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\par
\noindent {\bf Solution to Question 1.} 
\par\noindent
(a) [{\it Proof from lectures}]. 
We are given: $|f(a_0)| < |f'(a_0)|^2$\ $(*)$.
Define $f_j(x)$ by:
\par\noindent $f(x + y) = f_0(x) + f_1(x) y + f_2(x) y^2 + \ldots,$
so that $f_0(x) = f(x), f_1(x) = f'(x)$. Define
$b_0 = -f(a_0)/f'(a_0)$. By $(*)$, $|b_0| < 1$.
\par Define $a_1 = a_0 + b_0 = a_0 - f(a_0)/f'(a_0)$. Then:
\par\noindent $| f'(a_1) - f'(a_0) | = | f'(a_0+b_0) - f'(a_0) |
= | (\hbox{poly in }a_0) b_0 + (\hbox{poly in }a_0) b_0^2 + \ldots |$
\par \ \ \ \ \ \ \ \ \ \ \ $\leqslant | b_0 | < | f'(a_0) |$
\ \ (by $(*)$), so that $| f'(a_1) | = | f'(a_0) |$.
\par\noindent Also,
$| f(a_1) | = | f(a_0 + b_0) | =
| f_0(a_0) + f_1(a_0)b_0 + f_2(a_0)b_0^2 + \ldots |$
\par\ \ \ $= | f_2(a_0)b_0^2 + \ldots |$\ \
[since $f_0(a_0) + f_1(a_0)b_0 = 0$]
% N.B. I might omit the following line:
%\par\ \ \ $\leqslant \hbox{max}_{j\geqslant 2} |f_j(a_0)| |b_0|^j
$\leqslant | b_0 |^2 = \frac{ |f(a_0 )|^2}{|f'(a_0 )|^2}$
\par\ \ \ $= \rho | f(a_0) | < | f(a_0) |$,
where $\rho = \frac{ |f(a_0 )|}{|f'(a_0 )|^2} < 1$.
Summarising: 
\par\noindent $ |f'(a_1) | = |f'(a_0)|$ and
$| f(a_1) | \leqslant \rho | f(a_0) | < | f(a_0) |$, where
$\rho = \frac{ |f(a_0 )|}{|f'(a_0 )|^2} < 1$.
% \par\noindent Given $a_n\in R$, define
\par\noindent For all~$n$, given $a_n\in R$, define
$b_n = -f(a_n)/f'(a_n)$ and $a_{n+1} = a_n + b_n$
\par\noindent $= a_n -f(a_n)/f'(a_n)$.
%\par As above: $|f'(a_{n+1})| = \ldots |f'(a_1)| = |f'(a_0)|$.
%\ \ \ \ \ \ (1)
Assume, as induction hypothesis, that:
\par\noindent $|f'(a_n)| = \ldots = |f'(a_0)|$
and $ | f(a_n) | \leqslant \rho |f(a_{n-1})| \leqslant \ldots
\leqslant \rho^n |f(a_0)|$.\ \ \ \ \ \ (1)
\par\noindent Then, as above: $|f'(a_{n+1})| = \ldots = |f'(a_1)| = |f'(a_0)|$.
\par\noindent Then $| f(a_{n+1}) | \leqslant |b_n|^2$ \ \ [justified as
for~$n=0$ above]
$= \frac{ | f(a_n) |^2 }{ | f'(a_n) |^2 }$
\par\noindent
$= \frac{ | f(a_n) |^2 }{ | f'(a_0) |^2 }$\ [by~(1)]
$\leqslant \frac{ | f(a_0) | }{ | f'(a_0) |^2 } |f(a_n)|$
 \ [by~(1)]
$= \rho | f(a_n) | \leqslant \rho^{n+1} | f(a_0) |$
\ [by~(1)].
\par\noindent By induction, 
\par\noindent $\forall n$, $|f'(a_n)| = |f'(a_0)|$ and
$ | f(a_{n}) | \leqslant \rho^n | f(a_0) |$ which $\rightarrow 0$
as $n\rightarrow \infty$. \ \ \ \ \ \ (2)
\par\noindent Now, $| b_n | = | f(a_n) |/|f'(a_n)|
= | f(a_n) |/|f'(a_0)| \rightarrow 0$, so
[by thm from lectures that a series converges
in a non-Archimedean field iff its terms converge to~$0$]:
\par $a_n = a_0 + b_0 + b_1 + \ldots + b_n$
converges to~$a$, say.
\par\noindent By continuity of polynomials,
$f(a) = \lim f(a_n) = 0$\ \ [by~(2)].
Furthermore: 
\par $| a - a_0 | = | \sum b_n | \leqslant
\max | b_n | = \max \frac{| f(a_n) |}{|f'(a_n)|}
= \max \frac{| f(a_n) |}{|f'(a_0)|}
= \frac{| f(a_0) |}{|f'(a_0)|}$ [by~(2)].
