I'm trying to read the proof of
LEMMA 6.1 (Nagell)
Let $u_n$ be defined by $u_0=0$, $u_1=1$ and
$u_n=u_{n-1}-2u_{n-2} \hspace{20pt} (n\geq 2)$.
Then $u_n=\pm1$ only for $n=1,2,3, 5 $ and 13.
I get stuck at (6.13), but I'd better give some context. Sorry that I have written so much below, you can probably just skip to (6.11) and check back for information as you need it.
The first few values of $u_n$ are
$\begin{array}{c| c c c c c c c c c c} n & 0 & 1 &2&3&4&5&6&7&8&9 \\ \hline u_n & 0 & 1 & 1 & -1 & 3 & -1 & 5 & 7 & -3 & -17 \end{array}.$
Solving the recurrence relation by sixth-form mathematics, we have
$u_n=\frac{\alpha^n-\beta^n}{\alpha-\beta}$, (6.2)
where $\alpha$, $\beta$ are the roots of
$F(X)=X^2-X+2.$ (6.3)
We work out that $F(X)$ splits in $\mathbb{Q}_{11}$, and
we find that there is a root $\alpha\in \mathbb{Z}_{11}$ with
$\alpha \equiv 16 (11^2)$. (6.4)
The other root is
$\beta=1-\alpha\equiv 106 (11^2)$. (6.5)
Our first thought is to expand $u_n$ as a power series in $n$ and apply Strassmann's theorem.
To do this we set
$A=\alpha^{10}\equiv 1 \mod 11$
$B=\beta^{10} \equiv 1 \mod 11$ (6.6)
We therefore write
$n=r+10s$ $0 \leq r \leq 9$
so $u_{r+10s}=\frac{\alpha^rA^s-\beta^rB^s}{\alpha-\beta}$ (6.7)
We note that
$u_{r+10s} \equiv u_r (11)$ (6.8)
and so the only $r$ which we need consider are $r=1,2,3,5.$
We now write
$\alpha^{10}=A=1+a$, $\beta^{10}=B=1+b$, (6.9)
so
$a \equiv 99 (11^2)$, $b\equiv 77(11^2)$ (6.10)
and develop
$(\alpha-\beta)(u_{r+10s} \mp 1)=\alpha^r(1+a)^s-\beta^r(1+b)^s \mp (\alpha-\beta)$ (6.11)
as a power series
$c_0+c_1s+c_2s^2+...$ (6.12)
using Lemma 5.2. Here the upper sign is correct for $r=1,2$ and the lower for $r=3,5$. In every case $c_0=0$. It is easy to see that
$c_j \equiv 0 (11^2)$ (all $j \geq 2$). (6.13)
I am not finding this easy to see! I know $c_j\equiv 0 (11^2)$ for sufficiently large $j$ because the power series converges in $\mathfrak{o}$ (this is what Lemma 5.2 says). Usually I would differentiate $k$ times and evaluate at zero to check $j=k$, but this makes no sense at all in this context. Thanks for any help.
For reference, Lemma 5.2:
LEMMA 5.2 Let $b \in \mathbb{Q}_p$ and suppose that
$|b|\leq 2^{-2}$ ($p=2$)
$|b|\leq p^{-1}$ (otherwise).
where |.| is the $p$-adic valuation. Then there is a power series
$\Phi_b(X)=\sum_{n=0}^\infty\gamma_nX^n$, (5.2)
where
$\gamma_n \in \mathbb{Q}_p$, $\gamma_n \rightarrow 0$
such that
$(1+b)^r=\Phi_b(r)$
for all $r \in \mathbb{Z}$.