The system of diophantine equations $$\{x^2y^2+z^2u^2+q^2t^2=0,\,xy+ztuq=0 \}$$ is given. Do the formulas $$x:=(j(p^24ps+3s^2)(ps)(3p^24ps+s^2))k^2+2(j2(ps))(ps)kn+(jp+s)n^2, $$ $$y:=(ps)(4j(ps)3p^2+4pss^2)k^2+2(ps)(j2(ps))kn(ps)n^2, $$ $$z:=(j(p^24ps+3s^2)+s(3p^24ps+s^2))k^2+2(ps)(2s+j)kn+(j+s)n^2, $$ $$t:=(j(p+s)3p^2+4pss^2)(ps)k^2+2[jp2(ps)(ps)]kn+(jp+s)n^2, $$ $$q:=(j(5p^28ps+3s^2)(ps)(3p^24ps+s^2))k^2+2(j(2ps)2(ps)(ps))kn+(jp+s)n^2, $$ $$u:=(j(p^24ps+3s^2)+(2sp)(3p^24ps+s^2))k^2+2(ps)(j+2(2sp))kn+(j+2sp)n^2, $$ where $p:=a^23b^2,\,s=2ab4b^2,\, j:=3b^24ab+a^2,$ produce all its solutions?
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The equations $x^2y^2+z^2u^2+q^2t^2=0$, $xy+ztuq=0$ are the real and imaginary parts of $w_1^2 + w_2^2 + w_3^2 = 0$ where $(w_1,w_2,w_3) = (x+iy,z+it,qiu)$. 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. 

