Integral points on a curve parametrized by quadratics.

Question

I would like to know the integer solutions to the non-linear equations

$$x=\frac{\sqrt7}{7}\left(\frac{t^2+\frac{3}{7}}{t^2-\frac{3}{7}}\right)$$ and

$$y=\frac{2\sqrt{7}}{7}\left(\frac{t}{t^2 - \frac{3}{7}} \right)$$.

where $t \in \mathbb{R}$ and $t^2 \neq \frac{3}{7}$.

I was wondering if there are general techniques for finding integer values of such equations. Finding separate integer values for $x$ and $y$ is not that hard. But making sure both $x$ and $y$ are integers is rather difficult. With the aid of computer, I have been able to find values for $x$ and $y$ by solving the equation for integers values of $y$ ranging between $1$ to $1000$, and checking if the corresponding $x$ values would be an integer. Those values are $(2,3)$ and $(218, 333)$ given by $t=\frac{\sqrt{7}}{21}+\frac{2}{3}$ and $t=\frac{\sqrt{7}}{2331}+\frac{218}{333}$.

I have not been able to come up with a general pattern. Do we even know if there are infinitely many solutions?

Answer 1

With the aid of computer, I have been able to find values for x and y by solving the equation for integers values of y ranging between 1 to 1000, and checking if the corresponding x values would be an integer.

If you use Python, there is a Sympy module specially dedicated to Diophantine equations: https://docs.sympy.org/latest/modules/solvers/diophantine.html

Answer 2

The first thing to notice is that $t$ cannot just be any real number; the condition that $x$ and $y$ are integers forces $t$ to be an algebraic number of rather low degree, by the looks of it at most degree $4$, possibly just degree $2$. Let's try to find an integer polynomial that $t$ must be a root of by manipulating the expressions for $x$ and $y$ a bit:

Dividing $x$ by $y$ cancels a lot of the mess, and we see that $$\frac{x}{y}=\frac{t^2+\tfrac37}{2t},$$ where $x$ and $y$ are integers. Clearing denominators then shows that$$7yt^2-14xt+3y=0,$$ and so $t$ is a root of a quadratic polynomial with integral coefficients, so $t$ lies in a quadratic extension of $\Bbb{Q}$. The discriminant of this polynomial is $$\Delta=196x^2-84y^2=28(7x^2-3y^2),$$ and plugging the expressions for $x$ and $y$ in terms of $t$ back in shows that $$7x^2-3y^2=1.\tag{1}$$ In particular $t\in\Bbb{Q}(\sqrt{7})$, and $(x,y)$ is a solution to a Pell equation. Or more traditionally, the pair $(7x,y)$ is a solution to the Pell equation $$X^2-21Y^2=7.$$ Of course Pell equations have been studied for millennia, their solutions are well understood and effectively computable, and thousands of near-identical questions on them exist on this site. Expressing $t$ in terms of either $x$ or $y$ is then a matter of routine algebra.