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!

Your equation, $$a^2 + b^2 = c^4$$, is the same as $$a^2 + b^2 = c^2$$, except that $$c$$ has been replaced with $$c^2$$. Thus you can find solutions by choosing $$m$$ and $$n$$ so that \begin{align} a&=2mn \\ b&=m^2-n^2 \\ c^2&=m^2+n^2. \end{align} But how can you make sure that $$c$$ is an integer? Easy: that last equation, $$c^2 = m^2 + n^2$$ says that $$m,n$$ and $$c$$ also form a Pythagorean triple. Thus you can find solutions with two new parameters, $$r$$ and $$s$$ such that \begin{align} m&=2rs \\ n&=r^2-s^2 \\ c&=r^2+s^2. \end{align} Substituting these in the first set of equations to find $$a$$ and $$b$$, you have \begin{align} a&=4rs(r^2-s^2) \\ b&=6r^2 s^2-r^4-s^4 \\ c&=r^2 + s^2. \end{align} For example, choosing $$r=2$$ and $$s=1$$ gives you $$a = 24$$, $$b=7$$, $$c=5$$ and, indeed, $$24^2 + 7^2 = 5^4$$.
Whenever you have $$a^2+b^2=c^2$$ you also get $$(ca)^2+(cb)^2=c^4$$ -- does that count for you?