Are Diophantine equations in four variables solvable?


Re: Are Diophantine equations in four variables solvable?

Postby Guest » Fri Jan 25, 2019 2:16 am



Given the Diophantine equation,

1. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex],

of four nonzero integer variables, w, x, y, and z with known nonzero integer constants, [tex]a_{1 }[/tex], [tex]a_{2 }[/tex], [tex]a_{3}[/tex], and [tex]a_{4 }[/tex] with

known nonzero integer exponents, [tex]e_{1 }[/tex], [tex]e_{2 }[/tex], [tex]e_{3}[/tex], and [tex]e_{4 }[/tex] and with known integer constant, n,

can we develop an algorithm which generally solves equation one?
Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Fri Jan 25, 2019 9:11 pm

Guest wrote:


Given the Diophantine equation,

1. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex],

of four nonzero integer variables, w, x, y, and z with known nonzero integer constants, [tex]a_{1 }[/tex], [tex]a_{2 }[/tex], [tex]a_{3}[/tex], and [tex]a_{4 }[/tex] with

known nonzero integer exponents, [tex]e_{1 }[/tex], [tex]e_{2 }[/tex], [tex]e_{3}[/tex], and [tex]e_{4 }[/tex] and with known integer constant, n,

can we develop an algorithm which generally solves equation one?


The answer is yes! We can develop an algorithm which computes a system of 8 equations with 8 variables according to equation one.

However, we can also develop a much more
complicated equation than equation one involving more terms or expressions of two or more (maximum is four) variables which may make our quest (an algorithm) much more difficult or impossible according to the insolvability of Hilbert's Tenth Problem.

Hmm. Employing just four integer variables in our Diophantine equation, we may be able to develop a cipher which defeats any algorithm developed to crack it. Wow!

Relevant Reference Link:

https://en.m.wikipedia.org/wiki/Hilbert%27s_tenth_problem

Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Wed Feb 13, 2019 2:57 am

The eight variables are w, x y, z, [tex]n_{1 }[/tex], [tex]n_{2 }[/tex], [tex]n_{3 }[/tex], and [tex]n_{4 }[/tex] where each [tex]n_{j }[/tex] is the sum of a distinct combination of three terms of equation one. The eight equations are obvious. Right?

Note: The integer constant, n, of equation one could be zero (very interesting), positive, or negative.
Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Sat Mar 02, 2019 4:32 am

"Zero hides data..."

Keywords: Algorithm, Numerical Analysis, and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];
...

4. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

...









Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Sun Mar 03, 2019 3:26 pm

Guest wrote:"Zero hides data..."

Keywords: Algorithm, Numerical Analysis, and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]w = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]w = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]w = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }x^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}[/tex].

Step O: Guess n_{10}, n_{20}, n_{30 }, and n_{40 };

Step 1: Compute the corresponding w_{0 }, x_{0},

y_{0}, and z_{0};

...











Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Sun Mar 03, 2019 3:28 pm

"Zero hides data..."

Keywords: Algorithm, Numerical Analysis, and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }x^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}[/tex].

Step O: Guess n_{10}, n_{20}, n_{30 }, and n_{40 };

Step 1: Compute the corresponding w_{0 }, x_{0},

y_{0}, and z_{0};

...











Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Sun Mar 03, 2019 3:38 pm

Guest wrote:"Zero hides data..."

Keywords: Algorithm, Numerical Analysis, and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }x^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}[/tex].

Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],

[tex]n_{30}[/tex], and [tex]n_{40}[/tex];

Step 1: Compute the corresponding [tex]w_{0}[/tex],

[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];











Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Wed Mar 06, 2019 11:43 pm

Guest wrote:
Guest wrote:"Zero hides data..."

Keywords: Algorithm, Numerical Analysis, and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}[/tex].

Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],

[tex]n_{30}[/tex], and [tex]n_{40}[/tex];

Step 1: Compute the corresponding [tex]w_{0}[/tex],

[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];











Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Tue Mar 12, 2019 1:11 pm

"Zero hides data..."

Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}[/tex].

Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],

[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each term of equations, 5 - 7.

Step 1: Compute the corresponding [tex]w_{0}[/tex],

[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];

...











Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Tue Mar 12, 2019 4:33 pm

Guest wrote:"Zero hides data..."

Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}[/tex].

Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],

[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each term of equations, 5 - 7.

Step 1: Compute the corresponding [tex]w_{0}[/tex],

[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];

...







Relevant Reference Link (Newton's Method):

https://en.wikipedia.org/wiki/Newton%27s_method



Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Tue Mar 12, 2019 4:37 pm

Guest wrote:
Guest wrote:"Zero hides data..."

Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{3}y^{e_{3 }}[/tex].

Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],

[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each term of equations, 5 - 7.

Step 1: Compute the corresponding [tex]w_{0}[/tex],

[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];

...







Relevant Reference Link (Newton's Method):

https://en.wikipedia.org/wiki/Newton%27s_method



Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Tue Mar 12, 2019 4:40 pm

Guest wrote:
Guest wrote:"Zero hides data..."

Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }z^{e_{3 }}[/tex].

Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],

[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each term of equations, 5 - 7.

Step 1: Compute the corresponding [tex]w_{0}[/tex],

[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];

...







Relevant Reference Link (Newton's Method):

https://en.wikipedia.org/wiki/Newton%27s_method



Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Tue Mar 12, 2019 4:40 pm

"Zero hides data..."

Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}[/tex].

Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],

[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each term of equations, 5 - 7.

Step 1: Compute the corresponding [tex]w_{0}[/tex],

[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];

...







Relevant Reference Link (Newton's Method):

https://en.wikipedia.org/wiki/Newton%27s_method



Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Wed Mar 13, 2019 4:37 am

Guest wrote:"Zero hides data..."

Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}[/tex].

Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],

[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each variable of every term of equations, 5 - 8.

Step 1: Compute the corresponding [tex]w_{0}[/tex],

[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];

...







Relevant Reference Link (Newton's Method):

https://en.wikipedia.org/wiki/Newton%27s_method



Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Sat Mar 16, 2019 1:33 pm

x_{1 }"Zero hides data..."

Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}[/tex].

Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],

[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each variable of every term of equations, 5 - 8.

Step 1: Compute the corresponding [tex]w_{0}[/tex],

[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];

...

Remark: This is a challenging and beautiful problem.

We assume a system of nonlinear equations which means there exist some [tex]e_{i } > 1[/tex] where [tex]i\in[/tex] {1, 2, 3, 4}.

Important Derived Equations/Functions:

5. [tex]n_{1 } = 3n -( n_{2 } + n_{3 } + n_{4 } )[/tex];

6. [tex]n_{2 } = 3n -( n_{1 } + n_{3 } + n_{4 } )[/tex];

7. [tex]n_{3 } = 3n -( n_{1 } + n_{2 } + n_{4 } )[/tex];

8. [tex]n_{4 } = 3n -( n_{1 } + n_{2 } + n_{3} )[/tex];

Notes:

[tex]\frac{\partial n_j }{\partial w} = a_1 * e_1 * w^{e_1 - 1} = \frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}[/tex] or 0 if j =1;

[tex]\frac{\partial n_j }{\partial x} = a_2 * e_2 * x^{e_2 - 1} = \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}[/tex] or 0 if j =2;

[tex]\frac{\partial n_j }{\partial y} = a_3 * e_3 * y^{e_3 - 1} = \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}[/tex] or 0 if j =3;

[tex]\frac{\partial n_j }{\partial z} = a_4 * e_4 * z^{e_4 - 1} = \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}[/tex] or 0 if j =4;

And of course,

[tex]\frac{e_{j }(n - n_{j })}{((n - n_{j })/a_{j })^{1/e_{j }}} = \frac{e_{j }(n - n_{j })}{(|(n - n_{j })/a_{j }|)^{1/e_{j }}}[/tex] if [tex]e_{j }[/tex] is even.

Central Equation (CE) of our Algorithm:

We have,

(CE): [tex]n_{j(k+1) } = n_{j(k) } - J^{-1} * n_{j }[/tex],

where J (Jacobian Matrix) and the function, [tex]n_{j }[/tex], are evaluated at the point, ([tex]n_{1(k)}, n_{2(k)}, n_{3(k)}, n_{4(k)}[/tex]).

And of course, we seek the right the point, ([tex]n_{1(k+1)}, n_{2(k+1)}, n_{3(k+1)}, n_{4(k+1)}[/tex]), after the kth iteration such that we solve our main equation E.

We shall review our work here, and define our initial point, ([tex]n_{1(0)}, n_{2(0)}, n_{3(0)}, n_{4(0)}[/tex]), J, and its inverse, [tex]J^{-1}[/tex], next time. And hopefully, we can soon complete our algorithm and solve an example too.

Dave,

https://www.researchgate.net/profile/David_Cole29



Relevant Reference Link (Newton's Method):

https://en.wikipedia.org/wiki/Newton%27s_method



Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Sat Mar 16, 2019 3:20 pm

Guest wrote:x_{1 }"Zero hides data..."

Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}[/tex].

Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],

[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each variable of every term of equations, 5 - 8.

Step 1: Compute the corresponding [tex]w_{0}[/tex],

[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];

...

Remark: This is a challenging and beautiful problem.

We assume a system of nonlinear equations which means there exist some [tex]e_{i } > 1[/tex] where [tex]i\in[/tex] {1, 2, 3, 4}.

Important Derived Equations/Functions:

5. [tex]n_{1 } = 3n -( n_{2 } + n_{3 } + n_{4 } )[/tex];

6. [tex]n_{2 } = 3n -( n_{1 } + n_{3 } + n_{4 } )[/tex];

7. [tex]n_{3 } = 3n -( n_{1 } + n_{2 } + n_{4 } )[/tex];

8. [tex]n_{4 } = 3n -( n_{1 } + n_{2 } + n_{3} )[/tex];

Notes:

[tex]\frac{\partial n_j }{\partial w} = a_1 * e_1 * w^{e_1 - 1} = \frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}[/tex] or 0 if j =1;

[tex]\frac{\partial n_j }{\partial x} = a_2 * e_2 * x^{e_2 - 1} = \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}[/tex] or 0 if j =2;

[tex]\frac{\partial n_j }{\partial y} = a_3 * e_3 * y^{e_3 - 1} = \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}[/tex] or 0 if j =3;

[tex]\frac{\partial n_j }{\partial z} = a_4 * e_4 * z^{e_4 - 1} = \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}[/tex] or 0 if j =4;

And of course,

[tex]\frac{e_{j }(n - n_{j })}{((n - n_{j })/a_{j })^{1/e_{j }}} = \frac{e_{j }(n - n_{j })}{(|(n - n_{j })/a_{j }|)^{1/e_{j }}}[/tex] if [tex]e_{j }[/tex] is even.

Central Equation (CE) of our Algorithm:

We have,

(CE): [tex]n_{j(k+1) } = n_{j(k) } - J^{-1} * n_{j }[/tex],

where the inverse of J (Jacobian Matrix), [tex]J^{-1}[/tex], and the function, [tex]n_{j }[/tex], are evaluated at the point, ([tex]n_{1(k)}, n_{2(k)}, n_{3(k)}, n_{4(k)}[/tex]).

And of course, we seek the right the point, ([tex]n_{1(k+1)}, n_{2(k+1)}, n_{3(k+1)}, n_{4(k+1)}[/tex]), after the kth iteration such that we solve our main equation E.

We shall review our work here, and define our initial point, ([tex]n_{1(0)}, n_{2(0)}, n_{3(0)}, n_{4(0)}[/tex]), J, and its inverse, [tex]J^{-1}[/tex], next time. And hopefully, we can soon complete our algorithm and solve an example too.

Dave,

https://www.researchgate.net/profile/David_Cole29



Relevant Reference Link (Newton's Method):

https://en.wikipedia.org/wiki/Newton%27s_method

Here's our tentative definition of the Jacobian, J:

[tex]J = \frac{\partial ( n_1, n_2, n_3, n_4 ) }{\partial (w, x, y, z)} = \begin{bmatrix}0& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& 0& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& 0& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}&0\end{bmatrix}[/tex].


Tentatively, we have calculated the determinant of J:

[tex]|J| = \frac{3e_{1 } * e_{2 } * e_{3 } * e_{4 } *(n - n_{1 })(n - n_{2 })(n - n_{3 })(n - n_{4 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}((n - n_{2 })/a_{2 })^{1/e_{2 }}((n - n_{3 })/a_{3 })^{1/e_{3 }}((n - n_{4 })/a_{4 })^{1/e_{4 }}}[/tex].

...

Dave.

Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Sat Mar 16, 2019 3:40 pm

x_{1 }"Zero hides data..."

Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}[/tex].

Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],

[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each variable of every term of equations, 5 - 8.

Step 1: Compute the corresponding [tex]w_{0}[/tex],

[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];

...

Remark: This is a challenging and beautiful problem.

We assume a system of nonlinear equations which means there exist some [tex]e_{i } > 1[/tex] where [tex]i\in[/tex] {1, 2, 3, 4}.

