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Solving Diophantine system of degree three that contains 4 equations with 16 unknowns over $mathbb Z_n$ [closed]

-1

$begingroup$

The following Diophantine system

$25333-123,a_2a_1-a_2a_1a_3-478,a_2b_1-a_
2b_1c_3-223,c_2a_1-c_2a_1b_3-589,c_
2b_1-c_2b_1d_3-a_4=0
$
$
29151-123,b_2a_1-b_2a_1a_3-478,b_2b_1-b_
2b_1c_3-223,d_2a_1-d_2a_1b_3-589,d_
2b_1-d_2b_1d_3-b_4=0
$
$
54507-123,a_2c_1-a_2c_1a_3-478,a_2d_1-a_
2d_1c_3-223,c_2c_1-c_2c_1b_3-589,c_
2d_1-c_2d_1d_3-c_4=0
$
$
62645-123,b_2c_1-b_2c_1a_3-478,b_2d_1-b_
2d_1c_3-223,d_2c_1-d_2c_1b_3-589,d_
2d_1-d_2d_1d_3-d_4=0
$

has a solution

$
[a_1=1,b_1=2,c_1=3,d_1=4,a_2=5,b_2=7,c_2=
13,d_2=14,a_3=11,b_3=21,c_3=31,d_3=41,a_4=21,b
_4=31,c_4=41,d_4=51]
$

It is known that in 1900, David Hilbert proposed the solvability of all Diophantine equations as the tenth of his fundamental problems. In 1970, Yuri Matiyasevich solved it negatively, by proving that a general algorithm for solving all Diophantine equations cannot exist.

Is there any mathematical way to find the solution of the above system without doing loop over $mathbb Z_n$?

How many solutions in $mathbb Z_n$?

number-theory systems-of-equations diophantine-equations

share|cite|improve this question

edited Mar 20 at 16:13

Ahmad Al-Dweik

asked Mar 20 at 7:46

Ahmad Al-DweikAhmad Al-Dweik

11

$endgroup$

closed as off-topic by José Carlos Santos, Thomas Shelby, Eevee Trainer, egreg, Parcly Taxel Mar 21 at 0:43

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." – José Carlos Santos, Thomas Shelby, Eevee Trainer, egreg, Parcly Taxel

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

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You might consider formatting this to be legible: as it stands, I suspect most people won't take the trouble to try and figure out what you're asking.
  $endgroup$
  – postmortes
  Mar 20 at 8:44
  

  
  
  
  

add a comment | 

-1

$begingroup$

The following Diophantine system

$25333-123,a_2a_1-a_2a_1a_3-478,a_2b_1-a_
2b_1c_3-223,c_2a_1-c_2a_1b_3-589,c_
2b_1-c_2b_1d_3-a_4=0
$
$
29151-123,b_2a_1-b_2a_1a_3-478,b_2b_1-b_
2b_1c_3-223,d_2a_1-d_2a_1b_3-589,d_
2b_1-d_2b_1d_3-b_4=0
$
$
54507-123,a_2c_1-a_2c_1a_3-478,a_2d_1-a_
2d_1c_3-223,c_2c_1-c_2c_1b_3-589,c_
2d_1-c_2d_1d_3-c_4=0
$
$
62645-123,b_2c_1-b_2c_1a_3-478,b_2d_1-b_
2d_1c_3-223,d_2c_1-d_2c_1b_3-589,d_
2d_1-d_2d_1d_3-d_4=0
$

has a solution

$
[a_1=1,b_1=2,c_1=3,d_1=4,a_2=5,b_2=7,c_2=
13,d_2=14,a_3=11,b_3=21,c_3=31,d_3=41,a_4=21,b
_4=31,c_4=41,d_4=51]
$

It is known that in 1900, David Hilbert proposed the solvability of all Diophantine equations as the tenth of his fundamental problems. In 1970, Yuri Matiyasevich solved it negatively, by proving that a general algorithm for solving all Diophantine equations cannot exist.

Is there any mathematical way to find the solution of the above system without doing loop over $mathbb Z_n$?

