Describe the set of rational points on the curve

Question

Describe the set of rational points on the curve

$x^2-7y^2=2$

Given that $(3,1$) is on the curve

Answer 1

Yes, your comments show you have the right idea. It is a generalization of the standard rational parametrization of the unit circle. Look at the general line through $(3,1)$ with slope $t$. This has equation $y-1=t(x-3)$.

Find out where this line meets the hyperbola again. We substitute $t(x-1)+1$ for $y$.

The result is $x^2-7(t(x-3)+1)^2=2$. Solve for $x$ in terms of $t$. Then use $y=t(x-3)+1$ to express $y$ in terms of $t$. Now you have a rational parametrization of the curve $x^2-7y^2=2$. As $t$ ranges over the rationals, your $(x(t),y(t))$ ranges over all rational points on the curve except $(3,1)$. Actually, $(3,1)$ is not missed either. It corresponds to $t=\infty$.

Technical note: The required root of the quadratic is easy to compute. For from the quadratic you know the product of the roots. But one of the roots is $3$, so the other root is fairly simple to write down. Or else you can use the fact that you know the sum of the roots, and one root.

Remark: A similar idea is of great importance in the theory of elliptic curves.