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Just to add another method to the collection: Let $\Delta = \sigma + 4 m$, be the fundamental discriminant of a quadratic field, where $\sigma \in { 0, 1 }$. Let $(x / z , y / z)$ be a rational solution of the Pell conic $x^2 + \sigma x y - m y^2 = 1$, with $\text{gcd}(x, y) = 1$, and let $\beta \equiv x \cdot y^{-1} \pmod{z}$. Assume that $q_1 = (t_1, u_1)$, $q_2 = (t_2, u_2)$ satisfy $z t^2 + (2 \beta + \sigma ) t u + \frac{\beta^2 + \sigma \beta - m}{z} u^2 = 1$ in rational integers, then $\nu (q_2 , q_1 ) = $ I am trying to write a matrix times a vector: $\begin{pmatrix}{z (z t_1 + (\beta + \sigma ) u_1}&{\beta , \beta t_1 + \frac{\beta^2 + \sigma \beta - m}{z} u_1}\{u_1,- u_1}&{t_1}\end{pmatrix} \begin{pmatrix}{t_2}\{u_2}\end{pmatrix}$ u_1,t_1 ) (t_2, u_2)$ satisfies the Pell conic in rational integers. I learned about the map $\nu$ from Franz Lemmermeyer, his articles and book `Binary quadratic forms'. There is a bijection between the integer points $(t, u)$ and the primitive integer points $(T, U)$ of $z T^2 + (2 x + \sigma y) T U + z U^2 = y^2 $. Given a primitive integer point $(T, U)$, we also have a primitive integer point $(U, T)$. Using this bijection, given an integer point $(t_1, u_1)$, we obtain another point $(t_2, u_2) = (\kappa t_1 + \kappa' u_1, y t_1 - \kappa u_1)$, where $\kappa = \frac{x - \beta y}{z} $, and $\kappa' = \frac{(2 \beta + \sigma ) x - (\beta^2 + m ) y}{z^2}$. The points $(t_1, u_1), (t_2, u_2) $ are used with $\nu$ to obtain an integer point of the Pell conic. In particular, letting $\gamma = \frac{\beta^2 + \sigma \beta - m}{z}$, $( x t^2 + (z \kappa' + \sigma \kappa + y \gamma ) t u + ((\beta + \sigma ) \kappa' - \kappa \gamma ) u^2, y t^2 - 2 \kappa t u - \kappa' u^2 )$. One must check that $T \not= U$, equivalently that $y t \not= (\kappa + 1) u$, for otherwise this method will not work. This follows from "Arithmetic of Pell surfaces'surfaces", Acta Arith., 146, (2011), no. 1, 1--12.

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Just to add another method to the collection: Let $\Delta = \sigma + 4 m$, be the fundamental discriminant of a quadratic field, where $\sigma \in { 0, 1 }$. Let $(x / z , y / z)$ be a rational solution of the Pell conic $x^2 + \sigma x y - m y^2 = 1$, with $\text{gcd}(x, y) = 1$, and let $\beta \equiv x \cdot y^{-1} \pmod{z}$. Assume that $q_1 = (t_1, u_1)$, $q_2 = (t_2, u_2)$ satisfy $z t^2 + (2 \beta + \sigma ) t u + \frac{\beta^2 + \sigma \beta - m}{z} u^2 = 1$ in rational integers, then $\nu (q_2 , q_1 ) = $ $\begin{pmatrix}{z t_1 + (\beta + \sigma ) u_1}&{\beta t_1 + \frac{\beta^2 + \sigma \beta - m}{z} u_1}\{- u_1}&{t_1}\end{pmatrix} \begin{pmatrix}{t_2}\{u_2}\end{pmatrix}$ satisfies the Pell conic in rational integers. I learned about the map $\nu$ from Franz Lemmermeyer, his articles and book Binary quadratic forms'. There is a bijection between the integer points $(t, u)$ and the primitive integer points $(T, U)$ of $z T^2 + (2 x + \sigma y) T U + z U^2 = y^2 $. Given a primitive integer point $(T, U)$, we also have a primitive integer point $(U, T)$. Using this bijection, given an integer point $(t_1, u_1)$, we obtain another point $(t_2, u_2) = (\kappa t_1 + \kappa' u_1, y t_1 - \kappa u_1)$, where $\kappa = \frac{x - \beta y}{z} $, and $\kappa' = \frac{(2 \beta + \sigma ) x - (\beta^2 + m ) y}{z^2}$. The points $(t_1, u_1), (t_2, u_2) $ are used with $\nu$ to obtain an integer point of the Pell conic. In particular, letting $\gamma = \frac{\beta^2 + \sigma \beta - m}{z}$, $( x t^2 + (z \kappa' + \sigma \kappa + y \gamma ) t u + ((\beta + \sigma ) \kappa' - \kappa \gamma ) u^2, y t^2 - 2 \kappa t u - \kappa' u^2 )$. One must check that $T \not= U$, equivalently that $y t \not= (\kappa + 1) u$, for otherwise this method will not work. This follows fromArithmetic of Pell surfaces', Acta Arith., 146, (2011), no. 1, 1--12.