# Are Diophantine equations in four variables solvable?

Guest

### Re: Are Diophantine equations in four variables solvable?

Given the Diophantine equation,

1. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$,

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

known nonzero integer exponents, $$e_{1 }$$, $$e_{2 }$$, $$e_{3}$$, and $$e_{4 }$$ 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?

Guest wrote:

Given the Diophantine equation,

1. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$,

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

known nonzero integer exponents, $$e_{1 }$$, $$e_{2 }$$, $$e_{3}$$, and $$e_{4 }$$ 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!

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

Guest

### Re: Are Diophantine equations in four variables solvable?

The eight variables are w, x y, z, $$n_{1 }$$, $$n_{2 }$$, $$n_{3 }$$, and $$n_{4 }$$ where each $$n_{j }$$ 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?

"Zero hides data..."

Keywords: Algorithm, Numerical Analysis, and Convergence

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

4. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

...

Guest

### Re: Are Diophantine equations in four variables solvable?

Guest wrote:"Zero hides data..."

Keywords: Algorithm, Numerical Analysis, and Convergence

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }x^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}$$.

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?

"Zero hides data..."

Keywords: Algorithm, Numerical Analysis, and Convergence

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }x^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}$$.

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?

Guest wrote:"Zero hides data..."

Keywords: Algorithm, Numerical Analysis, and Convergence

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }x^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }x^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}$$.

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?

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

Keywords: Algorithm, Numerical Analysis, and Convergence

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}$$.

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?

"Zero hides data..."

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

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}$$.

Step O: Guess $$n_{10}$$, $$n_{20}$$,

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

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

$$x_{0}$$ , $$y_{0}$$, and $$z_{0}$$;

...

Guest

### Re: Are Diophantine equations in four variables solvable?

Guest wrote:"Zero hides data..."

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

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{3 }z^{e_{3 }}$$.

Step O: Guess $$n_{10}$$, $$n_{20}$$,

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

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

$$x_{0}$$ , $$y_{0}$$, and $$z_{0}$$;

...

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

Guest

### Re: Are Diophantine equations in four variables solvable?

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

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

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }w^{e_{1}} + a_{2}y^{e_{2}} + a_{3}y^{e_{3 }}$$.

Step O: Guess $$n_{10}$$, $$n_{20}$$,

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

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

$$x_{0}$$ , $$y_{0}$$, and $$z_{0}$$;

...

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

Guest

### Re: Are Diophantine equations in four variables solvable?

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

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

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }z^{e_{3 }}$$.

Step O: Guess $$n_{10}$$, $$n_{20}$$,

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

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

$$x_{0}$$ , $$y_{0}$$, and $$z_{0}$$;

...

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

Guest

### Re: Are Diophantine equations in four variables solvable?

"Zero hides data..."

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

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}$$.

Step O: Guess $$n_{10}$$, $$n_{20}$$,

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

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

$$x_{0}$$ , $$y_{0}$$, and $$z_{0}$$;

...

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

Guest

### Re: Are Diophantine equations in four variables solvable?

Guest wrote:"Zero hides data..."

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

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}$$.

Step O: Guess $$n_{10}$$, $$n_{20}$$,

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

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

$$x_{0}$$ , $$y_{0}$$, and $$z_{0}$$;

...

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

Guest

### Re: Are Diophantine equations in four variables solvable?

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

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

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}$$.

Step O: Guess $$n_{10}$$, $$n_{20}$$,

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

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

$$x_{0}$$ , $$y_{0}$$, and $$z_{0}$$;

...

Remark: This is a challenging and beautiful problem.

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

Important Derived Equations/Functions:

5. $$n_{1 } = 3n -( n_{2 } + n_{3 } + n_{4 } )$$;

6. $$n_{2 } = 3n -( n_{1 } + n_{3 } + n_{4 } )$$;

7. $$n_{3 } = 3n -( n_{1 } + n_{2 } + n_{4 } )$$;

8. $$n_{4 } = 3n -( n_{1 } + n_{2 } + n_{3} )$$;

Notes:

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

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

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

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

And of course,

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

Central Equation (CE) of our Algorithm:

We have,

(CE): $$n_{j(k+1) } = n_{j(k) } - J^{-1} * n_{j }$$,

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

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

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

Dave,

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

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

Guest

### Re: Are Diophantine equations in four variables solvable?

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

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

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}$$.

