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.

This is one of those CW answers. Country and Western.
I did Gerry's recipe and I quite like how it works. Educational, you might say. 
I took the slope $t = \frac{q}{r}$ and starting rational point $(2,1).$ The other point works out to be
$$  x = \frac{2 t^2 - 2 t - 2}{t^2 + 1}, \; \; \; y =  \frac{- t^2 - 4 t + 1}{t^2 + 1},   $$ so multiply everything by $r^2$ to arrive at
$$  x = \frac{2 q^2 - 2qr  - 2r^2}{q^2 + r^2}, \; \; \; y =  \frac{- q^2 - 4 qr + r^2}{q^2 + r^2}.   $$
So far $x^2 + y^2 = 5.$  Multiply through by $q^2 + r^2$ to get 
$$ a = 2  q^2 - 2 q  r - 2  r^2 $$  
$$ b = -q^2 - 4  q  r + r^2 $$ 
$$ c = q^2 + r^2 $$
$$ a^2 + b^2 =  5 q^4 + 10 r^2 q^2 + 5 r^4 $$
and
$$   c^2 =  q^4 + 2 r^2 q^2 + r^4 $$
and
$$ a^2 + b^2 = 5  c^2 $$
