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% Possible shortenings to Question 1:
% remove second part [done]; change Q_p to Z_p in part (b) [done].
\par
\noindent {\bf Solution to Question 1.} 
\par\noindent
(a) [{\it Proof from lectures}]. 
For any two points~$\aaa,\bb$ on~$\C$, let~$\ell_{\aaa,\bb}$
denote the line which meets~$\C$ at~$\aaa,\bb$ [if~$\aaa,\bb$ are distinct
then~$\ell_{\aaa,\bb}$ is the unique line through~$\aaa,\bb$;
if~$\aaa=\bb$ 
then $\ell_{\aaa,\bb}$ is the line tangent to~$\C$ at~$\aaa=\bb$].
Then~$\ell_{\aaa,\bb}$ and~$\C$ have~$3$ points of intersection (B\'ezout).
Let~$\dd$ be the third point of intersection between~$\C$
and~$\ell_{\aaa,\bb}$.
Let~$\cc$ be the third point of intersection between~$\C$
and~$\ell_{\oo,\dd}$. We define $\aaa + \bb = \dd$.
\par
We recall from lectures the following lemma.
\par\noindent {\bf Lemma.} Let~$P_1,\ldots , P_8$ be such that 
no~$4$ points lie on a line and no~$7$ points lie on a conic. Then 
there exists a unique point~$P_9$ which is
a 9th point of intersection of any two cubics passing
through~$P_1,\ldots ,P_8$.
\par
In order to prove associativity, consider the following diagram.
\par
\hskip 50 pt \vrule width 1 pt depth 2 pt height 240 pt \hskip 60 pt
\vrule width 1 pt depth 2 pt height 240 pt \hskip 60 pt
\vrule width 1 pt depth 2 pt height 240 pt 
%\reallynopagebreak
\vskip-230pt \hskip 127 pt {\lower 3.5pt \hbox{\bf w}} \vskip -5 pt
\hskip 53 pt {\bf a} \hskip 67 pt {\lower 3pt\hbox{$|$}} 
\hskip 40 pt {\bf v} \hskip 56 pt $r$ \vskip -2.5 pt
\hrule width 240 pt depth 1 pt height 1 pt \vskip 15 pt
\hskip 98 pt {\bf f} \hskip -2 pt --- \vskip 28 pt
\hskip 52.5 pt {\bf b} \hskip 50 pt {\bf c} \hskip 54 pt {\bf u}
\hskip 56 pt $s$ \vskip 2pt
\hrule width 240 pt depth 1 pt height 1 pt \vskip 65 pt
\hskip 52.5 pt {\bf d} \hskip 50 pt {\bf e} \hskip 55 pt \oo
\hskip 58 pt $t$ \vskip 1pt
\hrule width 240 pt depth 1 pt height 1 pt \vskip 50 pt
\hskip 43 pt $\ell$ \hskip 51 pt $m$ \hskip 51 pt $n$
\medskip
\par
Here, $r,s,t,\ell,m,n$ are lines. On each line, the labelled points
are the points of intersection between~$\C$ and that line.
From our construction of the law for adding points:
$\aaa + \bb = \ee$, and so: $(\aaa + \bb) + \cc$ is the 3rd point of 
intersection on~$\ell_{\oo,\ff}$.
Similarly, $\bb + \cc = \vv$,
and $\aaa + (\bb + \cc) =$ is the 3rd point of intersection 
on~$\ell_{\oo,\ww}$.
To show $(\aaa + \bb) + \cc = \aaa + (\bb + \cc)$, it
is sufficient to show that~$\ff = \ww$. Let $F_1 = \ell m n$
and $F_2 = rst$, both of which are cubic curves.
\par $\C$ and~$F_1$ have~$8$ common points: 
$\aaa,\bb,\cc,\dd,\ee,\uu,\vv,\oo$.
\par $\C$ and~$F_2$ also have these~$8$ 
common points: $\aaa,\bb,\cc,\dd,\ee,\uu,\vv,\oo$.
From the above lemma, the 9th point of
intersection of~$\C$ and~$F_1$ must be the same as the 9th
point of intersection of~$\C$ and~$F_2$; that is, $\ff = \ww$,
as required.
\hfill {\bf [10~marks]}
%\bigskip
%
%\par\noindent
%{\bf (b)} Using $2YY' = 3X^2-8$ gives $Y' = (3\cdot 1^2 - 8)/2 = -5/2$
%at $P = (1,1)$, so the line tangent to $\E$ at~$P$
%is $Y = (-5/2)X + 7/2$. Substituting: $((-5/2)X + 7/2)^2 = X^3-8X+8$
%gives $X^3 - (25/4)X^2 + \ldots$, so the $x$-coordinate
%of $2P$ is: $25/4 - 1 - 1 = 17/4$, and the $y$-coordinate
%is $-( (-5/2)(17/4) + 7/2 ) = 57/8$. Hence $2P = (17/4, 57/8)$.
