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\begin{document}

\noindent
\centerline{{\bf Elliptic Curves MT25: Solutions to Sheet 2}}
\bigskip

\begin{enumerate}[{\bf (1)}]

\item
{\bf (a).}
Let $x,y\in K$ be such that $|x| \not= |y|$, without loss of generality,
say that $|x| < |y|$. Since $K$ is non-Archimedean, we
know that $|x \pm y | \leqslant \hbox{max}( |x|, |y| ) = |y|$,
and so it is sufficient to show that 
$|x \pm y | \not< |y|$. Imagine $|x \pm y | < |y|$; then
$|y| = |(x \pm y) - x| \leqslant \hbox{max}(|x \pm y |, |x|) < |y|$,
a contradiction, as required. [Note that an immediate
consequence is the implication:
$| u \pm v | < | u | \Rightarrow |u| = |v|$].
\par\noindent {\bf (b).}
Let $x_1, \ldots , x_n \in K$
be such that
$|x_\ell| > |x_i|$ for all $i\not= \ell$.
Let $x = x_1 + \ldots + x_{\ell-1} + x_{\ell+1} + \ldots + x_n$
and let $y = x_\ell$.
Then $|x| \leqslant 
\hbox{max}(|x_i| : 1\leqslant i \leqslant n, i\not= \ell\} < |y|$
and so: 
$|x_1 + \ldots + x_n| = |x + y| = |y| = |x_\ell|$, by part~(a).
\smallskip
\par\noindent {\bf (c).} 
Since $K,|\ |$ satisfies the triangle inequality $|x+y| \leqslant |x|+|y|$,
we can apply the standard trick from first year Analysis:
$|x| \leqslant |(x - y) + y| \leqslant |x-y| + |y|$
to give: $|x| - |y| \leqslant |x-y|$; similarly
$|y| - |x| \leqslant |x-y|$, and so: $|\ |x| - |y|\ |_\infty 
\leqslant | x - y |$.
If $s_n \rightarrow s$
in $K,|\ |$ then $|s_n - s| \rightarrow 0$ in $\R$,
and so $|\ |s_n| - |s|\ |_\infty \leqslant | s_n - s | \rightarrow 0$,
giving $|s_n| \rightarrow |s|$
in $\bbR, |\ |_\infty$, as required. 
\par
Suppose $s \not= 0$; then $|s| > 0$ and taking $\epsilon = |s|$,
there exists $N$ such that $|s_n - s| < |s|$ for all $n > N$,
so that $|s_n| = |s|$ for all $n > N$ (since if $|s_n| \not= |s|$
then~(a) would give $|s_n - s| = \hbox{max}(|s_n|,|s|)
\geqslant |s|$, a contradiction).
\medskip

