Conference rusure::math

    If you consider your function f(x,y) as a plane algebraic curve,
    then a rational parameterization only exists if the curve has
    genus zero.  This is the case for the conic sections (circles,
    hyperbolas, parabolas), for cubics with a self intersection
    (or double point) or for quartics with a triple point.

    In implicit form you have a quartic:

	f(x,y) = x^2*y^2 - (x^2+y^2) - 4*E*x*y + 1 = 0

    Is there a point where this curve intersects itself twice?  If so,
    a line drawn thru that point will only meet the curve at one other
    point and the slope of that line will be a rational parameterization.

    I may have to look this up at home since I don't remember the
    exact details, but it's in Semple and Roth's (no relation) book on
    algebraic geometry.

    [genus zero means the curve as a Riemann surface is topologically a
    sphere;  the most general functions on the sphere are the rational
    functions.  A general cubic is genus one, or a torus and elliptic
    functions are required for parameterization.  It rapidly gets
    hairy for higher degrees...]

    later...

        - Jim

    I sketched some examples of this curve for various values of E in the
    range [-2,2] and it looks like you're out of luck for a rational
    parameterization. (It makes a nice symmetric design, actually...)

    - Jim

>   Is there a point where this curve intersects itself twice?
    
    I thought 'certainly there is'.  But I solved for y, and set the
    discriminant to 0.
    
    	y = (2 E x � sqrt(x^4 + (4 E� - 2) x� + 1))/(x� - 1)
    	                  x^4 + (4 E� - 2) x� + 1 = 0
    
    But that's (x� - 1)� + 4 E�x�, which is never 0.  Sigh, no luck.
    
    
    (BTW, is there some 'easy' way of recognizing an expression as the sum
    of two squares?  Will MAPLE do this?)
    
    
>   and the slope of that line will be a rational parameterization.
    
    This sounds very strange and interesting.  I hope it's in Semple and Roth.
    
    
    Well, how about:
    
    	(w - 1/w) (x - 1/x) = (y - 1/y) (z - 1/z)
    
    Does this have non-trivial parameterizations?

>    (BTW, is there some 'easy' way of recognizing an expression as the sum
>    of two squares?  Will MAPLE do this?)

I think the answers are {No, No}. FWIW, every *integer* is the sum of 4 
integer squares, and they can be found by a simple backtrack scheme, 
largest first. But I don't see any way of dealing formally with something
like 
                       2  2       2          2      2      2
                    5 x  y  - 12 x  y - 6 x y  + 9 x  + 9 y

(original form appears after the spoiler!) 




                                2              2
                   (3 x - 2 x y)  + (x y - 3 y)

But we can always write an expression E as the *difference* of two squares.

		      E + w�  2     E - w�  2
		E = (--------)  - (--------)
		       2 w           2 w

Where w is a free variable or expression.

re .-1: Yes, but that's like the moral equivalent of saying "For any 
object, one can find a set of objects of the same type of which the given
object is a non-member." Not a lot of help in attacking the problem!

        re .3,
        
>    (BTW, is there some 'easy' way of recognizing an expression as the sum
>    of two squares?  Will MAPLE do this?)

        The first order theory of real closed fields in the
        language of fields {0,1,+,*} is complete.  So it can be
        proven whether or not a polynomial in x,y,z,... with
        rational coefficients ever takes on a negative value for
        real x,y,z,....  If it does, then it cannot be a sum of
        squares.  If it never takes on negative values, then I
        think it may be possible to write as a sum of squares of
        rational functions (quotients of polynomials), but I
        haven't been able to track down where I read that to see
        a precise statement of the theorem (such as, does it
        apply here).
        
        Dan