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\begin{document}
\newfont{\Bbs}{msbm10 scaled\magstep0}
\def\Oh{\mathcal{O}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\Q}{\mathbb{Q}}
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\newcommand{\D}{\mathcal{D}}
\newcommand{\G}{\mathcal{C}(\Q)}
\newcommand{\HH}{\mathcal{D}(\Q)}
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\newcommand{\hp}{{\hat \phi}}
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\newcommand{\oo}{{\bf o}}

\noindent {\bf Solution 1}
\par\noindent
{\bf (a)} 
Let $S_N = \sum_{n=1}^N x_n$.
Assume that~$x_n \rightarrow 0$ in~$K$.
Then:
\par\noindent $| S_N - S_M | = | x_{M+1} + \ldots + x_N |
\leqslant \max\bigl( |x_{M+1}|, \ldots , |x_N| \bigr)
\rightarrow 0$ as $M,N\rightarrow \infty$.
\par\noindent
Hence $S_N$ is Cauchy and so convergent
(since~$K$ is complete), giving that
$\sum x_n$ is convergent.
Conversely, assume that $\sum x_n$ is convergent, that is,
$S_N \rightarrow \ell$ for some $\ell \in K$. Then: 
\par\noindent
$| x_n - 0 | = | x_n | = | S_n - S_{n-1} |
= | S_n - \ell + \ell - S_{n-1} |
\leqslant | S_n - \ell | + | S_{n-1} - \ell | \rightarrow 0$
as $n \rightarrow \infty$,
so that $x_n \rightarrow 0$ in~$K$, ${{|\ \, |}}$.
\par
Let $x,y\in K$ satisfy $|x| \not= |y|$; without loss of generality,
say $|x| > |y|$. Since ${{|\ \, |}}$
is non-Archimedean, $|x + y| \leq \hbox{max}(|x|,|y|)$.
Imagine $|x + y| < \hbox{max}(|x|,|y|) = |x|$;
then $|x| = | x + y + (-y) | \leq \hbox{max}( |x+y|, |y|) < |x|$,
a contradiction; hence $|x + y| = \hbox{max}(|x|,|y|)$.

[{\bf 5 marks} Bookwork.]
\bigskip

\par\noindent
{\bf (b)} Note that $1 + 2p + 2p^2 + 2p^3 + \ldots
= 1 + 2p(1 + p + p^2 + \ldots) = 1 + 2p/(1-p) = (1+p)/(1-p)$,
which equals $-9/8$ exactly when $8(1+p) = -9(1-p)$,
that is: $p = 17$.

[{\bf 2 marks}; Unseen.]
\bigskip

\par\noindent
{\bf (c)} Let
$x_0 = a_0 = 2$. Then $| x_0^2 + 3 |_7 = 7^{-1}$.
Look for $x_1 = a_0 + 7a_1$ such that $| x_1^2 + 3 |_7 \leqslant 7^{-2}$.
This is satisfied if: $(2 + 7a_1)^2 \equiv -3$ (mod $7^2$)
$\iff 2\cdot 2\cdot 7 a_1 \equiv -3 - 2^2\ (\hbox{mod } 7^2)
\iff 4 a_1 \equiv -1\ (\hbox{mod } 7) \iff a_1 \equiv 5\ (\hbox{mod } 7)$.
So, now define: $a_1 = 5$ and $x_1 = a_0 + 7a_1 = 37$,
which satisfies $|x_1^2 + 3 |_7 < 7^{-1}$.
\par
For any $x\in \Z$, $| x^2 |_7 = 7^{2r}$, for some $r\in\Z$,
whereas $|14|_7 = 7^{-1}$, so that $| x^2 |_7 \not= |14|_7$.
By~(a), $| x^2 + 14 |_7 = \hbox{max}( |x^2|_7, |14|_7 ) \geq 7^{-1}$,
and so we cannot have $| x^2 + 14 |_7 < 7^{-1}$.

