seeking an integer parameterization for A^2+B^2=C^2+D^2+1 I'm looking for a complete [integer] parameterization of all integer solutions to the Diophantine equation
$A^2+B^2=C^2+D^2+1$,
analogous to the classical parameterization of the Pythagorean equation, i.e.
$A^2+B^2=C^2 \implies t,m,n \text{ such that } (A,B,C)=t(m^2-n^2,2mn,m^2+n^2)$.
Dickson's History contains many references and examples, but most appear to be inadequate, incomplete, or simply incorrect. Barnett and Bradley independently reached almost the same parameterization of the more general equation
$A^2+B^2+C^2=D^2+E^2+F^2$,
but I have so far been unable to reduce their parameterization(s) to one which solves the first equation I posted.
Any help or further references would be greatly appreciated.
Thanks!
Kieren.
 A: This is completely unrelated to my other answer. This class of problems is considered by L. N. Vaserstein in his 2006 Annals paper (preprint here): Polynomial parametrization
for the solutions of Diophantine equations and arithmetic groups. Vaserstein appears to show that there is a polynomial parametrization (or at least a decomposition into polynomially parametrized sets) of integer solutions for this class of problems, but it ain't going to be pretty.
A: Such a parametrization is not possible.
Proof. Suppose A,B,C,D are polynomials with integer coefficients (in any number of variables) and A^2+B^2 = C^2+D^2+1. 
Then we have a parametrization for the congruence subgroup  H of SL(2,Z) reducing to
the permutation matrices modulo 2 (see Noam D. Elkies Aug 13 '12 at 15:49 above).
This H appears in Example 14 in my Annals paper. The first rows in H are (A+C, B+D).
Modulo 2, (A+C)(B+D) = 0, hence either  A+C or C+D is always even. But those rows contain both (1,0) and (0,1). This is a contradiction. QED.
A: This question (and answers/comments) is extremely relevant: integer solutions to quadratic forms
A: Though what any solution of the equation: $X^2+Y^2=Z^2+R^2+1$
Need to write. Though such.
$X=2t+1$
$Y=t-1$
$Z=2t$
$R=t+1$
$t,a,b$ any integer.
$X=b(a(a-1)b-1)$
$Y=a(a-1)b^2+1$
$Z=(a-1)b(ab+1)$
$R=ab((a-1)b-1)$