%To calculate $2P$, we would first need to find the slope
%of the tangent at $2P$, $Y' = (3\cdot 47^2 - 8)4/{57}$ which
%would require finding the inverse of~$57$ mod~$N = 57$, impossible.
%For $N = p_1 p_2$, the Elliptic Curve Method is successful
%when $kP = \oo$ in ${\widetilde \E}(\F_{p_1})$
%and $kP \not= \oo$ in ${\widetilde \E}(\F_{p_2})$,
%so that the denominator that arises when computing $kP$
%will have an hcf with~$N$ which gives a proper factor.
%Here $p_1 = 3$ and $p_2 = 19$ and $P$ has order~$4$ in
%both ${\widetilde \E}(\F_{p_1})$ and ${\widetilde \E}(\F_{p_2})$.
%When $4 | k$ the hcf is~$N$ and when $4 \nmid k$ the
%hcf is~$1$; so, for any~$k$, the method with this choice
%of~$\E$ and~$P$ will never give a proper factor.
%[{\bf 5 marks}; Unseen.]
\bigskip

\par\noindent
{\bf (b)} 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
%Imagine $x,y \in \Q_2$ satisfied $x^2 - 2y^2 = 3$.
%Note that $|x^2|_2 = 2^{\hbox{\ssrm even}}$ and 
%$|2y^2|_2 = 2^{\hbox{\ssrm odd}}$,
%so that $|x^2|_2 \not= |2y^2|_2$, giving that
%$1 = |3|_2 = |x^2 - 2y^2|_2 = \hbox{max}( |x^2|_2, |2y^2|_2 )$
%[using thm: $|\alpha|_p \not= |\beta|_p \implies 
%|\alpha \pm \beta |_p = \hbox{max}( |\alpha|_p , |\beta|_p )$].
%Hence $|x|_2^2, 2^{-1} |y|_2^2 \leqslant 1$, giving
%$|x|_2, |y|_2 \leqslant 1$ and so $x,y\in \Z_2$.
%Then $x^2 - 2y^2 \equiv 3$ (mod~$8$). This gives
%$|x|_2 = 1$ (otherwise $1 = |3|_2 = |x^2 - 2y^2|_2 < 1$),
%and so $x^2 = (1 + 2k)^2 \equiv 1$ (mod~$8$),
%and $y^2 \equiv 0,1$ or~$4$ (mod~$8$), so that
%$x^2 - 2y^2 \equiv 1\hbox{ or }7 \not\equiv 3$ (mod~$8$),
%a contradiction. Hence there are no solutions $x,y\in \Q_2$. 
%\par
%Imagine $x,y \in \Q_3$ satisfied $x^2 - 2y^2 = 3$.
%Note that $|x|_3 = |y|_3$ (otherwise $3^{-1} = |3|_3 
%= |x^2 - 2y^2|_3 = \hbox{max}(|x^2|_3, |2y^2|_3) = 3^{\hbox{\ssrm even}}$).
%Also $|x|_3 = |y|_3 \geqslant 1$ (otherwise
%$3^{-1} = |3|_3 = |x^2 - 2y^2|_3 \leqslant 
%\hbox{max}(|x^2|_3, |2y^2|_3) \leqslant 3^{-2}$).
%Hence $x = x'/3^k, y = y'/3^k$ for some $k \geqslant 0$
%and $|x'|_3 = |y'|_3 = 1$, so that $x',y' \in \Z_3$.
%This gives: $x'^2 - 2y'^2 = 3$, so that $x'^2 \equiv 2y'^2$
%and $(x'/y')^2 \equiv 2$ (mod~$3$), impossible, since
%$\bigl( \frac{2}{3} \bigr) = -1$. Hence there are
%no such $x,y \in \Z_3$.
Imagine $x,y \in \Z_2$ satisfy $x^2 - 2y^2 = 3$.
Then $x^2 - 2y^2 \equiv 3$ (mod~$8$). This gives
$|x|_2 = 1$ (otherwise $1 = |3|_2 = |x^2 - 2y^2|_2 < 1$),
and so $x^2 = (1 + 2k)^2 \equiv 1$ (mod~$8$),
and $y^2 \equiv 0,1$ or~$4$ (mod~$8$), so that
$x^2 - 2y^2 \equiv 1\hbox{ or }7 \not\equiv 3$ (mod~$8$),
a contradiction. Hence: no solutions $x,y\in \Z_2$. 
\par
Imagine $x,y \in \Z_3$ satisfy $x^2 - 2y^2 = 3$.
Then $|y|_3 = 1$ (otherwise $|2y^2|_3 \leqslant 3^{-2}$
and $3^{\hbox{\ssrm even}} = |x^2|_3 = |3 + 2y^2|_3 
= \hbox{max}( |3|_3, |2y^2|_3 ) = 3^{-1}$, using
the theorem: $|\alpha|_p \not= |\beta|_p \implies
|\alpha \pm \beta |_p = \hbox{max}( |\alpha|_p , |\beta|_p )$.
