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Often seen similar systems of equations. Usually consider such systems in which decisions no. Such as there. Is there $a,b,c,d\in \mathbb N$ so that $a^2+b^2=c^2$, $b^2+c^2=d^2$?

I think it would be more interesting to solve the system in which there are solutions. For example to find out whether such a system solution?

$$\left\{\begin{aligned}&a^2+b^2+c^2=q^2\\&c^2+q^2=k^2\end{aligned}\right.$$

What is the right approach? And how to solve it?

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individ
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  • It's easy to write down all the solutions of $c^2+q^2=k^2$. Then, given any such solution, you just want to know whether $q^2-c^2$ is a sum of two squares. But it's well-understood which numbers are sums of two squares. – Gerry Myerson Feb 14 '15 at 08:56
  • @GerryMyerson it is clear, but the challenge comes to solving systems of nonlinear equations. Are all wondering what approach is needed to address them? – individ Feb 14 '15 at 09:03
  • If you're asking for a general method for solving systems of Diophantine equations, I think you're asking too much; there isn't even a general method for solving single Diophantine equations. – Gerry Myerson Feb 14 '15 at 09:06
  • @GerryMyerson General method exists. But can not solve all the equations. The question is - how else can you facilitate the calculation? I did not understand, and which equation can be tricky? – individ Feb 14 '15 at 09:11
  • Are you familiar with the work on Hilbert's 10th problem? – Gerry Myerson Feb 14 '15 at 09:14
  • @GerryMyerson Yes. All equations to solve the impossible, but a very large class of equations is easily solved. – individ Feb 14 '15 at 09:17
  • So then what's your question? If your question is, "Tell me methods for solving systems of Diophantine equations", then you are asking us to write a book. If that's the question, it should be closed as "too broad". – Gerry Myerson Feb 14 '15 at 09:20
  • @GerryMyerson the question is quite specific. Moreover, the system is not that difficult. What we need to do the conversion in order to write a formula solution? – individ Feb 14 '15 at 09:25
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    So, wait a minute --- you are just asking about the one system in the question? Not about all possible systems? I think my first comment tells you what you need to do to solve that system in the question. – Gerry Myerson Feb 14 '15 at 09:30
  • @GerryMyerson It is always interesting to write a formula. Numerical search can be done easily. And to write a formula task is much complicated. – individ Feb 14 '15 at 09:33

2 Answers2

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The $2.n.m$ Diophantine equations have well-known complete solutions (a.k.a. parameterizations) for all $n,m \ge 1$ (for example, see Bradley's paper or Barnett's paper).

In particular, the solutions for the $2.1.2$ and $2.1.3$ equations you are asking about have been known for centuries. If $q$ is odd (which is essentially equivalent to “all primitive integer solutions”), you’re looking for \begin{align} k &= r^2+s^2 \\ c &= 2rs \\ q &= r^2-s^2 \end{align} where $$q = t^2+u^2+v^2+z^2$$ and either $$c = t^2+u^2-v^2-z^2 \tag{$\star$}$$ or $$c = 2(tz-uv)\tag{$\star\star$}$$ With respect to $q$, you are are ultimately finding $r,s,t,u,v,z$ such that \begin{align} r^2 &= s^2+t^2+u^2+v^2+z^2; \end{align} this is the $2.1.5$ Diophantine equation, for which the complete solution (a.k.a. parameterization) is known. You then simply need to intersect that result with the solutions for ($\star$) and ($\star\star$) to obtain the complete solution to your original system of equations.

Kieren MacMillan
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  • The links that were given I can't open. There is no access to these logs. I know what you're talking about. But I use a different approach there. Below is the formula - where you just need to substitute the numbers and not think. You can also write another formula. I just gave an example of using a standard method for solving nonlinear systems. – individ Apr 15 '21 at 05:45
  • JSTOR allows 100 free article views per month, so you can access the linked articles by simply registering your email address. As for your method, you have not proved that the equations give all solutions, so it’s somewhat useless for practical applications. – Kieren MacMillan Apr 15 '21 at 18:30
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If you solve the system of equations:

$$\left\{\begin{aligned}&a^2+b^2+c^2=q^2\\&c^2+q^2=w^2\end{aligned}\right.$$

When the standard approach solution and using a replacement.

$$p=9t^2-10tk+5k^2$$

$$s=5t^2-10tk+9k^2$$

$$x=7t^2-10tk+7k^2$$

$$y=4(t^2-k^2)$$

$$z=5t^2-14tk+5k^2$$

Then the solution can be written as :

$$a=2zx$$

$$b=z^2+y^2-x^2$$

$$c=2yx$$

$$q=z^2+y^2+x^2$$

$$w=p^2+s^2$$

$t,k$ - integers which we ask.

individ
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  • This clearly isn’t a complete solution, since $(a,b,c,q,w)=(4,5,20,21,29)$ is a solution to the original system of equations, but no $x,y,z$ satisfy your “solution”. – Kieren MacMillan Apr 15 '21 at 19:01
  • There is often a formula that looks like you need to take into account mutually simple solutions. That is, after substituting the coefficients, reduce by the common divisor. And it often happens that the formula of all solutions can break down into several. And there's nothing wrong with that. It's just that these equations have this property. Or rather, the formula of their decisions. – individ Apr 16 '21 at 06:04
  • Okay… What $(x,y,z)$ [with or without common divisor] yields the solution $(a,b,c,q,w)=(4,5,20,21,29)$? – Kieren MacMillan Apr 17 '21 at 00:39
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    Eh... another arithmetic assignment. I don't like such questions. Okay. Substitute it.. $t=1$ ; $k=3$ $$p=83$$ $$s=87$$ $$x=85$$ $$y=84$$ $$z=81$$ Let's reduce all the numbers by 8. Then we'll set it up. $$a=52$$ $$b=-24$$ $$c=220$$ $$q=221$$ $$w=229$$ Then we will reduce it by 2 again. And we will get what we need. Will we check the other numbers again???? – individ Apr 17 '21 at 09:03
  • No, that gives $(a,b,c,q,w) = (5,-4,-20,21,29)$, which is not what I asked for. You may not like “such questions”, but that’s what math is all about. – Kieren MacMillan Apr 18 '21 at 02:33