The system of diophantine equations $$\{x^2-y^2+z^2-u^2+q^2-t^2=0,\,xy+zt-uq=0 \}$$ is given. Do the formulas
$$x:=(j(p^2-4ps+3s^2)-(p-s)(3p^2-4ps+s^2))k^2+2(j-2(p-s))(p-s)kn+(j-p+s)n^2, $$
$$y:=(p-s)(4j(p-s)-3p^2+4ps-s^2)k^2+2(p-s)(j-2(p-s))kn-(p-s)n^2, $$
$$z:=(j(p^2-4ps+3s^2)+s(3p^2-4ps+s^2))k^2+2(p-s)(2s+j)kn+(j+s)n^2, $$
$$t:=(j(p+s)-3p^2+4ps-s^2)(p-s)k^2+2[jp-2(p-s)(p-s)]kn+(j-p+s)n^2, $$
$$q:=(j(5p^2-8ps+3s^2)-(p-s)(3p^2-4ps+s^2))k^2+2(j(2p-s)-2(p-s)(p-s))kn+(j-p+s)n^2, $$
$$u:=(j(p^2-4ps+3s^2)+(2s-p)(3p^2-4ps+s^2))k^2+2(p-s)(j+2(2s-p))kn+(j+2s-p)n^2,
 $$ where $p:=a^2-3b^2,\,s=2ab-4b^2,\, j:=3b^2-4ab+a^2,$ produce all its solutions?



Solutions of system of diophantine equations
The equations $x^2-y^2+z^2-u^2+q^2-t^2=0$, $xy+zt-uq=0$
are the real and imaginary parts of $w\_1^2 + w\_2^2 + w\_3^2 = 0$
where $(w\_1,w\_2,w\_3) = (x+iy,z+it,q-iu)$. So we have a Pythagorean triple
over the Gaussian numbers (with the hypotenuse multiplied by $i$,
making it more symmetrical), and can just use the standard
method for parametrizing a conic. Remember at the end to multiply
by an arbitrary scalar.


