Generating Pythagorean triples for $a^2+b^2=5c^2$? Just trying to figure out a way to generate triples for $a^2+b^2=5c^2$. The wiki article shows how it is done for $a^2+b^2=c^2$ but I am not sure how to extrapolate.

Here is a cute way to construct a family of solutions for the given diophantine:
Key Theorem: Let $a, b, c, d \in \mathbb Z$. Then, $$(a^2 + b^2)(c^2 + d^2) = (ac + bd)^2 + (ad - bc)^2$$
Proof: Expand.
Consider a pythagorean triple $(a, b, c)$, which you can easily generate. A simple application of the key theorem yields:
$a^2 + b^2 = c^2$, $\implies$ $5(a^2 + b^2) = 5c^2$, so $(2^2 + 1^2)(a^2 + b^2) = 5c^2$ $\implies$ $(2a + b)^2 + (2b - a)^2 = 5c^2$.
Therefore, if $(a,b,c)$ is a pythagorean triple, then $(2a + b, 2b - a, c)$ is a solution of $x^2 + y^2 = 5z^2$.
$\blacksquare$.
