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.
 A: I. The edit of the OP suggests that the context of the problem is to find 4 numbers $a>b>c>d$ such that the difference of any two is a square,
$$a-b=x^2\\ a-c =z^2\\ a-d=w^2\\ b-c=y^2\\ b-d=t^2\\ c-d=s^2$$
Solving the first 3 eqns for $b,c,d$ and plugging those into the last 3 eqns, we recover the OP's system,
$$x^2+y^2=z^2\\ x^2+t^2=w^2 \\ s^2+z^2=w^2$$
Solving for $s,t,y$,
$$s = \sqrt{w^2-z^2}\\ t = \sqrt{w^2-x^2}\\ y = \sqrt{z^2-x^2}$$
Thus, the real problem is to find three squares $w^2,z^2,x^2$ such that the difference between any two is also a square.

This is one version of the centuries-old problem called Mengoli-Six Square Problem, or MSP. A parametric solution can be given as,
$$w = (e^2+f^2)(g^2-h^2)\\ z =(e^2-f^2)(g^2+h^2)\\ x=  (e^2-f^2)(g^2-h^2)$$
where,
$$e = 4q\\ \quad\quad f = \sqrt{p^2+(3q)^2}\\ g = 4pq\\ h=p^2+5q^2$$
It is easy to find rational $f$ since it is just a Pythagorean triple in disguise. So one small solution is $p = 4, q=1$ which yields,
$$w,z,x = 7585, 6273, 1665$$
$$s,t,y = 4264, 7400, 6048$$
and infinitely many more.
A: 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:
$$x=s^2+k^2-t^2+2(t-k-s)q$$
$$y=t^2+k^2-s^2+2ks-2tk$$
$$z=s^2+k^2-t^2$$
$$e=t^2+k^2+s^2-2kt-2ts$$
$$j=t^2+s^2-k^2+2ks-2ts$$
$$p=3s^2+3k^2+3t^2-6kt-6st+8ks+2(t-k-s)q$$
$$b=\frac{x^2+y^2-e^2}{2}$$
$$a=\frac{e^2+x^2-y^2}{2}$$
$$c=\frac{e^2+y^2-x^2}{2}$$
$$f=\frac{2z^2+e^2-x^2-y^2}{2}$$
