Are Diophantine equations in four variables solvable?

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An Update:

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/vector, ([tex]n_{1(k)}, n_{2(k)}, n_{3(k)}, n_{4(k)})[/tex], with [tex]N_{j(k) } = [n_{1(k)}, n_{2(k)}, n_{3(k)}, n_{4(k)}][/tex] (transposed). We are dealing with a column matrix (4x1) and a 4x4 matrix in CE...

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"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/vector, ([tex]n_{1(k)}, n_{2(k)}, n_{3(k)}, n_{4(k)})[/tex], with [tex]N_{j(k) } = [n_{1(k)}, n_{2(k)}, n_{3(k)}, n_{4(k)}][/tex] (transposed). We are dealing with a column matrix (4x1) and a 4x4 matrix in CE...

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 (kth iteration):

[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_{2k })}{((n - n_{2k })/a_{2 })^{1/e_{2 }}}& \frac{e_{3 }(n - n_{3k })}{((n - n_{3k })/a_{3 })^{1/e_{3 }}}& \frac{e_{4 }(n - n_{4k })}{((n - n_{4k })/a_{4 })^{1/e_{4k }}}\\\frac{e_{1 }(n - n_{1k })}{((n - n_{1k })/a_{1 })^{1/e_{1 }}}& 0 & \frac{e_{3 }(n - n_{3k })}{((n - n_{3k })/a_{3 })^{1/e_{3 }}}& \frac{e_{4 }(n - n_{4k })}{((n - n_{4k })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1k })}{((n - n_{1k })/a_{1 })^{1/e_{1 }}}& \frac{e_{2 }(n - n_{2k })}{((n - n_{2k })/a_{2 })^{1/e_{2 }}}& 0& \frac{e_{4 }(n - n_{4k })}{((n - n_{4k })/a_{4 })^{1/e_{4 }}}\\\frac{e_{1 }(n - n_{1k })}{((n - n_{1k })/a_{1 })^{1/e_{1 }}}& \frac{e_{2 }(n - n_{2k })}{((n - n_{2k })/a_{2 })^{1/e_{2 }}}& \frac{e_{3 }(n - n_{3k })}{((n - n_{3k })/a_{3 })^{1/e_{3 }}}&0\end{bmatrix}[/tex].


Tentatively, we have calculated the determinant of J (kth iteration):

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

[tex]J^{-1} = \begin{bmatrix} \frac{-2((n - n_{1k })/a_{1 })^{1/e_{1 }}}{3e_{1 }(n - n_{1k })}& \frac{((n - n_{1k })/a_{1 })^{1/e_{1 }}}{3e_{1 }(n - n_{1k })}& \frac{((n - n_{1k })/a_{1 })^{1/e_{1 }}}{3e_{1 }(n - n_{1k })}& \frac{((n - n_{1k })/a_{1 })^{1/e_{1 }}}{3e_{1 }(n - n_{1k })}& \\\frac{((n - n_{2k })/a_{2 })^{1/e_{2 }}}{3e_{2 }(n - n_{2k })}& \frac{-2((n - n_{2k })/a_{2 })^{1/e_{2 }}}{3e_{2 }(n - n_{2k })}& \frac{((n - n_{2k })/a_{2 })^{1/e_{2 }}}{3e_{2 }(n - n_{2k })}& \frac{((n - n_{2k })/a_{2 })^{1/e_{2 }}}{3e_{2 }(n - n_{2k })}& \\\frac{((n - n_{3k })/a_{3 })^{1/e_{3 }}}{3e_{3 }(n - n_{3k })}& \frac{((n - n_{3k })/a_{3 })^{1/e_{3 }}}{3e_{3 }(n - n_{3k })}& \frac{-2((n - n_{3k })/a_{3 })^{1/e_{3 }}}{3e_{3 }(n - n_{3k })}& \frac{((n - n_{3k })/a_{3 })^{1/e_{3 }}}{3e_{3 }(n - n_{3k })}& \\\frac{((n - n_{4k })/a_{4 })^{1/e_{4 }}}{3e_{4 }(n - n_{4k })}& \frac{((n - n_{4k })/a_{4 })^{1/e_{4 }}}{3e_{4 }(n - n_{4k })}& \frac{((n - n_{4k })/a_{4 })^{1/e_{4 }}}{3e_{4 }(n - n_{4k })}& \frac{-2((n - n_{4k })/a_{4 })^{1/e_{4 }}}{3e_{4 }(n - n_{4k })}\end{bmatrix}[/tex].
...

Dave.

