What types of Diophantine equations are unsolvable?

[Guest]

Update:

On Q2: [tex]y(\beta_{1}^{3} + 4\beta_{1}^{2} \beta_{2} -8\beta_{1} - 9 \beta_{2}^{3} - 44 \beta_{1} \beta_{2}^{3}y^{2})^{1/3} \approx 201,412,490.4[/tex] ...

Remark: We need to answer Q1 before we can answer Q2. Right?

Hmm. [tex]|y| < (201,412,490.4)^{\frac{3}{5}} \approx 96, 039[/tex]. Right?

[Guest]

Remark: Machine Learning (AI) and human learning can benefit in the quest to solve Hilbert's great tenth problem.

[Guest]

Hmm. [tex]|y| < (201,412,490.4)^{\frac{3}{5}} \approx 96, 039[/tex]. Right?

The true sign of [tex]y[/tex] and the lower bound for [tex]y[/tex] may be deduced from answering Q1. Right?

[Guest]

Remark: A good answer for Q1 determines how x and [tex]z[/tex] are related. Right?

[Guest]

Remark: A good answer for Q1 determines how x and [tex]z[/tex] are related. Right?

[Guest]

'Rabbit Hole' Analysis:

Is [tex]\beta_{1} \ne \beta_{2}[/tex]? Why?

Hmm. We are clueless.

[Guest]

'Rabbit Hole' Analysis:

Is [tex]\beta_{1} \ne \beta_{2}[/tex]? Why?

Hmm. We are clueless.

If [tex]\beta_{1} = \beta_{2}[/tex], then [tex]x = z[/tex]. Right?

And we can significantly simplify matters. How?

[Guest]

An Optimistic Remark: If our analysis works well for our latest example, then we will be confident we can solve Diophantine equations with more than three variables. However, the complexity ([tex]\beta_{1}[/tex], [tex]\beta_{2}[/tex], [tex]\beta_{3}[/tex], ..., [tex]\beta_{k > 3}[/tex]) ... will grow, and the problems will generally become more difficult to solve.

But we optimistically believe a polynomial-time can be constructed to solve Hilbert's Tenth Problem.

We remain open to the possibility we may fail too, and that the difficult MRDP (Matiyasevich's) theorem is sound.

Go Blue!

[Guest]

An Optimistic Remark: If our analysis works well for our latest example, then we will be confident we can solve Diophantine equations with more than three variables. However, the complexity ([tex]\beta_{1}[/tex], [tex]\beta_{2}[/tex], [tex]\beta_{3}[/tex], ..., [tex]\beta_{k > 3}[/tex]) ... will grow, and the problems will generally become more difficult to solve.

But we optimistically believe a polynomial-time algorithm can be constructed to solve Hilbert's Tenth Problem.

We remain open to the possibility we may fail too, and that the difficult MRDP (Matiyasevich's) theorem is sound.

Go Blue!

[Guest]

Question: What value of [tex]k[/tex] (number of unknown variables ...) will stop us?

"We must know. We will know!" -- David Hilbert.

[Guest]

'Rabbit Hole' Analysis:

Is [tex]\beta_{1} \ne \beta_{2}[/tex]? Why?

Hmm. We are clueless.

If [tex]\beta_{1} = \beta_{2}[/tex], then [tex]x = z[/tex]. Right?

And we can significantly simplify matters. How?

From [tex]x^{3} + 4x^2z -8xy^{2}-44xyz^{3}-9z^{3} = 8,170,698,744,133,625,080,294,398[/tex] for [tex]x = z , y \in \mathbb{Z}[/tex] with [tex]x = \beta_{1} y[/tex],

we generate the equation,

[tex](4 \beta_{1}^{2} + 8 + 44\beta_{1}^{3}y^{2})\beta_{1} y^{3} = - 8,170,698,744,133,625,080,294,398[/tex] with [tex]-96,039 < y < 96,039[/tex].

Does [tex]\beta_{1} \in \mathbb{Q}[/tex]?

