,x_4)$. Since this mapping is injective, we know that there are more or equal number of odd solutions than that of even solutions. But this mapping is not surjective since e.g. $(3,1,5,5)$ is not in the image. Thus we conclude there are more odd solutions than even solutions.

share|cite|improve this answer

edited Jan 11 at 21:46

answered Jan 11 at 20:09

SongSong

12.2k630

$endgroup$

add a comment | 

2

$begingroup$

Assuming 'natural' means 'positive integer':

The total number of solutions is:$$binom{14-1}{4-1}=286$$

Even solutions are positive integer solutions to $y_1+y_2+y_3+y_4=7$, where $y_i=frac{x_i}{2}$

The number of even solutions is:$$binom{7-1}{4-1}=20$$

Odd solutions are positive integer solutions to $z_1+z_2+z_3+z_4=9$, where $z_i=frac{x_i+1}{2}$

The number of odd solutions is:$$binom{9-1}{4-1}=56$$

share|cite|improve this answer

answered Jan 11 at 21:03

Daniel MathiasDaniel Mathias

1,10418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the odd part, why is it $9$ and not $7$?
  $endgroup$
  – kickstart
  Jan 12 at 11:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because it is $(14+4)/2$
  $endgroup$
  – Daniel Mathias
  Jan 12 at 11:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you add $4$?
  $endgroup$
  – kickstart
  Jan 12 at 14:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
  $endgroup$
  – Daniel Mathias
  Jan 12 at 15:56
  
  
  

  
  
  
  

add a comment | 

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

7

$begingroup$

By letting $x_1'=x_1+1$, $x_2'=x_2+1$, $x_3'=x_1-1$ and $x_4'=x_4-1$, we can produce an odd solution $(x_1',x_2',x_3',x_4')$ from an even solution $(x_1,x_2,x_3,x_4)$. Since this mapping is injective, we know that there are more or equal number of odd solutions than that of even solutions. But this mapping is not surjective since e.g. $(3,1,5,5)$ is not in the image. Thus we conclude there are more odd solutions than even solutions.

share|cite|improve this answer

edited Jan 11 at 21:46

answered Jan 11 at 20:09

SongSong

12.2k630

$endgroup$

add a comment | 

7

$begingroup$

By letting $x_1'=x_1+1$, $x_2'=x_2+1$, $x_3'=x_1-1$ and $x_4'=x_4-1$, we can produce an odd solution $(x_1',x_2',x_3',x_4')$ from an even solution $(x_1,x_2,x_3,x_4)$. Since this mapping is injective, we know that there are more or equal number of odd solutions than that of even solutions. But this mapping is not surjective since e.g. $(3,1,5,5)$ is not in the image. Thus we conclude there are more odd solutions than even solutions.

share|cite|improve this answer

edited Jan 11 at 21:46

answered Jan 11 at 20:09

SongSong

12.2k630

$endgroup$

add a comment | 

7

7

7

$begingroup$

By letting $x_1'=x_1+1$, $x_2'=x_2+1$, $x_3'=x_1-1$ and $x_4'=x_4-1$, we can produce an odd solution $(x_1',x_2',x_3',x_4')$ from an even solution $(x_1,x_2,x_3,x_4)$. Since this mapping is injective, we know that there are more or equal number of odd solutions than that of even solutions. But this mapping is not surjective since e.g. $(3,1,5,5)$ is not in the image. Thus we conclude there are more odd solutions than even solutions.

share|cite|improve this answer

edited Jan 11 at 21:46

answered Jan 11 at 20:09

SongSong

12.2k630

$endgroup$

By letting $x_1'=x_1+1$, $x_2'=x_2+1$, $x_3'=x_1-1$ and $x_4'=x_4-1$, we can produce an odd solution $(x_1',x_2',x_3',x_4')$ from an even solution $(x_1,x_2,x_3,x_4)$. Since this mapping is injective, we know that there are more or equal number of odd solutions than that of even solutions. But this mapping is not surjective since e.g. $(3,1,5,5)$ is not in the image. Thus we conclude there are more odd solutions than even solutions.

share|cite|improve this answer

edited Jan 11 at 21:46

answered Jan 11 at 20:09

SongSong

12.2k630

share|cite|improve this answer

share|cite|improve this answer

edited Jan 11 at 21:46

edited Jan 11 at 21:46

edited Jan 11 at 21:46

answered Jan 11 at 20:09

SongSong

12.2k630

answered Jan 11 at 20:09

SongSong

12.2k630

answered Jan 11 at 20:09

SongSong

12.2k630

12.2k630

add a comment | 

add a comment | 

2

$begingroup$

Assuming 'natural' means 'positive integer':

The total number of solutions is:$$binom{14-1}{4-1}=286$$

Even solutions are positive integer solutions to $y_1+y_2+y_3+y_4=7$, where $y_i=frac{x_i}{2}$

The number of even solutions is:$$binom{7-1}{4-1}=20$$

Odd solutions are positive integer solutions to $z_1+z_2+z_3+z_4=9$, where $z_i=frac{x_i+1}{2}$

