question on rational varieties and searching for parametric solutions
American Citizen <[email protected]> Wed, 25 Feb 2026 18:18:06 -0800
| Newsgroups | gmane.comp.mathematics.pari.user |
|---|---|
| Message-ID | <[email protected]> |
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