Is it possible that $(a,b,c)$, $(x,y,a)$, $(p,q,b)$ are Pythagorean triples simultaneously? Do there exist postive integers $a,b,c,x,y,p,q$ such $(a,b,c)$, $(x,y,a)$, $(p,q,b)$ are all Pythagorean triples? That is, does the system
$$\begin{cases}
a^2+b^2=c^2\\
x^2+y^2=a^2\\
p^2+q^2=b^2
\end{cases}$$
have a postive integer solution?

The system of equations.
$$\left\{\begin{aligned}&a^2+b^2=c^2\\&x^2+y^2=a^2\\&f^2+q^2=b^2\end{aligned}\right.$$
Equivalent to the need to solve the following system of equations.
$$\left\{\begin{aligned}&a=2ps=z^2+t^2\\&b=p^2-s^2=j^2+v^2\\&c=p^2+s^2\\&x=2zt\\&y=z^2-t^2\\&f=2jv\\&q=j^2-v^2\end{aligned}\right.$$
To ease calculations we change.
$$B=k^2+2n^2-r^2$$
$$A=k^2+2n^2+r^2-4nr$$
$$W=2k(r-2n)$$
$$Q=2n^2+r^2-k^2-2rn$$
Then we need the number to obtain Pythagorean triples can be found by the formula.
$$p=B^2+A^2$$
$$s=2Q^2$$
$$z=2BQ$$
$$t=2AQ$$
$$j=B^2+A^2-2Q^2$$
$$v=2WQ$$
At any stage of the computation can be divided into common divisor.