The number of odd solutions is:$$binom{9-1}{4-1}=56$$

share|cite|improve this answer

answered Jan 11 at 21:03

Daniel MathiasDaniel Mathias

1,10418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the odd part, why is it $9$ and not $7$?
  $endgroup$
  – kickstart
  Jan 12 at 11:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because it is $(14+4)/2$
  $endgroup$
  – Daniel Mathias
  Jan 12 at 11:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you add $4$?
  $endgroup$
  – kickstart
  Jan 12 at 14:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
  $endgroup$
  – Daniel Mathias
  Jan 12 at 15:56
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Assuming 'natural' means 'positive integer':

The total number of solutions is:$$binom{14-1}{4-1}=286$$

Even solutions are positive integer solutions to $y_1+y_2+y_3+y_4=7$, where $y_i=frac{x_i}{2}$

The number of even solutions is:$$binom{7-1}{4-1}=20$$

Odd solutions are positive integer solutions to $z_1+z_2+z_3+z_4=9$, where $z_i=frac{x_i+1}{2}$

The number of odd solutions is:$$binom{9-1}{4-1}=56$$

share|cite|improve this answer

answered Jan 11 at 21:03

Daniel MathiasDaniel Mathias

1,10418

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the odd part, why is it $9$ and not $7$?
  $endgroup$
  – kickstart
  Jan 12 at 11:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because it is $(14+4)/2$
  $endgroup$
  – Daniel Mathias
  Jan 12 at 11:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you add $4$?
  $endgroup$
  – kickstart
  Jan 12 at 14:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
  $endgroup$
  – Daniel Mathias
  Jan 12 at 15:56
  
  
  

  
  
  
  

add a comment | 

2

2

2

$begingroup$

Assuming 'natural' means 'positive integer':

The total number of solutions is:$$binom{14-1}{4-1}=286$$

Even solutions are positive integer solutions to $y_1+y_2+y_3+y_4=7$, where $y_i=frac{x_i}{2}$

The number of even solutions is:$$binom{7-1}{4-1}=20$$

Odd solutions are positive integer solutions to $z_1+z_2+z_3+z_4=9$, where $z_i=frac{x_i+1}{2}$

The number of odd solutions is:$$binom{9-1}{4-1}=56$$

share|cite|improve this answer

answered Jan 11 at 21:03

Daniel MathiasDaniel Mathias

1,10418

$endgroup$

Assuming 'natural' means 'positive integer':

The total number of solutions is:$$binom{14-1}{4-1}=286$$

Even solutions are positive integer solutions to $y_1+y_2+y_3+y_4=7$, where $y_i=frac{x_i}{2}$

The number of even solutions is:$$binom{7-1}{4-1}=20$$

Odd solutions are positive integer solutions to $z_1+z_2+z_3+z_4=9$, where $z_i=frac{x_i+1}{2}$

The number of odd solutions is:$$binom{9-1}{4-1}=56$$

share|cite|improve this answer

answered Jan 11 at 21:03

Daniel MathiasDaniel Mathias

1,10418

share|cite|improve this answer

share|cite|improve this answer

answered Jan 11 at 21:03

Daniel MathiasDaniel Mathias

1,10418

answered Jan 11 at 21:03

Daniel MathiasDaniel Mathias

1,10418

answered Jan 11 at 21:03

Daniel MathiasDaniel Mathias

1,10418

1,10418

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the odd part, why is it $9$ and not $7$?
  $endgroup$
  – kickstart
  Jan 12 at 11:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because it is $(14+4)/2$
  $endgroup$
  – Daniel Mathias
  Jan 12 at 11:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you add $4$?
  $endgroup$
  – kickstart
  Jan 12 at 14:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
  $endgroup$
  – Daniel Mathias
  Jan 12 at 15:56
  
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  For the odd part, why is it $9$ and not $7$?
  $endgroup$
  – kickstart
  Jan 12 at 11:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because it is $(14+4)/2$
  $endgroup$
  – Daniel Mathias
  Jan 12 at 11:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why do you add $4$?
  $endgroup$
  – kickstart
  Jan 12 at 14:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
  $endgroup$
  – Daniel Mathias
  Jan 12 at 15:56
  
  
  

  
  
  
  

$begingroup$
For the odd part, why is it $9$ and not $7$?
$endgroup$
– kickstart
Jan 12 at 11:45

$begingroup$
For the odd part, why is it $9$ and not $7$?
$endgroup$
– kickstart
Jan 12 at 11:45

$begingroup$
Because it is $(14+4)/2$
$endgroup$
– Daniel Mathias
Jan 12 at 11:50

$begingroup$
Because it is $(14+4)/2$
$endgroup$
– Daniel Mathias
Jan 12 at 11:50

$begingroup$
Why do you add $4$?
$endgroup$
– kickstart
Jan 12 at 14:29

$begingroup$
Why do you add $4$?
$endgroup$
– kickstart
Jan 12 at 14:29

$begingroup$
@kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
$endgroup$
– Daniel Mathias
Jan 12 at 15:56

$begingroup$
@kickstart Starting with $x_1+x_2+x_3+x _4=14$ with all $x_i$ odd, we add $1$ to each $x_i$ to get $(x_1+1)+(x_2+1)+(x_3+1)+(x _4+1)=14+4$. Divide by $2$ and let $z_i=frac{x_i}{2}$ to get $z_1+z_2+z_3+z_4=9$
$endgroup$
– Daniel Mathias
Jan 12 at 15:56

add a comment | 

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