Rational solutions of the equation $2xy^2=kx^2+1$

Question

I am looking for the rational solutions of the equation: $$2xy^2=kx^2+1$$

where $k$ is a fixed positional natural number. For $k=6$, Maple shows it is irreducible and doesn't produces the rational parametrization.

Answer 1

Look at this as a degree $2$ equation in $x$, i.e., $kx^2-2y^2x+1=0$. The solutions are $x_{1,2}=\frac{y^2}{k}\pm\frac{\sqrt{y^4-k}}{k}$. Since $y,k\in\mathbb{Q}$, these solutions are rational if and only if $\sqrt{y^4-k}\in\mathbb{Q}$.

So to get all solutions, you just need to find $y\in\mathbb{Q}_+$ such that $\sqrt{y^4-k}\in\mathbb{Q}$ (restricted to $\mathbb{Q}_+$ since if $(x,y)$ is a solution, then so is $(x,-y)$).

Letting $y=\frac{r}{s}$, this means $\big(\frac{r}{s}\big)^4=k+\big(\frac{p}{q}\big)^2$, where $r,s,p,q$ are naturals and $r,s$ and $p,q$ are coprime. It is not difficult to see that it is necessary that $q=s^2$, so we are left with the equation \begin{equation*} r^4-ks^4=p^2, \quad\quad (*) \end{equation*} for some natural $p$ under the restriction that $r$ and $s$ are coprime.

Consider the 'weaker' equation \begin{equation*} R^2-kS^2=p^2, \quad\quad (**) \end{equation*} which is a quadratic Diophantine equation, and note that if $(r,s,p)$ is a solution for $(*)$, then clearly $(r^2,s^2,p)$ is a solution for $(**)$ and if $(R,S,p)$ is a solution for $(**)$, it gives a solution for $(*)$ if and only if $R$ and $S$ are both squares.

So it suffices to parametrize the solutions for $(**)$ and we can recover those of $(*)$ as a subset. There are known ways of listing solutions for homogeneous degree $2$ Diophantine equations (see e.g. https://en.wikipedia.org/wiki/Diophantine_equation).

It is also maybe interesting to point out that $(**)$ will always have non-trivial solutions. If $k$ is not a square, we can for example find $(R,S)$ as the solutions of the Pell equation $R^2-kS^2=1$ by setting $p=1$. If $k=\ell^2$, then we can find Pythagorean solutions for the equation $R^2=\ell^2+p^2$ by just taking $S=1$.