Important Derived Equations/Functions:

5. [tex]n_{1 } = 3n -( n_{2 } + n_{3 } + n_{4 } )[/tex];

6. [tex]n_{2 } = 3n -( n_{1 } + n_{3 } + n_{4 } )[/tex];

7. [tex]n_{3 } = 3n -( n_{1 } + n_{2 } + n_{4 } )[/tex];

8. [tex]n_{4 } = 3n -( n_{1 } + n_{2 } + n_{3} )[/tex];

Notes:

[tex]\frac{\partial n_j }{\partial w} = a_1 * e_1 * w^{e_1 - 1} = \frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}[/tex] or 0 if j =1;

[tex]\frac{\partial n_j }{\partial x} = a_2 * e_2 * x^{e_2 - 1} = \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}[/tex] or 0 if j =2;

[tex]\frac{\partial n_j }{\partial y} = a_3 * e_3 * y^{e_3 - 1} = \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}[/tex] or 0 if j =3;

[tex]\frac{\partial n_j }{\partial z} = a_4 * e_4 * z^{e_4 - 1} = \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}[/tex] or 0 if j =4;

And of course,

[tex]\frac{e_{j }(n - n_{j })}{((n - n_{j })/a_{j })^{1/e_{j }}} = \frac{e_{j }(n - n_{j })}{(|(n - n_{j })/a_{j }|)^{1/e_{j }}}[/tex] if [tex]e_{j }[/tex] is even.

Central Equation (CE) of our Algorithm:

We have,

(CE): [tex]n_{j(k+1) } = n_{j(k) } - J^{-1} * n_{j }[/tex],

where the inverse of J (Jacobian Matrix), [tex]J^{-1}[/tex], and the function, [tex]n_{j }[/tex], are evaluated at the point, ([tex]n_{1(k)}, n_{2(k)}, n_{3(k)}, n_{4(k)}[/tex]).

And of course, we seek the right the point, ([tex]n_{1(k+1)}, n_{2(k+1)}, n_{3(k+1)}, n_{4(k+1)}[/tex]), after the kth iteration such that we solve our main equation E.

We shall review our work here, and define our initial point, ([tex]n_{1(0)}, n_{2(0)}, n_{3(0)}, n_{4(0)}[/tex]), J, and its inverse, [tex]J^{-1}[/tex], next time. And hopefully, we can soon complete our algorithm and solve an example too.

Dave,

https://www.researchgate.net/profile/David_Cole29



Relevant Reference Link (Newton's Method):

https://en.wikipedia.org/wiki/Newton%27s_method

Here's our tentative definition of the Jacobian, J:

[tex]J = \frac{\partial ( n_1, n_2, n_3, n_4 ) }{\partial (w, x, y, z)} = \begin{bmatrix}0& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& 0& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& 0& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}&0\end{bmatrix}[/tex].


Tentatively, we have calculated the determinant of J:

[tex]|J| = \frac{3e_{1 } * e_{2 } * e_{3 } * e_{4 } *(n - n_{1 })(n - n_{2 })(n - n_{3 })(n - n_{4 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}((n - n_{2 })/a_{2 })^{1/e_{2 }}((n - n_{3 })/a_{3 })^{1/e_{3 }}((n - n_{4 })/a_{4 })^{1/e_{4 }}}[/tex].

Thus, the inverse of J is,

[tex]J^{-1} = [m'_{kl }][/tex]

where

[tex]m'_{kl } = \frac{(-1)^{l + k} * M_{lk }}{|J|}[/tex]

and where [tex]M_{lk }[/tex] are the determinants of the appropriate minor matrices of J.

...

Dave.

Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Mon Mar 18, 2019 10:52 am

Guest wrote:x_{1 }"Zero hides data..."

Keywords: Algorithm, Numerical Analysis (Newton's Method, etc.), and Convergence

E. [tex]a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n[/tex].

1. [tex]w = ( ( n - n_{1 } ) / a_{1 } )^{1/e_{1 }}[/tex];

2. [tex]x = ( ( n - n_{2 } ) / a_{2} )^{1/e_{2 }}[/tex];

3. [tex]y = ( ( n - n_{3 } ) / a_{3} )^{1/e_{3}}[/tex];

4. [tex]z = ( ( n - n_{4} ) / a_{4} )^{1/e_{4 }}[/tex];

5. [tex]n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

6. [tex]n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}[/tex];

7. [tex]n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}[/tex];

8. [tex]n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}[/tex].

Step O: Guess [tex]n_{10}[/tex], [tex]n_{20}[/tex],

[tex]n_{30}[/tex], and [tex]n_{40}[/tex] while being mindful of the coefficients and powers associated with each variable of every term of equations, 5 - 8.

Step 1: Compute the corresponding [tex]w_{0}[/tex],

[tex]x_{0}[/tex] , [tex]y_{0}[/tex], and [tex]z_{0}[/tex];

...

Remark: This is a challenging and beautiful problem.

