Just edited the posted question to correct a dead link to one of the pdf files.
Apologies if this question has been asked before - there are many "pell" entries in this forum.
Given:
(a) $\;D \in \mathbb{Z^+}\;$ is not a perfect square.
(b) The continued fraction expansion of $\sqrt{D}\;$ is
$\;[a_1, \overline{a_2, \cdots, a_n, 2a_1}].$
(c) Coprime $\;\frac{p_r}{q_r}\;$ denotes the $r^{th}$ convergent of $\sqrt{D}.$
(d) $\;\forall k \in \mathbb{Z^+}, \;(P_k,Q_k)$ is defined by
$\;(p_n + q_n\sqrt{D})^k = P_k + Q_k\sqrt{D}.$
To prove: (e) $\;\forall k \in \mathbb{Z^+},$ $\;(p_{nk},q_{nk}) = (P_k, Q_k).$
My research: Let $E_1$ denote the pell equation $\;x^2 - Dy^2 = \pm 1.\;$ I reviewed pages 1-7 (through theorem 1.19) of pdf-1, chapter 1 (only) of pdf-2, this pdf-3, and html-4. From these references, I concluded:
(1) $\;\{(p_n,q_n), (p_{2n},q_{2n}), (p_{3n},q_{3n}), \cdots \}\;$ is the complete set of all positive integer solutions to $E_1.$
(2) $\;\{(P_1,Q_1), (P_2,Q_2), (P_3,Q_3), \cdots \}\;$ is also the complete set of all positive integer solutions to $E_1.$
(3) The 4 sequences $\;\{p_n, p_{2n}, p_{3n}, \cdots \}, \;\{q_n, q_{2n}, q_{3n}, \cdots \}, \;\{P_1, P_2, P_3, \cdots \},\;$ and $\;\{Q_1, Q_2, Q_3, \cdots \}\;$ are each strictly increasing.
From these conclusions I indirectly deduced result (e) above. However, as I have elaborated in my Partial Work section (below), I would like to establish this result directly, through algebra.
Partial Work: I decided to attempt to algebraically demonstrate that
coprime $\;(p_{2n}, q_{2n}) = ([{p_n}^2 + D{q_n}^2], 2p_nq_n) = (P_2, Q_2).\;$
If successful, I then hoped to find a pattern in the demonstration that would allow
me to show that
$[F_1]\;\; \forall k\in\mathbb{Z^+},
\;(p_{n(k+1)}, q_{n(k+1)}) = (p_{nk}p_n + Dq_{nk}q_n, p_{nk}q_n + q_{nk}p_n).\;$
This would allow me to inductively conclude result (e) above.
$a_1 = $ the floor of $\sqrt{D}.$
Let $A$ denote the continued fraction $\;[a_1, a_2, \cdots, a_n] = \frac{p_n}{q_n}.$
Let $B$ denote the continued fraction $\;[2a_1, a_2, \cdots, a_n].$
Then, $\;B = a_1 + \frac{p_n}{q_n}\;$ and
$\;\frac{p_{2n}}{q_{2n}} = \;$ the continued fraction represented by
$\;[A, B].$
I can now generate 2 non-linear equations ($F_2$ and $F_3,$ below)
for $\;p_{2n}, q_{2n}\;$
that depend on the computation of coprime $p_{n-1}$ and $q_{n-1}.$
I may have to give special consideration for when $\;n=1\;$
(e.g. $\sqrt{2} = [1, \overline{2}]$),
since then $(p_{n-1}, q_{n-1}) = \;$ the artificially contrived $(1, 0).$
$[F_2]\;\; \dfrac{Bp_n + p_{n-1}}{Bq_n + q_{n-1}} = \;$ the ratio $\dfrac{p_{2n}}{q_{2n}}.$
$[F_3]\;\; {p_{2n}}^2 - D{q_{2n}}^2 = (-1)^{2n} = 1.$
Assuming that $p_{2n}$ and $q_{2n}$ are each expressed in terms of
$\;p_n, q_n, p_{n-1}, q_{n-1}, a_1\;$ and $\;D,\;$
I now have to compute $p_{n-1}$ and $q_{n-1}$ in terms of
$\;p_n, q_n, a_1\;$ and $\;D.\;$
Let $\beta $ denote $[\overline{2a_1, a_2, \cdots, a_n}] = a_1 + \sqrt{D}.$
Then $\sqrt{D}$ may be represented by the continued fraction $\;[A,\beta].$
Therefore, (I suspect that) the following two equations will generate unique
values for $p_{n-1}$ and $q_{n-1}:$
$[F_4]\;\; \dfrac{\beta p_n + p_{n-1}}{\beta q_n + q_{n-1}} = \sqrt{D}.$
$[F_5]\;\; p_nq_{n-1} - q_np_{n-1} = (-1)^n\;$
(since the first element of the continued fraction is $a_1,$ rather than $a_0$).
Unfortunately, $F_2, F_3, F_4,\;$ and $F_5\;$ seem very ugly to me, and there is no way to predict the challenges in $F_1.$ I guess that it is not suprising that an indirect proof was chosen for the last theorem in html-4.
Anyway, has any professional mathematician wrestled with this? For example, would any of the later chapters in Khinchin's pdf-2 be pertinent? Alternatively, does anyone have any ideas about making the attack less crude?
My knowledge of field theory is fairly weak, so please be merciful.