\item
{\bf (a).} $3/50 = 5^{-2}\cdot 3/2$ (where $5$ has no
common factor with either the numerator and denominator
of $3/2$)
and so 
$| 3/50 |_5 = 5^2$. Similarly, $3/50 = 3^{1}\cdot 1/50$ and so 
$| 3/50 |_3 = 3^{-1}$. Similarly, $3/50 = 7^0 \cdot 3/50$,
and so $| 3/50 |_7 = 7^0 = 1$.
\par
$d_5(2/3 , 1/5) = | 2/3 - 1/5 |_5 = | 7/15 |_5 = 5$,
$d_7(2/3 , 1/5) = | 7/15 |_7 = 7^{-1}$, $d_{11}(2/3, 1/5)
= | 7/15 |_{11} = 1$.
\par\noindent {\bf (b).} $ |3/7|_3 = 3^{-1}$, $ |3/7|_7 = 7$.
For all other primes~$p$, $\hbox{gcd}(p,3) = \hbox{gcd}(p,7) = 1$
and so $ |3/7|_p = 1$. Note also that $ |3/7|_{\infty} = 3/7$.
The given product $\prod | 3/7 |_i$ has all factors equal to~1
apart from the factors: $ |3/7|_3 = 3^{-1}$, $ |3/7|_7 = 7$
and $ |3/7|_3 = 3/7$, whose product is~$1$. Hence the given
product $\prod | 3/7 |_i$ is~1.
For any $x\in\Q$, write $x = \pm n/d$, where $n$ and $d$ are integers
with $\hbox{gcd}(n,d) = 1$, and write $n,d$ in terms of
their primes factorisations: $n = p_1^{s_1}\cdot \ldots p_k^{s_k}$
and $d = q_1^{t_1}\cdot \ldots q_\ell^{t_\ell}$, where $p_1,\ldots p_k,
q_1,\ldots q_\ell$ are distinct primes. For any prime~$p$,
if $p = p_i$ for some~$i$, then $|x|_p = p_i^{-s_i}$.
If $p = q_j$ for some~$j$, then $|x|_p = q_j^{t_j}$.
If $p=\infty$, then $|x|_p = n/d = (p_1^{s_1}\cdot \ldots p_k^{s_k})/
( q_1^{t_1}\cdot \ldots q_\ell^{t_\ell})$. For all other primes~$p$,
we have $|x|_p = 1$. It follows that
the product
$\prod | x |_i$ is equal to $(p_1^{-s_1}\cdot \ldots p_k^{-s_k})\cdot
 q_1^{t_1}\cdot \ldots q_\ell^{t_\ell}\cdot (p_1^{s_1}\cdot \ldots p_k^{s_k})/
( q_1^{t_1}\cdot \ldots q_\ell^{t_\ell})$, which is again equal
to~$1$.
\medskip

\newpage
\item
{\bf (a).} $| 1/5^n |_5 = 5^n \rightarrow \infty$; this means
that $| 1/5^n |_5$ does not converge, and so
$1/5^n$ does not converge in~$\Q_5$ (since, if $a_n \rightarrow \ell$
then $| a_n |_p \rightarrow | \ell |_p$). 
\par\noindent {\bf (b).} {\it Method 1.} 
$ | n |_5 \le 5^{-1}$ for $n = 5,10,15,\ldots$
and $ | n |_5  = 1$ otherwise; this means that $| n |_5$ does not converge
(since there is a subsequence $| n |_5$ for $5 \not | n$ converging to~$1$
and a subsequence $| n |_5$ for 
$5 | n$ and $5^2 \not | n$ converging to~$1/5$), 
and so $n$ does not converge in~$\Q_5$ (since, if $a_n \rightarrow \ell$
then $| a_n |_p \rightarrow | \ell |_p$).
\par\noindent {\bf (b).} {\it Method 2.}
$n = \sum 1$, which does not converge, since $| 1 |_5 = 1$
which does not converge to~0 (using the result from lectures
which says that $\sum c_i$ converges iff $| c_i |_p \rightarrow 0$).
\par\noindent {\bf (c).} $n! = 5^r\cdot k$, where $r > n/5 - 1$,
since every fifth factor of $1\cdot 2\cdot \ldots n$ 
(i.e. the factors $5,10,15,\ldots $) contributes at least
one new factor of~5. Hence $ | n!  - 0 |_5 = | n! |_5 < 5^{-(n/5 - 1)}
\rightarrow 0$, and so $n!$ converges to~0 in $\Q_5$. 
\par
\noindent
{\bf (d).} 
$| (3+10^n) - 3 |_5 = |10^n|_5 = |5^n\cdot 2^n|_5 = 5^{-n} \rightarrow
0$, and so $3+10^n$ converges to~$3$ in $\Q_5$.
\par
\noindent {\bf (e).} $|10^n|_5 = |5^n\cdot 2^n|_5 = 5^{-n} \rightarrow 
0$ in $\Q_5$ and so $\sum 10^n$ converges in $\Q_5$
(using the result from lectures
which says that $\sum c_i$ converges iff $| c_i |_p \rightarrow 0$).  
\par
\noindent
{\bf (f).}
$| 7^n |_5 = 1$ which does not converge to~0,
and so $\sum 7^n$ does not converge in $\Q_5$ (using the
same result from lectures as used in part (e)).
\medskip

