# Diophantine equation in four variables

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?

• Could you write this equation in this form? $q^2=(z^2+d^2-x^2-y^2)^2-4(d^2-x^2)(d^2-y^2)$ And look for a solution in integers. Because this equation has a solution. May 30, 2014 at 17:19
• Yes indeed, this amounts to a rescaling of my variables. Where can I find a solution to that equation? May 30, 2014 at 17:23
• Nowhere. It is necessary to solve this equation. May 30, 2014 at 17:24
• Ok, I see. Are there somewhat systematic methods for solving equations like this? Perhaps it is better to start with a simpler version of this. E.g., the following eq. is also relevant for me, $(1 - b2^2) (1 - b3^2) - 4 (1 - b1^2) = 0$. It has one variable less. Jun 2, 2014 at 13:29
• For such equations the more unknown so easily solved. It is better not to reduce the number of unknowns. And what is the equation you need? Jun 2, 2014 at 14:00

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.