Hi, I'm trying to find all rational points on the surface of the title, in connection with the Euler Brick (AKA Rational Box) problem.
This surface is equivalent to $ x^2 z^2 - 1 = (x^2 - z^2) y^2 $, and it is easy to find an addition formula for rational points by jogging the second into a form conformable with the equations defining Jacobian elliptic functions:
A rational point on the second form clearly implies rational $a, b, .. e$ (and conversely) satisfying the pair $ (a b)^2 - 1 = e (a b c)^2 $ and $ (a b)^2 - b^4 = e (a b d)^2 $ and dividing each by $ (a b)^2 $, one can express the variables as follows with modulus $ k = b^2 $ :
$ sn(k,u) = 1 / (a b) $
$ cn(k,u) = \sqrt e . c $
$ dn(k,u) = \sqrt e . d $
Because in the addition formulae for Jacobian elliptic functions, $cn$ and $dn$ arise only in products of pairs, and "cross products" $cn . dn$ occur only in the resulting $sn$, this means that a pair of rational solutions with $e = e_1$ and $e = e_2$ implies a rational solution with $E = e_1 e_2$ to the pair:
$ C^2 + E S^2 = 1 $
$ D^2 + b^4 E S^2 = 1 $
Treating this as a Diophantine pair in its own right, without reference to Elliptic functions, one can absorb $ S^2 $ into E, and plug the expression for $E$ given by the first into the second to obtain:
$ D^2 - b^4 C^2 = 1 - b^4 $
For any given $b$ this is a conic curve, with a rational point $ C, D = 1, 1 $ and hence has a rational parametrization over Q, in b^4 and some parameter t say.
Now I'm sure the original surface is not rational (I gather it is a K3 surface). But I wonder if the above can be reversed, to some extent, to at least give a procedure for finding all rational points on the surface from those of the pencil of conics parametrized by $b$.
