How does one solve the diophantine equation $x^2 + y^2 = z^2 + w^2 $? Can solutions be parameterized in three variables analogously to the Pythagorean triples case?



integer solutions to quadratic forms
Here is the standard geometric argument: after extracting common factors, you are asking for rational points on the quadric $Q\colon x^2-w^2=z^2-y^2$ in $\mathbb{P}^3$, which is isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$ (as Matt Young's comment explains). Projection from a point on $Q$ gives a birational map $p:Q\to \mathbb{P}^2$ hence a rational parameterisation of $Q$. Explicitly, projecting from $P\_0=(0:1:0:1)$ to the $(w=0)$-plane gives the parameterisation 
$$
p : (a:b:c)\in\mathbb{P}^2 \mapsto (2ab:c^2-a^2+b^2:2bc:c^2-a^2-b^2) \in Q
$$
of all the points except for the two lines on $Q$ passing through $P\_0$ - i.e. the intersection of $Q$ with the tangent plane at $P\_0$, which are $z=w$, $x=\pm y$. Notice also that the points $\{b=0,a\ne\pm c\}$ all go to $P\_0$, for the same reason.


The advantage of this method is that there is nothing particularly special about the quadratic form $x^2+y^2-z^2-w^2$ here; the same argument parameterises the rational zeros of any nonsingular quadratic form in 3 or more variables, as soon as it has one non-trivial zero. 