Then: $x^2 \equiv 2y^2$ and
and so $(x/y)^2 \equiv 2$ (mod~$3$), impossible, since
$\bigl( \frac{2}{3} \bigr) = -1$. Hence:
no solutions $x,y \in \Z_3$.
\par
For $p \not= 2,3$, the set $\{ x^2 : x \in \F_p \}$
has size $(p+1)/2$, since $x \mapsto x^2$ is $2$-to-$1$
on $\F_p^*$ and maps $0 \mapsto 0$. Similarly the
set $\{ 3 + 2y^2 : y \in \F_p \}$ has size $(p+1)/2$.
These cannot be disjoint, since $(p+1)/2 + (p+1)/2 > p = \# \F_p$,
so there exist $x_0, y_0 \in \Z$ such that 
$x_0^2 \equiv 3 + 2y_0^2$ (mod~$p$), and cannot have
both $|x_0|_p < 1$ and $|y_0|_p < 1$ (since $|3|_p = 1$),
so $|x_0|_p = 1$ or $|y_0|_p = 1$, wlog $|x_0|_p = 1$.
For $f(x) = x^2 - (3 + 2y_0^2)$, we have $|f(x_0)|_p < 1$
and $|f'(x_0)|_p = |3x_0^2|_p = 1$, so by Hensel's lemma,
there exists $x_1 \in \Z_p$ such that $f(x_1) = 0$,
that is: $x_1,y_0 \in \Z_p$ such that $x_1^2 - 2 y_0^2 = 3$.
[{\bf 7 marks}; Unseen]
\bigskip

\par\noindent
{\bf (c)} For $p=3$, set $z=1$; then 
$y^3 - z^3 + b - a \equiv y^3 - 1$ (mod~27). Let $f(y) = y^3 - 1^3 + b - a$.
Let $y_0 = 1$. Then $|f(y_0)|_3 = |b-a|_3 \leqslant 3^{-3}$
and $|f'(y_0)|_3 = |3|_3 = 3^{-1}$, so that $|f(y_0)|_3 < |f'(y_0)|^2_3$.
By Hensel's lemma, there exists $y_1 \in \Z_3$ such that $f(y_1) = 0$.
Then: $y=y_1, z=1$ and~$x$ arbitrary gives a solution in~$\Z_3$
to $y^3 - z^3 + b - a = 0$ and so to the given equation.
\par
Now let $p \not= 3$. If $a \equiv 0$ (mod~$p$) then take $y=1$
and $f(x) = x^3 - 1^3 + a$ and $x_0 = 1$. Then $|f(x_0)|_p < 1$
and $|f'(x_0)|_p = |3|_p = 1$, so by HL there exists $x_1 \in \Z_p$
such that $f(x_1) = 0$. Then $x_1 = x_1, y = 1$ and $z$ arbitrary
gives a solution in $\Z_p$ to $x^3 - y^3 + a = 0$ and
so to the given equation. Similarly if $b \equiv 0$ (mod~$p$)
then there is a solution in $\Z_p$ to $x^3 - z^3 + b$;
similary if $b - a \equiv 0$ (mod~$p$) then there is a solution
in $\Z_p$ to $y^3 - z^3 + b - a = 0$. 
So, now assume that $a \not\equiv 0$, $b \not\equiv 0$, 
$b - a \not\equiv 0$ (mod~$p$). The map $x \mapsto x^3$ 
is at most $3$-to-$1$ on $\F_p^*$ (since $x^3 - \alpha$ has
at most~$3$ solutions) and $0 \mapsto 0$. So,
$\# \{ x^3 : x \in \F_p \} \geqslant (p-1)/3 + 1 = (p+2)/3$,
and similarly $\{ y^3 - a : y \in \F_p \}$ and 
$\{ z^3 - b : z \in \F_p \}$.
Hence these sets cannot all be disjoint, since
$(p+2)/3 + (p+2)/3 + (p+2)/3 > \# \F_p$.
So, there exist $x_0,y_0,z_0 \in \Z$
such that $x_0^3 - y_0^3 + a \equiv 0$ or $x_0^3 - z_0^3 + b \equiv 0$
or $y_0^3 - z_0^3 + b - a \equiv 0$ (mod~$p$),
wlog say that $x_0^3 - y_0^3 + a \equiv 0$ (mod~$p$).
Since $a \not\equiv 0$ (mod~$p$), $x_0\not\equiv 0$ or $y_0 \not\equiv 0$,
wlog $x_0\not\equiv 0$ (mod~$p$).
Let $f(x) = x^3 - y_0^3 + a$. Then $|f(x_0)|_p < 1$
and $|f'(x_0)|_p = |3x_0^2|_p = 1$, so by HL there exists $x_1 \in \Z_p$
such that $f(x) = 0$. Then $x=x_1, y=y_0$ and $z$ arbitrary
is a solution in $\Z_p$ to $x^3 - y^3 + a = 0$ and so
to the given equation. Hence, there are solutions in $\Z_p$
for all~$p$.