\item
{\bf (a).} We are looking for $x$ such that $|x^2 +1|_5 \leqslant 5^{-4}$.
Take $x_0 = a_0 = 2$ as the initial approximation for which
$|a_0^2 + 1|_5 = |5|_5 = 5^{-1}$. Take $x_1 = a_0 + 5a_1 = 2+5a_1$. Then
we want $(2 + 5a_1)^2 \equiv -1$ mod~$5^2$, and so $20 a_1 \equiv -5$
mod~$5^2$, which is the same as $4a_1 \equiv -1$ mod~$5$, and
so $a_1 = 1$. This gives a better approximation: $x_1 = 2 + 1\cdot 5 = 7$,
which satisfies $|x_1^2 + 1|_5 = 5^{-2}$. Simlarly, we then
find $a_2 = 2$ and so $x_2 = 7 + 2\cdot 5^2 = 57$. Then, finally,
we similarly find $a_3 = 1$ and so $x_3 = 57 + 1\cdot 5^3 = 182$,
satisfying $|x^2 + 1|_5 = 5^{-4}$, as required.  
\par
\noindent {\bf (b).} If
there were an integer~$x$ such that $|x^2 - 7/8|_3 \leqslant
3^{-7}$, then we would have $x^2 \equiv 7/8$ mod~$3^7$ and
so $x^2 \equiv 7/8$ mod~$3$. But $7/8 \equiv 2$ mod~$3$, which is
not a quadratic residue (since $0^2 \equiv 0, 1^2 \equiv 1, 2^2\equiv 1$
mod~3); this means that no such~$x$ can exist.
\par\noindent {\bf (c).} If there
were an integer~$x$ such that $|x^2 - 5/4 |_5
\leqslant 5^{-4}$ then consider the possibilities for $|x^2|_5$.
If $|x^2|_5 > |5/4|_5$ then (using the result
that $|a+b|_p = \hbox{max}( |a|_p, |b|_p)$
when $|a|_p \not= |b|_p$) we have 
$|x^2 - 5/4 |_5 = \hbox{max}(|x^2|_5 , |5/4|_5)
=
|x^2|_5 > |5/4|_5
= 5^{-1} > 5^{-4}$, contradicting $|x^2 - 5/4 |_5 
\leqslant 5^{-4}$.
If $|x^2|_5 < |5/4|_5$ then (using the same result)
we have
$|x^2 - 5/4 |_5 = \hbox{max}(|x^2|_5 , |5/4|_5)
=
|5/4|_5
= 5^{-1} > 5^{-4}$, again contradicting $|x^2 - 5/4 |_5
\leqslant 5^{-4}$. It must then be that $|x^2|_5 = |5/4|_5 = 5^{-1}$,
but this is again impossible since $|x^2|_5 = 5^r$ where
$r$ is an even integer. Hence no such~$x$ exists.
\medskip
 
\item
$200$ mod $7$ is~$4$, so write: $200 = 4 + 7\cdot 28
= $ (similarly) $ 4 + 7\cdot (0 + 7\cdot 4) = 4 + 0\cdot 7^1 + 4\cdot 7^2
= 4,04$. 
\par For $3/14$, it is easier first to write:
$3/14 = 7^{-1}\cdot 3/2$. First find the $7$-adic expansion of~$3/2$,
with the idea that the $7$-adic expansion of $3/14$ will then
just be the $7$-adic expansion of $3/2$ shifted one place
to the left). Now, $|3/2|_7 = 1$ and so $3/2 = a_0,a_1a_2\ldots$.
Using $2( a_0 + 7a_1 + 7^2a_2 + \ldots) = 3$, we first find
$2a_0 \equiv 3$ mod~$7$ and so $a_0 = 5$. Continuing as usual,
we find that $a_1 = 3, a_2=3, a_3=3\ldots$. At this point,
we suspect that $3/2 = 5,\bar{3}$. We can prove this rigorously
as follows. Let $\alpha = 5,\bar{3}$. Then $\alpha - 5 = 0,\bar{3}$,
and so $7^{-1}(\alpha - 5) - 3 = 3,\bar{3} - 3 = 0,\bar{3} = \alpha - 5$.
Hence, $\alpha - 5 - 21 = 7\alpha - 35$ and so $\alpha = 3/2$,
proving that $3/2 = 5,\bar{3}$ in $\Q_7$. Finally,
$3/14 = 7^{-1}\cdot 3/2 = 53,\bar{3}$. 
\par 
Let $\beta = 2,\overline{34} \in \Q_5$. Then $\beta - 2 = 0,\overline{34}$
and so $5^{-2}(\beta - 2) = 34,\overline{34}$, giving:
$5^{-2}(\beta - 2) - 3\cdot 5^{-1} - 4\cdot 5^0 = 0,\overline{34}
= \beta - 2$, and so: $\beta = -67/24$.
\medskip

