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Jul 30, 2010 at 3:01 vote accept john mangual
Jul 29, 2010 at 22:35 comment added Pietro Majer Also (especially for rational solutions): make the substitutions w:=x+h, and the equation becomes linear in x. Use y,z,h as parameters and solve in x. Normalizing gives a parametrization for integers solution also (with some care about parity).
Jul 29, 2010 at 22:33 answer added Tony Scholl timeline score: 15
Jul 29, 2010 at 22:14 comment added Qiaochu Yuan Matt's approach is the way I would go. It's worth noting that a similar approach works for Pythagorean triples and, with a little effort, for Pythagorean quadruples x^2 + y^2 + z^2 = w^2 as well.
Jul 29, 2010 at 21:22 comment added Daniel Litt +1 Eric Tressler. Expanding on is comment, the relevant equations on the mathworld page are labeled 16-18. Basically, your question boils down to finding the splittings of the prime factors of $n=x^2+y^2$ that are equal to 1 mod 4.
Jul 29, 2010 at 21:18 comment added Matt Young By rearranging your formula, you get to the equation $(x-z)(x+z)=(w-y)(w+y)$. Basically $x-z$ and $x+z$ can be viewed as independent variables (the slight catch is that they have the same parity) and similarly for the right hand side. So in effect you want to solve the equation $ab = cd$ where $a = b \pmod{2}$ and $c = d \pmod{2}$. By letting $a$ and $b$ vary you then want to find all the different factorizations of $ab$ with a constraint modulo $2$.
Jul 29, 2010 at 21:08 comment added Eric Tressler A good starting point is research.att.com/~njas/sequences/A004018, which will tell you the number of ways to write $n$ as the sum of two squares, and give you some references. See also mathworld.wolfram.com/SumofSquaresFunction.html
Jul 29, 2010 at 20:59 history asked john mangual CC BY-SA 2.5