\par
Imagine there were a solution $x,y,z\in \Z$. Then
$x^3 - y^3 + a = 0$ or $x^3 - z^3 + b = 0$ or $y^3 - z^3 + b - a = 0$.
But $x^3 - y^3 + a$ $\equiv$ 
$(0 \hbox{ or } \pm 1) - (0 \hbox{ or } \pm 1) + 4$ $\not\equiv 0$ (mod~$9$),
so $x^3 - y^3 + a \not= 0$.
Similarly $x^3 - z^3 + b \not\equiv 0$ (mod~$9$) so 
$x^3 - z^3 + b \not= 0$.
Also, $y^3 - z^3 + b - a$ $\equiv$ 
$(0 \hbox{ or } \pm 1) - (0 \hbox{ or } \pm 1) + 3$ $\not\equiv 0$ (mod~$7$),
so $y^3 - z^3 + b - a \not= 0$. Hence there are no solutions in~$\Z$.
[{\bf 8 marks}; Unseen]

\bigskip
\hrule
\newpage
% Possible shortening to Question 2: remove (a)(ii),
% in which case (a)(i) would just become (a).
\noindent {\bf Solution to Question 2}
\par\noindent
{\bf (a)(i)} [{\it Proof from lectures}].
Let $P(T) = F_X(0, T)^{-1}$, so that $\omega(T) = P(T)\ddd T$.
Note that $F_X(0, T) = 1 + \ldots$ is invertible, so
that $P(T)$ is indeed a member of~$R[[T]]$.
Furthermore, $P(0) = 1$, so that it is normalised.
\par We need to show that $\omega$ is an invariant differential;
that is, $\omega \circ F(T,S) = \omega(T)$, which is equivalent to:
$P\bigl( F(T,S) \bigr) F_X (T,S) = P(T)$ so, in our case,
it is sufficient to show:
$$ F_X\bigl( 0, F(T,S) \bigr)^{-1} F_X(T,S) = F_X(0,T)^{-1}, $$ 
which is true iff:
$$ F_X\bigl( 0, F(T,S) \bigr) = F_X(T,S) F_X(0,T). $$
But this last statement is immediate from differentiating
the associativity axiom
$F\bigl( U, F(T,S) \bigr) = F\bigl( F(U,T), S \bigr)$
with respect to~$U$ to get: 
$F_X\bigl( U, F(T,S) \bigr) = F_X\bigl( F(U,T), S\bigr) F_X(U,T)$
and setting~$U = 0$. Hence~$\omega$ is an invariant differential. 
\par Suppose that $\hat \omega = Q(T)\ddd T \in R[[T]] \ddd T$ is also an 
invariant differential, so that $Q(T)$ satisfies
$Q\bigl( F(T,S) \bigr) F_X(T,S) = Q(T)$.
Substituting~$T=0$ gives $Q(S) F_X(0,S) = Q(0)$, so
that $Q(S) = Q(0) F_X(0,S)^{-1}$. It follows that
$\hat \omega = a \omega$, where $a = Q(0)$.
\hfill {\bf [7~marks]}
\bigskip

\par\noindent
{\bf (ii)} [{\it Proof from lectures}].
First, note that $\omega_G \circ f\bigl( F(T,S) \bigr)
= \omega_G \bigl( G( f(T), f(S) \bigr) = \omega_G \circ f (T)$,
so that $\omega_G \circ f$ is an inviariant differential
on~$G$. From part~(a), it follows
that $\omega_G \circ f = a\ \omega_F$, for some~$a\in R$.
Since~$\omega_F,\omega_G$ are normalised,
$(1 + \ldots ) \ddd f(T)  = a (1 + \ldots )\ddd T$,
and so $(1 + \ldots ) f'(T) \ddd T  = a (1 + \ldots )\ddd T$; 
equating constant terms gives $a = f'(0)$, as required.
\par Let~$\omega$ be the normalised invariant differential on~$F$.
Since $[p](T) = p T + \dots$, it satisfies $[p]'(0) = p$.
Applying the previous result to~$[p]$, a homomorphism from~$F$
to itself, gives: $\omega \circ [p] = [p]'(0) \omega = p \omega$,
and so
$$ p \omega(T) = \omega \circ [p](T)
= (1 + \ldots ) \ddd ( [p](T) ) = (1 + \ldots ) [p]'(T) \ddd T.$$
Hence $[p]'(T) \in p\, R_T$. Each term $a_n T^n$ in $[p](T)$
must then satisfy $ p | n a_n$ and so $p | n$ or $p | a_n$,
as required.
\hfill {\bf [6~marks]}
\bigskip

\par\noindent
{\bf (b)(i)} 
Differentiating: $2YY' = 3X^2 + 2(\alpha + \beta)X + \alpha\beta$,
so the tangent to~$\C$ at~$(x,y)$ is $Y = \lambda X + \mu$, where 
$\lambda = \bigl( 3x^2 + 2(\alpha + \beta)x + \alpha\beta \bigr)/(2y)$.