How many solutions in $mathbb Z_n$?

number-theory systems-of-equations diophantine-equations

share|cite|improve this question

edited Mar 20 at 16:13

Ahmad Al-Dweik

asked Mar 20 at 7:46

Ahmad Al-DweikAhmad Al-Dweik

11

$endgroup$

closed as off-topic by José Carlos Santos, Thomas Shelby, Eevee Trainer, egreg, Parcly Taxel Mar 21 at 0:43

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." – José Carlos Santos, Thomas Shelby, Eevee Trainer, egreg, Parcly Taxel

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

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You might consider formatting this to be legible: as it stands, I suspect most people won't take the trouble to try and figure out what you're asking.
  $endgroup$
  – postmortes
  Mar 20 at 8:44
  

  
  
  
  

add a comment | 

$begingroup$

The following Diophantine system

$25333-123,a_2a_1-a_2a_1a_3-478,a_2b_1-a_
2b_1c_3-223,c_2a_1-c_2a_1b_3-589,c_
2b_1-c_2b_1d_3-a_4=0
$
$
29151-123,b_2a_1-b_2a_1a_3-478,b_2b_1-b_
2b_1c_3-223,d_2a_1-d_2a_1b_3-589,d_
2b_1-d_2b_1d_3-b_4=0
$
$
54507-123,a_2c_1-a_2c_1a_3-478,a_2d_1-a_
2d_1c_3-223,c_2c_1-c_2c_1b_3-589,c_
2d_1-c_2d_1d_3-c_4=0
$
$
62645-123,b_2c_1-b_2c_1a_3-478,b_2d_1-b_
2d_1c_3-223,d_2c_1-d_2c_1b_3-589,d_
2d_1-d_2d_1d_3-d_4=0
$

has a solution

$
[a_1=1,b_1=2,c_1=3,d_1=4,a_2=5,b_2=7,c_2=
13,d_2=14,a_3=11,b_3=21,c_3=31,d_3=41,a_4=21,b
_4=31,c_4=41,d_4=51]
$

It is known that in 1900, David Hilbert proposed the solvability of all Diophantine equations as the tenth of his fundamental problems. In 1970, Yuri Matiyasevich solved it negatively, by proving that a general algorithm for solving all Diophantine equations cannot exist.

Is there any mathematical way to find the solution of the above system without doing loop over $mathbb Z_n$?

How many solutions in $mathbb Z_n$?

number-theory systems-of-equations diophantine-equations

share|cite|improve this question

edited Mar 20 at 16:13

Ahmad Al-Dweik

asked Mar 20 at 7:46

Ahmad Al-DweikAhmad Al-Dweik

11

$endgroup$

share|cite|improve this question

edited Mar 20 at 16:13

Ahmad Al-Dweik

asked Mar 20 at 7:46

Ahmad Al-DweikAhmad Al-Dweik

11

share|cite|improve this question

edited Mar 20 at 16:13

Ahmad Al-Dweik

asked Mar 20 at 7:46

Ahmad Al-DweikAhmad Al-Dweik

11

asked Mar 20 at 7:46

Ahmad Al-DweikAhmad Al-Dweik

11

asked Mar 20 at 7:46

Ahmad Al-DweikAhmad Al-Dweik

11

1 Answer
1

active

oldest

votes

1

$begingroup$

Applying the condition, $3(a+c)=2(b+d)$ and substituting as below in the given equations we have:

($a_1,a_2,a_3,a_4$)=$(a,e,j,n)$

($b_1,b_2,b_3,b_4$)=$(b,f,k,p)$

($c_1,c_2,c_3,c_4$)=$(c,g,L,q)$

($d_1,d_2,d_3,d_4$)=$(d,h,m,r)$

$(s,t,u,v)=(25333,29151,54507,62645)$

$(x,y,z,w)=((L+478),(k+223),(j+123),(m+589))$

Because of common factor's the simultaneous equation given

by 'OP' simplifies to as shown below:

$2(s+t+u+v)=(a+c)[(e+f)(3x+2z)+(g+h)(2y+3w)]+2(n+p+q+r)$

But because the equation is cubic in nature & the number

of unknowns are sixteen which is more than number of equations

the problem is difficult. Unless the number of unknown's are reduced.

share|cite|improve this answer

edited Mar 23 at 6:03

Community♦

1

answered Mar 20 at 17:31

SamSam

111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have solved the problem. There many solutions. One can just simply solve the four equations for the unknowns a4, b4, c4, d4. Then the number of solutions is n^12.
  $endgroup$
  – Ahmad Al-Dweik
  Mar 22 at 13:00
  