\par For uniqueness, imagine that $\hat a \not= a$ also satisfied
$f(\hat a) = 0$ and that:
\par\noindent $| \hat a - a_0 | \leqslant | f(a_0) |/ | f'(a_0) |$.
Let $\hat b = \hat a - a \not= 0$.
\par Then $0 = f(\hat a) - f(a) = f(a + \hat b) - f(a)
= {\hat b} f_1(a) + {\hat b}^2 f_2(a) + \ldots$ \ \ \ \ \ \ (3)
\par\noindent But $| \hat b | = | \hat a - a_0 + a_0 - a |
\leqslant \max( | \hat a - a_0 |, | a - a_0 | )
\leqslant | f(a_0) |/ | f'(a_0) | $
\par \ \ \ $< | f'(a_0) \hbox{ [by (*)] }|
= | f_1(a_0) | = | f_1(a) |$ \ \ [by~(1) and continuity of $|f'(x)|$].
\par\noindent This gives $|{\hat b}^j f_j(a)| \leqslant
|{\hat b}^j| \leqslant |{\hat b}^2| < |{\hat b} f_1(a)|$
(since $|{\hat b}| \not= 0$ \& $|{\hat b}| < |f_1(a)|$)
for~$j\geqslant 2$, so that the leading term of the sum in~(3)
has valuation strictly greater than the valuations of the other terms,
which is inconsistent with the sum being~$0$. Hence~$a$
is unique.
\hfill {\bf [10~marks]}
\smallskip

\par\noindent
{\bf (b)} 
For any $x \in \Z$, if $11 | x$ then $x^{10} \equiv 0$ (mod~$11$),
and if $11 \nmid x$ then $x^{10} \equiv 1$ (mod~$11$) by
Fermat's Little Theorem. So always $x^{10} \not\equiv 2$ (mod~$11$).
Hence $| x^{10} - 2 |_{11} = 1$ and so
$| x^{10} - 2 |_{11} \not\leqslant 11^{-3}$.
\par
For any $x \in \Z$, $|x|_p = p^{-n}$, some $n\in \Z$, $n\geqslant 0$.
Hence $|x^p|_p = p^{pn}$. Also $|p|_p = p^{-1}$,
so $|x^p|_p \not= |p|_p$ and
$|x^p - p|_p = \hbox{max}(|x^p|_p, |p|_p) \geqslant |p|_p = p^{-1}$,
using the theorem: $|\alpha|_p \not= |\beta|_p \implies
| \alpha \pm \beta |_p = \hbox{max}( |\alpha|_p, |\beta|_p )$.
Hence $|x^p - p|_p \not\leqslant p^{-2}$.
\par
Let $f(x) = x^2 + 2$, and take $x_0 = a_0 = 1$. Then
$|x_0^2 + 2|_3 = 3^{-1}$. 
We want $x_1 = a_0 + a_1 3^1$
such that $|x_1^2 + 2|_3 \leqslant 3^{-2}$, that is:
$(a_0 + a_1 3^1)^2 \equiv -2$ (mod~$3^2$) $\iff$
$1 + 2a_1 3^1 \equiv -2$ (mod~$3^2$) $\iff$
$2a_1 3^1 \equiv -3$ (mod~$3^2$) $\iff$
$2a_1 \equiv -1$ (mod~$3$). So take $a_1 = 1$ and 
$x_1 = a_0 + a_1 3^1 = 1 + 1\cdot 3^1 = 4$.
\par
We next want $x_2 = a_0 + a_1 3^1 + a_2 3^2$
such that $|x_2^2 + 2|_3 \leqslant 3^{-3}$, that is:
$(a_0 + a_1 3^1 + a_2 3^2)^2 \equiv -2$ (mod~$3^3$) $\iff$
$4^2 + 2\cdot 4 a_2 3^2 \equiv -2$ (mod~$3^3$) $\iff$
$8 a_2 3^2 \equiv -18$ (mod~$3^3$) $\iff$
$8 a_2 \equiv -2$ (mod~$3$). So take $a_2 = 2$ and 
$x_2 = a_0 + a_1 3^1 + a_2 3^2 = 22$.
Note that already $|x_2^2 + 2|_3 = |484 + 2|_3 = |486|_3 
= | 2 \cdot 3^5 |_3 = 3^{-5}$, so we can take $x = 22$
as our solution.
\par
Let $f(x) = x^2 + 7$, and take $x_0 = 1$, so
that $| f(x_0) |_2 = |1^2 + 7|_2 = 2^{-3}$
and $f'(x) = 2x$, so that $| f'(x_0) |_2 = 2^{-1}$.
We use the method in the proof of Hensel's lemma.
Let $x_1 = x_0 - f(x_0)/f'(x_0) = 1 - (1^2 + 7)/(2\cdot 1) = 1 - 4 = -3$.
Then $|f(x_1)|_2 = |(-3)^2 + 7|_2 = |16|_2 = 2^{-4}$.