This meets~$\C$ at points whose $x$-coordinates satisfy
$(\lambda X + \mu)^2 = X(X+\alpha)(X+\beta)$,
so that $0 = X^3 - (\lambda^2 - \alpha - \beta)X^2 + \ldots$.
Hence the $x$-coordinate of $2(x,y)$ equals
$-(\hbox{coeff of }X^2) - 2x$ $=$ $\lambda^2 - \alpha - \beta - 2x$,
which equals
\begin{equation*}
\frac{\bigl( 3x^2 + 2(\alpha + \beta)x + \alpha\beta \bigr)^2}
{4x(x+\alpha)(x+\beta)}
- \alpha - \beta - 2x
= \frac{(x^2 - \alpha \beta)^2}{4x(x+\alpha)(x+\beta)}.
\end{equation*}
The points of order~2 in $\C(\Q)$ are $(0,0),(\alpha,0),(\beta,0)$,
and so:
\begin{equation*}
\C(\Q) \hbox{ has a point of order }4 \iff 
(0,0)\hbox{ or }(-\alpha,0)
\hbox{ or }(-\beta,0)\in 2\C(\Q).\ \ \ (*)
\end{equation*}
If $(0,0)\in2\C(\Q)$ then $(0,0) = 2(x,y)$ for some $(x,y)\in\C(\Q)$,
and so using the above duplication formula on the $x$-coordinate,
we have: $(x^2 - \alpha\beta)^2 = 0$, so that $\alpha\beta\in(\Q^*)^2$
and $x^2 = \alpha\beta$. Then
$y^2 = x(x+\alpha)(x+\beta)$ $=$ $x(x+\alpha)(x + x^2/\alpha)$
$=$ $x^2 (x + \alpha)^2/\alpha$. Hence
$\alpha = \bigl( x(x+\alpha)/y \bigr)^2 \in (\Q^*)^2$,
and so also $\beta = x^2/\alpha \in (\Q^*)^2$.
\par
Conversely, if $\alpha, \beta \in (\Q^*)^2$, say $\alpha = a^2$
and $\beta = b^2$ for some $a,b\in\Q^*$. Then
$(ab,ab(a+b)) \in \C(\Q)$ [since $ab(ab+a^2)(ab+b^2) = a^2 b^2 (a+b)^2$],
and from the above duplication formula, the $x$-coordinate
of $2(ab,ab(a+b))$ is~$0$, so that $2(ab,ab(a+b)) = (0,0)$.
Hence $(0,0)\in 2\C(\Q)$. In summary:
\begin{equation*}
(0,0) \in 2\C(\Q) \iff \alpha\in (\Q^*)^2 \hbox{ and } \beta\in (\Q^*)^2.
\ \ \ \ (1)
\end{equation*}
The map $(x,y) \mapsto (x+\alpha,y)$ is a birational transformation
from~$\C$ to the curve $\C' : Y^2 = X(X - \alpha)(X + \beta - \alpha)$,
so that:
\begin{equation*}
(-\alpha,0)\in 2\C(\Q) \iff (0,0)\in 2\C'(\Q)
\iff -\alpha \in (\Q^*)^2 \hbox{ and } \beta - \alpha \in (\Q^*)^2.
\ \ (2)
\end{equation*}
The map $(x,y) \mapsto (x+\beta,y)$ is a birational transformation
from~$\C$ to the curve $\C'' : Y^2 = X(X - \beta)(X + \alpha - \beta)$,
so that:
\begin{equation*}
(-\beta,0)\in 2\C(\Q) \iff (0,0)\in 2\C''(\Q) 
\iff -\beta \in (\Q^*)^2 \hbox{ and } \alpha - \beta \in (\Q^*)^2.
\ \ (3)
\end{equation*}
Substituting $(1),(2),(3)$ into~$(*)$ gives the required result.
\hfill [{\bf 6~marks}; Unseen]
\bigskip

\par\noindent
{\bf (ii)} Here 
$\C : Y^2 = X(X+c^4)(X+d^4) = X(X^2 + (c^4 + d^4)X + c^4 d^4)$
and the $2$-isogenous curve is 
$\D : V^2 = U\bigl( U^2 - 2(c^4 + d^4)U + (c^4 - d^4)^2 \bigr)$
where, from lectures, the $2$-isogenies are
$\phi : \C \mapsto \D : (x,y) \mapsto 
\bigl( (y/x)^2, y - c^4 d^4 y/x^2)$
and $\hat\phi : \D \mapsto \C : (u,v) \mapsto
\bigl( (1/4)(v/u)^2, (1/8)( v - (c^4-d^4)^2 v/u^2 ) \bigr)$,
satisfying $\hat\phi \circ \phi = [2]$.
By the previous part, we have that 
$P = ( c^2 d^2, c^2 d^2 (c^2 + d^2) ) \in \C(\Q)$ is 
a point of order~$4$. The $x$-coordinate of~$P$ is $c^2 d^2 \in (\Q^*)^2$
so by a theorem from lectures, $P \in \hat\phi( \D(\Q))$,
so that $P = \hat\phi( (u,v) )$, for some $(u,v)\in \D(\Q)$.
