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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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1  
I've added added page Google Books link to your post - I guess this might be useful for some users. (Of course, I am aware that they will not be displayed to everyone.) I hope you don't mind.\\ You have typeset your post very nicely, so I believe there is sufficient context for the question even without seeing the text in the book. –  Martin Sleziak May 7 '12 at 11:54
    
Thanks very much. Anything that might encourage someone to help me is much appreciated. –  Cruiskeen Lawn May 7 '12 at 13:15
2  
I don't have Cassels at hand, so I don't know what, for example, Lemma 5.2. is saying. But could it be something like the following. The element $a$ is in the maximal ideal $\mathfrak{p}=11\mathbb{Z}_{11}$. Therefore $\log (1+a)=a-a^2/2+\cdots\in \mathfrak{p}$. Now if everything works out, we get as the power series in $s$ $$(1+a)^s=exp(s\log(1+a))=\sum_n\frac{(s\log(1+a))^n}{n!}.$$ Here $\log(1+a)\in\mathfrak{p}$, so the claim follows from this (together with the usual estimate of the 11-adic value of $n!$). Similarly for $(1+b)^s$? –  Jyrki Lahtonen May 7 '12 at 18:03
    
I've added Lemma 5.2 to my question. I didn't know that $(1+a)^s=\exp (s\log(1+a))$ for $\mathbb{Q}_p$. –  Cruiskeen Lawn May 7 '12 at 21:20
1  
Thanks, @Martin. MA390, I'm not positive about the identity. For integer $s$ it should follow by induction from the convergence of the series involved and from the equation $exp(x+y)=exp(x) exp(y)$ for all $x,y\in\mathfrak{p}$. It should hold for all $a\in\mathfrak{p}, s\in \mathbb{Z}_{11}$. Heck, I dare guess that one might define $(1+a)^s$ this way. It looks like you need to delve into the proof of Lemma 5.2. to get more explicit information about the coefficients of the series. For now, my comment just adds plausibility to the claim. A busy day grading the last midterm coming up... –  Jyrki Lahtonen May 8 '12 at 6:06
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1 Answer

up vote 3 down vote accepted

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.

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