\item
First we can simplify the question by dividing out the square factor $4$ from $-28$. So it suffices to determine the primes $p$ such that $-7$ has a square root in $\Q_p$. Write $f(x) = x^2 + 7$. If $p \ne 2, 7$ it follows from Hensel's lemma that $f(x)$ has a root in $\Q_p$ if and only if it has a root in $\bbF_p$. This holds if and only if the Legendre symbol $\left(\frac{-7}{p}\right) = +1$. We apply quadratic reciprocity and Euler's formula $\left(\frac{-1}{p}\right) = (-1)^{(p-1)/2}$ to tell us that  $\left(\frac{-7}{p}\right) = \left(\frac{p}{7}\right)$. This is equal to $+1$ when $p = 1, 2$ or $4$ mod $7$. 

It remains to deal with $p=2,7$. For $p=2$, we can again apply Hensel's lemma with $x_0 = 1$. Since $|f(x_0)|_2 = 2^{-3}$ and $|f'(x_0)|_2 = 2^{-1}$, the conditions for Hensel's lemma are satisfied and there is a square root of $-7$ in $\Q_2$. 

For $p=7$, there are no solutions to $x^2 = -7$, since the $7$-adic valuation of the right hand side is $7^{-1}$, but the valuation of the left hand side lies in $7^{2\Z}$. 

To conclude, $-28$ has a square root in $\Q_p$ if and only if $p = 1, 2$ or $4$ mod $7$. 

\item In~$\R$, we have for example the
root $\sqrt{2}$.
In $\Q_2$, note that $17 \equiv 1$ mod~8, and so $17$
is a square in $\Q_2$. In $\Q_{17}$, note that $|2|_{17} = 1$
and $2$ is a quadratic residue mod~$17$, so that $2$ is a square
in~$\Q_{17}$.
\par Finally, let $p \not= 2,17$. Then each of $({2\over p}), ({17\over p}),
({34\over p})$ is $1$ or $-1$. They cannot all be $-1$
since $({34\over p}) = ({2\over p}) ({17\over p})$ [and $-1 \not= (-1)(-1)$],
so at least one of them must be~1.
Wlog, say that  $({2\over p}) = 1$. Then $|2|_p = 1$ and $2$ is
a quadratic residue mod~$p$, so that $2$ must be a square in $\Q_p$.
\medskip

\item
Suppose there were an $x\in \Q_3$
such that $x^3 = 4$ in $\Q_3$. Then $|x|_3^3 = |4|_3 = 1$
and so $|x|_3 = 1$, which means that
$x$ can be written $x = a_0,a_1a_2\ldots$.
Reducing $x$ modulo~9 would then give
the integer $a_0 + 3a_1$ whose cube is~$4$ modulo~9. But $0,\ldots 8$
square to give: $0,1,8,0,1,8,0,1,8$, respectively, so that
$x^3 = 4$ is an impossible congruence mod~9. Hence~$4$ is
not a cube in $\Q_3$.
\par
Let $f(x) = x^3 - 28$ and let $x_0 = 1$. Then $| f(x_0) |_3
= |-27 |_3 = 3^{-3}$, whereas $| f'(x_0) |_3 = | 3 |_3 = 3^{-1}$.
Hence $| f(x_0) |_3 < | f'(x_0) |_3^2$, and so by Hensel's
Lemma there must be a root
of $f(x)$ in $\Q_3$; that is $28$ must be a cube in $\Q_3$.
\par
Let $f(x) = x^3 - 13$ and let $x_0 = -1$.
Then $| f(x_0) |_7 
= |-14 |_3 = 7^{-1}$, whereas $| f'(x_0) |_7 = | 3 |_7 = 1$. 
Hence $| f(x_0) |_7 < | f'(x_0) |_7^2$, and so by Hensel's
Lemma again there must be a root 
of $f(x)$ in $\Q_7$; that is $13$ must be a cube in $\Q_7$.
\medskip

