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\begin{question}{1}


Find an integer $n\in\mathbb{Z}$ such that $|n^3-5|_2<2^{-3}$.  Show
that there is no integer $n$ for which $|n^3-5|_5<5^{-3}$. 
\bigskip

Let $K$ be a field, complete with respect to a discrete
non-Archimedean valuation 
$|\; |$, with valuation ring $R = \{x\in K : |x| \le 1\}$. Prove 
that if $f(x) \in R[x]$ and $a_0 \in R$ satisfies the inequality
$|f(a_0)| < |f'(a_0)|^2$, then one can  produce an infinite sequence
$(a_n)_0^{\infty}$ in $R$, starting at $a_0$, such that
\begin{enumerate}
\item[(i)] $|a_{i+1}-a_i|\le |f(a_i)|/|f'(a_i)|$ for $i\ge 0$;
\item[(ii)] $|f'(a_0)|=|f'(a_1)|=|f'(a_2)|=\ldots\;$; and
\item[(iii)] $|f(a_0)|>|f(a_1)|>\ldots$
\end{enumerate}
Deduce that there exists an element $a \in R$ such that 
$|a - a_0| \le |f(a_0)|/|f'(a_0)|$ and $f(a) = 0$ .
\bigskip

Show that the above conclusion remains true if $|f(a_0)| = |f'(a_0)|^2$,
providing that one has $0<|f'(a_0)|<1$ and $|\tfrac12 f''(b)|<1$ for all $b\in
R$.
\bigskip

For which values $n\in\mathbb{Z}$ does the equation $x^3+3x+y^3+3y=n$
have solutions $x,y\in\mathbb{Z}_3$ ?
\end{question}


\begin{question}{2}




Let ${\mathcal{E}} : y^2 = x^3 + A x + B$, be an elliptic curve,
where $A,B\in \mathbb{Z}_p$.
Show that any $(x,y) \in {\mathcal{E}}_{\mathrm{tors}}(\mathbb{Q}_p)$ 
satisfies $|x|_p\leqslant 1, |y|_p\leqslant 1$. 
\bigskip

[You may assume that if $F(X,Y)$ is a formal group defined over
  $\mathbb{Z}_p$, and that if $z\in p\mathbb{Z}_p$ is a non-trivial
torsion point for  the group operation $x \oplus y = F(x,y)$ on $
p\mathbb{Z}_p$, then it has order $p^n$, and
  $|z|_p\ge p^{-1/(p^n-p^{n-1})}$.]
\bigskip

Let ${\widetilde {\mathcal{E}}}$
denote the reduction of ${\mathcal{E}}$ modulo $p$.
If ${\widetilde {\mathcal{E}}}$ is 
non-singular, show that ${\mathcal{E}}_{\mathrm{tors}}(\mathbb{Q}_p)$
is isomorphic to a subgroup of ${\widetilde {\mathcal{E}}}(\mathbb{F}_p)$.
\bigskip

Let $y^2=x^3+ax+b$ be an elliptic curve defined over $\mathbb{Z}$.
Show that if $(x_1,y_1)\in\mathbb{Q}^2$ is a non-trivial torsion
point, then $x_1,y_1\in\mathbb{Z}$, and either $y_1=0$ or
$y_1^2|4a^3+27b^2$.
\bigskip

[You may use without proof the polynomial identity
\[\phi_1(X) \psi_1(X) + \phi_2(X) \psi_2(X) = 4a^3 + 27b^2,\]
where $\phi_1(X)= 3X^2+4a$, $\psi_1(X) = (3X^2+a)^2$,
$\phi_2(X)= -27(X^3 + aX - b)$ and $\psi_2(X) = X^3 + aX + b$.]
\bigskip

\begin{enumerate}
\item[(a)] If $b=a^2$, find a point of infinite order on 
$\mathcal{E}(\mathbb{Q})$.
\item[(b)] Find the torsion elements over $\mathbb{Q}$, and their
  group structure, in the case $a=1, b=2$.
\item[(c)] Prove that if $a\equiv 1\pmod{10}$ and $b\equiv 5\pmod{10}$ then the
point at infinity is the only rational torsion point. 
\end{enumerate}


\end{question}

\begin{question}{3}



Let  $\mathcal{E}: y^2=x(x^2+ax+b)$ be an elliptic curve, with
$a,b\in\mathbb{Z}$, and let $\mathcal{C}=\mathcal{E}(\mathbb{Q})$.
Define the map $q:\mathcal{C}\rightarrow
\mathbb{Q}^{\times}/(\mathbb{Q}^{\times})^2$ by setting
$q(x,y)=x(\mathbb{Q}^{\times})^2$ if $x\ne 0$ and 
$q(x,y)=b(\mathbb{Q}^{\times})^2$ otherwise.  Show that the image of
$q$ is finite.
\bigskip

Now let $\mathcal{E}'$ be the elliptic curve $y^2=x(x^2+a_1x+b_1)$,
with $a_1=-2a$ and $b_1=a^2-4b$, and let $\mathcal{D}=\mathcal{E'}(\mathbb{Q})$.
Stating clearly any properties you may require concerning the
2-isogeny $\phi:\mathcal{E}\rightarrow
\mathcal{E'}$ and its dual $\hat{\phi}$, show that
$\mathcal{C}/2\mathcal{C}$ is finite.
\bigskip

[You may assume that the map $q$ above is a homomorphism, with kernel
  $\hat{\phi}(\mathcal{D})$.] 
\bigskip

In the special case $\mathcal{E}: y^2=x(x^2-p^2)$, where $p$ is an odd
prime, show that the rank is at most 2.
\bigskip

[You may use without proof the fact that the order of
  $\mathcal{C}/2\mathcal{C}$ is $2^{R+s}$ where $R$ is the rank of
  $\mathcal{C}$ and $2^s$ is the order of its rational 2-torsion
  subgroup.]
\bigskip

Show that if $x^4+p^2y^4=2z^2$ has a non-trivial integer
solution, then $(\tfrac{2}{p})=+1$, so that $p\equiv\pm 1\pmod{8}$.
\bigskip

Deduce that, for $\mathcal{E}: y^2=x(x^2-p^2)$ as before, the rank is at most
1 if $p\equiv 3$ or $5\pmod {8}$.  For $p=5$ display a non-integral
point $(x,y)$ on the curve. [It may help to consider points for which
  $x$ is a square.]  Deduce that for $p=5$ the rank is
exactly 1.



\end{question}

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