Given a Pell solution $(u_k,v_k)$, is there a closed form "descent" to $(u_{k-1},v_{k-1})$? Given: a solution $(u_k,v_k)$ to the Pell equation
$$U^2-dV^2=1, \qquad(\star)$$
where $d$ is a non-square integer, and $k \ge 1$ is an arbitrary integer.
There are well-known recurrences to ascend to $(u_{k+1}, v_{k+1})$, given also the fundamental solution $(u_1,v_1)$. I'm wondering if there's a closed form “reverse recurrence” which gives $(u_{k-1},v_{k-1})$, without knowing $(u_1,v_1)$ explicitly.
As a concrete numerical example, when $d=2$, we can compute $v_{k-1} = u_k-v_k$, and then compute $u_{k-1}=v_k-v_{k-1}$, which gives a solution to the negative version of $(\star)$; applied twice, you get back to the next smaller solution for the positive version. It's this kind of “descent”, in a general case, that I'm interested in.
 A: The answer to your question is no. Knowledge of how to perform your descent is equivalent to knowledge of the fundamental solution. Given $d > 0$ not a square, the previous solution is given by multiplying the column vector by
$$
\left(
\begin{array}{rr}
s & -d t \\
-t & s
\end{array}
\right)
$$
where $$ s^2 - d t^2 = 1  $$
is the fundamental solution.
Given $(x,y)$ that solves $x^2 - d y^2 = 1,$ the previous solution is
$$ (s x - d t y, \; \;  -t x + s y); $$ but the next solution is
$$ (s x + d t y, \; \;  t x + s y); $$
Meanwhile, the matrix (either) has trace $2s$ and determinant $1.$ If you know two consecutive solutions, Cayley-Hamilton says 
$$ x_{n+2} = 2 \, s \;  x_{n+1} - x_n,  $$
$$ y_{n+2} = 2 \, s \; y_{n+1} - y_n.  $$
Once again, knowledge of $2s$ tells you the fundamental solution.
A: If we have any solution of the equation Pell: $$x^2-dy^2=1$$
   Having a form: $(x_1;y_1)$    The following solution we can always be obtained by the formula:  $$x_2=x_0x_1+dy_0y_1$$   $$y_2=y_0x_1+x_0y_1$$    This formula gives all solutions. In order to choose the direction of growth is necessary to change the sign of the substitution solution. Instead " $+$ " write " $-$ "
$(x_0;y_0)$ - the first solution of the Pell equation.