\item
Since $p \equiv 2$~(mod~$3$),
we have $\hbox{gcd}(3,p-1) = 1$,
and so there exist $\lambda, \mu \in \bbZ$ such that
$3\lambda + (p-1)\mu = 1$. For any $x,y\in\bbF_p^*$,
we have $x^{p-1} = y^{p-1} = 1$ [by Fermat's Little Theorem],
and so:
$$ x^3 = y^3 \Rightarrow 
\bigl( x^3 \bigr)^\lambda \cdot 1^\mu
= \bigl( y^3 \bigr)^\lambda \cdot 1^\mu
\Rightarrow
\bigl( x^3 \bigr)^\lambda \cdot \bigl(x^{p-1}\bigr)^\mu
= \bigl( y^3 \bigr)^\lambda \cdot \bigl(y^{p-1}\bigr)^\mu
$$
$$
\ \ \ \ \ \Rightarrow x^{ 3\lambda + (p-1)\mu } 
= y^{ 3\lambda + (p-1)\mu } 
\Rightarrow x = y.
$$
Thus the map $x \mapsto x^3$ is injective on $\bbF_p^*$
and so is a bijection.
\par Now, let 
$a\in \Z$ be such that $p\notdiv a$.
From above, there must exist $x_0 \in \Z$ such
that $x_0^3 \equiv a$ (mod~$p$); clearly
$p\notdiv x_0$. Let $f(x) = x^3 - a$.
Then $|f(x_0)|_p = |x_0^3 - a|_p < 1$,
since $x_0^3 - a \equiv 0$ (mod~$p$).
Also, $|f'(x_0)|_p = |3 x_0^2|_2 = 1$,
since $|3|_p = 1$ and $p\notdiv x_0$.
Hence $|f(x_0)|_p < |f'(x_0)|_p^2$, and so by Hensel's Lemma,
there exists
$x\in \Z_p$ with $f(x) = 0$, that is, $x^3 = a$, as required.

\item We use Hensel's lemma again. First note that any root of unity lies in $\Z_p^\times$, since $|x^m|_p = 1$ implies $|x|_p = 1$. Suppose $\alpha$ is a primitive $m$th root of unity. Then the reduction $\bar{\alpha}$ is a $m$th root of unity in the residue field $\bbF_p$. Let $m'$ be the order of $\bar{\alpha}$ in $\bbF_p^\times$. Since $\bbF_p^\times$ is cyclic of order $p-1$, $m'$ divides $p-1$. We also have $m'|m$, since $\bar{\alpha}^m = 1$. 

Consider the polynomials $f_{n}(x) = x^n-1$ for $n \ge 1$. It follows from Hensel's lemma that $x^{m}-1$ has a unique root which is $= \bar{\alpha}$ mod $p$. This root must be $\alpha$. But we can also apply Hensel's lemma to the polynomial $x^{m'}-1$. This also has a unique root $\alpha'$ which is $= \bar{\alpha}$ mod $p$. Since $\alpha'^{m'} = 1$, we also have $\alpha'^m = 1$. We deduce from uniqueness that $\alpha' = \alpha$, so $\alpha$ is an $m'$th root of unity. This means that $m' = m$ (since $\alpha$ was a \emph{primitive} $m$th root). We deduce that $m$ must divide $p-1$. 

Conversely, if $m$ divides $p-1$ then we have a primitive $m$th root of unity $\bar{\alpha} \in \bbF_p$. Applying Hensel's lemma as above, we find a root of $x^m - 1$ in $\bbQ_p$ which gives the desired primitive $m$th root of unity.  

\url{https://kconrad.math.uconn.edu/blurbs/gradnumthy/hensel.pdf} Theorem 3.1 proves the precise description of roots of unity in $\Q_p$.

