Exercise 5.6.2 (Parametrization of $x^2 - 2y^2$, starting from $(3,2)$)

Answers

Proof. The line with slope $t$ passing by the point $(3, 2)$ has for equation

y = 2 + t (x - 3) .

Let $(x, y) \in ℚ^{2}$ . Then

\begin{array}{l} x^{2} - 2 y^{2} = 1 & ⟺ {\begin{matrix} (x, y) = (3, 2) or \\ x \neq 3 and \exists t \in ℚ, {\begin{matrix} x^{2} - 2 {(2 + t (x - 3))}^{2} = 1, \\ y = 2 + t (x - 3), \end{matrix} \end{matrix} \\ ⟺ {\begin{matrix} (x, y) = (3, 2) or \\ x \neq 3 and \exists t \in ℚ, {\begin{matrix} (x - 3) [x + 3 - 8 t - 2 t^{2} (x - 3)] = 0, \\ y = 2 + t (x - 3), \end{matrix} \end{matrix} \\ ⟺ (x, y) = (3, 2) or \exists t \in ℚ, {\begin{matrix} x (1 - 2 t^{2}) + 6 t^{2} - 8 t + 3 = 0, \\ y = 2 + t (x - 3), \end{matrix} \\ ⟺ (x, y) = (3, 2) or \exists t \in ℚ, {\begin{matrix} x = \frac{6 t^{2} - 8 t + 3}{2 t^{2} - 1}, \\ y = - \frac{2 (2 t^{2} - 3 t + 1)}{2 t^{2} - 1} . \end{matrix} \end{array}

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With Sagemath:

var(’x,y,t’)
solve([x^2 - 2*y^2 == 1, y == 2 + t*(x-3)],x,y)

[[x = 3, y = 2], [x = \frac{6 t^{2} - 8 t + 3}{2 t^{2} - 1}, y = - \frac{2 (2 t^{2} - 3 t + 1)}{2 t^{2} - 1}]]

2025-05-03 10:39