Parameterize the solutions $x^2 + 5y^2 = z^2$ Parameterize the solutions in integers to $$x^2 + 5y^2 = z^2$$
To make it easier, consider only the solutions such that the GCD of x, y, and z is 1. Also
assume that x, y, and z are positive, and that x is odd. 
The analysis will probably still give two cases, but we can combine them into one by use of absolute values. Basically I have to prove that the analysis is correct.
 A: Another way to think about such problems is to convert this to 
$$\left(\frac{a}{c}\right)^2+5\left(\frac{b}{c}\right)^2=1,$$
and then think about rational points (both coordinates rational) on the curve
$$x^2+5y^2=1.$$
One obvious point is $(1,0)$. Let us consider a line that passes through this point $y=m(x-1)$ with rational slope $m$. Observe that all the points of intersection of this line and the curve (given above) will be rational. To find the intersection we solve
$$x^2+5m^2(x-1)^2=1.$$
This is equivalent to
$$x^2(5m^2+1)-10m^2x+(5m^2-1)=0.$$
Since one of the roots of this equation has to be $x=1$ (our trivial rational point) so the other root will be 
$$\alpha=\frac{5m^2-1}{5m^2+1}.$$
Thus we obtain the following point as the other point of intersection:
$$\left(\frac{5m^2-1}{5m^2+1}, \, \frac{-2m}{5m^2+1}\right).$$
So if we take $a=5m^2-1, b=-2m$ and $c=5m^2+1$ (with $m \in \mathbb{Z}$) then we can obtain integer solutions to the equation $a^2+5b^2=c^2$. 
More generally, if we take $m=\frac{r}{s}$ (with $r,s \in \mathbb{Z}$ and $s \neq 0$) then we can have $a=5r^2-s^2, b=-2rs$ and $c=5r^2+s^2$ as the solutions.