[{\bf 5 marks}; Seen similar.]
\bigskip

\par\noindent
{\bf (d)} Statement of Hensel's Lemma: 
Let~$K$ be a field, complete with respect 
to a non-Archimedean valuation~${{|\ \, |}}$, with valuation 
ring~$R = \{ x\in K : |x| \leq 1\}$. Let $f(x) \in R[x]$ 
and let $a_0\in R$ satisfy: $| f(a_0) | < | f'(a_0) |^2$.
Then there exists a unique $a\in R$ such that
$f(a) = 0$ and $| a - a_0 | \leq | f(a_0) |/ | f'(a_0) |$.
\par
We are given that $p\equiv 1$~(mod~$3$). Then~$p\not= 2$, so
$p$ is odd, and so satisfies either $p \equiv 1$~(mod~$4$)
or $p \equiv 3$~(mod~$4$). In the first case, we have
that $\bigl( {-3\over p} \bigr)
= \bigl( {3\over p} \bigr) \bigl( {-1\over p} \bigr)
= \bigl( {p\over 3} \bigr) \bigl( {-1\over 3} \bigr)$
[by quadratic reciprocity]
$= \bigl( {1\over p} \bigr) \bigl( {-1\over 3} \bigr)
= 1\cdot 1$ [using the result from notes that 
$\bigl( {-1\over p} \bigr) = 1$ when $p\equiv 1$~(mod~$4$)].
In the second case, $\bigl( {-3\over p} \bigr)
= \bigl( {3\over p} \bigr) \bigl( {-1\over p} \bigr)
= -\bigl( {p\over 3} \bigr) \bigl( {-1\over p} \bigr)$
[by quadratic reciprocity]
$= -\bigl( {1\over 3} \bigr) \bigl( {-1\over p} \bigr)
= -1\cdot (-1)$ [using the result from notes that
$\bigl( {-1\over p} \bigr) = 1$ when $p\equiv 1$~(mod~$4$)].
In either case, we have that $\bigl( {-3\over p} \bigr) = 1$,
as required.
\par
We now have that there exists $a_0 \in \Z$ such that
$a_0^2 \equiv -3$ (mod~$p$), and so $|f(a_0)|_p \leq p^{-1}$,
where $f(x) = x^2 + 3$; clearly $|a_0|_p = 1$
(since $|a_0^2|_p \equiv |-3|_p = 1$) and so 
$|f'(a_0)|_p = 2 |a_0|_p = 1$; hence $|f(a_0)|_p < |f(a_0)|_p^2$;
it follows from Hensel's Lemma that $f(x)$ has a root in~$\Z_p$;
hence $-3$ is a cube in~$\Z_p$ and so in~$\Q_p$.

