Triple Pythagorean with $a^2+b^2=c^4$ It is well known that there exist integer solutions to the equation $a^2+b^2=c^2$.
For example, an explicit formula for integer values of $a$ , $b$ , and $c$ is
\begin{align}a&=2mn \\ b&=m^2-n^2 \\ c&=m^2+n^2.\end{align}
I want to find solutions to the equation $a^2+b^2=c^4$. However, I got $a,b,c$ in a general form that substitutes $x$ in integer and get $a,b,c$ in the integer. What should I do first? I'm very stuck!

Whenever you have $a^2+b^2=c^2$ you also get $(ca)^2+(cb)^2=c^4$ -- does that count for you? 