Let $x_2 = x_1 - f(x_1)/f'(x_1) 
= -3 - \bigl( (-3)^3 + 7 \bigr)/\bigl( 2\cdot (-3) \bigr)$
$= -3 - 16/(-6) = -1/3$. Then
$|f(x_2)|_2 = | (-1/3)^2 + 7 |_2 = | 64/9 |_2 = 2^{-6}$.
Let $x_3 = x_2 - f(x_2)/f'(x_2)$
$= -1/3 - \bigl( (-1/3)^2 + 7 \bigr)/\bigl( 2\cdot (-1/3) \bigr)
= -1/3 + (64/9)/(2/3) = 31/3$. Then
$|f(x_3)|_2 = | (31/3)^2 + 7 |_2 
= | (31 + 96)/9 |_2 = | 1024/9 |_2 = 2^{-10}$.
So, we can take $x = 31/3$ as a solution (there are also
integer solutions: $\pm 181, \pm 331, \pm 693, \pm 843$).
[{\bf 8 marks}; Seen similar types of examples]

\par\noindent
{\bf (c)} 
Let $\C : Y^2 = f(X) = p_1 X^4 - p_1 p_2^2 p_3^2 
= p_1 (X^2 + p_2 p_3)(X^2 - p_2 p_3)$,
so $f'(X) = 4 p_1 X^3$.
Any repeated root~$x_0$ of~$f(X)$ must satisfy
$f(x_0) = f'(x_0) = 0$, and so
$0 = 4(p_1 x_0^4 - p_1 p_2^2 p_3^2) - x_0(4 p_1 x_0^3) 
= -4 p_1 p_2^2 p_3^2$. So $\tC$ over $\F_p$ (for $p \not= 2,p_1,p_2,p_3$)
has ${\tilde f}(x)$ with no repeated roots, so by a theorem
from lectures, $\tC(\F_p)$ has at least $p - 1 - 2\sqrt{p}$
affine points; if also $p \geqslant 7$ this is positive,
so there is an affine point $(x_1,y_1)\in \tC(\F_p)$, which
is nonsingular (since ${\tilde f}(x)$ has no repeated roots),
and so lifts to a point $(x_2,y_2) \in \C(\Q_p)$ (by the theorem
that any nonsingular point on $\tC(\F_p)$ lifts to a point on $\C(\Q_p)$).
This shows the result for all $p \not\in \{ 2,3,5,p_1,p_2,p_3 \}$.
\par
For $p=2$, note that $p_2 p_3 \equiv 1$ (mod~$8$), so 
$p_2 p_3 \in (\Q_2^*)^2$ [by Cor to HL], so there exists
$x_3 \in \Q_2$ such that $x_3^2 = p_2 p_3$, and we can
take $x=x_3, y=0$ as a solution. For $p=3$, note that
$p_2 p_3$ or $- p_2 p_3 \equiv 1 \in (\F_3^*)^2$, so by Cor to HL,
$-p_2 p_3$ or $p_2 p_3 \in (\Q_3^*)^2$, so there exists
$x_4 \in \Q_3$ such that $x_4^2 = -p_2 p_3$ or $p_2 p_3$,
and we can take $x=x_4, y=0$ as a solution.
\par
For $p = p_1$, we are given $p_2,p_3 \in (\F_{p_1}^*)^2$
and so $p_2 p_3 \in (\F_{p_1}^*)^2$, so by Cor to HL,
$p_2 p_3 \in (\Q_{p_1}^*)^2$. Then there exists $x_5 \in \Q_{p_1}^*$
such that $x_5^2 = p_2 p_3$ and we can take $x = x_5, y = 0$
as a solution.
\par
For $p = p_2$, let $x_6 = 1 \in \Q_{p_2}^*$.
Then $f(x_6) = p_1( 1 + p_2 p_3 )( 1 - p_2 p_3)
\equiv p_1 \in (\F_{p_2}^*)^2$,
so by Cor to HL, 
$p_1 (1 + p_2 p_3)(1 - p_2 p_3) \in (\Q_{p_2}^*)^2$.
Then there exists $\beta \in \Q_{p_2}^*$ such
that $\beta^2 = p_1 (1 + p_2 p_3)(1 - p_2 p_3)$,
and $x = x_6 = 1, y = \beta$ is a solution.
The argument for $p = p_3$ is the same.
\par
For $p=5$, if some $p_i = 5$ then we are already covered
by the above cases $p = p_1,p_2$ or~$p_3$,
so assume $p_1 \not= 5$, $p_2 \not= 5$, $p_3 \not= 5$.
Then $p_2 p_3 \not\equiv 0$ (mod~$5$). If $p_2 p_3 \equiv 1\hbox{ or }4$
(mod~$5$) $\in (\F_5^*)^2$ then (by Cor to HL) there
exists $x_7 \in \Q_5^*$ such that $x_7^2 = p_2 p_3$
and $x=x_7,y=0$ is a solution. So assume $p_2 p_3 \equiv 2\hbox{ or }3$,
so that $p_2^2 p_3^2 \equiv 4$ (mod~$5$) and 
$f(x) \equiv p_1(x^4 + 1)$ (mod~$5$). 
