I'm looking for a complete [integer] parameterization of all integer solutions to the Diophantine equation


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


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.

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    $\begingroup$ $A^2 + B^2 = C^2 + D^2 + 1$ is equivalent to $$ 1 = (A^2-C^2) - (D^2-B^2) = (A+C)(A-C) - (D+B)(D-B). $$This identifies the set of solutions with the congruence subgroup of $\mathop{\rm SL}_2({\bf Z})$ consisting of matrices that reduce mod $2$ to either the identity or $({0\phantom.1\atop1\phantom.0})$. I don't know if there's a parametrization of this group available, but maybe enough is known about its elements for your needs. $\endgroup$ – Noam D. Elkies Aug 13 '12 at 15:49
  • $\begingroup$ What exactly do you need a parameterization for? Writing down solutions is straightforward using the Euclidean algorithm, so if that's all you want to do... $\endgroup$ – Qiaochu Yuan Aug 13 '12 at 17:19
  • $\begingroup$ Do you have a reference for Barnett or Bradley's parameterizations? $\endgroup$ – Zack Wolske Aug 13 '12 at 18:41
  • $\begingroup$ Bradley: jstor.org/stable/3620159 Barnett: jstor.org/stable/2302941 $\endgroup$ – Kieren MacMillan Aug 13 '12 at 20:32

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.

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  • $\begingroup$ This is interesting -- thanks! The solutions I've been working towards (as near as I've gotten them) are definitely not pretty, so this is at least validation, if not particularly encouraging. Thanks! Kieren. $\endgroup$ – Kieren MacMillan Aug 13 '12 at 20:28

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.

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  • $\begingroup$ So the proofs (and included parameterizations) by Bradley (jstor.org/stable/3620159) and Barnett (jstor.org/stable/2302941) are false? Or is it just not possible to reduce their general parameterization(s) $A^2+B^2=C^2+D^2+E^2$ to the special case $e = \pm1$? $\endgroup$ – Kieren MacMillan Jun 14 '14 at 21:29

This question (and answers/comments) is extremely relevant: integer solutions to quadratic forms

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    $\begingroup$ I'm afraid it only looks similar: $A^2+B^2=C^2+D^2$ is a homogeneous equation, so solutions amount to rational points on a surface (up to scaling), which are easily parametrized in this case (the surface is birational to ${\bf P}^1 \times {\bf P}^1$; whereas $A^2+B^2=C^2+D^2+1$ is integral points on a smooth threefold, so the underlying geometry is quite different. $\endgroup$ – Noam D. Elkies Aug 13 '12 at 16:58
  • $\begingroup$ @Noam: I am fully aware of what you say, but notice that transforming this (essentially as you did, and as Matt Young did in the referenced question) into $(D+B)(D-B) = (A+C)(A-C) - 1$ allows one to generate all the solutions by letting $u=A+C, v = A-C$ and then $(D+B), (D-B)$ factors of $uv - 1$ (modulo some parity considerations) gives a quick way to generate solutions (but does not quite answer the question, I admit). $\endgroup$ – Igor Rivin Aug 13 '12 at 17:19

Though what any solution of the equation: $X^2+Y^2=Z^2+R^2+1$

Need to write. Though such.





$t,a,b$ any integer.





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  • 2
    $\begingroup$ Why are these all of the solutions? $\endgroup$ – Alex Degtyarev Nov 1 '14 at 19:40

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