<s>[INST] Solutions to a system of three equations with Pythagorean triples Is there any solution to this system of equations where $x,y,z,s,w,t\in\mathbb{Z}$, none are $0$.
\begin{align*}
x^2+y^2=z^2\\\
s^2+z^2=w^2\\\
x^2+t^2=w^2
\end{align*}
EDIT: Thank you zwim for the answer. Maybe I should explain where this came from. Well these six numbers correspond to another four numbers $a,b,c,d\in\mathbb{Z}$ in such a way that if you take any two numbers among $a,b,c,d$ their difference produces a perfect square. There is six pairs so it corresponds to finding six perfect squares that satisfy the system of equations above. In the end we have a set of numbers that all differ between each other by a perfect square. I now wonder if there is a set of five numbers with such a property.
 [/INST] Time... began to write, and I will add a few words....
Do there exist four distinct integers such that the sum of any two of them is a perfect square?
This is equivalent to solving the following system of equations:
$$\left\{\begin{aligned}& b+a=x^2 \\&b+c=y^2\\&b+f=z^2\\&a+c=e^2\\&a+f=j^2\\&c+f=p^2\end{aligned}\right.$$
Let: $F,T,R,D$ - any asked us integers.
For ease of calculation, let's make a replacement.
$$q=(8F^2+4FT-T^2)R^2+2(T+2F)RD-D^2$$
$$k=(8F^2+8FT+2T^2)R^2+2(T+2F)RD$$
$$s=-T^2R^2+2(T+2F)RD-D^2$$
$$t=(8F^2+12TF+3T^2)R^2+2(T+2F)DR-D^2$$
Then the solutions are of the form:
$$x=s^2+k^2-t^2+2(t-k-s)q$$
$$y=t^2+k^2-s^2+2ks-2tk$$
$$z=s^2+k^2-t^2$$
$$e=t^2+k^2+s^2-2kt-2ts$$
$$j=t^2+s^2-k^2+2ks-2ts$$
$$p=3s^2+3k^2+3t^2-6kt-6st+8ks+2(t-k-s)q$$
$$b=\frac{x^2+y^2-e^2}{2}$$
$$a=\frac{e^2+x^2-y^2}{2}$$
$$c=\frac{e^2+y^2-x^2}{2}$$
$$f=\frac{2z^2+e^2-x^2-y^2}{2}$$
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