If $p_1 \in (\F_5^*)^2$ then (by Cor to HL) there exists
$\gamma \in \Q_5^*$ such that $\gamma^2 = p_1$ so that $x=0,y=\gamma$
is a solution; if $p_1 \not\in (\F_5^*)^2$
then $2p_1 \in (\F_5^*)^2$ (since $2\not\in (\F_5^*)^2$)
so by Cor to HL, there exists $\delta \in \Q_5^*$
such that $\delta^2 = 2 p_1$, so $x=1, y=\delta$
is a solution.
[{\bf 7 marks}; Unseen]
\smallskip

\hrule
\newpage
\noindent {\bf Solution to Question 2}
\par\noindent
{\bf (a)} [{\it Proof from lectures}].
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
on the exam paper, $y^2 | (\phi_1(x)\psi_1(x) + \phi_2(x)\psi_2(x))
= \Delta$, as required.
\hfill {\bf [8~marks]}
\bigskip

\par\noindent
{\bf (b)} 
For $\E : Y^2 = X^3 - 3$, we have $\Delta = 4\cdot 0^3 + 27\cdot (-3)^2 
= 3^5$. Let $(x,y) \in \ETQ$. By Nagell-Lutz, $x,y\in \Z$
and $y = 0$ or $y^2 | \Delta$,
so $y \in \{ 0, \pm 1, \pm 3, \pm 9 \}$.
Hence $x^3 = y^2 + 3 \in \{ 3,4,12,84 \}$, a contradiction,
since these are not cubes. Hence $\ETQ$ is the trivial group $\{ \oo \}$.
\par
Let $p > 3$. Then $p \equiv 1$ or $2$ (mod~$3$).
\par
When $p \equiv 1$ (mod~$3$), then:
\par\smallskip $p \equiv 1$ (mod~$4$) $\implies
\bigl( \frac{-3}{p} \bigr) = 
\bigl( \frac{-1}{p} \bigr) \bigl( \frac{3}{p} \bigr)
= \bigl( \frac{3}{p} \bigr) = \bigl( \frac{p}{3} \bigr)$
[by QR] $= \bigl( \frac{1}{3} \bigr) = 1$.
\par\smallskip $p \equiv 3$ (mod~$4$) $\implies
\bigl( \frac{-3}{p} \bigr) = 
\bigl( \frac{-1}{p} \bigr) \bigl( \frac{3}{p} \bigr)
= -\bigl( \frac{3}{p} \bigr) = \bigl( \frac{p}{3} \bigr)$
[by QR] $= \bigl( \frac{1}{3} \bigr) = 1$.
\par\smallskip\noindent In all cases $\bigl( \frac{-3}{p} \bigr) = 1$,
so there exists $\alpha \in \F_p$ such that $\alpha^2 = -3$.
Then $(0,\alpha)\in \tE(\F_p)$ and the line $Y = \alpha$
substituted into~$\tE$ gives: $\alpha^2 = X^3 - 3$,
so $X^3 = 0$, so that $Y = \alpha$ intersects $\tE$
three times at~$(0,\alpha)$. Hence $3(\alpha,0) = \oo$
(and $(\alpha,0) \not= \oo$), so $(\alpha, 0)$ has
order~$3$, giving: $3 | \# \tE(\F_p)$. 
\par
When $p \equiv 2$ (mod~$3$), then:
$\hbox{gcd}(3,p-1) = 1$, so there exist $\lambda, \mu \in \Z$
such that $3\lambda + (p-1)\mu = 1$. For any $x_1,x_2\in \F_p^*$
we have:
\par $x_1^3 = x_2^3 \implies 
x_1^3 = x_2^3 \hbox{ and } x_1^{p-1} = x_2^{p-1}$ (by FLT)
\par
$\implies x_1^{3\lambda} x_1^{(p-1)\mu} = x_1^{3\lambda} x_1^{(p-1)\mu}$
$\implies x_1^{3\lambda + (p-1)\mu} = x_1^{3\lambda + (p-1)\mu}$
$\implies x_1 = x_2$.
\par\noindent Hence the map $x \mapsto x^3$ is injective and
so surjective $ : \F_p^* \longrightarrow \F_p^*$.
Also: $0 \mapsto 0$, so that $x \mapsto x^3$ is
a bijection $: \F_p \longrightarrow \F_p$. Then, for
all $y = 0,1,\ldots ,p-1 \in \F_p$ there exists a unique
$x \in \F_p$ such that $x^3 = y^2 + 3$.
Hence there are precisely~$p$ affine points, and 
so $\tE(\F_p)$ (including~$\oo$, the point at infinity)
satisfies $\tE (\F_p) = p+1 \equiv 0$ (mod~$3$),
so that again: $3 | \# \tE(\F_p)$.
\hfill [{\bf 7~marks}; Unseen]
\bigskip

\par\noindent
{\bf (c)} 
Let $\E : Y^2 = X^3 + X^2 - X$.
