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Solution 1
\bigskip

$(-3)^3\equiv 5\pmod{32}$ so $n=-3$ is acceptable.

[2 marks]
\bigskip

If $|n^3-5|_5<5^{-3}$ then $n^3\equiv 5 \pmod{625}$, whence $5|n$ and
$125|n$, a contradiction.

[2 marks]
\bigskip

Suppose we are given a finite sequence $a_0,\ldots,a_n$ in $R$
satisfying (i), (ii) and (iii) up to the $n$-th term.
We show how to extend the sequence with an $(n+1)$-th term.  We 
define $a_{n+1}=a_n+b_n$, where $b_n=-f(a_n)/f'(a_n)$.  Note that
$f'(a_n)\not=0$, since $|f'(a_n)|^2=|f'(a_0)|^2>|f(a_0)|$, whence we
cannot have $|f'(a_n)|=0$.  Thus (i) will hold if $n$ is
replaced by $n+1$.

[2 marks]
\bigskip

Note that
\begin{equation}\label{e1}
|b_n|^2=|f(a_n)|^2/|f'(a_n)|^2\le |f(a_n)|\frac{|f(a_0)|}{|f'(a_0)|^2}<|f(a_n)|
\end{equation}
by (ii) and (iii), whence
\begin{equation}\label{e2}
|b_n|<\sqrt{|f(a_n)|}\le\sqrt{|f(a_0)|}<|f'(a_0)|=|f'(a_n)|\le 1,
\end{equation}
by a further application of (ii) and (iii).

A Taylor series expansion gives
$f'(a_{n+1})=f'(a_n+b_n)=f'(a_n)+b_nf''(a_n)+\ldots$, in which the
$k$-th term takes the form $b_n^kf_k(a_n)$ with a polynomial
$f_k(x)\in R[x]$.  It follows that $|f_k(a_n)|\le 1$ for $k\ge 1$.
Moreover (\ref{e2}) show that $|b_n^k|<|f'(a_n)|$ for $k\ge 1$.  It 
follows that 
$|b_nf''(a_n)+b_n^2f_2(a_n)+\ldots|<|f'(a_n)|$, and hence that
$|f'(a_n)+b_nf''(a_n)+b_n^2f_2(a_n)+\ldots|=|f'(a_n)|$.  We therefore
deduce that $|f'(a_{n+1})|=|f'(a_{n})|$.  Thus (ii) above remains true
up to $n+1$.

[4 marks]
\bigskip

Finally we have another Taylor expansion $f(a_{n+1})=f(a_n+b_n)=
f(a_n)+b_nf'(a_n)+\ldots$, in which the
$k$-th term now takes the form $b_n^kF_k(a_n)$ with a polynomial
$F_k(x)\in R[x]$. Our choice of $b_n$ ensures that
$f(a_{n+1})=b_n^2F_2(a_n)+\ldots $.  As before we have $|F_k(a_n)|\le
1$ for $k\ge 2$. Moreover (\ref{e1}) yields
$|b_n^kF_k(a_n)|<|f(a_n)|$ for $k\ge 2$, whence $|f(a_{n+1})|<|f(a_n)|$
as required for (iii).

[3 marks]
\bigskip

We therefore see that we can produce an infinite sequence satisfying
(iii), and since the valuation is discrete this implies that
$|f(a_n)|\rightarrow 0$. Parts (i) and (ii) then imply that
$|a_{n+1}-a_n|\rightarrow 0$, which suffices to ensure that the
sequence $(a_n)$ is convergent, with limit $a^*$, say.  By continuity
of polynomials $f(a^*)=0$. Moreover
\[|a^*-a_0|=\lim |a_n-a_0|\le \sup|a_{i+1}-a_i|\le\sup
|f(a_i)|/|f'(a_i)|\le |f(a_0)|/|f'(a_0)|.\]
Thus $a=a^*$ fulfills the conditions of the question.