[{\bf 6 marks}; Unseen]
\bigskip

\par\noindent
{\bf (e)} 
When $p \equiv 2$~(mod~$3$), note that $3$ is coprime to
$p-1$, and so there exist $\lambda , \mu \in \Z$ such
that $(p-1) \lambda  + 3 \mu  = 1$. Now, suppose that $v^3 = w^3$ in
$\F_p^*$.
Then $v^{3 \mu} = w^{3 \mu}$. Also $(p-1) \lambda$ is a multiple of
the order of the group $\F_p^*$ and so $v^{(p-1) \lambda}
= w^{(p-1) \lambda}$ (since both are equal to~1). Multiplying these
last two equations gives: $v^{(p-1) \lambda  + 3 \mu} = 
w^{(p-1) \lambda  + 3 \mu}$, and so $v = w$. We have shown that
$v^3 = w^3$ implies $v=w$, and so the map $v \mapsto v^3$ is
injective, and hence surjective (and hence bijective),
from $\F_p^*$ to $\F_p^*$. 
Since~$q$ is not divisible by~$p$ (since $q\equiv 1$ mod~27), 
it follows that there exists~$x_0 \in \Z$
such that~$x_0^3 \equiv q$~(mod~$p$). This is the same
as $| f(x_0) |_p < 1$, where $f(x) = x^3 - q$.
But from the fact that $q$ is not divisible by~$p$
and $x_0^3 \equiv q$~(mod~$p$) it follows that
$x_0$ is not divisible by~$p$, and so $|x_0|_p = 1$,
giving that $| f'(x_0) |_p = | 3 x_0^2|_p = 1$.
Therefore, $| f(x_0) |_p < | f'(x_0) |_p^2$ and so,
by Hensel's Lemma, there exists a solution in~$\Q_p$
to $x^3 - q = 0$, which is also a solution to $(x^2+3)(x^3-q)=0$.
\par When $p\equiv 1$~(mod~3), we have already established
in~(d) that $x^2 + 3 = 0$ has a solution~$\Q_p$,
which will also be a solution to $(x^2+3)(x^3-q)=0$.
When $p=3$, note
that we are given $q \equiv 1$~(mod~$27$), so that
$ | f(x_0) |_3 \le 3^{-3}$, where $x_0=1$ and $f(x) = x^3 - q$.
Also, $| f'(x_0) |_3 = | 3\cdot 1^2 | = 3^{-1}$,
so that $ | f(x_0) |_3 < | f'(x_0) |_3^2$. By Hensel's
Lemma, it follows that there exists a solution in~$\Q_3$
to $x^3 - q = 0$, and so to $(x^2+3)(x^3-q)=0$. Since $q>0$,
there also exists a solution to $x^3 - q = 0$ in~$\R$,
and so to $(x^2+3)(x^3-q)=0$.
Hence there is a solution to $(x^2 + 3)(x^3 - q) = 0$
in~$\R$ and every~$\Q_p$, as required. Finally note that
$x^2 + 3$ has no solution in~$\Q$ (since none in $\R$)
and $x^3 - q$ has no solution in~$\Q$ (since none in $\Q_q$,
since $|x^3|_q = q^{3r}$, whereas $|q|_q = q^{-1}$).

[{\bf 7 marks}; Unseen, although they have previously seen the idea of
$x \mapsto x^3$ being bijective on~$\F_p^*$, when $p \equiv 2$~(mod~$3$), 
in a different problem on a sheet.]


\bigskip
\hrule
\newpage
\noindent {\bf Solution 2}
\bigskip

\noindent
{\bf (a)} 
Let~$\e$ be the elliptic curve~$y^2 = x^3 + 4x$.
Then~{\bf o} and~$(0,0)$ [which has order~$2$] are both members
of the torsion subgroup. Furthermore, consider the point
$(2,4)\in \e(\Q)$; the tangent to~$\e$ at~$(2,4)$
has slope~$(3\cdot 2^2 + 4)/ (2\cdot 4) = 2$ and has equation~$y = 2x$.
So, the $x$-coordinate of $2(2,4)$ is $2^2 - 2 - 2 = 0$, with corresponding
$y$-coordinate~$0$, giving: $2(2,4) + (0,0) = {\bf o}$
and so $2(2,4) = -(0,0) = (0,0)$.
Hence~$4(2,4) = 2(0,0) = {\bf o}$; similarly~$4(2,-4) = {\bf o}$.
We have found~$4$ torsion elements: ${\bf o}, (0,0), (2,4), (2,-4)$,
and so the torsion subgroup is of order at least~$4$. 
The discriminant of $x^3 + 4x$ is~$4\cdot 4^3 + 27\cdot 0^2
= 2^8$, and so $\widetilde \e : y^2 = x^3 + 4x$ is an elliptic curve
for all $p\not= 2$. The elements of
$\widetilde \e (\F_3 )$ are:
${\bf o}, (0, 0), (2,\pm 1)$,
and so $\widetilde \e (\F_3 )$ has order~$4$.
Since the torsion subgroup 
of $\e (\Q)$ injects into this group, its order must 
divide~$4$. Hence the torsion subgroup 
of $\e (\Q)$ consists precisely of the~$4$ torsion
elements already found,
namely:~$\{ {\bf o}, (0,0), (2, \pm 4) \}$.

