find all rational points on $x^2+7 y^2 = 1$

Question

I am familiar with the solution of "how to find all rational points on $x^2+y^2=1$ ?".

I would like to know if i can solve: "how to find all rational points on $x^2+7y^2=1$ ?" using the same technique? and if so, can you please write the solution for me (thanks).

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NOTE: the solution I'm speaking about is choosing a rational point on the ellipse and using a rational line ($ax+by = c $ for $ a,b,c \in \mathbb{Z}$) that pass trough my chosen point, and so finding all others rational points (you can find many examples for this in this cite)

Answer 1

WLOG

$x=\cos2p=\dfrac{1-t^2}{1+t^2}$

and $\sqrt7y=\sin2p=\dfrac{2t}{1+t^2}$ where $t=\tan p$

$\implies y=\dfrac{2t/\sqrt7}{1+t^2}$

So, it is sufficient to make $t/\sqrt7$ rational

Answer 2

The point $(1,0)$ is on the curve. We consider a general line through this point, it has the equation $y=t(x-1)$, $t\in \Bbb Q$, and intersect them, with the given quadratic equation, computer supported by sage, here for the case "with the $7$", compared with the one "with the $1$" instead:

sage: solve( [x^2 + 7*y^2 == 1, y == t*(x-1)], [x,y] )
[[x == 1, y == 0], [x == (7*t^2 - 1)/(7*t^2 + 1), y == -2*t/(7*t^2 + 1)]]

sage: solve( [x^2 + y^2 == 1, y == t*(x-1)], [x,y] )
[[x == 1, y == 0], [x == (t^2 - 1)/(t^2 + 1), y == -2*t/(t^2 + 1)]]

(Now eventually replace $t=m/n$, irreducible fraction.)