[3 marks]
\bigskip

Under the new hypothesis we may proceed as before, to get a sequence
satisfying (i) with $n$ replace by $n+1$.  Instead of (\ref{e1}) we
have
\[|b_n|^2=|f(a_n)|^2/|f'(a_n)|^2\le
|f(a_n)|\frac{|f(a_0)|}{|f'(a_0)|^2}\le |f(a_n)|,\]
and instead of (\ref{e2}),
\[|b_n|\le\sqrt{|f(a_n)|}\le\sqrt{|f(a_0)|}\le |f'(a_0)|=|f'(a_n)|< 1.\]
This suffices to show that $|b_n^kf_k(a_n)|<|f'(a_n)|$ for $k\ge 2$,
while for $k=1$ we have $f_1(x)= f''(x)$, whence
$|b_nf_1(a_n)|\le|f'(a_n)f_1(a_n)|<|f'(a_n)|$.  The treatment of (ii)
then goes through as before. 

For (iii) we see from the above estimate that $|b_n^k|<|f(a_n)|$ if
$k\ge 3$, so that $|b_n^kF_k(a_n)|<|f(a_n)|$.  For $k=2$ we have
$F_2(x)=\tfrac12 f''(x)$, whence $|b_n^2F_2(a_n)|\le|f(a_n)F_2(a_n)|<|f(a_n)|$.
The proof may now be completed as before.

[5 marks]
\bigskip

Let $f(x)=x^3+3x+y^3+3y-n$, so that $f'(x)=3x^2+3$ and $\tfrac12
f''(x)=3x$. It follows that $|f'(b)|=3^{-1}$ and $|\tfrac12
f''(b)|<1$ for any $b\in\mathbb{Z}_3$. Thus if there exist $a_0$
and $y$ such that $|f(a_0)|\le 3^{-2}$, then $f=0$ is solvable. Hence
it is sufficient (and clearly necessary) that $x^3+3x+y^3+3y\equiv
n\pmod 9$ should be solvable.  The required $n$ are therefore those
congruent to 0, 1, 4, 5 or 8 $\pmod{9}$.

[4 marks]
\newpage

Solution 2
\bigskip

Either $|x|_p\leqslant 1, |y|_p\leqslant 1$ or $|x|_p,|y|_p>1$.
Assume the latter, for a contradiction.  Then the parameters $z=-x/y$
and $w=-1/y$ satisfy $|z|_P=|x|_p^{-1/2}<1$ and $|w|_p<1$, by the
equation for ${\mathcal{E}}$. If $(x,y)$ were torsion, then $z$ 
would be a torsion point in $F_{\mathcal{E}}(p\mathbb{Z}_p)$. Thus it has order
$p^n$ for some $n\ge 1$, and $|z|_p\ge p^{-1/(p^n-p^{n-1})}$. However,
$p^n-p^{n-1}>1$ for $n\ge 1$, unless $p=2$ and $n=1$. If
$(x,y)$ has order 2 we have $y=0$, contradicting the assumption that
$|y|_p>1$. Otherwise $1>|z|_p>p^{-1}$, which is also impossible.

[4 marks]
\bigskip

When ${\widetilde {\mathcal{E}}}$ is non-singular the group
$\widetilde{\mathcal{E}_{\mathrm{ns}}}$ is the whole of ${\widetilde
{\mathcal{E}}}$, and the reduction map is a group homomorphism
from $\mathcal{E}(\mathbb{Q}_p)$ to ${\widetilde
  {\mathcal{E}}}(\mathbb{F}_p)$. The result just proved shows that no
non-trivial torsion points of $\mathcal{E}(\mathbb{Q}_p)$ are mapped
to zero by the reduction map, whence it is an injection on
${\mathcal{E}}_{\mathrm{tors}}(\mathbb{Q}_p)$.  Thus
${\mathcal{E}}_{\mathrm{tors}}(\mathbb{Q}_p)$
is isomorphic to a subgroup of ${\widetilde
  {\mathcal{E}}}(\mathbb{F}_p)$.

[4 marks]
\bigskip

If $y^2=x^3+ax+b$ is defined over $\mathbb{Z}$ and
$(x_1,y_1)\in\mathbb{Q}^2$ is a non-trivial torsion
point, then $|x_1|_p\le 1,\,|y_1|_p\le 1$ for every prime $p$, whence
$x_1,y_1\in\mathbb{Z}$.