[{\bf 5 marks}; Seen similar.]
\bigskip

\noindent
{\bf (b)} 
Let~$\e$ be the elliptic 
curve~$y^2 = x(x+1)(x+n^2) = x^3 + (n^2+1)x^2 + n^2 x$.
The tangent to~$\e$ at~$(n, n(n+1))$
has slope~$(3n^2 + 2(n^2+1)n + n^2)/ (2n(n+1)) = n+1$, and
so equation~$y = (n+1)x +c$ for some~$c$;
the line must pass through~$(n, n(n+1))$ so that~$c=0$ and
the equation of the tangent is: $y = (n+1)x$.
The $x$-coordinate of~$P = 2(n, n(n+1))$ is $(n+1)^2 - (n^2+1) - 2n = 0$,
with $y$-coordinate~$(n+1)\cdot 0 = 0$.
Hence $2(n, n(n+1)) + (0,0) = {\bf o}$, 
so that $2(n, n(n+1)) = -(0,0) = (0,0)$,
giving $4(n, n(n+1)) = {\bf o}$. So $(n, n(n+1))$ is torsion,
as must be $(n, -n(n+1)) = -(n, n(n+1))$.
Replacing every occurrence of~$n$ with~$-n$
gives that the tangent to~$\e$ at~$(-n, n(n-1))$
has equation~$y = (-n+1)x$ and~$2(-n, n(n-1)) = (0,0)$; we deduce,
as before, that $(-n, \pm n(n-1))$ are torsion elements.
Therefore there are at least~$8$ members of the torsion subgroup
of~$\e(\Q)$, namely: ${\bf o}, (0,0), (-1,0), (-n^2,0)$ and
the above~$4$ points of order~$4$.
Finally, let~$n \equiv 2$~(modulo~$5$), so that~$n^2 \equiv 4$~(modulo~$5$).
Then $\widetilde \e$ modulo~$5$ is: $y^2 = x(x+1)(x+4)$, and 
the cubic has no repeated roots, so that~$\e$ 
has good reduction at~$5$.
The elements of
$\widetilde \e (\F_5 )$ are:
${\bf o}, (0, 0), (1, 0), (2, \pm 1), (3, \pm 2), (4,0)$,
and so $\widetilde \e (\F_5 )$ has order~$8$.
Since the torsion subgroup
of $\e (\Q)$ injects into this groups, its order must
divide~$8$. Hence the torsion subgroup
of $\e (\Q)$ consists precisely of the~$8$ torsion
elements already found, 
namely:~$\{ {\bf o}, (0,0), (-1,0), (-n^2,0), 
(n, \pm n(n+1)) , (-n, \pm n(n-1)) \}$.

[{\bf 7 marks}; Unseen.]
\bigskip

\noindent
{\bf (c)} 
The Nagell-Lutz Theorem
states that, if~$(x,y)$ is a $\Q$-rational torsion point 
on $\e : y^2 = x^3 + Ax + B$, where $A,B\in \Z$, then
$x,y\in\Z$ and $y = 0$ or $y^2 | \Delta$, where $\Delta = 4A^3 + 27B^2$. 
\par\noindent {\bf Proof.} We are given the result that
$x,y\in \Z$. If $y = 0$ then the result is satisfied; otherwise,
$(x,y)$ is not $2$-torsion and we consider $(x_2,y_2) = 2(x,y)$,
with $(x_2,y_2) \not= {\bf o}$, and so $x_2,y_2\in \Q$. But~$(x_2,y_2)$
is also a torsion point, so $x_2 , y_2 \in \Z$. Now, the line
tangent to~$\e$ at~$(x,y)$ has slope $(3x^2+A)/(2y)$, from
which we immediately get: $x_2 = \bigl( (3x^2+A)/(2y) \bigr)^2 - 2x$.
Now, we know $x_2, x\in \Z$ and so $\bigl( (3x^2+A)/(2y) \bigr)^2\in \Z$.
It follows that $4y^2 | (3x^2+A)^2$ and so $y^2 | (3x^2+A)^2 = \psi_1(x)$.
Also, $y^2 = x^3 + Ax + B = \psi_2(x)$. Using the identity given
in the question, $y^2 | (\phi_1(x)\psi_1(x) + \phi_2(x)\psi_2(x))
= \Delta$, as required.

