I later discovered a couple of solutions, (g,h,q) = (15,17,4)/12 and (12,17,12)/15 --- apologies for earlier false claim, and thanks to those who contributed! The upside to this fresh clanger was that, thinking there were no solutions, I generalised the construction to incorporate a new parameter. Instead of an outermost cuboid, there is a trapezohedron with 2 congruent e x f (empty) rectangles above and below, a band of 4 congruent isosceles trapezium faces of vertical depth g around the outside, and 16 interior triangular faces as before connecting to a quadrangle of height h and radius d (was q). This (homogeneous) 5-parameter polytore is isometric / origami / planar just when vanishes the quartic (2*d^2+3*e^2+3*f^2+2*e*f-4*e*d-4*f*d)*h^2 + (2*d^2-e^2-f^2-2*e*f)*g^2 + 2*(e-f)*(-e+2*d-f)*g*h + (d^2-e^2-f^2)*(-e-f+d)^2 ; conditions for embedding are unfortunately not as obvious as before, involving 3 new quadratic inequalities.

From a pedagogic point of view, we can now construct examples which are both rational, and easier to visualise than the cuboid special cases --- (d,e,f,g,h) = (9,16,15,8,16) is particularly nicely balanced.

Fred Lunnon On 9/1/09, Warut Roonguthai <warut822@gmail.com> wrote:

On Wed, Sep 2, 2009 at 4:24 AM, Fred W. Helenius<fredh@ix.netcom.com> wrote:

...

While sitting through a seminar this afternoon, I found that g = 20/9, h = 7/3, q = 8/9 works.

This proves that my parametric solution does not cover all the rational solutions! :P

Warut

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I've thought off and on about the info from John Cremona reported by Victor; but I must confess that I don't really understand its implications. I'm used to thinking about extensions by irrationals, but not by rationals ... What can be said in general about the structure of the rational solution sets of these equations (the cube case with e = f = g (= 1), the cuboid with e = f (= 1), or the (most general) trapezoidal with all 5 (homogeneous) parameters? For example, what's their dimension? Current presentation (with old q -> new d) of "planar_trapeze" constraint : + ((e-f)^2+2*(e+f-d)^2)*h^2 - ((e+f)^2-2*d^2)*g^2 - 2*(e-f)*(e+f-2*d)*g*h - (e^2+f^2-d^2)*(e+f-d)^2 = 0 . Wlog, we can assume that all of d,e,f,g,h are in |N. A necessary condition on the embedding region is 0 < 2*d < e+f [more required for sufficiency, but not relevant --- yet]. To establish the sufficiency of planar_trapeze constraint, I need to bound the "right_angle" polynomial e^2 + f^2 - (e+f)*d + g^2 - h^2 + e*f away from zero in this region, subject to constraint. I can do this for the cuboid case. For the general trapezoidal case, I can show that right_angle > 1/2 on the boundary in the same way, by expressing it as a sum of squares --- but I cannot see how to establish that there are no turning points in the interior. Any ideas, anybody? [Lagrange lambda shows only that there is a maximum outside, on the boundary of the immersion region at e = f = 1 (say), d = 2+2sqrt2, g = h = 0.] Fred Lunnon On 9/4/09, 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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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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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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