Step O: Guess $$n_{10}$$, $$n_{20}$$,

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

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

$$x_{0}$$ , $$y_{0}$$, and $$z_{0}$$;

...

Remark: This is a challenging and beautiful problem.

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

Important Derived Equations/Functions:

5. $$n_{1 } = 3n -( n_{2 } + n_{3 } + n_{4 } )$$;

6. $$n_{2 } = 3n -( n_{1 } + n_{3 } + n_{4 } )$$;

7. $$n_{3 } = 3n -( n_{1 } + n_{2 } + n_{4 } )$$;

8. $$n_{4 } = 3n -( n_{1 } + n_{2 } + n_{3} )$$;

Notes:

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

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

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

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

And of course,

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

Central Equation (CE) of our Algorithm:

We have,

(CE): $$n_{j(k+1) } = n_{j(k) } - J^{-1} * n_{j }$$,

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

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

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

Dave,

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

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

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

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

Tentatively, we have calculated the determinant of J:

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

...

Dave.

Guest

### Re: Are Diophantine equations in four variables solvable?

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

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

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}$$.

Step O: Guess $$n_{10}$$, $$n_{20}$$,

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

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

$$x_{0}$$ , $$y_{0}$$, and $$z_{0}$$;

...

Remark: This is a challenging and beautiful problem.

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

Important Derived Equations/Functions:

5. $$n_{1 } = 3n -( n_{2 } + n_{3 } + n_{4 } )$$;

6. $$n_{2 } = 3n -( n_{1 } + n_{3 } + n_{4 } )$$;

7. $$n_{3 } = 3n -( n_{1 } + n_{2 } + n_{4 } )$$;

8. $$n_{4 } = 3n -( n_{1 } + n_{2 } + n_{3} )$$;

Notes:

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

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

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

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

And of course,

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

Central Equation (CE) of our Algorithm:

We have,

(CE): $$n_{j(k+1) } = n_{j(k) } - J^{-1} * n_{j }$$,

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

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

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

Dave,

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

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

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

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

Tentatively, we have calculated the determinant of J:

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

Thus, the inverse of J is,

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

where

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

and where $$M_{lk }$$ are the determinants of the appropriate minor matrices of J.

...

Dave.

Guest

### Re: Are Diophantine equations in four variables solvable?

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

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

E. $$a_{1 }w^{e_{1 }} + a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }} = n$$.

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

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

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

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

5. $$n_{1} = a_{2 }x^{e_{2}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

6. $$n_{2} = a_{1 }w^{e_{1}} + a_{3}y^{e_{3}} + a_{4 }z^{e_{4 }}$$;

7. $$n_{3} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{4 }z^{e_{4 }}$$;

8. $$n_{4} = a_{1 }w^{e_{1}} + a_{2}x^{e_{2}} + a_{3 }y^{e_{3 }}$$.

Step O: Guess $$n_{10}$$, $$n_{20}$$,

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

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

$$x_{0}$$ , $$y_{0}$$, and $$z_{0}$$;

...

Remark: This is a challenging and beautiful problem.

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

Important Derived Equations/Functions:

5. $$n_{1 } = 3n -( n_{2 } + n_{3 } + n_{4 } )$$;

6. $$n_{2 } = 3n -( n_{1 } + n_{3 } + n_{4 } )$$;

7. $$n_{3 } = 3n -( n_{1 } + n_{2 } + n_{4 } )$$;

8. $$n_{4 } = 3n -( n_{1 } + n_{2 } + n_{3} )$$;

Notes:

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

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

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

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

And of course,

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

Central Equation (CE) of our Algorithm:

We have,

(CE): $$n_{j(k+1) } = n_{j(k) } - J^{-1} * n_{j(k) }$$,

where the inverse of J (Jacobian Matrix), $$J^{-1}$$, is evaluated at the point, ($$n_{1(k)}, n_{2(k)}, n_{3(k)}, n_{4(k)}$$). 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, ($$n_{1(k+1)}, n_{2(k+1)}, n_{3(k+1)}, n_{4(k+1)}$$), after the kth iteration such that we solve our main equation E.

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

Dave,

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

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

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

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

Tentatively, we have calculated the determinant of J:

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

Thus, the inverse of J is,

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

where

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

and where $$M_{lk }$$ are the determinants of the appropriate minor matrices of J.

...

Dave.

Guest

### Re: Are Diophantine equations in four variables solvable?

Example: We shall solve the following equation:

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

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