[{\bf 7 marks}; Bookwork.]
\bigskip

\noindent
{\bf (d)} The disciminant of $x^3 - k^2 x + k^3$
is $4\cdot (-k^2)^3 + 27 \cdot (k^3)^2 = 23 k^6$. If
$(x,y)$ is a torsion point in $\e (\Q)$,
then we know by Nagell-Lutz that $x,y\in \Z$ with $y^2 | 23 k^6$ and so
$| y |^2 \le 23 |k|^6$; that is:
$| y | \le \sqrt{23} |k|^3 \le 5 |k|^3$.
\par
Since $k\in \Z$, and we are given $k \not= 0$, we have 
$|k| \geqslant 1$ and so:
$| k |^2 \leqslant | k |^4$ and $| k |^3 \leqslant | k |^6$.
\par\noindent Imagine $ | x | > 3 | k |^2$.
\par\noindent Then:
$| x^2 - k^2 | \ge | x^2 | - | k^2 | > 9 | k |^4 - | k^2 |
\geqslant 9 | k |^4 - | k^4 | = 8 | k |^4$, so:
\par\noindent
$ | y |^2 = 
%| x^3 - k^2 x + k^3 | = 
| x(x^2 - k^2) + k^3 |
\ge | x(x^2 - k^2) | - | k^3 | = | x | | x^2 - k^2 | - | k^3 |
> 3 | k |^2 8 | k |^4 - | k |^6 = 23 | k |^6$,
contradicting $| y |^2 \le 23 |k|^6$ (above).
Hence $| x | \le 3 | k |^2$, as required. 

[{\bf 6 marks}; Unseen]

