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Question 1.
\bigskip

(a)\medskip

We require $m^2\equiv 2\pmod{7^3}$. We have $3^2\equiv 2\pmod{7}$, and
$(3+7k)^2\equiv 9+42k\pmod{7^2}$.  Thus $(3+7k)^2\equiv 2\pmod{7^2}$
iff $49\mid 7+42k$ iff $7\mid 1+6k$ iff $k\equiv 1\pmod{7}$.  Hence
$10^2\equiv 2\pmod{49}$, and $(10+7^2k)^2\equiv 100+980k\pmod{7^3}$.  
Thus $(10+49k)^2\equiv 2\pmod{7^3}$ iff $7^3\mid 98+980k$ iff 
$7\mid 2+20k$ iff $k\equiv 2\pmod{7}$.  Hence  $m=10+49.2=108$ is a
solution. 

[3 marks; S]\medskip

If $|n^3-2|_7\le 7^{-3}$ then $n^3\equiv 2\pmod{7^3}$ and hence 
$n^3\equiv 2\pmod{7}$. But $0^3,1^3,2^3\ldots,6^3\equiv
0,1,1,6,1,6,6\pmod{7}$ so that we never have $n^3\equiv 2\pmod{7}$.

[2 marks; S]\medskip

Fact: an infinite $p$-adic series converges iff the individual terms
tend to zero. Since $p\mid k$ we have $|k|_p\le p^{-1}$ and hence 
$|k^n|_p\le p^{-n}\rightarrow\infty$. The series therefore converges.

[2 marks; S]\medskip

We claim the limit is $(1-k)^{-1}$.  Note that $1-k\not=0$ since
$p\mid k$.  We have $\sum_{n<N}k^n=(1-k^N)/(1-k)$, whence
\[\left|\sum_{n<N}k^n-\frac{1}{1-k}\right|_p=\left|\frac{k^N}{1-k}\right|_p
\le\frac{p^{-N}}{|1-k|_p}\rightarrow 0\]
as required.

[3 marks; S]\medskip

We claim that $3^{2^n}\equiv 1\pmod{2^n}$ for all $n\in\mathbb{N}$,
which will suffice.  This is true for $n=1$, and if $3^{2^n}=1+2^nk$
with $n\ge 1$ then 
\[3^{2^{n+1}}=1+2^{n+1}(k+2^{n-1}k^2)\equiv 1\pmod{2^{n+1}}. \]
The claim therefore follows by induction.  Hence $|3^{2^n}-1|_2\le
2^{-n}$, which tends to zero as $n$ tends to infinity, as required.

[3 marks; N]\medskip

The sequence $b+(a-b)3^{2^n}$ therefore converges to $a$ in $\mathbb{Q}_2$
and to $b$ in $\mathbb{Q}_3$.

[2 marks; N]\newpage

(b)\medskip

One can determine $p$ and $q$ if the calculation involves a multiple
$jP$ for which $jP$ is the point at infinity modulo one of the primes
but not modulo the other.

[2 marks; B]\medskip

We start by computing $2P$ modulo 391, using the tangent line $y=mx+c$
at $(3,4)$. Thus $m=(3.3^2)/(2.4)\equiv 3.9.49\equiv 150\pmod{391}$, since
$8.49=392\equiv 1\pmod{391}$; and $c=4-3m$. This line intersect
the curve when $x^3-11-(mx+c)^2$=0, giving three roots $x=3,3$ and
$x_2$ say, whose sum is $m^2$.  Thus $x_2\equiv 150^2-6\equiv
207\pmod{391}$, and the corresponding $y$ value is
$y_2=mx_2+c=4+m(x_2-3)\equiv 4+150.204\equiv 106\pmod{391}$.  Hence
$2P=(207,-106)$. 

[5 marks; S]\medskip

We now compute $3P=P+2P$, using the line $y=nx+d$ between $P$ and
$2P$.  Thus $n=(4-(-106))/(3-207)=-110/204$. However 204 does not have
an inverse modulo 391. Indeed the highest common factor is found to be
17, and we then find $391=17\times 23$.