Then $(1/4)(v/u)^2 = c^2 d^2$ and so $v = \pm 2cdu$.
Substituting this into~$\D$ gives 
$(\pm 2cdu )^2 = u(u^2 - 2(c^4 + d^4)u + (c^4 - d^4)^2)$, so that
\begin{equation*} 
\begin{split}
0 &= 
u\bigl( u^2 - 2(c^2 + d^2)^2 u + (c+d)^2 (c^2 + d^2)(c-d)^2 (c^2 + d^2) \bigr)\\
&=
u\bigl( u - (c+d)^2(c^2 + d^2) \bigr)
 \bigl( u - (c-d)^2(c^2 + d^2) \bigr).
\end{split}
\end{equation*}
But $\hat\phi : (0,0) \mapsto \oo$, so that
$u = (c\pm d)^2 (c^2 + d^2) \in (\Q^*)^2$, since we
are given that $c^2 + d^2 \in (\Q^*)^2$.
So, by the same theorem from lectures, $(u,v) \in \phi(\C(\Q))$,
so that $(u,v) = 2(x,y)$, for some $(x,y) \in \C(\Q))$.
Then $P = \hat\phi \circ \phi ( (x,y) ) = 2(x,y)$;
already know that~$P$ is a point of order~$4$
so that $(x,y)$  is a point of order~$8$ in $\C(\Q))$, as required.
\par
As an example, take $c=3,d=4$, so that $c^2 + d^2 = 25 \in (\Q^*)^2$,
and then $\C : Y^2 = X(X + 3^4)(X + 4^4)$ has a point of
order~$8$ in~$\C(\Q)$.
\hfill [{\bf 6~marks}; Unseen]

\bigskip
\hrule
\bigskip

\newpage

\noindent
{\bf Solution to Question 3}
\par\noindent
{\bf (a)} [{\it Proof from lectures}].
Let~$(u_1,v_1),
(u_2,v_2),(u_3,v_3)$ be~3 points on~$\D(\Q)$ which sum to~$\oo$,
so that $(u_1,v_1) + 
(u_2,v_2)= (u_3,-v_3)$.
Then these are the~3 points of intersection between~$\D$
and some line defined over~$\Q$: 
$Y = \ell X + m$, say. Substituting~$Y = \ell X + m$
into $\D$ gives: $X(X^2 + a_1X + b_1) - (\ell X + m)^2$,
whose 3~roots must be~$u_1,u_2,u_3$.  
That is: $X(X^2 + a_1X + b_1) - (\ell X + m)^2
= (X-u_1)(X-u_2)(X-u_3)$. Equating constant terms gives:
$u_1 u_2 u_3 = m^2 = 1$ in $\qmods$, and so $u_1 u_2 = 1/u_3 = u_3$
in $\qmods$.
Therefore, by the definition of~$q$ we have:
$q\bigl( (u_1,v_1) \bigr) q\bigl( (u_2,v_2) \bigr)
= q\bigl( (u_3,v_3) \bigr)$, as required.
\hfill {\bf [5~marks]}
\bigskip

\noindent
{\bf (b)} 
Let $\CC : Y^2 = X(X^2 - pX + p^2)$.
Here, $a=-p, b=p^2$ and so $a_1 = -2a = 2p,
b_1 = a^2 - 4b = -3p^2$, giving
$\D : V^2 = U(U^2 + 2pU - 3p^2) = U(U - p)(U + 3p)$. 
%The isogeny
%$\phi : \CC \rightarrow \D$ is given by
%$\phi(x,y) = \Bigl( \bigl( \frac{y}{x} \bigr)^2, y - \frac{by}{x^2} \Bigr)
%= \Bigl( \bigl( \frac{y}{x} \bigr)^2, y - \frac{p^2y}{x^2} \Bigr)$.
%The isogeny $\hp : \D \longrightarrow \CC$
%is given by
%$\hp(u,v) =
%\Bigl( \frac{1}{4} \bigl( \frac{v}{u} \bigr)^2,
%\frac{1}{8} \bigl( v - \frac{b_1 v}{u^2}\bigr) \Bigr)
%= \Bigl( \frac{1}{4} \bigl( \frac{v}{u} \bigr)^2,
%\frac{1}{8} \bigl( v + \frac{3 p^2 v}{u^2}\bigr) \Bigr)$.
\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 $r | b_1 = -3p^2, r\in \Z$, $r$ square free,
that is,
$\im q \leqslant \{ \pm 1, \pm 3, \pm p, \pm 3p \}$.
But $q(\oo) = 1$, $q(0,0) = b_1 = -3$, $q(p,0) = p$, $q(-3p,0) = -3p$,
so that $\{ 1,-3,p,-3p \} \leqslant \im q$.
There is just one nontrivial coset: take~$-1$ as the representative.
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:
$W_{-1} : - \ell^4 + 2 p \ell^2 m^2 + 3 p^2 m^4 = n^2$.
