Rational solutions of $x^4(k^2+4)-4kx^2=z^2$, where $k>0$ is given rational.

Question

How to generate some parametric family of rational solutions to $x^4(k^2+4)-4kx^2=z^2$, where $k>0$ is given rational.

Answer 1

For $k=4$, Above equation is shown below:

$20x^4=(4x)^2+z^2$ ------(A)

Equation $(A)$ has parametric solution:

$x=2p(5m^2-4m+1)$

$z=16(p^2)(25m^4-45m^3+30m^2-9m+1)$

Where, $p=[1/(5m^2-1)]$

For $m=3$ we have:

$(x,z,k)=[(17/11),(1054/121),(4)]$

Answer 2

The equation can be written \begin{equation*} x^2((k^2+4)x^2-4k)=z^2 \end{equation*} and, so we need \begin{equation*} (k^2+4)x^2-4k=y^2 \end{equation*}

If $x=1$, then $y=\pm (k-2)$, so the line $y=k-2+m(x-1)$ will meet the quadric at the further point \begin{equation*} x=\frac{-(k^2-2km+m^2+4m+4)}{k^2-m^2+4} \end{equation*}

For example, if $k=1/2$, putting $m=4$ gives \begin{equation*} z^2=\frac{3^2*5^2*43^2*103^2}{2^2*47^4} \end{equation*}

Allan Macleod