[3 marks; S]


\newpage

Question 2.
\bigskip

(a)\medskip

We require
\begin{enumerate}
\item[(i)]
$F(X,Y)=X+Y+$ higher order terms;
\item[(ii)]
$F(X,Y)=F(Y,X)$; and
\item[(iii)]
$F(F(X,Y)Z)=F(X,F(Y,Z))$.
\end{enumerate}

[3 marks; B]\medskip

We will prove by induction that there is a unique series of
polynomials $Z_n(T)$ of degree $\le n$ such that (a) $Z_n$ and
$Z_{n+1}$ agree up to degree $n$, and (b) $F(X,Z_n)$ has only terms
of degree $>n$.  This suffices since we may take $i(T)$ as the formal 
limit of the
polynomials $Z_n(T)$; and this is clearly the only admissible choice,
since if $F(T,i(T))=0$ we may take $Z_n(T)$ as the terms of $i(T)$ of 
degree $\le n$, giving an admissible sequence $Z_n$.

[3 marks; B]\medskip

For the induction we see that $Z_1(T)=-T$ is admissible by
(i). Generally suppose $F(T,Z_n)=cT^{n+1}+\ldots$, and consider
$Z_{n+1}=Z_n+dT^{n+1}$. We have $F(T,Z+W)=F(T,Z)+WF_Y(T,Z)+$terms
involving higher powers of $W$, and $F_Y(T,Z)=1+$higher order terms,
by (i). Hence $F(T,Z_n+dT^{n+1})=F(T,Z_n)+dZ^{n+1}+$higher order terms, so that
$F(T,Z_n+dT^{n+1})=(c+d)Z^{n+1}+$higher order terms.  We therefore see
that there is exactly one possible choice for $d$, namely $d=-c$.

[3 marks; B]\medskip

We require that $P(F(T,S))F_X(T,S)=P(T)$ identically in $T$ and $S$.

[1 mark; B]\medskip

For $P(T)=F_X(0,T)^{-1}$ we first note that $F_X(0,T)=1+$higher order
terms, by (i) above, whence $F_X(0,T)^{-1}$ exists as a power series.
It then suffices to check that
$F_X(0,F(T,S))^{-1}F_X(T,S)=F_X(0,T)^{-1}$, or equivalently that
$F_X(T,S)F_X(0,T)=F_X(0,F(T,S))$.  However $F(X,F(T,S))=F(F(X,T),S)$
by (iii) above, and the result follows on differentiating with respect
to $X$ and setting $X=0$.

[4 marks; B]\newpage

(b)\medskip

Theorem: The kernel of the reduction map $\mathcal{E}(\mathbb{Q}_p)\rightarrow
\tilde{\mathcal{E}}(\mathbb{F}_p)$ is torsion-free.

[2 marks; B]\medskip

Hence if we restrict the reduction map to the subgroup 
$\mathcal{E}_{\rm tors}(\mathbb{Q}_p)$ the kernel is trivial, giving
us an injection into $\tilde{\mathcal{E}}(\mathbb{F}_p)$, which is
finite.  It follows that $\mathcal{E}_{\rm tors}(\mathbb{Q}_p)$ must
also be finite.

[3 marks; B]\medskip

(c)\medskip

The discriminant is $4n^6$ and hence there is good reduction for any
prime $\ge 5$ which does not divide $n$. For $p\equiv 5\pmod{6}$ 
the congruence $x^3\equiv
y^2-n^2\pmod{p}$ has exactly one solution $x$ for each $y$ modulo
$p$.  Thus
the elliptic curve has $p+1$ points modulo $p$, allowing for the point
at infinity.  

[3 marks; S]\medskip

It follows that $\#\mathcal{E}_{\rm tors}(\mathbb{Q}_p)$ divides
$p+1$.  Since $\mathcal{E}_{\rm tors}(\mathbb{Q})$ is a subgroup of
$\mathcal{E}_{\rm tors}(\mathbb{Q}_p)$ we deduce that its order, $t$ say, also
divides $p+1$.   If $k=(p+1)/6$ there is a prime $p_1\equiv
5\pmod{6}$ for which $(p_1+1)/6$ is coprime to $k$.  But then $t$
divides both $p+1$ and $p_1+1$, and hence divides their highest
common factor, which is 6.

[3 marks; N]


\newpage

Question 3.
\bigskip

One defines $q:\mathcal{H}\rightarrow\mathbb{Q}^\times/(\mathbb{Q}^\times)^2$
by taking $q(\mathbf{0})$ to be the identity coset, by taking
$q(0,0)$ to be the coset of $b_1$, and by taking $q(u,v)$ to be the
coset of $u$ for all other elements $(u,v)$ of $\mathcal{H}$.

