Diophantine equation in four variables

Question

I would like to find a parametric solution for the following diophantine equation:

$-4 (1-a_1^2)(1-a_2^2) + (1+a_3^2 -a_1^2 -a_2^2)^2 = a_4^2$

Does such a solution exist? How does one go about solving such questions systematically?

Answer 1

Equation:

$q^2=(z^2+d^2-x^2-y^2)^2-4(d^2-x^2)(d^2-y^2)$

Has the solution:

$x=(p-s)sa^2+(p^2-ps+s^2)t^2-psk^2$

$y=-(p-s)sa^2+(p^2-ps+s^2)t^2-psk^2+2psak-2(p^2-ps+s^2)at$

$d=(p-s)sa^2-(p^2-ps+s^2)t^2-psk^2-2(p-s)sat+2pskt$

$z=(p-s)sa^2+(p^2-ps+s^2)t^2+psk^2-2(p^2-ps+s^2)kt-2(p-s)sak$

$q=4((p^2-ps+s^2)((p^2-s^2)k-(p^2-2ps)a)t^3+ps((p^2-s^2)t+as^2)k^3+$

$+s((p^3-3sp^2+2ps^2)t+k(p-s)s^2)a^3-(p^4-s^4)t^2k^2+p(p^3-4sp^2+6ps^2-4s^3)t^2a^2-$

$-(2p-s)s^3a^2k^2+(2p^3-3sp^2-ps^2+s^3)sakt^2-(p^3-2ps^2+2s^3)satk^2-$

$-(p^3-3sp^2+ps^2-s^3)skta^2)$

$p,s,a,t,k$ - Any integers.