Can the fundamental solution of a Pell equation be “triangulated” given multiple known solutions? In this question about “descent” given a single Pell solution, Will Jagy gave the [accepted] answer that, for a Pell equation
$$
  U^2 - dV^2 = \pm 1,  \tag{$\star$}
$$
there is no way to determine the fundamental solution $(u_1,v_1)$ — or even the next smaller solution $(u_{k-1},v_{k-1})$ — given only a single solution $(u_k,v_k)$ for [unknown] integer $k \ge 1$.
My followup question is this:
Would it be possible to determine the fundamental solution of ($\star$) given $m \ge 2$ solutions $(u_{k_1},v_{k_1}) \le (u_{k_2},v_{k_2}) \le \dots \le (u_{k_m},v_{k_m})$, without knowing exactly which solutions these were (e.g., in particular, it cannot be assumed that they are adjacent solutions)? If so, is there a minimum $m$ for which such a method/algorithm would always work?
For example, given only the three [ordered] solutions
$$(7,5),\quad(1393,985),\quad(9369319, 6625109),$$
could one determine that $(u_1,v_1)=(1,1)$?
EDIT: As individ pointed out, the numerical example is trivial to calculate from. I'm actually thinking about more general equations with variables involved, e.g., given the three ordered, but not necessarily consecutive, solutions
$$(u_{k_1},v_{k_1}),\quad (u_{k_2},v_{k_2}),\quad (u_{k_3},v_{k_3}),$$
is there an algebraic method to determine $(u_1,v_1)$ as an algebraic identity in terms of [one or more of] the variables $u_{k_1},v_{k_1},u_{k_2},v_{k_2},u_{k_3},v_{k_3}$, or a way to calculate $d$ as an integer explicitly (say, given some divisibility properties or factors of the variables)?
 A: The question is poorly phrased, but if you represent the solutions as $u+v\sqrt{d}$, where $u^2-dv^2 = \pm 1$, then every solution (in positive integers) is a power of the fundamental solution.
So, different question: given a bunch of powers $z^a, z^b, z^c,\ldots$, can I determine $z$?  Not necessarily!  I might have been unlucky and all of $a,b,c$ are even.  Then I can never directly construct any non-even powers.  Same with multiples of $3$, and so on.
On the other hand, if I have just two solutions where the powers are relatively prime, then I can find the primitive solution using the Euclidean algorithm.
For example, if I'm given $a=7+5\sqrt{2}$ and $b=1393 + 985\sqrt{2}$, I can compute $a/b = 99+70\sqrt{2}$, $a/b^2 = 7+5\sqrt{2}$... and uh-oh, it looks like my two solutions were both third powers, and $7+5\sqrt{2}$ is as low as I can go.  But if I throw $c= 9369319 + 6625109\sqrt{2}$ into the mix, I can find the fundamental solution, e.g. $c/b^2 = 1+\sqrt{2}$.
EDIT (Kieren): See comments for details on using the g.c.d. of the powers.