[Guest]

And does [tex]x = \beta_{1} y \in \mathbb{Z}[/tex]?

[Guest]

Remark: We confidently assume [tex]\beta_{1} \ne \beta_{2}[/tex] or [tex]x \ne z[/tex] and that [tex]x, y, z[/tex] are unequal.

[Guest]

Remark: We must complete our 'Rabbit Hole' analysis and solve our equation for x, y, and z.

Good Luck!

[Guest]

Remark: There are minor/non-serious gaps or mistakes for sure in our evolving theory/computations to solve Diophantine equations of the type suggested by Hilbert's great tenth problem. That's okay. We will do better and close those gaps and correct our mistakes too.

So far so good!

[Guest]

Important Remark: Our approach to solving problems here is to reduce complexity/complications by whatever means (established theory, cleverness, methods, heuristics, etc.) necessary.

If we have an equation with four variables, we will try to reduce it to an equivalent equation with three variables, etc.

[Guest]

Remark: We confidently assume [tex]\beta_{1} \ne \beta_{2}[/tex] or [tex]x \ne z[/tex] and that [tex]x, y, z[/tex] are unequal.

We let [tex]d_{\beta_{1} \beta_{2} }= |\beta_{1} - \beta_{2}|[/tex] .

Moreover, [tex]\beta_{1} \ne \beta_{2}[/tex] implies Rabbit Hole 1:[tex]\beta_{1} > \beta_{2}[/tex] or Rabbit Hole 2: [tex]\beta_{1} < \beta_{2}[/tex].

Which one is correct? We are clueless.

[Guest]

Remark: We confidently assume [tex]\beta_{1} \ne \beta_{2}[/tex] or [tex]x \ne z[/tex] and that [tex]x, y, z[/tex] are unequal.

We let [tex]d_{\beta_{1} \beta_{2} }= |\beta_{1} - \beta_{2}|[/tex] .

Moreover, [tex]\beta_{1} \ne \beta_{2}[/tex] implies Rabbit Hole 1:[tex]\beta_{1} > \beta_{2}[/tex] or Rabbit Hole 2: [tex]\beta_{1} < \beta_{2}[/tex].

Which one is correct? We are clueless.

We recall the equations,

[tex]x^{3} + 4x^2z -8xy^{2}-44xyz^{3}-9z^{3}

= (\beta_{1} y)^{3} + 4(\beta_{1} y)^2( \beta_{2} y) -8(\beta_{1} y)y^{2}-44(\beta_{1} y)y( \beta_{2} y)^{3}-9( \beta_{2} y)^{3}

= (\beta_{1}^{3} + 4\beta_{1}^{2} \beta_{2} -8\beta_{1} - 9 \beta_{2}^{3})y^{3} - 44 \beta_{1} \beta_{2}^{3}y^{5} = 8,170,698,744,133,625,080,294,398[/tex] with

[tex]-96,039 < y < 96,039[/tex].

We solve for [tex]\beta_{1}[/tex]...

[Guest]

Hmm. Solving for [tex]\beta_{1}[/tex] is unworkable. Right?

Can we estimate roughly the range of possible values for [tex]\beta_{1}[/tex] or [tex]\beta_{ 2}[/tex] by holding [tex]y[/tex] as a constant and by employing the Newton's Method?

It's worth a try. Good luck!

[Guest]

Hmm. Solving for [tex]\beta_{1}[/tex] is unworkable. Right?

Can we estimate roughly the range of possible values for [tex]\beta_{1}[/tex] or [tex]\beta_{ 2}[/tex] by holding [tex]y[/tex] as a constant and by employing the Newton's Method?

It's worth a try.

We hold [tex]y[/tex] as a constant and vary the values of [tex]\beta_{ 2}[/tex] and determine the range of possible values for [tex]\beta_{1}[/tex].

Or we hold [tex]y[/tex] as a constant and vary the values of [tex]\beta_{ 1}[/tex] and determine the range of possible values for [tex]\beta_{2}[/tex].

Let's generate some results. Good Luck!