[2 marks; B]\medskip

For any non-zero rational number $u$ we may write $u=rw^2$ with $r$
being a square-free integer, and $w\in\mathbb{Q}$.  Then if
$v^2=u(u^2+a_1u+b_1)$ we must have $u^2+a_1u+b_1=rs^2$ for some
rational number $s$, and $v=rsw$. Suppose there is a prime with $p\mid
r$ for which $p\nmid b_1$. We consider two cases.  If $|u|_p<1$ then
$|u^2|_p<1$ and $|a_1u|_p<1$, since $|a_1|_p\le 1$.  Thus
$|u^2+a_1u+b_1|_p=1$, since $|b_1|_p=1$. This is impossible, since
$rs^2$ is either zero, or contains $p$ to an odd power.  On the
other hand if $|u|_p\ge 1$ then in fact $|u|_p>1$, since $u=rw^2$ 
contains $p$ to an odd power.  Thus
$|u^2|_p>|u|_p\ge |a_1u|_p$, and
$|u^2|_p>1=|b_1|_p$, whence $|u^2+a_1u+b_1|_p=|u|^2_p$.  Again this 
is impossible, since
$rs^2$ is either zero, or contains $p$ to an odd power.
It follows that $p|b_1$ whenever $p|r$, so that $r|b_1$.

[8 marks; S]\medskip

The maps $\phi$ and $\hat{\phi}$ are group homomorphisms, and 
$\hat{\phi}\phi=[2]$.

[1 mark; B]\medskip

Let $g_1,\ldots,g_m$ be distinct coset representatives for 
$\mathcal{G}/\hat{\phi}(\mathcal{H})$, and $h_1,\ldots,h_n$
similarly for $\mathcal{H}/\phi(\mathcal{G})$.  Then for any $g\in G$
we have $g=g_i+\hat{\phi}(h)$ for some $h\in\mathcal{H}$, and $h=h_j+
\phi(g')$ for some $g'\in\mathcal{G}$.  Hence $g=g_i+\hat{\phi}(h_j)+
\hat{\phi}(\phi(g'))=g_i+\hat{\phi}(h_j)+2g'$.  It follows that the set
of all combinations $g_i+\hat{\phi}(h_j)$ form a set of (not
necessarily distinct) coset representatives for 
$\mathcal{G}/2\mathcal{G}$.  Hence
\[\#\left(\frac{\mathcal{G}}{2\mathcal{G}}\right)\le mn=
\#\left(\frac{\mathcal{G}}{\hat{\phi}(\mathcal{H})}\right)
\#\left(\frac{\mathcal{H}}{\phi(\mathcal{G})}\right).\]

[5 marks; B]\medskip

The image of $q$ consists of cosets $r(\mathbb{Q}^\times)^2$ with
integers $r$ dividing $b_1$, for which the equation
\[r\ell^4+a_1\ell^2m^2+b_1r^{-1}m^4=n^2\]
has an integral solution with $\ell,m,n$ not all zero.  
Such a class corresponds to
a point $(u,v)$ with $u=r\ell^2/m^2$. There is an entirely analogous
map from $\mathcal{G}/\hat{\phi}(\mathcal{H})$ to
$\mathbb{Q}^\times/(\mathbb{Q}^\times)^2$, where the corrsponding
integers $r$ divide $b$, and the equation
\[r\ell^4+a\ell^2m^2+br^{-1}m^4=n^2\]
is solvable.

For $\mathcal{C}$ this latter equation is $r\ell^4+5\ell^2m^2+3r^{-1}m^4=n^2$
with $r|3$.  The point $(0,0)$ produces $r=b=3$, while the points
$(1,3)$ and $(-3,3)$ produce $r=1$ and $r=-3$ respectively.  Thus these
3 points, together with the point at infinity, form a complete set of
coset representatives for $\mathcal{G}/\hat{\phi}(\mathcal{H})$.

[4 marks; S/N]\medskip

Similarly for $\mathcal{D}: v^2=u(u^2-10u+13)$ we have equations
$r\ell^4-10\ell^2m^2+13r^{-1}m^4=n^2$ with $r|13$.  Clearly there are
no solutions with when $r<0$, and $r=1$ and $r=13$ arise from the
point at infinity and $(0,0)$ respectively.  These two points are
therefore coset representatives for $\mathcal{H}/\phi(\mathcal{G})$.

[3 marks; S/N]\medskip

Since $\hat{\phi}(0)=\hat{\phi}(0,0)=\mathbf{0}$ the argument above shows that
$\mathcal{G}/(2\mathcal{G})$ consists of the cosets of $\mathbf{0}, (0,0),
  (1,3)$ and $(-3,3)$, so that the group has order 4.

[2 marks; S/N]\medskip

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