Exercise 5.6.1 (Parametrization of $x^2 - 2y^2$, starting from $(1,0)$)

Question

Find a parametrization of the rational points on the hyperbola $x^2 - 2y^2 = 1$, starting from the point $(1,0)$.

Richard Ganaye · Accepted Answer

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

\begin{proof} The line with slope $t$ passing by the point $(1,0)$ has for equation
$$y = t(x-1).$$
Let $(x,y) \in \Q^2$. Then
\begin{align*}
x^2 - 2y^2 = 1 &\iff 
\left\{
\begin{array}{l}
(x, y) = (1,0)  	ext{ or } \
x 
e 1 	ext{ and } \exists t \in \Q,\ 
    \left\{
    \begin{array}{l}
    x^2 - 2t^2(x-1)^2 = 1,\
    y = t(x-1),
    \end{array}
    ight.
\end{array} 
ight.\
&\iff
\left\{
\begin{array}{l}
(x, y) = (1,0) 	ext{ or } \
x 
e 1 	ext{ and } \exists t \in \Q,\ 
    \left\{
    \begin{array}{l}
    (x-1)[x+1 - 2t^2(x-1)] = 0,\
    y = t(x-1),
    \end{array}
    ight.
\end{array} 
ight.\
&\iff
\begin{array}{l}
(x, y) = (1,0)  	ext{ or } \exists t \in \Q,\ 
    \left\{
    \begin{array}{l}
    x(1-2t^2) + 2t^2 + 1 = 0,\
    y = t(x-1),
    \end{array}
    ight.
\end{array} \
&\iff
\begin{array}{l}
(x, y) = (1,0)  	ext{ or } \exists t \in \Q,\ 
    \left\{
    \begin{array}{l}
    x = \frac{2t^2+1}{2t^2 - 1},\
    y = \frac{2t}{2t^2 - 1}.
    \end{array}
    ight.
\end{array} 
\end{align*}

\end{proof}
With Sagemath:
\begin{verbatim}
var('x,y,t')
solve([x^2 - 2*y^2 == 1, y == t*(x-1)],x,y)
\end{verbatim}
$$
\left[\left[x = \frac{2 \, t^{2} + 1}{2 \, t^{2} - 1}, y = \frac{2 \, t}{2 \, t^{2} - 1}ight], \left[x = 1, y = 0ight]ight]
$$

Exercise 5.6.1 (Parametrization of $x^2 - 2y^2$, starting from $(1,0)$)

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Exercise 5.6.1 (Parametrization of $x^2 - 2y^2$, starting from $(1,0)$)

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