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?

5$\begingroup$ why do you want to know? Or, more to the point, why would I want to know? $\endgroup$ – Will Jagy Nov 2 '13 at 20:46

$\begingroup$ @ Will Jagy: Thank you for the interest to the question. You might find the method of the solutions of such hard problems of interest. $\endgroup$ – user64494 Nov 2 '13 at 21:05

6$\begingroup$ user64494: what I think Will Jagy was suggesting is that you should include some motivation with your question, saying where this specific system of equations came from, and how you came up with your solutions. Without any such motivation, your question just looks like some random pair of equations. Such a question becomes much more interesting if the equations are "nice" in some way  which is the case for your equations, being the real and imaginary parts of $a^3+b^3+c^3=0$. Why not say that, to help the reader appreciate your question, and also to point the way to a solution? $\endgroup$ – Michael Zieve Nov 3 '13 at 3:38
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.

$\begingroup$ Could you give a reference to the standard method for parametrizing a conic over the Gaussian numbers? $\endgroup$ – user64494 Nov 2 '13 at 21:42

1$\begingroup$ Same as over any field. Here you can just adapt the standard formula for Pythagorean triples: $(w_1,w_2,w_3) = c(m^2n^2,2mn,i(m^2+n^2))$. For my take on it, see the first few pages of my article "The ABC's of Number Theory" (thehcmr.org/issue1_1/elkies.pdf). $\endgroup$ – Noam D. Elkies Nov 2 '13 at 21:49

$\begingroup$ Sorry, but the Gaussian numbers do not form a field. $\endgroup$ – user64494 Nov 2 '13 at 21:54

6$\begingroup$ Gaussian numbers = complex numbers with rational real and imaginary parts. Yes, they form a field. The Gaussian integers in that field are a subring with unique factorization, so you can take $c,m,n$ to be relatively prime Gaussian integers if you know that the $w_j$ will be Gaussian integers. $\endgroup$ – Noam D. Elkies Nov 2 '13 at 21:59