Then $(0,0) \in \E(\Q)$ has order~$2$, so $(0,0)\in \ETQ$.
Calculate $2(1,1)$, as follows. $2YY' = 3X^3 + 2X - 1$,
so that $Y' = 4/2 = 2$ at $(1,1)$, and so $Y = 2X - 1$
is the line tangent to~$\E$ at~$(1,1)$.
Substituting $Y = 2X - 1$ into~$\E$ gives:
$(2X - 1)^2 = X^3 + X^2 - X$,
and so: $X^3 - 3X^2 + \ldots = 0$. 
Hence the $x$-coordinate of $2(1,1)$ is: $3 - 1 - 1 = 1$.
So, the third point of intersection is~$(1,1)$,
giving $2(1,1) = (1,-1)$ and $3(1,1) = \oo$.
So $(1,1)$ is of order~$3$ and $(1,1) \in \ETQ$.
\par
Hence:
\par\noindent
$\{ \oo, (1,1), (1,-1), (0,0), (0,0)+(1,1) = (-1,1),
(0,0)+(1,-1) = (-1,-1) \}$
\par
$\leqslant \ETQ$.
\par\noindent
Also, $\E : Y^2 = X(X^2 + X - 1)$ and the discriminant
of $X^2 + X - 1$ is~$5$, so there are no repeated roots
of~${\tilde f}(x)$ when $p \not= 5$ and~$\tE$ is nonsingular
when $p \not= 2,5$. Hence~$3$ is a prime of good reduction,
and we find that:
\par $\tE(\F_3) = \{ \oo, (1,1), (1,2), (0,0), (2,1), (2,2) \}$,
\par\noindent
so that $\# \tE (\F_p) = 6$. By a theorem from lectures,
$\# \ETQ | \# \tE (\F_p) = 6$. Hence:
$\ETQ = \{ \oo, (1,1), (1,-1), (0,0), (-1,1), (-1,-1) \}$.
\hfill [{\bf 6~marks}; Seen similar]
\bigskip

\par\noindent
{\bf (d)} 
Imagine $|x| > \hbox{max}( 2|A|, |\Delta| + |B| )$. Then:
\par $|y^2| = |x^3 + Ax + B| \geqslant |x|^3 - |A||x| - |B|
= |x|^2 |x| - |A| |x| - |B|$
\par
$> 4|A|^2|x| - |A||x| - |B| \geqslant 
4|A||x| - |A||x| - |B|$ [since $A\in\Z$]
\par $= 3|A||x| - |B| \geqslant |x| - |B|
\geqslant (|\Delta| + |B|) - |B| = |\Delta|$,
\par\noindent a contradiction with Nagell-Lutz.
\hfill [{\bf 4~marks}; Unseen]

\bigskip
\hrule
\bigskip

\newpage

\noindent
{\bf Solution to Question 3}
\par\noindent
{\bf (a)} [{\it Proof from lectures}].
Let $r \in \qmods, r \in \hbox{im}\, q,
r \in \Z, r$ square free. We want to prove that $r | b_1$.
Suppose $r = q(u,v)$, where $(u,v)\in \D (\Q)$, which must
exist since $r \in \hbox{im}\, q$. Then: $r = q(u,v) = u = u^2 + a_1 u + b_1$
in $\qmods$ [since $u(u^2 + a_1 u + b_1) = v^2$]. 
So, $r, u, u^2 + a_1 u + b_1$ are all the same modulo squares,
which means we can write:
$u^2 + a_1 u + b_1 = rs^2, u = rt^2,\hbox{ for some } s,t \in \Q.$
Hence: $ (rt^2)^2 + a_1 (rt^2) + b_1 = rs^2$. Let $t = \ell/m$,
where $\ell, m \in \Z$, gcd$(\ell,m) = 1$.
Then: $ r^2 \ell^4 / m^4 + a_1 r \ell^2 / m^2 + b_1 = rs^2$,
and so: $ r^2 \ell^4 + a_1 r \ell^2 m^2 + b_1 m^4 = r(m^2s)^2$.
Now, $a_1, b_1, r,\ell, m \in \Z$, so the LHS of this last equation
is in~$\Z$, and so the RHS is also in~$\Z$; that is: $r(m^2s)^2\in \Z$.
Since $r$ is square free, we must therefore have $m^2 s \in \Z$.
Define: $n = m^2 s \in \Z$. Then our equation becomes:
\par\smallskip\par
$r^2 \ell^4 + a_1 r \ell^2 m^2 + b_1 m^4 = r n^2,\ \ 
\hbox{ for some } \ell, m, n \in \Z, \hbox{ gcd}(\ell,m) = 1.
\ \ (\dagger )$.
We want to show that $r | b_1$, and we know that $r$ is square free.
It is sufficient to show, for any prime~$p$, that $p | r \Rightarrow p | b_1$.
\par
Imagine $p | r$ and $p \nmid b_1$, for some prime~$p$.