  
  
  
  

add a comment | 

1

$begingroup$

Applying the condition, $3(a+c)=2(b+d)$ and substituting as below in the given equations we have:

($a_1,a_2,a_3,a_4$)=$(a,e,j,n)$

($b_1,b_2,b_3,b_4$)=$(b,f,k,p)$

($c_1,c_2,c_3,c_4$)=$(c,g,L,q)$

($d_1,d_2,d_3,d_4$)=$(d,h,m,r)$

$(s,t,u,v)=(25333,29151,54507,62645)$

$(x,y,z,w)=((L+478),(k+223),(j+123),(m+589))$

Because of common factor's the simultaneous equation given

by 'OP' simplifies to as shown below:

$2(s+t+u+v)=(a+c)[(e+f)(3x+2z)+(g+h)(2y+3w)]+2(n+p+q+r)$

But because the equation is cubic in nature & the number

of unknowns are sixteen which is more than number of equations

the problem is difficult. Unless the number of unknown's are reduced.

share|cite|improve this answer

edited Mar 23 at 6:03

Community♦

1

answered Mar 20 at 17:31

SamSam

111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have solved the problem. There many solutions. One can just simply solve the four equations for the unknowns a4, b4, c4, d4. Then the number of solutions is n^12.
  $endgroup$
  – Ahmad Al-Dweik
  Mar 22 at 13:00
  

  
  
  
  

add a comment | 

1

$begingroup$

Applying the condition, $3(a+c)=2(b+d)$ and substituting as below in the given equations we have:

($a_1,a_2,a_3,a_4$)=$(a,e,j,n)$

($b_1,b_2,b_3,b_4$)=$(b,f,k,p)$

($c_1,c_2,c_3,c_4$)=$(c,g,L,q)$

($d_1,d_2,d_3,d_4$)=$(d,h,m,r)$

$(s,t,u,v)=(25333,29151,54507,62645)$

$(x,y,z,w)=((L+478),(k+223),(j+123),(m+589))$

Because of common factor's the simultaneous equation given

by 'OP' simplifies to as shown below:

$2(s+t+u+v)=(a+c)[(e+f)(3x+2z)+(g+h)(2y+3w)]+2(n+p+q+r)$

But because the equation is cubic in nature & the number

of unknowns are sixteen which is more than number of equations

the problem is difficult. Unless the number of unknown's are reduced.

share|cite|improve this answer

edited Mar 23 at 6:03

Community♦

1

answered Mar 20 at 17:31

SamSam

111

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have solved the problem. There many solutions. One can just simply solve the four equations for the unknowns a4, b4, c4, d4. Then the number of solutions is n^12.
  $endgroup$
  – Ahmad Al-Dweik
  Mar 22 at 13:00
  

  
  
  
  

add a comment | 

$begingroup$

Applying the condition, $3(a+c)=2(b+d)$ and substituting as below in the given equations we have:

($a_1,a_2,a_3,a_4$)=$(a,e,j,n)$

($b_1,b_2,b_3,b_4$)=$(b,f,k,p)$

($c_1,c_2,c_3,c_4$)=$(c,g,L,q)$

($d_1,d_2,d_3,d_4$)=$(d,h,m,r)$

$(s,t,u,v)=(25333,29151,54507,62645)$

$(x,y,z,w)=((L+478),(k+223),(j+123),(m+589))$

Because of common factor's the simultaneous equation given

by 'OP' simplifies to as shown below:

$2(s+t+u+v)=(a+c)[(e+f)(3x+2z)+(g+h)(2y+3w)]+2(n+p+q+r)$

But because the equation is cubic in nature & the number

of unknowns are sixteen which is more than number of equations

the problem is difficult. Unless the number of unknown's are reduced.

share|cite|improve this answer

edited Mar 23 at 6:03

Community♦

1

answered Mar 20 at 17:31

SamSam

111

$endgroup$

share|cite|improve this answer

edited Mar 23 at 6:03

Community♦

1

answered Mar 20 at 17:31

SamSam

111

edited Mar 23 at 6:03

Community♦

1

edited Mar 23 at 6:03

Community♦

1

answered Mar 20 at 17:31

SamSam

111

answered Mar 20 at 17:31

SamSam

111