[1 mark]
\bigskip

If $y_1 = 0$ then the result is immediate.  Otherwise
$(x_1,y_1)$ is not $2$-torsion and we consider $(x_2,y_2) = 2(x,y)$,
with $(x_2,y_2) \not=\mathbf{0}$, whence $x_2,y_2\in \mathbb{Q}$.  However 
$(x_2,y_2)$ is also a torsion point, so that $x_2 , y_2 \in
\mathbb{Z}$.  The line
tangent to $\mathcal{E}$ at $(x_1,y_1)$ has slope $(3x_1^2+a)/(2y_1)$, from
which we have $x_2 = \bigl( (3x_1^2+a)/(2y_1) \bigr)^2 - 2x_1$.
Since $x_2, x\in \mathbb{Z}$ we have $\bigl( (3x_1^2+a)/(2y_1)
\bigr)^2\in \mathbb{Z}$. 
It follows that $4y_1^2 | (3x_1^2+a)^2$ and so $y_1^2 | (3x_1^2+a)^2 =
\psi_1(x_1)$. 
Also, $y_1^2 = x_1^3 + ax_1 + b = \psi_2(x_1)$. Using the identity given
we conclude that $y_1^2 | (\phi_1(x_1)\psi_1(x_1) + \phi_2(x_1)\psi_2(x_1))
= \Delta$, as required.

[7 marks]
\bigskip

\begin{enumerate}
\item[(a)] $(1/4\,,\,a+1/8)$ is on the curve, but is non-integral, and
hence cannot be a torsion point.

[2 marks]

\item[(b)] $\Delta=4.1^3+27.2^2=112$, so there is good reduction at 5,
  where the points in $\mathbb{F}_5$ are $(1,\pm 2), (4,0)$ and
  $\mathbf{0}$.  Only one of these has order 2 (with $y=0$) so this is
  $C_4$.  Globally the point $(-1,0)$ has order 2, and if $P=(1,2)$
  calculation shows that $x(2P)=-1$, so that $2P$ must be $(-1,0)$.
  Hence the $\mathbb{Q}$-torsion is
  $\{(1,2),(1,-2),(-1,0),\mathbf{0}\}$ of type $C_4$.

[3 marks]

\item[(c)] 
$\Delta=4a^3+27b^2\not\equiv 0\pmod{5}$ so there is good reduction at
5.  Moreover ${\widetilde {\mathcal{E}}}$ is given by $y^2=x^3+x$,
which has the points $\mathbf{0}, (\pm 2,0)$ and $(0,0)$ over
$\mathbb{F}_5$.  Thus the rational torsion group has order dividing
4. Hence if there is non-trivial torsion there would be a rational
point $(\alpha,0)$ of order 2. Then $\alpha$ would be an integer root
of $x^3+ax+b\equiv x^3+x+1\pmod{2}$.  This is impossible as $x^3+x+1$
is irreducible modulo $2$.

[4 marks]

\end{enumerate}

\newpage

Solution 3
\bigskip

As coset representatives for $\mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^2$
we may take non-zero square-free values $r\in\mathbb{Z}$. Such an $r$
arises from $q(x,y)$ if and only if $x=ru^2$ and $x^2+ax+b=rv^2$ for
some rational $u,v$.  If $u=l/m$ in lowest terms, then
$r^2l^4+ral^2m^2+bm^4=rn^2$ with integers $l,m$ for which $(l,m)=1$
and $m\ne 0$.  Since $rn^2$ is an integer and $r$ is square-free it
follows that $n$ is also an integer.  We claim that $r|b$.  Otherwise 
there would be a prime $p|r$ with $p\nmid b$. Then
$p|r(n^2-rl^4-al^2m^2)=bm^4$, whence $p|m$.  This entails
$p^2|r^2l^4+ral^2m^2+bm^4$, whence $p|rn^2$, from which we deduce that
$p|n$.  Finally we would have $p^3|rn^2-ral^2m^2-bm^4=r^2l^4$, which
imples that $p|l$.  This contradicts our assumption that $(l,m)=1$,
and therefore shows that no such $p$ can exist.  It therefore follows
that $r|b$, so that $q$ must have finite image.

