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.
 A: Consider the circle $$x^2+y^2=5$$ Find a rational point on it (that shouldn't be too hard). Then imagine a line with slope $t$ through that point. It hits the circle at another rational point. So you get a family of rational points, parametrized by $t$. Rational points on the circle are integer points on $a^2+b^2=5c^2$. 
A: 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
$$ a^2 + b^2 = 5  c^2 $$
A: 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 $x^2 + y^2 = 5z^2$.
$\blacksquare$.