We assume a system of nonlinear equations which means there exist some [tex]e_{i } > 1[/tex] where [tex]i\in[/tex] {1, 2, 3, 4}.

Important Derived Equations/Functions:

5. [tex]n_{1 } = 3n -( n_{2 } + n_{3 } + n_{4 } )[/tex];

6. [tex]n_{2 } = 3n -( n_{1 } + n_{3 } + n_{4 } )[/tex];

7. [tex]n_{3 } = 3n -( n_{1 } + n_{2 } + n_{4 } )[/tex];

8. [tex]n_{4 } = 3n -( n_{1 } + n_{2 } + n_{3} )[/tex];

Notes:

[tex]\frac{\partial n_j }{\partial w} = a_1 * e_1 * w^{e_1 - 1} = \frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}[/tex] or 0 if j =1;

[tex]\frac{\partial n_j }{\partial x} = a_2 * e_2 * x^{e_2 - 1} = \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}[/tex] or 0 if j =2;

[tex]\frac{\partial n_j }{\partial y} = a_3 * e_3 * y^{e_3 - 1} = \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}[/tex] or 0 if j =3;

[tex]\frac{\partial n_j }{\partial z} = a_4 * e_4 * z^{e_4 - 1} = \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}[/tex] or 0 if j =4;

And of course,

[tex]\frac{e_{j }(n - n_{j })}{((n - n_{j })/a_{j })^{1/e_{j }}} = \frac{e_{j }(n - n_{j })}{(|(n - n_{j })/a_{j }|)^{1/e_{j }}}[/tex] if [tex]e_{j }[/tex] is even.

Central Equation (CE) of our Algorithm:

We have,

(CE): [tex]n_{j(k+1) } = n_{j(k) } - J^{-1} * n_{j(k) }[/tex],

where the inverse of J (Jacobian Matrix), [tex]J^{-1}[/tex], is evaluated at the point, ([tex]n_{1(k)}, n_{2(k)}, n_{3(k)}, n_{4(k)}[/tex]). We are dealing with a column matrix (4x1) and a 4x4 matrix in CE. Please forgive the current notation...

And of course, we seek the right the point, ([tex]n_{1(k+1)}, n_{2(k+1)}, n_{3(k+1)}, n_{4(k+1)}[/tex]), after the kth iteration such that we solve our main equation E.

We shall review our work here, and define our initial point, ([tex]n_{1(0)}, n_{2(0)}, n_{3(0)}, n_{4(0)}[/tex]), J, and its inverse, [tex]J^{-1}[/tex], next time. And hopefully, we can soon complete our algorithm and solve an example too.

Dave,

https://www.researchgate.net/profile/David_Cole29



Relevant Reference Link (Newton's Method):

https://en.wikipedia.org/wiki/Newton%27s_method

Here's our tentative definition of the Jacobian, J:

[tex]J = \frac{\partial ( n_1, n_2, n_3, n_4 ) }{\partial (w, x, y, z)} = \begin{bmatrix}0& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& 0& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& 0& \frac{e_{4 }(n - n_{4 })}{((n - n_{4 })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}}& \frac{e_{2 }(n - n_{2 })}{((n - n_{2 })/a_{2 })^{1/e_{2 }}}& \frac{e_{3 }(n - n_{3 })}{((n - n_{3 })/a_{3 })^{1/e_{3 }}}&0\end{bmatrix}[/tex].


Tentatively, we have calculated the determinant of J:

[tex]|J| = \frac{3e_{1 } * e_{2 } * e_{3 } * e_{4 } *(n - n_{1 })(n - n_{2 })(n - n_{3 })(n - n_{4 })}{((n - n_{1 })/a_{1 })^{1/e_{1 }}((n - n_{2 })/a_{2 })^{1/e_{2 }}((n - n_{3 })/a_{3 })^{1/e_{3 }}((n - n_{4 })/a_{4 })^{1/e_{4 }}}[/tex].

Thus, the inverse of J is,

[tex]J^{-1} = [m'_{kl }][/tex]

where

[tex]m'_{kl } = \frac{(-1)^{l + k} * M_{lk }}{|J|}[/tex]

and where [tex]M_{lk }[/tex] are the determinants of the appropriate minor matrices of J.

...

Dave.

Guest
 

Re: Are Diophantine equations in four variables solvable?

Postby Guest » Tue Mar 19, 2019 1:21 pm

Example: We shall solve the following equation:

[tex]129w^{5} + 879x^{3} - 11,013y^{7} + 87,345,889z = -1,853,116,821,365,070,995,917,491[/tex].
Guest
 

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