# System of quadratic Diophantine equations 2

I am looking for a way to simultaneously transform the following four expressions into perfect squares, $1+x_1^2, 1+x_2^2, 1+x_3^2, x_1^2+x_2^2+x_3^2$, i.e. I want to find a rational parametrization of $x_1,x_2,x_3$, and some $a_1,a_2,a_3,a_4$, such that

$1+x_1^2 = a_1^2\quad$ $1+x_2^2 = a_2^2\quad$ $1+x_3^2 = a_3^2\quad$ $x_1^2+x_2^2+x_3^2 = a_4^2$

E.g. the first three equations can be satisfied by setting $x_{1} = 1/(2 y_1)-y_1/2$ and $a_1 = (1+y_1^2)/2/y_1$, etc., but then the forth equation still needs to be satisfied.

I have found similar cases on https://sites.google.com/site/tpiezas/007, but I don't know how to approach such questions systematically.

• for what purpose? – Will Jagy May 29 '14 at 18:51
• This question arises in the context of particle scattering in theoretical physics. The $x_1, x_2, x_3$ are related to geometric quantities. In the equations one encounters square roots of the expressions above, and removing the latter would be very desirable. – Johannes May 29 '14 at 18:57
• Take a look at math.stackexchange.com/questions/660143/… let me know what you think. Your task has some sorts of answers but not others, what you really want may not be possible. – Will Jagy May 29 '14 at 19:12
• Yes, what I am looking for is exactly a parametrization as in your answer to the question there, i.e. $x = 2 r p, y = 2 r q, z = p^2 + q^2 - r^2, t = p^2 + q^2 + r^2$ for $x^2+y^2+z^2=t^2$, except that here I would like to solve all equations simultaneously. Knowing whether or not such a solution might exist would of course also be interesting! – Johannes May 29 '14 at 19:34
• Solve such a system difficult. In this form it has no solutions. There is one option. Increase the number of unknowns. Look at this topic. Maybe what that formula handy. math.stackexchange.com/questions/757664/… – individ May 30 '14 at 5:28