Solving Cubic Systems of Diophantine Equations

Question

What techniques are there for solving systems of Cubic Diophantine equations? I know there is no general purpose technique and looking at some papers it can quickly go over my head even for just a single equation. Any good resources for someone who has not worked with such equations very often and the foundations? When looking I mainly find papers which are not great for learning from or seems like a lot the other beginner resources mainly cover linear Diophantine systems which I can deal with via linear algebra, and also not what I am looking for since I am dealing with cubics.

For example I can brute force the following with a computer program I quickly wrote:

$$\left\{\!\begin{aligned} & z^3 = w^3 + x^3 - y^3 \\ & z^3 = w^3 + u^3 - v^3 \end{aligned}\right. $$

One such solution I get is:

$$ \left\{\!\begin{aligned} & 9^3 = 16^3 + 33^3 - 34^3\\ & 9^3 = 16^3 + 2^3 - 15^3 \end{aligned}\right. $$

The problem is if I start to have more variables and more equations to the system brute force becomes infeasible quickly. So some better methods to attack the problem to know if even a solutions exists and even find them would be great.

Also I notice when looking at just an equation in isolation a lot times I can find someone who has already found a parametrization of a given equation out the various systems I am messing with. I have tried combining 2 or more parametrizations, but this also seems quite hard.

For example the equations I gave as example above, I have found a paramizations for $ a^3 + b^3 = c^3 + d^3 $ which both equations in the example are a form of. Since the example is a set of simultaneous equations with repeated variables means I can I set parts of the parametrization equal to each other, but this just causes the number of variables to explode if the system has lots more equations and often again falls back to brute force unless I can rewrite the parametrization in the same variables as the other one. However, when I try this I often end up with an equation that no long gives integer solutions mainly because I end up with having cuberoots/squareroots and such. So it would seems like this method would not work.

So I guess back to my first paragraph, I am just mainly looking for information/resource and techniques for working with such systems. Ideally with the goal parametrization of the system or determining if a solution exists.

--Edited-- Individ asked for an example to work on. I guess we can use the example posted above to see what techniques one may use to maybe solve such a system. Solutions exists since I found some via brute force so at least so a parametric solution should exist I think?

Answer 1

Why is this task so difficult? Formally, it is necessary to solve such a system of equations.

$$A^3+B^3=C^3+Q^3=R^3+F^3$$

It is impossible to solve such a system of equations directly. Firstly, there are few variables for a power of 3. The more variables, the easier it is to solve. Secondly, a direct solution leads to an algebraic equation of degree 9. That is, the formula cannot be written.

Therefore, we will need some kind of trick to deceive nature. And the idea here is simple. In fact, we are dealing with the problem of 4 cubes. If we have parametrization, then it will be enough for us to solve a system of 2 diophantine equations in which $A,B -$ are expressed in 2 sets of different parameters.

And then I decided to use the well-known parametrization formulas for the 4-cube problem. https://ru.wikipedia.org/wiki/Задача_о_четырёх_кубах

I can write to a large extent myself. Only in the solution it will not help.

Find all integer solutions to Diophantine equation $x^3+y^3+z^3=w^3$

https://artofproblemsolving.com/community/c3046h1872888_five_dice

And then I was disappointed. Parameterization is usually of the 3rd degree and higher.

Instead of one complex system, it will be necessary to solve another even more complex system. I'm a lazy person, so I thought and decided that I needed to find some simple parametrization. The second degree is best. From three variables.

Because quadratic parametrization from 2 variables does not give all solutions. That's where the problem arose. Quadratic parametrization is possible if we solve a quadratic equation. That is, the equation of the 3rd degree must somehow be converted into a square. And I can solve the quadratic equation well.

As soon as this chain of conclusions and the direction of decisions became clear. Everything else turned out to be very simple. And in the equation...

$$A^3+B^3=C^3+Q^3$$

You can write such a parametrization.

$$A=28a^2+12ab-68ac+2b^2-16bc+42c^2$$

$$B=21a^2+ab-43ac-b^2+bc+21c^2$$

$$C=42a^2+16ab-100ac+2b^2-20bc+60c^2$$

$$Q=-35a^2-15ab+85ac-b^2+17bc-51c^2$$

Well, then you will need to solve a system of quadratic diophantine equations.I'll write about it later.... So far, I have solved the problem of 4 cubes.

P.S. - I would like to say a few words about the formula. When I wrote it for 4 cubes. I decided to use it to solve the system in finding coefficients. And there a rather funny pattern turned out. Which naturally did not allow to solve this system. Maybe someone will have ideas how to get around it?

If we take and set any numbers - as parameters --- $a,b,c$

And if we substitute new coefficients in the formula instead of the old ones - as a single number added to them.

$a \Longrightarrow p+a $

$b \Longrightarrow p+b $

$c \Longrightarrow p+c $

Then the same numbers will always be obtained... $-- A,B,C,Q$ It's so funny.

Note 2....

It turned out that parameterization of such a form can be made up a lot. For example, this. Which can save us from the symmetry of the formula. But so far it has been possible to obtain a formula with large coefficients...

$$A^3+B^3=C^3+Q^3$$

$$A=1144a^2-2024ab-11176ac+900b^2+9896bc+27300c^2$$

$$B=-559a^2+989ab+5461ac-435b^2-4826bc-13335c^2$$

$$C=273a^2-547ab-2731ac+269b^2+2726bc+6825c^2$$

$$Q=1092a^2-1928ab-10664ac+856b^2+9424bc+26040c^2$$

If there are other formulas, then we must try to find one that is not symmetric and with small coefficients.

Square parameters are reduced to two anyway.

$$A=k(3(k^3-2t^3)(k^2+kt+t^2)x^2+3(k+t)(k^3-2t^3)xy+(k+t)(k^2-t^2)y^2)$$

$$B=t(k+t)(3(k^4+k^2t^2+t^4)x^2+3t^3xy+(t^2-k^2)y^2)$$

$$C=k(k+t)(3(k^4+k^2t^2+t^4)x^2+3k^3xy+(k^2-t^2)y^2)$$

$$Q=t(-3(2k^3-t^3)(k^2+kt+t^2)x^2-3(k+t)(2k^3-t^3)xy+(k+t)(t^2-k^2)y^2)$$

Further, to solve it, it is necessary to solve a system of nonlinear equations.

Another approach of the solution - led anyway to two quadratic parameters.

$$A=(3k-t)((81k^5-108k^4t+63k^3t^2-27k^2t^3+6kt^4-t^5)x^2+3(9k^3-9k^2t+3kt^2-t^3)xy+(3k-2t)y^2)$$

$$B=t(3k(27k^4-36k^3t+21k^2t^2-6kt^3+t^4)x^2+t^3xy-(3k-2t)y^2)$$

$$C=(3k-t)(3k(27k^4-36k^3t+21k^2t^2-6kt^3+t^4)x^2+(3k-t)^3xy+(3k-2t)y^2)$$

$$Q=t((-162k^5+216k^4t-126k^3t^2+45k^2t^3-9kt^4+t^5)x^2-3(18k^3-18k^2t+6kt^2-t^3)xy-(3k-2t)y^2)$$