Let $n>3$ be an integer. What are the rational solutions of


$$y^2 = 4x^n + z^{n-1}$$


?



What are the rational solutions of $y^2 = 4x^n + z^{n-1}$?
We consider two cases: $n$ is odd or $n$ is even. If $n=2k+1$ then put
$$
x=uz, \quad y=vz^{k},
$$
where $u, v$ are non-zero rational parameters. Thus $v^2z^{2k}=4u^{2k+1}z^{2k+1}+z^{2k}$ and after division by $z^{2k}$ we get linear equation in $z$ which can be easily solved, i.e., $z=(v^2-1)/(4u^{2k+1})$. We thus get
$$
x=\frac{v^2-1}{4u^{2k}},\quad y=v\left(\frac{v^2-1}{4u^{2k+1}}\right)^{k}. 
$$


If $n=2k$ one can use exactly the same type of reasoning, i.e., we put
$$
y=ux^{k},\quad z=vx.
$$
After necessary simplifications, we get $x=v^{2k-1}/(4-u^2)$ and thus
$$
y=u\left(\frac{v^{2k-1}}{4-u^2}\right)^{k},\quad z=v^{2k}/(4-u^2).
$$