\bigskip
\hrule
\bigskip

\newpage

\noindent
{\bf Solution 3}
\bigskip

{\bf (a)} 
Let $\CC : Y^2 = X(X^2 + X + 2)$.
Here, $a=1, b=2$ and so $a_1 = -2a = -2,
b_1 = a^2 - 4b = -7$, giving
$\D : V^2 = U(U^2 - 2U - 7)$. The isogeny
$\phi : \CC \rightarrow \D$ is given by
$\phi(x,y) = \Bigl( \bigl( \frac{y}{x} \bigr)^2, y - \frac{by}{x^2} \Bigr)
= \Bigl( \bigl( \frac{y}{x} \bigr)^2, y - \frac{2y}{x^2} \Bigr)$.
The isogeny $\hp : \D \longrightarrow \CC$
is given by
$\hp(u,v) =
\Bigl( \frac{1}{4} \bigl( \frac{v}{u} \bigr)^2,
\frac{1}{8} \bigl( v - \frac{b_1 v}{u^2}\bigr) \Bigr)
= \Bigl( \frac{1}{4} \bigl( \frac{v}{u} \bigr)^2,
\frac{1}{8} \bigl( v + \frac{7 v}{u^2}\bigr) \Bigr)$.
\par\noindent {\bf Step 1.} Find $\HH /\phi(\G)$.
We need to consider $r | b_1 = -7, r\in \Z$, $r$ square free,
that is, $r = \pm 1, \pm 7$, so that
$\im q \leqslant \{ \pm 1, \pm 7\}$.
But $q(\oo) = 1,\ q(0,0) = b_1 = -7,\ q(-1,2) = -1,\ q(7,14) = 7$
[note that $(7,14)$ can be found as $(0,0) + (-1,2)$ without
needing to search],
so that $\im q = \{ \pm 1, \pm 7\}$
and $\HH / \phi(\G) = \{ \oo, (0,0), (-1,2), (7,14) \}
= \langle (0,0), (-1,2) \rangle$.
\par\noindent {\bf Step 2.} Find $\G /\hp(\HH)$.
We need to consider $r | b = 2, r\in \Z$, $r$ square free,
that is, $r = \pm 1, \pm 2$.
Also, $\hq(\oo) = 1,\ \hq(0,0) = b = 2$, so
that $\{ 1,2\} \leqslant \im \hq \leqslant \{ \pm 1, \pm 2\}$.
Note that $-1 \in \im \hq \iff -2 \in \im \hq$,
and so it is only necessary to check one member of
the coset $\{ -1, -2\}$. 
\par Choose $r=-1$. Then ${\widehat W}_{-1}$,
$r \ell^4 + a \ell^2 m^2 + (b/r) m^4 = n^2$ becomes:
$$
{\widehat W}_{-1} : - \ell^4 + \ell^2 m^2 - 2 m^4 = n^2,\ \
\hbox{ for some } \ell, m, n \in \Z, 
\hbox{ not all $0$, with gcd}(\ell,m) = 1.
$$
On multiplying both sides by~$4$ and completing the square, we obtain:
$$ -( 2\ell^2 - m^2 )^2 - 7 m^4 = 4 n^2.\ \ \ \ \ (1)$$
This gives $-( 2\ell^2 - m^2 )^2 \equiv 4 n^2$~(mod~$7$).
\par Imagine $7 \nmid ( 2\ell^2 - m^2 )$; then $2\ell^2 - m^2$ 
would have an inverse~$\alpha$ mod~$7$, and so
$-1 \equiv (2\alpha n)^2$~(mod~$7$), contradicting the fact
that $-1$ is not a quadratic residue mod~$7$.
\par Hence, by reductio, $7 | ( 2\ell^2 - m^2 )$
and so $7 | n$ [since $7 | 4n^2$ and $7 \nmid 4$], giving also
that $7^2 | ( 2\ell^2 - m^2 )^2$ and $7^2 | 4 n^2$,
so that, from~(1), $7^2 | 7 m^4$, and so $7 | m$.
But combining $7 | m$ with $7 | 2\ell^2 - m^2$ gives
$7 | 2\ell^2$, so that $7 | \ell$. We have shown
that $7 | \ell$ and $7 | m$, contradicting $\gcd(\ell,m) = 1$.
Hence there are no solutions to ${\widehat W}_{-1}$, giving that
$-1 \not\in \im \hq$. 
%[indeed, we have shown that there are no solutions in $\Q_7$].
\par We conclude that $\im \hq = \{ 1, 2\}$
and $\G / \hp(\HH) = \{ \oo, (0,0) \}
= \langle (0,0) \rangle$.
\par\noindent
{\bf Step 3.} Find $\G/2\G$. This is by
$\G / \hp (\HH) = \{ \oo, (0,0) \}$,
together with $\hp\bigl( \HH/\phi(\G) \bigr)
= \{ \hp( \oo), \hp (0,0), \hp(-1,2), \hp(7,14) \}$
$$ = \{ \oo, \oo, \Bigl( \frac{1}{4} \bigl( \frac{2}{-1} \bigr)^2,
\frac{1}{8} \bigl( 2 + \frac{7 \cdot 2}{(-1)^2}\bigr) \Bigr),
\Bigl( \frac{1}{4} \bigl( \frac{14}{7} \bigr)^2,
\frac{1}{8} \bigl( 14 + \frac{7 \cdot 14}{7^2}\bigr) \Bigr)\}
$$
$ = \{ \oo, \oo, (1,2), (1,2) \} = \{ \oo, (1,2) \}$.
Therefore $\G/2\G$ is:
\par\noindent $\{ \oo , (0,0), (1,2), (0,0) + (1,2) = (2,-4)\}
= \langle (0,0), (1,2) \rangle \cong C_2 \times C_2$.
\par Finally, use the result: $\G/2\G \cong \CC(\Q)[2] \times C_2^r$,
and $\CC(\Q)[2] = \{ \oo , (0,0) \} \cong C_2$
(since the other points of order~2 are $( (-1 \pm \sqrt{-7})/2, 0 )$),
to deduce that the rank $r = 1$.

