# Primitive solutions to $a^2 + 4b^2 = c^2$

I am trying to generate primitive solutions (GCD is 1 for $a, b, c$) to the equation $a^2+4b^2=c^2$. I attempted to do this by modifying the usual Pythagorean triplet $(m^2-n^2)^2 + (2mn)^2 = (m^2+n^2)^2$ but was unable to get anywhere with that approach.

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$$a^2+4b^2=z^2\iff a^2+(2b)^2=z^2\iff \\ a=m^2-n^2, \ b=mn , \ z=m^2+n^2 , \ \ (m,n)=1 , m-n>0.$$
@WhatsInAName: To generate positive values $m$ must be larger than $n$. –  P.. Jan 13 '13 at 17:47