The system of Diophantine equations. 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?

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

What is the right approach? And how to solve it?
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 &amp;= r^2+s^2  \\
  c &amp;= 2rs  \\
  q &amp;= 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 &amp;= 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.
If you solve the system of equations:

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

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.