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?
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5$\begingroup$ why do you want to know? Or, more to the point, why would I want to know? $\endgroup$– Will JagyCommented Nov 2, 2013 at 20:46
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$\begingroup$ @ Will Jagy: Thank you for the interest to the question. You might find the method of the solutions of such hard problems of interest. $\endgroup$– user64494Commented Nov 2, 2013 at 21:05
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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 ZieveCommented Nov 3, 2013 at 3:38
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$\begingroup$ MSE is a right place for such type questions ... $\endgroup$– dohmatobCommented Nov 21, 2020 at 6:01
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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.
answered Nov 2, 2013 at 21:37
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$\begingroup$ Could you give a reference to the standard method for parametrizing a conic over the Gaussian numbers? $\endgroup$ Commented Nov 2, 2013 at 21:42
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1$\begingroup$ Same as over any field. Here you can just adapt the standard formula for Pythagorean triples: $(w_1,w_2,w_3) = c(m^2-n^2,2mn,i(m^2+n^2))$. 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). $\endgroup$ Commented Nov 2, 2013 at 21:49
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$\begingroup$ Sorry, but the Gaussian numbers do not form a field. $\endgroup$ Commented Nov 2, 2013 at 21:54
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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. $\endgroup$ Commented Nov 2, 2013 at 21:59