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.
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.