On completing the square: $-(\ell^2 - pm^2)^2 + 4p^2 m^4 = n^2$.
Imagine $2 | \ell$. Then $3 p^2 m^4 \equiv n^2$ (mod~$4$)
and so $2 | m$ (ow LHS $\equiv 3$ and RHS $\equiv 1$ mod~$4$),
a contradiction. Hence $2 \nmid \ell$. Imagine $2 | m$.
Then $-\ell^4 \equiv n^2$ (mod~$4$) and so $2 | \ell$
(ow LHS $\equiv 3$ and RHS $\equiv 1$ mod~$4$),
a contradiction. Hence $2 \nmid m$.
\par
Hence $\ell^2 \equiv 1$ and $m^2 \equiv 1$ (mod~$8$), so
$\ell^2 - p m^2 \equiv 1 - 5 \equiv 4$ (mod~$8$) (since $p\equiv 5$ mod~$8$),
so that $2 | n$, so write $n = 2n'$ for some $n' \in \Z$.
Then $-\bigl( (\ell^2 - pm^2)/2 \big)^2 + p^2 m^2 = n'^2$.
But $(\ell^2 - pm^2)/2 \equiv 2$ (mod~$4$) and so is 
$\equiv 2\hbox{ or }6$ (mod~$8$), so that 
$\bigl( (\ell^2 - pm^2)/2 \big)^2 \equiv 4$ (mod~$8$).
Then LHS $= -\bigl( (\ell^2 - pm^2)/2 \big)^2 + p^2 m^2$ 
$\equiv 4 + 1 \equiv 5$ (mod~$8$),
whereas RHS $= n'^2 \equiv 0,1\hbox{ or }4$ (mod~$8$), a contradiction.
Hence $-1\not\in \im q$, giving $\im q = \{ 1, -3, p, -3p \}$
and $\D(\Q) /\phi(\C(\Q)) = \{ \oo, (0,0), (p,0), (-3p,0) \}$
$= \langle (0,0), (p,0) \rangle$ $\cong C_2 \times C_2$.
\par\smallskip\noindent {\bf Step 2.} Find $\C(\Q) /\hat\phi(\D(\Q))$.
Here $\hat q : \C(\Q) /\hat\phi(\D(\Q)) \longrightarrow \qmods$.
We need to consider $r | b = p^2, r\in \Z$, $r$ square free,
that is, $\im \hat q \leqslant \{ \pm 1, \pm p \}$.
But $\hat q(\oo) = 1$, $\hat q(0,0) = b = 1$,
so that $\{ 1 \} \leqslant \im \hat q$.
There are three nontrivial cosets: $-1,-p,p$.
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 $0$, with $\hbox{gcd}(\ell,m) = 1$. For $r < 0$,
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. Hence $-1,-p \not\in \im \qhat$. For $r = p$,
we have:
${\widehat W}_{p} : p \ell^4 - p \ell^2 m^2 + p m^4 = n^2$,
giving $p | n$ so write $n = p n'$.
Then $\ell^4 - \ell^2 m^2 + m^4 = p n'^2$. Multiplying by~$4$
and completing the square gives: $(2\ell^2 - m^2)^2 + 3 m^4 = 4p n'^2$.
Hence $(2\ell^2 - m^2)^2 \equiv - 3 m^4$ (mod~$p$).
Imagine $p \nmid m$. Then 
$\bigl( (2\ell^2 - m^2)/m^2 \bigr)^2 \equiv -3$ (mod~$p$),
giving $\bigl( \frac{-3}{p} \bigr) = 1$. but in fact
$\bigl( \frac{-3}{p} \bigr) 
= \bigl( \frac{-1}{p} \bigr)\bigl( \frac{3}{p} \bigr)$
$ = \bigl( \frac{3}{p} \bigr) = \bigl( \frac{p}{3} \bigr)$
[by QR, since $p \equiv 1$ (mod~$4$)]
$= \bigl( \frac{2}{3} \bigr)$ [since $p \equiv 2$ (mod~$3$)] $= -1$,
a contradiction. Hence $p | m$; but then from
$\ell^4 - \ell^2 m^2 + m^4 = p n'^2$ we see also that $p | \ell$,
impossible. So $p \not\in \im \qhat$.
Hence $\im \qhat = \{ 1 \}$ and $\C(\Q) /\hat\phi(\D(\Q)) = \{ \oo \}$,
the trivial group.
\par\smallskip\noindent {\bf Step 3.} Find $\C(\Q) /2 \C(\Q))$,
which is generated by $\C(\Q) /\hat\phi(\D(\Q)) = \{ \oo \}$
and $\hat\phi\bigl( \D(\Q) /\phi(\C(\Q)) \bigr)$
$= \hat\phi \bigl( \langle (0,0), (p,0) \rangle \bigr)$
$= \langle (0,0) \rangle$, since $\hat\phi : (0,0) \mapsto \oo$,
$(p,0) \mapsto (0,0)$. 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
$( (p \pm p\sqrt{-3})/2, 0 ) \not\in \C(\Q)$. Hence the rank is~$0$.