[{\bf 11 marks}; Seen similar.]
\bigskip

\noindent
{\bf (b)}
We are given that $A/2A$ is finite,
so let $A/2A = S = \{ Q_1,\ldots Q_r \} \subset A$.
Let~$P$ be any element of~$A$. Then $P = Q_{i_1}$ in $A/2A$
for some $Q_{i_1} \in S$ and so we can write:
$P = 2P_1 + Q_{i_1}$, for some $P_1 \in A$.
Inductively, continue to write:
$P_1 = 2P_2 + Q_{i_2}, P_2 = 2P_3 + Q_{i_3}, \ldots$,
where each $P_j \in A$
and each $Q_{i_j}\in S$. Now:
\par
\noindent $h(P_j) \le \frac{1}{4}\bigl( h(2P_j) + C_2 \bigr)$ [by~(2)]
$= \frac{1}{4}\bigl( h(P_{j-1}-Q_{i_j}) + C_2 \bigr)
\le \frac{1}{4}\bigl( 2h(P_{j-1}) + C_1' + C_2 \bigr)$ [by~(1)], where:
\par
\noindent $C_1' = \hbox{max}\{ C_1(-Q) : Q\in S\}$.
So, if $h(P_{j-1}) > (C_1' + C_2)/2$ then:
\par
$h(P_j) < \frac{1}{4} \bigl( 2h(P_{j-1}) + 2h(P_{j-1}) \bigr) 
= h(P_{j-1})$.
\par
Imagine that $h(P) > (C_1' + C_2)/2$ and 
$h(P_j) > (C_1' + C_2)/2$ for all~$j$. Then
the sequence $h(P), h(P_1), h(P_2),\ldots$ would be strictly decreasing,
giving infinitely many distinct members of~$A$
with height $\le h(P)$, which would contradict~(3).
This contradiction shows that there must exist an~$n$
such that $h(P_n) \le (C_1' + C_2)/2$. So, we can write:
$P = 2P_1 + Q_{i_1} = 2(2P_2 + Q_{i_2}) + Q_{i_1} = \ldots$,
and after~$n$ steps $P$ will be written as a linear combination
of~$P_n$ and members of~$S$. Let
$T = \{ Q \in A : h(Q) \le (C_1' + C_2)/2\}$. We have
shown (since $P_n\in T$) that any $P\in A$ is a linear
combination of members of~$S \cup T$. Furthermore,
$T$ is finite, by~(3). In conclusion: $A$ is generated
by the finite set $S \cup T$,
and so is finitely generated.

[{\bf 9 marks}; Bookwork.]
\bigskip

\noindent
{\bf (c)}
Imagine that $h(P) > {1\over 3}C_2$.
Then $C_2 < 3h(P)$ and so by~(2):
$h(2P) \ge 4h(P) - C_2 > 4h(P) - 3h(P) = h(P)$.
Inductively: $h(P) < h(2P) < h(4P) < \ldots$, so that
$P,2P,4P,\ldots$ are all distinct, which forces~$P$ to have
infinite order (since otherwise $P$ would generate a finite group).
Hence, if $P$ is a torsion element
then $h(P) \le {1\over 3}C_2$.

[{\bf 5 marks}; Unseen.]
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