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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 $r \in \mathbb{Z}$.

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    @Jyrki Thanks very much, $I$'ll work on it.2012-05-08

1 Answers 1

3

I think I have an answer now. In the proof of Lemma 5.2 we have

$(1+b)^r=\sum_{s=0}^\infty r(r-1)...(r-s+1)(b^s/s!)$

$=\sum_{s=0}^\infty \sum_{n=0}^s r^n (b^s/s!) g(n,s)$

where $g(n,s)$ is the coefficient of $r^n$ in $r(r-1)...(r-s+1)$ (in particular, $g$ is integer valued). Since $|b^s/s!|\rightarrow 0$, we can rearrange the order of summation to get

$=\sum_{n=0}^\infty \sum_{s=n}^\infty r^n (b^s/s!) g(n,s)$.

Using $|b^s/s!|\leq 11^{-2}$ for $s \geq 2$ tells us the coefficient of $r^2$ in the expansion of $(1+b)^r$ is divisible by $11^2$. Doing the same with $a$ gives the result.