\item
{\bf (a).} The number of positive multiples of an integer $k>0$ which are
$\leqslant n$ is clearly $\bigl[ {n\over k} \bigr]$. To count the
power of $p$ dividing $n!$, since $p$ is prime, it is enough to
count the powers of $p$ dividing $1,2,3,\ldots ,n$ and add these
powers up. Now, the number of multiples of $p$ among $1,2,3,\ldots ,n$
is $[ {n\over p} ]$. Each multiple of $p^2$ among
$1,2,3,\ldots ,n$ gives an additional power of $p$ dividing into $n!$,
giving $\bigl[ {n\over p} \bigr] + \bigl[ {n\over p^2} \bigr]$
so far. Continuing in this way we get that the total power of $p$
is as in the given formula. 

{\bf (b).} We have shown that $|n!|_p = p^{-M}$,
where $M = \bigl[ {n\over p} \bigr] + \bigl[ {n\over p^2} \bigr] + \ldots
\leqslant {n\over p} + {n\over p^2} + \ldots = {n\over p-1}$.
Hence $|n!| \geqslant p^{-{n\over p-1}}$, and so
$|1/n!| \leqslant p^{n\over p-1}$. Suppose that $|x| < p^{-{1\over p-1}}$
and let $\tau = |x|/p^{-{1\over p-1}} < 1$.
Then $\bigl| {x^n \over n!} \bigr| = \tau^n p^{-{n\over p-1}} 
\bigl| {1 \over n!} \bigr| \leqslant \tau^n p^{-{n\over p-1}} 
p^{n\over p-1} = \tau^n \rightarrow 0$. Hence, $\hbox{exp}_p(x)$
converges (using the result from lectures
which says that $\sum c_i$ converges iff $| c_i |_p \rightarrow 0$).
\par
On the other hand, if $|x| \geqslant p^{-{1\over p-1}}$,
note that, for any $n=p^\ell$, the above ${n\over p}, {n\over p^2},\ldots
{n\over p^\ell} = p^{\ell-1}, p^{\ell-2}, \ldots , 1 \in \Z$ and
so $M = p^{\ell-1} + p^{\ell-2}, \ldots + 1 = {p^\ell - 1\over p - 1}$; 
this means that the subsequence
$\bigl| {x^{p^\ell}\over (p^\ell)!}\bigr| \geqslant 
(p^{-{1\over p-1}})^{p^\ell} p^{p^\ell - 1\over p - 1}
= p^{-1 \over p-1}$, and so $\bigl| {x^n \over n!} \bigr| \not\rightarrow 0$.
Hence $\hbox{exp}_p(x)$ does not converge when
$|x| \geqslant p^{-{1\over p-1}}$.
%\par When $K = \Q_p$ ($p\not= 2$), any $|x|_p < 1$
%must satisfy $|x|_p \leqslant p^{-1}$ [since $|x|_p = p^r$
%for some $r\in \Z$] $< p^{-{1\over p-1}}$,
%since $p\geqslant 3$. Any $|x|_p \geqslant 1$
%satisfies $|x| \geqslant p^{-{1\over p-1}}$.
%When $K = \Q_2$, any $|x|_2 < 1/2$
%must satisfy $|x|_2 \leqslant 2^{-2}$ [since $|x|_p = p^r$
%for some $r\in \Z$] $< p^{-{1\over p-1}} = 2^{-1}$.
%Any $|x|_p \geqslant 1/2$
%satisfies $|x| \geqslant p^{-{1\over p-1}}= 2^{-1}$.

{\bf (c).} The idea is to define the $p$-adic logarithm map using the Taylor series $\log_p(1+x) = x -x^2/2 + \cdots$ and show that this gives an inverse to $\exp_p$. It also needs to be verified that $\exp_p$ is a group homomorphism. See Theorem 8.13 in \url{https://kconrad.math.uconn.edu/blurbs/gradnumthy/infseriespadic.pdf} for details, as well as the cautionary Example 8.14 which shows that care is needed when computing the composition of two functions defined by power series. 

\end{enumerate}

\end{document}