Then $p | r^2 \ell^4, a_1 r \ell^2 m^2, r n^2$ and so by $(\dagger)$,
$p | b_1 m^4$, which in turn gives: $p | m$ [since $p \nmid b_1$].
Hence, since now $p | r$ and $p | m$, 
we have: $p^2 | r^2 \ell^4, a_1 r \ell^2 m^2,
b_1 m^4$, and so by $(\dagger)$, $p^2 | r n^2$, which in turn
gives: $p | n$ [since~$r$ is square free]. Hence, since now  
$p | r,m,n$, we have: $p^3 | a_1 r \ell^2 m^2, b_1 m^4, r n^2$,
and so by $(\dagger)$, $p^3 | r^2 \ell^4$, which in turn
gives: $p | \ell$ [since~$r$ is square free]. This is a contradiction,
since $p | \ell$ and $p | m$ but $\hbox{ gcd}(\ell,m) = 1$.
\par
The above assumption that $p | r$ and $p \nmid b_1$ let to a contradiction,
and so it is impossible for any prime~$p$ 
to satisfy $p | r$ and $p \nmid b_1$.
This is the same as saying that $ p | r \Rightarrow p | b_1 $ for
any prime~$p$. Since $r$ is square free, we conclude
that $r | b_1$, as required
\hfill {\bf [9~marks]}
\bigskip

\noindent
{\bf (b)} 
Let $\CC : Y^2 = X(X^2 - X - 1)$.
Here, $a=-1, b=-1$ and so $a_1 = -2a = 2,
b_1 = a^2 - 4b = 5$, giving
$\D : V^2 = U(U^2 + 2U + 5)$. 
%The isogeny
%$\phi : \CC \rightarrow \D$ is given by
%$\phi(x,y) = \Bigl( \bigl( \frac{y}{x} \bigr)^2, y - \frac{by}{x^2} \Bigr)
%= \Bigl( \bigl( \frac{y}{x} \bigr)^2, y + \frac{y}{x^2} \Bigr)$.
%The isogeny $\hp : \D \longrightarrow \CC$
%is given by
%$\hp(u,v) =
%\Bigl( \frac{1}{4} \bigl( \frac{v}{u} \bigr)^2,
%\frac{1}{8} \bigl( v - \frac{b_1 v}{u^2}\bigr) \Bigr)
%= \Bigl( \frac{1}{4} \bigl( \frac{v}{u} \bigr)^2,
%\frac{1}{8} \bigl( v - \frac{5 v}{u^2}\bigr) \Bigr)$.
\par\medskip\noindent {\bf Step 1.} Find $\D(\Q) /\phi(\C(\Q))$.
Here $q : \D(\Q) /\phi(\C(\Q)) \longrightarrow \qmods$.
We need to consider $r | b_1 = 5, r\in \Z$, $r$ square free,
that is,
$\im q \leqslant \{ \pm 1, \pm 5 \}$.
But $q(\oo) = 1$, $q(0,0) = b_1 = 5$,
so that $\{ 1,5 \} \leqslant \im q$.
There is one nontrivial coset:~$-1$.
Then $W_r : r \ell^4 + a_1 \ell^2 m^2 + (b_1/r) m^4 = n^2$,
for some $\ell, m, n \in \Z$,
$\hbox{not all $0$, with gcd}(\ell,m) = 1$ becomes:
$W_{-1} : - \ell^4 + 2 \ell^2 m^2 - 5 m^4 = n^2$.
On completing the square: $-(\ell^2 - m^2)^2 - 4 m^4 = n^2$.
Then $\hbox{LHS} \leqslant 0$ and $\hbox{RHS} \geqslant 0$
and both sides are equal only when $\ell = m = n = 0$, impossible.
Hence $-1\not\in \im q$, giving $\im q = \{ 1, 5 \}$
and $\D(\Q) /\phi(\C(\Q)) = \langle (0,0) \rangle$ $\cong C_2$.
\par\smallskip\noindent {\bf Step 2.} Find $\C(\Q) /\hat\phi(\D(\Q))$.
Here $\hat q : \C(\Q) /\hat\phi(\D(\Q)) \longrightarrow \qmods$.
Consider $r | b = -1, r\in \Z$, $r$ square free,
that is, $\im \hat q \leqslant \{ \pm 1 \}$.
But $\hat q(\oo) = 1$, $\hat q(0,0) = b = -1$,
Hence $\im \qhat = \{ \pm 1 \}$ and 
$\C(\Q) /\hat\phi(\D(\Q)) = \langle (0,0) \rangle \cong C_2$.
\par\smallskip\noindent {\bf Step 3.} Find $\C(\Q) /2 \C(\Q))$,
which is generated by $\C(\Q) /\hat\phi(\D(\Q)) = \{ \oo, (0,0) \}$
and $\hat\phi\bigl( \D(\Q) /\phi(\C(\Q)) \bigr)$
$= \hat\phi \bigl( \langle (0,0) \rangle \bigr)$
$= \{ \oo \}$, since $\hat\phi : (0,0) \mapsto \oo$.