[7 marks]
\bigskip

The maps $\phi:\mathcal{C}\rightarrow\mathcal{D}$ and
$\hat{\phi}:\mathcal{D}\rightarrow\mathcal{C}$ are homomomorphisms with
$\phi\hat{\phi}$ and $\hat{\phi}\phi$ being $[2]$ on the respective
groups $\mathcal{D}$ and $\mathcal{C}$.  We are told that ${\rm
  ker}(q)=\hat{\phi}(\mathcal{D})$, and we have a similar map
$\hat{q}$ on $\mathcal{D}$ with ${\rm
  ker}(\hat{q})=\phi(\mathcal{C})$. We then obtain induced injective
homomorphisms from $\mathcal{C}/\hat{\phi}(\mathcal{D})$ and
$\mathcal{D}/\phi(\mathcal{C})$ into
$\mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^2$, each with finite image,
as above.  Thus $\mathcal{C}/\hat{\phi}(\mathcal{D})$ and
$\mathcal{D}/\phi(\mathcal{C})$ are finite.

Let $\{g_1,\ldots,g_k\}$ be coset representatives for 
$\mathcal{C}/\hat{\phi}(\mathcal{D})$, and similarly $\{h_1,\ldots,h_l\}$ for 
$\mathcal{D}/\phi(\mathcal{C})$.  Given $g\in
\mathcal{C}$ there is a $g_i$ and an $h\in\mathcal{D}$
such that $g=g_i+\hat{\phi}(h)$.  Similarly there is an $h_j$ and a $g'\in
\mathcal{C}$ so that $h=h_j+\phi(g')$.  Since $\hat{\phi}$
is a homomorphism, and $\hat{\phi}\phi=[2]$ we have
\[g=g_i+\hat{\phi}(h)=g_i+\hat{\phi}(h_j+\phi(g'))
=g_i+\hat{\phi}(h_j)+\hat{\phi}\phi(g')=g_i+\hat{\phi}(h_j)+2g'.\]
Thus the cosets $g_i+\hat{\phi}(h_j)+2\mathcal{C}(\mathbb{Q})$ cover
$\mathcal{C}$, whence $\mathcal{C}/2\mathcal{C}$ is finite.

[7 marks --- Bookwork to here]
\bigskip

For $y^2=x(x^2-p^2)$ we have $s=2$.  The image of $q$ corresponds to
solvable equations $r^2l^4-p^2m^4=rn^2$, with $r|p$; and since $\mathcal{E}'$ is
given by $y^2=x(x^2+4p^2)$ the image of $\hat{q}$ corresponds to
solvable equations $r^2l^4+4p^2m^4=rn^2$, with $r|2p$.  For
$r^2l^4-p^2m^4=rn^2$ there are solutions $(1,0,1),(0,1,p), (1,1,0)$
and $(1,1,0)$ for $r=1,-1,p$ and $-p$ respectively.  For $r^2l^4+4p^2m^4=rn^2$
we must clearly have $r>0$.  Thus there are at most 4 possibilities
for $r$, namely $1,2,p$ and $2p$.  Hence, from the above,
\[2^{R+2}=\#\mathcal{C}/2\mathcal{C}\le
\big(\#\mathcal{C}/\hat{\phi}(\mathcal{D})\big)
\big(\#\mathcal{D}/\phi(\mathcal{C})\big)\le 4\times 4,\]
so that the rank is always at most 2.

[5 marks]
\newpage

If $x^4+p^2y^4=2z^2$ has a non-trivial integral solution we may choose
a minimal one.  If $p|z$ we would have $p|x$ and so
$y^4+p^2(x/p)^4=2(z/p)^2$, contradicting minimality.  Hence $p\nmid
z$.  However $2z^2\equiv x^4\pmod{p}$, and therefore $(\tfrac{2}{p})=+1$.

[2 marks]
\bigskip

Thus if $r^2l^4+4p^2m^4=rn^2$ and $r=2$ then $p\equiv\pm 1\pmod{8}$.
Thus for $p\equiv 3$ or $5\pmod{8}$ the case $r=2$ cannot occur, giving
$\#\mathcal{D}/\phi(\mathcal{C})\le 3$, whence $R<2$.  

[2 marks]
\bigskip

For $p=5$ the curve $C$ has a point $(x,y)=(25/4,75/8)$, which is
non-integral and therefore of infinite order.  Thus $R\ge 1$, and
hence $R=1$.

[2 marks --- Unseen, but similar to previous exam questions]

\end{document}
