Here is how I parametrized rational solutions to the Diophantine equation: h^2/(2-q^2) = g^2/(2-q)^2 + 1/2. If q is rational, we can write sqrt(2-q^2) as z*sqrt(m), where z is rational and m is an integer of the form a^2 - 2*b^2 (a, b are integers). Let h/sqrt(2-q^2) + g/(2-q) = k, h/sqrt(2-q^2) - g/(2-q) = 1/(2*k). Thus, g = (2*k^2-1)/(4*k)*(2-q), h = (2*k^2+1)/(4*k)*z*sqrt(m). Since g, h are rational, we must have (1): (2*k^2-1)/(4*k) = u and (2): (2*k^2+1)/(4*k) = v*sqrt(m) for some rational numbers u, v.
From (1)+(2) and (1)-(2), we have k = u + v*sqrt(m) and -1/(2*k) = u - v*sqrt(m), respectively.
So u^2 - m*v^2 = -1/2, or (2*u)^2 - m*(2*v)^2 = -2. It follows that m is also a number of the form c^2 + 2*d^2 (c, d are integers). The remaining steps are to find a parametric form of m = a^2 - 2*b^2 = c^2 + 2*d^2 and to solve for rational points (q, z) on 2 - q^2 = m*z^2 and rational points (u, v) on (2*u)^2 - m*(2*v)^2 = -2. These steps are rather straightforward. It's interesting to note that this derivation involves only elementary number theory. BTW, my comment on http://www.research.att.com/~njas/sequences/A155562 is a by-product of this effort. :) Warut On Mon, Oct 19, 2009 at 12:33 AM, Warut Roonguthai <warut822@gmail.com> wrote:
Fred Lunnon asked me if there are rational solutions to
h^2/(2-q^2) = g^2/(2-q)^2 + 1/2
with g = 1. (Such solutions have special geometric meaning that I don't understand.)
The answer is "no" and I'd like to share my proof with the math-fun list.
If g = 1, we have h^2 = 2*(2+(2-q)^2)*(2-q^2)/(4*(2-q)^2).
So we want 2*(2+(2-q)^2)*(2-q^2) to be a perfect square rational.
Suppose that q = a/b with gcd(a, b) = 1 makes 2*(2+(2-q)^2)*(2-q^2) a perfect square rational. It follows that 2*(2*b^2+(2b-a)^2)*(2*b^2-a^2) is a perfect square integer.
If 2*((2b-a)^2+2*b^2)*(2*b^2-a^2) is a square modulo 16, then a, b must be both even, hence contradicting to our assumption that gcd(a, b) = 1. ###
I'll describe how I derived the complete parametric solutions later...
Warut
On Mon, Oct 12, 2009 at 11:26 PM, Warut Roonguthai <warut822@gmail.com> wrote:
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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