I came across this Diophantine equation $$4x^2+y^4=z^2$$ Primitive solutions of this equation can be found by \begin{align} \begin{split} x&=2ab(a^2+b^2)\\ y&=a^2-b^2\\ z&=a^4+6a^2b^2+b^4\\ \end{split} \end{align} where $a$, $b$ are relatively prime and $1 \leq b < a$. One of these two, one is odd, and the other one is even.
I would like to know the intermediate steps that are required to find such a parametrization. I tried to manipulate the identity. $$4x^2+y^4=z^2 \implies 4x^2=(z+y^2)(z-y^2)$$ and then using parity check to further simplify it. Another approach was $$4x^2+y^4=z^2 \implies 4xy^2=(2x+y^2+z)(2x+y^2-z)$$ I could not make any more progress. Some help will be appreciated.
As JCAA's question comment suggests, rewriting the equation as $(2x)^2 + (y^2)^2 = z^2$ shows $2x$, $y^2$ and $z$ form a Pythagorean triple. Primitive solutions are obtained from
$$2x = 2mn \implies x = mn \tag{1}\label{eq1A}$$
$$y^2 = m^2 - n^2 \implies y^2 + n^2 = m^2 \tag{2}\label{eq2A}$$
$$z = m^2 + n^2 \tag{3}\label{eq3A}$$
where $m$, $n$ are relatively prime and $1 \leq n < m$.
Note \eqref{eq2A} shows $y$, $n$ and $m$ form another Pythagorean triple. In this case, it's another primitive solution since $\gcd(m,n) = 1$ means $y$ is also relatively prime to $m$ and $n$. Also, since $y$ is odd (due to $2x$ being even in the original equation), you get
$$y = a^2 - b^2 \tag{4}\label{eq4A}$$
$$n = 2ab \tag{5}\label{eq5A}$$
$$m = a^2 + b^2 \tag{6}\label{eq6A}$$
where $a$, $b$ are relatively prime and $1 \leq b < a$. Plugging \eqref{eq5A} and \eqref{eq6A} into \eqref{eq1A} gives
$$x = 2ab(a^2 + b^2) \tag{7}\label{eq7A}$$
while plugging \eqref{eq5A} and \eqref{eq6A} into \eqref{eq3A} gives
$$\begin{equation}\begin{aligned} z & = (a^2 + b^2)^2 + (2ab)^2 \\ & = a^4 + 2a^2b^2 + b^4 + 4a^2b^2 \\ & = a^4 + 6a^2b^2 + b^4 \end{aligned}\end{equation}\tag{8}\label{eq8A}$$