# Prove a parametrization function is surjective

As a starting note, I would like to say that I haven't (yet) taken courses in Set Theory, so some higher-level notation may be lost on me (and I may not write everything conventionally), but I'll do my best.

My question is as follows: I have sets of numbers which satisfy a particular Diophantine equation. Take, for example, the Pythagorean Quadruplets:
$a^2 + b^2 + c^2 = d^2; a,b,c,d\in\mathbb{R}$
Wolfram Mathworld gives the following (unscaled) parametrization for this equation:
$a = 2mp$
$b = 2np$
$c = p^2-(m^2+n^2)$
$d = p^2+(m^2+n^2)$
However, this set of equations is not surjective, as it's missing the solution 362 + 82 + 32 = 372

Wikipedia offers the following (unscaled) solution:
$a = m^2+n^2-p^2-q^2$
$b = 2(mq+np)$
$c = 2(nq-mp)$
$d = m^2+n^2+p^2+q^2$
This solves the above identity ({a,b,c,d} = {3,36,8,37}) with {m,n,p,q} = {2,4,-1,4}. This solution also happens to be surjective.

Here's the question: Given a Diophantine equation (e.g. Pythagorean Quadruplets), and a particular parametrization for this equation (e.g. Wikipedia's solution), how do you prove that the parametrization is surjective? That is, how do you prove that every set of integers that satisfies the Diophantine equation is generated by integer values of the parametrization?

Note that this is just an example; I want to know how it can be done in a general case.
Thanks for the help!
-Gabriel Benamy

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There is no general technique. It's frequently quite difficult and requires a deep understanding of the geometry of the associated diophantine equation. – Andy Putman May 23 '11 at 21:46
+1: I've always wanted to ask this question as well, with the specific equation $x^2+y^2=17z^2$ - Bjorn Poonen once told me that there's no "surjective" integer parametrization, and I always wanted to understand why. – Dror Speiser May 23 '11 at 22:08
@Dror: you can easily find all the rational solutions to $x^2+y^2=17$; just draw a line through $(1,4)$ with slope $\lambda$ and it hits the circle at a second point $(a(\lambda),b(\lambda))$. Now clear denominators, set $\lambda=m/n$ and you have a pretty good start for an integer parametrization. Perhaps the problem is just now a tedious one of the form "these solutions might not be coprime"? – Kevin Buzzard May 23 '11 at 22:24
@Kevin: if you do not allow infinite slope, the parametrization will miss the point (1,-4). In fact, affine parametrizations of conics will always miss one point - only occasionally the missing point is at infinity. In general, looking long enough at the method of parametrization should give you an idea which points you are going to miss. – Franz Lemmermeyer May 24 '11 at 10:53
@Kevin, Franz: for the specific equation any integer parametrization seems to miss infinitely many points, hence Poonen's statement. Kevin, I'm not sure what you mean by "these solutions...", but it might be this: there are rational solutions that give in the integer parametrization non-coprime numbers, and you have to divide by the gcd to get the coprime solution, and it you can't get it otherwise. Why does this happen? Andy's comment suggests this question is too general. If the question would've been only on the $=17z^2$ one, would it have received more serious attention? – Dror Speiser May 24 '11 at 18:07