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?
On
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.
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.