Hence $\C(\Q) /2 \C(\Q)) = \langle (0,0) \rangle \cong C_2$. 
From lectures, $\C(\Q) /2 \C(\Q)) \cong \C(\Q)[2] \times C_2^r$.
Here $\C(\Q)[2] = \{ \oo, (0,0) \} \cong C_2$, since
$( (1 \pm \sqrt{5})/2, 0 ) \not\in \C(\Q)$. Hence the rank is~$0$.
[{\bf 7 marks}; Seen similar.]
\bigskip

\noindent
{\bf (c)} 
Let $\CC : Y^2 = X(X^2 + p)$.
Here, $a=0, b=p$ and so $a_1 = -2a = 0,
b_1 = a^2 - 4b = -4p$, giving
$\D : V^2 = U(U^2 - 4p)$. 
%The isogeny
%$\phi : \CC \rightarrow \D$ is given by
%$\phi(x,y) = \Bigl( \bigl( \frac{y}{x} \bigr)^2, y - \frac{by}{x^2} \Bigr)
%= \Bigl( \bigl( \frac{y}{x} \bigr)^2, y - \frac{py}{x^2} \Bigr)$.
%The isogeny $\hp : \D \longrightarrow \CC$
%is given by
%$\hp(u,v) =
%\Bigl( \frac{1}{4} \bigl( \frac{v}{u} \bigr)^2,
%\frac{1}{8} \bigl( v - \frac{b_1 v}{u^2}\bigr) \Bigr)
%= \Bigl( \frac{1}{4} \bigl( \frac{v}{u} \bigr)^2,
%\frac{1}{8} \bigl( v + \frac{4p v}{u^2}\bigr) \Bigr)$.
\par\medskip\noindent {\bf Step 1.} Find $\D(\Q) /\phi(\C(\Q))$.
Here $q : \D(\Q) /\phi(\C(\Q)) \longrightarrow \qmods$.
We need to consider $r | b_1 = -4p, r\in \Z$, $r$ square free,
that is,
$\im q \leqslant \{ \pm 1, \pm 2, \pm p, \pm 2p \}$.
But $q(\oo) = 1$, $q(0,0) = b_1 = -4p = -p$,
so that $\{ 1,-p \} \leqslant \im q$.
There are three nontrivial coset:~$-1,2,-2$.
Then $W_r : r \ell^4 + a_1 \ell^2 m^2 + (b_1/r) m^4 = n^2$,
for some $\ell, m, n \in \Z$,
$\hbox{not all $0$, with gcd}(\ell,m) = 1$ becomes:
$W_{-1} : - \ell^4 + 4p m^4 = n^2$.
Imagine $p \nmid \ell$. Then $-1 \equiv (n/\ell)^2$ (mod~$p$),
and so $\bigl( \frac{-1}{p} \bigr) = 1$, impossible since
we are given $p \equiv 3$ (mod~$4$), and so $p | \ell$.
Hence $p | n$ and so, $p^2 | \ell^4$, $p^2 | n^2$,
giving $p^2 | 4 p m^4$ and so $p | m$, a contradiction.
Hence $-1\not\in \im q$.
\par Now consider $W_2 : 2 \ell^2 - 2 p m^4 = n^2$
and $W_{-2} : -2 \ell^4 + 2 p m^4 = n^2$.
\par We are given that $p \equiv 3$ (mod~$4$), so
either $p \equiv 3$ (mod~$8$) or $p \equiv 7$ (mod~$8$).
\par\noindent {\bf Case} $p \equiv 3$ (mod~$8$).
Then $\bigl( \frac{2}{p} \bigr) = -1$. Consider $W_2$ above.
Imagine $p \nmid \ell$. Then $W_2$ gives
$2 \equiv (n/\ell^2)^2$ (mod~$p$), impossible, so
that $p | \ell$. Hence $p | n$ and so, $p^2 | \ell^4$, $p^2 | n^2$,
giving $p^2 | 2 p m^4$ and so $p | m$, and contradiction.
Hence $2\not\in \im q$. So 
$\langle -p \rangle \leqslant \im q  \leqslant \langle -p, -2 \rangle$.
\par\noindent {\bf Case} $p \equiv 7$ (mod~$8$).
Then $\bigl( \frac{2}{p} \bigr) = 1$
and $\bigl( \frac{-2}{p} \bigr) 
= \bigl( \frac{-1}{p} \bigr)\bigl( \frac{2}{p} \bigr) = -1\cdot 1 = -1$. 
Consider $W_{-2}$ above.
Imagine $p \nmid \ell$. Then $W_{-2}$ gives
$-2 \equiv (n/\ell^2)^2$ (mod~$p$), impossible, so
that $p | \ell$. Hence $p | n$ and so, $p^2 | \ell^4$, $p^2 | n^2$,
giving $p^2 | 2 p m^4$ and so $p | m$, and contradiction.
