I'm not sure if there is anyone still interested in the Diophantine equation: h^2/(2-q^2) = g^2/(2-q)^2 + 1/2. Anyway, now I think I can parametrize all of its rational solutions with 6 parameters. I need 4 integers: t, u, v, w (rational values are ok, but not necessary) and 2 rationals: r, s. It's more compact to describe the solutions in steps as follows: Let a = t^2+u^2+v^2+w^2, b = t*v+t*w+u*v-u*w, c = t^2+u^2-v^2-w^2, d = t*v-t*w-u*v-u*w, and m = a^2-2*b^2 = c^2+2*d^2. Let q, z be the rational solutions of q^2 + m*z^2 = 2. So q = ((a-2*b)*r^2+2*m*r-(a-2*b)*m)/((a-b)*(r^2+m)), z = (r^2-2*(a-2*b)*r-m)/((a-b)*(r^2+m)). Let x, y be the rational solutions of x^2 - m*y^2 = -2. So x = ((c-4*d)*s^2+6*m*s+(c-4*d)*m)/((2*c+d)*(s^2-m)), y = (3*s^2-8*d*s+2*c*s+3*m)/((2*c+d)*(s^2-m)). Finally, we have g = (2-q)*x/2, h = m*z*y/2. Warut On Sat, Sep 5, 2009 at 1:48 AM, Warut Roonguthai <warut822@gmail.com> wrote:
Here's another parametric solution:
q = 4r/(2*r^2+1),
h = (2*s^2+1)*(2*r^2-1)/((2*s^2-1)*(2*r^2+1)),
g = 4*s*(2*r^2-2*r+1)/((2*s^2-1)*(2*r^2+1)),
where r, s are rational numbers. This one includes Fred W. Helenius's solution, but not my previous solutions. Fred Lunnon's solution is not included in any of my parametric solutions.
Warut
On Fri, Sep 4, 2009 at 3:32 AM, victor miller<victorsmiller@gmail.com> wrote:
I gave this to John Cremona who has an algorithm for finding a point (if it exists) on a diagonal conic over a function field. The equation in question is an example of such -- if we clear denominators we get:
(2-q)^2*h^2 = (2-q^2)*g^2 + 1/2*(2-q)^2*(2-q^2) (*)
So this is a conic in (g,h) over the field Q(q).
It's well know that if you have one solution to a conic, all the others are generated by a rational parametrization from that one. However, John's program showed that (*) had no points in Q(q). Warut's parametrized solutions essentially amount to finding one point on (*) in a quadratic extension of Q(q) (his parameter, r, satisfies a quadratic over Q(q)), and then using the general recipe to generate all other solutions over that field. An alternative approach would be to look at (*) as an elliptic curve in (h,q) over Q(g), which must have positive rank since we know that there are an infinite number of solutions.
Victor
On Tue, Sep 1, 2009 at 11:13 AM, Fred lunnon <fred.lunnon@gmail.com> wrote:
It's always nice to have examples of geometrical configurations with rational parameters --- apart from anything else, they're useful for testing software.
A polytore with square section edge 2, outer cuboid depth g, inner quadrangle height h, inner radius q, is planar just when
h^2/(2-q^2) = g^2/(2-q)^2 + 1/2
but this equation appears to have no small rational solutions, at any rate when 0 < q < 1.
Is it soluble over the rationals? WFL
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