question on rational varieties and searching for parametric solutions

[American Citizen <website.reader3@gmail.com>]

Hello:

I have an equation which is satisfied for face cuboids.

(1) ((r*s)^2 + 1) * (r^2 + s^2) = t^2.

Rational solutions are possible, one is r=13, s=9 and t=1850

I have found a parametric solution, with a degree 2 rational variety, if 
I am using the math nomenclature correctly.

This solution is

r(u,v) = (3*u^2 + 2*v*u - v^2)/(3*u^2 - 2*v*u - v^2)
s(u,v) = (3*u^2 + v^2)/(2*u*v)
t(u,v) = ((9*u^4 - 6*v*u^3 + 2*v^2*u^2 + 2*v^3*u + v^4)*(9*u^4 + 6*v*u^3 
+ 2*v^2*u^2 - 2*v^3*u + v^4))/(4*v^2*u^2*(u - v)^2*(3*u + v)^2)

This afternoon I ran a C++ search changing the coefficients for both 
r(u,v) and s(u,v) to see if other degree 2 rational varieties would 
work, but found none over the search range -10 < coefficients < +10

I have two questions.

#1. I have two other two parameter solutions which look like degree 2

r2(u,v) = (v*(2*u + v))/((u - v)*(u + v))
s2(u,v) = (u*(u + 2*v))/(2*(u^2 + u*v + v^2))
t2(u,v) = ((u^4 + 2*v*u^3 + 7*v^2*u^2 + 6*v^3*u + 2*v^4)*(2*u^4 + 
2*v*u^3 + v^2*u^2 + 2*v^3*u + 2*v^4))/(4*(u - v)^2*(u + v)^2*(u^2 + v*u 
+ v^2)^2)

r6(u,v) = (u*(u + 2*v))/(2*(u^2 + u*v + v^2))
s6(u,v) = ((u - v)*(u + v))/(v*(2*u + v))
t6(u,v) = ((u^4 + 2*v*u^3 + 7*v^2*u^2 + 6*v^3*u + 2*v^4)*(2*u^4 + 
2*v*u^3 + v^2*u^2 + 2*v^3*u + 2*v^4))/(4*v^2*(2*u + v)^2*(u^2 + v*u + 
v^2)^2)

and the following which is only 1 parameter :

r9(u) = (1 - 2*u - 3*u^2)/((1 + 3*u)*(1 - u))
s9(u) = (1 + 3*u^2)/(2*u)
t9(u) = (9*u^4 + 6*u^3 + 2*u^2 - 2*u + 1)*(9*u^4 - 6*u^3 + 2*u^2 + 2*u + 
1)/(4*u^2*(u - 1)^2*(3*u + 1)^2)

can we recover a 2 parameter solution for r9,s9,t9 ?

Main question: can we discover all two parameter degree 2 parametric 
solutions? I found 3 apparently but I am sure that more exist.

#2. How would one go about setting up degree 3 varieties for parametric 
solutions to (1) ? ditto on degree > 3 ?

Randall