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We have,

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

= [tex]\begin{bmatrix} n_{1k } - (\frac{-2((n - n_{1k })/a_{1 })^{1/e_{1 }}}{3e_{1 }(n - n_{1k })} * n_{1k } + \frac{((n - n_{1k })/a_{1 })^{1/e_{1 }}}{3e_{1 }(n - n_{1k })} * n_{2k } + \frac{((n - n_{1k })/a_{1 })^{1/e_{1 }}}{3e_{1 }(n - n_{1k })} * n_{3k } + \frac{((n - n_{1k })/a_{1 })^{1/e_{1 }}}{3e_{1 }(n - n_{1k })} * n_{4k })& \\\ n_{2k } - (\frac{((n - n_{2k })/a_{2 })^{1/e_{2 }}}{3e_{2 }(n - n_{2k })} * n_{1k } + \frac{-2((n - n_{2k })/a_{2 })^{1/e_{2 }}}{3e_{2 }(n - n_{2k })} * n_{2k } + \frac{((n - n_{2k })/a_{2 })^{1/e_{2 }}}{3e_{2 }(n - n_{2k })} * n_{3k } + \frac{((n - n_{2k })/a_{2 })^{1/e_{2 }}}{3e_{2 }(n - n_{2k })} * n_{4k })& \\\ n_{3k } - (\frac{((n - n_{3k })/a_{3 })^{1/e_{3 }}}{3e_{3 }(n - n_{3k })} * n_{1k } + \frac{((n - n_{3k })/a_{3 })^{1/e_{3 }}}{3e_{3 }(n - n_{3k })} * n_{2k } + \frac{-2((n - n_{3k })/a_{3 })^{1/e_{3 }}}{3e_{3 }(n - n_{3k })} * n_{3k } + \frac{((n - n_{3k })/a_{3 })^{1/e_{3 }}}{3e_{3 }(n - n_{3k })} * n_{4k })& \\\ n_{4k } - (\frac{((n - n_{4k })/a_{4 })^{1/e_{4 }}}{3e_{4 }(n - n_{4k })} * n_{1k } + \frac{((n - n_{4k })/a_{4 })^{1/e_{4 }}}{3e_{4 }(n - n_{4k })} * n_{2k } + \frac{((n - n_{4k })/a_{4 })^{1/e_{4 }}}{3e_{4 }(n - n_{4k })} * n_{3k } + \frac{-2((n - n_{4k })/a_{4 })^{1/e_{4 }}}{3e_{4 }(n - n_{4k })} * n_{4k }) \end{bmatrix}[/tex].

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Now, we are now ready to organize our algorithm and to code it to solve our example problem. We highly recommend Wolfram Cloud,https://www.wolframcloud.com/?source=nav for solving math problems. And we will remember to guess a good initial vector, [tex]N_{j0 } = [n_{10 }, n_{20}, n_{30 }, n_{40 }][/tex] (transposed).

Futhermore, we recall our example problem involving the following Diophantine equation:

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

What are the correct integer values for the variables, w, x, y, and z?

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If all goes well with our work here, we shall further develop our algorithm to solve more complex Diophantine equations of four variables and use our results to help investigate David Hilbert's Tenth Problem since I believe that important problem is still open...

Dave.

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[tex]\begin{cases} x + y = 4 \\ x - y = 0 \end{cases}[/tex]

If all goes well with our work here, we shall further develop our algorithm to solve more complex Diophantine equations of four variables and use our results to help us investigate David Hilbert's Tenth Problem since I believe that important and sound problem is still open...

Dave.

And what works well (sound algorithms) for the solutions of 4-variable Diophantine equations could be adapted rather easily... to work well for the solutions of Diophantine equations with five or more variables.

Dave.

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FYI:. On Some Practical Applications of Diophantine Equations, please see pdf:


http://ijmaa.in/v5n2-b/217-222.pdf

[Guest]

FYI:. On Some Practical Applications of Diophantine Equations, please see pdf:


http://ijmaa.in/v5n2-b/217-222.pdf

Apart from the practical aspects of Diophantine equations, we should ponder the nature of Diophantine equations especially their geometric nature.

Relevant Reference Link:

https://en.m.wikipedia.org/wiki/Diophantine_geometry

[Guest]

FYI:. On Some Practical Applications of Diophantine Equations, please see pdf:


http://ijmaa.in/v5n2-b/217-222.pdf

FYI:

https://www.researchgate.net/post/what_are_the_applications_of_diophantine_equations_where_do_we_apply_the_higher_degree_diophantine_equations