Let $d$ be a square-free positive integer $d>1$. Then there are no integers $x,y,z,t,a,b,c$ with $x\neq \pm z$, and $xt-yz\neq 0$ such that:
$$\begin{cases} x^2+y^2=a^2 d \\ z^2+t^2=b^2 d \\ (x-z)^2+(y-t)^2=c^2d\end{cases}$$
I have proved that $d\equiv 1\pmod 4$, but nothing more...
Let $x$ be an arbitrary integer, and take $y=a=x$, $z=t=b=1$, $c=x-1$ and $d=2$. Then $$\begin{cases} x^2+y^2=2x^2=a^2 d \\ z^2+t^2=1^2+1^2=b^2 d \\ (x-z)^2+(y-t)^2=2(x-1)^2=c^2d.\end{cases}$$ So we can choose $x\neq \pm z=\pm 1$.