A triple of pythagorean triples with an extra property

I'm trying to prove the non-existance of three positive integers $x,y,z$ with $x\geq z$ such that\begin{align} (x-z)^2+y^2 &\text{ is a perfect square,}\\ x^2+y^2 &\text{ is a perfect square,}\\ (x+z)^2+y^2 &\text{ is a perfect square.} \end{align} I failed tying to find such triple numerically. I tried to look at the differences, but that didn't give me much. Also, I tried the general solution of $a^2+b^2=c^2$, that is, $a=2pq$, $b=q^2-p^2$, and $c=q^2+p^2$ for some $p<q$, but this got ugly quite fast. I am not sure whether or not such $x,y,z$ exist. If it exists, one counterexample suffices to disprove my conjecture. Thanks in advance!

• This issue was discussed there. dxdy.ru/post1092838.html#p1092838 – individ Jan 28 '16 at 16:38
• @vrugtehagel: Every solution to your system is a rational point on a certain elliptic curve, so there are infinitely many. There is also an infinite number of unscaled polynomial solutions, one of which is individ's, $$x,\,y,\,z = 4b^4-17b^2+4,\;12b(b^2-1),\;2b^4+5b^2+2$$ – Tito Piezas III Feb 2 '16 at 15:29
• @ТитоPiezasIIIв this system is not I decided. By reference. – individ Feb 2 '16 at 15:39

$$(1904-1040)^2+990^2=1314^2\\ 1904^2+990^2=2146^2\\ (1904+1040)^2+990^2=3106^2$$
• @AndréNicolas It's because of the program I made wasn't brute-forcing all solutions. Instead, I assumed that $y=2d$ is even and then I looked for $a,b,c\mid d$ such that taking $x-z=a^2-(d/a)^2$ etc. would give $x,z$ integers. It's easy to see that this method will find all primitive solutions or their doubles. – Wojowu Jan 28 '16 at 16:47
It is impossible that non-existance because there are infinitely many counterexamples. In fact, we must have by the Pythagorean triples $$x-z=t^2-s^2;\space y=2ts\qquad (*)$$ $$x=t_1^2-s_1^2; \space y=2t_1s_1$$ $$x+z=t_2^2-s_2^2;\space \space y=2t_2s_2\qquad (**)$$ so we have from $(*)$ and $(**)$ $$x=\frac{t^2-s^2+t_2^2-s_2^2}{2}=t_1^2-s_1^2\iff t^2+t_2^2-2t_1^2=s^2+s_2^2-2s_1^2$$ On the other hand the identity $$(m^2-n^2)^2+(2mn)^2=2(m^2+n^2)^2$$ shows infinitely many solutions to the equation $X^2+Y^2=2Z^2$.
Thus we can get an infinite set of possible $t_1,t_2,t_3$