Hence $-2\not\in \im q$. So 
$\langle -p \rangle \leqslant \im q  \leqslant \langle -p, 2 \rangle$.
\par
% Could alternatively, after showing $-1 \not\in \im q$, just
% note that then $\# \im q < 8$ and has order a power of~$2$,
% and so is at most 4. This is already enough to give rank at most 1.
% But then one would still need to go back and consider $W_2$
% and $W_{-2}$ for the very last part about $p \equiv 7$ (mod~$16$). 
In either case, 
$\D(\Q) /\phi(\C(\Q)) = \langle (0,0) \rangle$ 
or $\langle (0,0), R \rangle$, for some $R \in \D(\Q)$.
\par\smallskip\noindent {\bf Step 2.} Find $\C(\Q) /\hat\phi(\D(\Q))$.
Here $\hat q : \C(\Q) /\hat\phi(\D(\Q)) \longrightarrow \qmods$.
We need to consider $r | b = p, r\in \Z$, $r$ square free,
that is, $\im \hat q \leqslant \{ \pm 1, \pm p \}$.
But $\hat q(\oo) = 1$, $\hat q(0,0) = b = p$,
so that $\{ 1,p \} \leqslant \im \hat q$.
There is one nontrivial coset: $-1$.
Then ${\widehat W}_r : r \ell^4 + a \ell^2 m^2 + (b/r) m^4 = n^2$
for some $\ell, m, n \in \Z$,
not all $0$, $\hbox{gcd}(\ell,m) = 1$. 
Then for ${\widehat W}_{-1} : -\ell^4 - p m^4 = n^2$
we see that $\hbox{LHS} \leqslant 0$ and $\hbox{RHS} \geqslant 0$,
and both sides are equal only when $\ell = m = n = 0$,
a contradiction. It follows that $-1 \not\in \im \qhat$. 
Hence $\im \qhat = \{ 1,p \}$ and 
$\C(\Q) /\hat\phi(\D(\Q)) = \{ \oo, (0,0) \} = \langle (0,0) \rangle$.
\par\smallskip\noindent {\bf Step 3.} Find $\C(\Q) /2 \C(\Q))$,
which is generated by $\C(\Q) /\hat\phi(\D(\Q)) = \langle (0,0) \rangle$
and $\hat\phi\bigl( \D(\Q) /\phi(\C(\Q)) \bigr)$
$= \hat\phi \bigl( \langle (0,0) \rangle \bigr) = \{ \oo \}$
or $\hat\phi \bigl( \langle (0,0), R \rangle \bigr)
= \langle \hat\phi(R) \rangle$,
since $\hat\phi : (0,0) \mapsto \oo$.
Hence $\C(\Q) /2 \C(\Q)) \cong C_2$ or $C_2 \times C_2$. From lectures, 
$\C(\Q) /2 \C(\Q)) \cong \C(\Q)[2] \times C_2^r$.
Here $\C(\Q)[2] = \{ \oo, (0,0) \} \cong C_2$, since
$(\pm \sqrt{-p}, 0 ) \not\in \C(\Q)$. Hence the rank is~$0$ or~$1$.
\par
Now assume that $p \equiv 7$ (mod~$16$). Then $p \equiv 7$ (mod~$8$)
and by the above argument (using the second case),
$\im q \leqslant \langle -p, 2 \rangle$ and:
\par $\hbox{rank} = 0 \iff \D(\Q)/\phi(\C(\Q)) = \langle (0,0) \rangle
\iff \im q = \langle -p \rangle \iff 2 \not\in \im q$.
\par\noindent So, sufficient to show $2 \not\in \im q$.
Recall $W_2 : 2 \ell^2 - 2 p m^4 = n^2$. Then $2 | n$,
so write $n = 2 n'$, some $n' \in \Z$,
giving: $\ell^4 - p m^4 = 2 n'^2$, so:
$2 | \ell \iff 2 | m$; but $\hbox{gcd}(\ell,m) = 1$,
so $2 \nmid \ell$ and $2 \nmid m$.
Hence $\ell^2 \equiv 1$ (mod~$8$), so 
$\ell^2 \equiv 1\hbox{ or }9$ and $\ell^4 \equiv 1$ (mod~$16$);
similarly $m^4 \equiv 1$ (mod~$16$).
Then $\hbox{LHS} = \ell^4 - p m^4 \equiv 1 - p \equiv 1 - 7
\equiv -6 \equiv 10$ (mod~$16$),
giving: $10 \equiv 2 n'^2$ (mod~$16$), so $5 \equiv n'^2$ (mod~$8$).
But $n'^2 \equiv 0,1\hbox{ or }4$ (mod~$8$),
%(since $n'$ odd
%gives $n'^2 \equiv 1$ and $n'$ even gives $n'^2 \equiv 0\hbox{ or }4$), 
a contradiction. Hence $2 \not\in \im q$ and the rank is~$0$,
as required.
[{\bf 9 marks}; Unseen.]

\medskip
\hrule

\end{document}