[{\bf 10 marks}; Seen similar.]
\bigskip

\noindent
{\bf (c)} 
Let $\CC : Y^2 = X(X^2 - p_1 p_2)$.
Here, $a=0, b=-p_1 p_2$ so $a_1 = 0,
b_1 = 4 p_1 p_2$, giving
$\D : V^2 = U(U^2 + 4 p_1 p_2)$.
%The isogeny
%$\phi : \CC \rightarrow \D$ is given by
%$\phi(x,y) = \Bigl( \bigl( \frac{y}{x} \bigr)^2, y - \frac{by}{x^2} \Bigr)
%= \Bigl( \bigl( \frac{y}{x} \bigr)^2, y + \frac{p_1 p_2 y}{x^2} \Bigr)$.
%The isogeny $\hp : \D \longrightarrow \CC$
%is given by
%$\hp(u,v) =
%\Bigl( \frac{1}{4} \bigl( \frac{v}{u} \bigr)^2,
%\frac{1}{8} \bigl( v - \frac{b_1 v}{u^2}\bigr) \Bigr)
%= \Bigl( \frac{1}{4} \bigl( \frac{v}{u} \bigr)^2,
%\frac{1}{8} \bigl( v - \frac{4 p_1 p_2 v}{u^2}\bigr) \Bigr)$.
\par\noindent {\bf Step 1.} Find $\D(\Q) /\phi(\C(\Q))$.
Here $q : \D(\Q) /\phi(\C(\Q)) \longrightarrow \qmods$.
We need to consider square free $r | b_1 = 4 p_1 p_2$,
that is,
$\im q \leqslant \langle - 1, 2, p_1, p_2 \rangle$.
But $q(\oo) = 1$, $q(0,0) = b_1 = p_1 p_2$.
Also, for any $r < 0$, 
$W_r : r \ell^4 + a_1 \ell^2 m^2 + (b_1/r) m^4 = n^2$
(for some $\ell, m, n \in \Z,
\hbox{ not all $0$, with gcd}(\ell,m) = 1$)
has $\hbox{LHS} \leqslant 0$ and $\hbox{RHS} \geqslant 0$,
with both sides equal only when $\ell = m = n = 0$,
so that $r \not\in \im q$. Hence $\{ 1, p_1 p_2 \}
\leqslant$ $\im q \leqslant$
$\{ 1,2,p_1,2p_1,p_2,2p_2,p_1 p_2, 2 p_1 p_2 \}$.
There are three nontrivial cosets: take $2, p_1, 2p_1$
as representatives.
\par\noindent Lemma $(*)$. $W_r : r \ell^4 + (4 p_1 p_2/r) m^4 = n^2$,
$i \in \{ 1,2 \}$; if $p_i \nmid r$ then $p_i \nmid \ell$. 
Proof: $p_i | \ell$ gives $p_i | n$, so:
$p_i^2 | \ell^4$, $p_i^2 | n^2$,
so that $p_i^2 | (4 p_1 p_2/r) m^4$ and $p_i | m$, contradiction.
\par
$W_2 : 2\ell^4 + 2 p_1 p_2 m^4 = n^2$, so
$2\ell^4 \equiv n^2$ (mod~$p_1$). By $(*)$: $p_1 \nmid \ell$.
$2 \equiv (n/\ell^2)^2$, so $\bigl(\frac{2}{p_1}\bigr) = 1$,
impossible since $p_1 \equiv 3$ (mod~$8$).
Hence $2 \not\in \im q$.
\par
$W_{p_1} : p_1 \ell^4 + 4 p_2 m^4 = n^2$.
$p_1 \ell^4 \equiv n^2$ (mod~$p_2$). By $(*)$: $p_2 \nmid \ell$.
$p_1 \equiv (n/\ell^2)^2$, so
$\bigl( \frac{p_1}{p_2} \bigr) = 1$, impossible
since given $\bigl( \frac{p_1}{p_2} \bigr) = -1$. 
Hence $p_1 \not\in \im q$. 
\par
$W_{2p_1} : 2 p_1 \ell^4 + 2 p_2 m^4 = n^2$,
so $2 p_1 \ell^4 \equiv n^2$ (mod~$p_2$). By $(*)$: $p_2 \nmid \ell$.
$2 p_1 \equiv (n/\ell^2)^2$, so
$\bigl( \frac{2 p_1}{p_2} \bigr) = 1$, impossible
since $\bigl( \frac{2 p_1}{p_2} \bigr) = 
\bigl( \frac{2}{p_2} \bigr) \bigl( \frac{p_1}{p_2} \bigr)
= 1\cdot (-1) = -1$.
Hence $2 p_1 \not\in \im q$.
\par
So: $\im q = \{ 1, p_1 p_2 \}$ and $\D(\Q) /\phi(\C(\Q))
= \{ \oo, (0,0) \} = \langle (0,0) \rangle \cong C_2$.
\par\noindent {\bf Step 2.} Find $\C(\Q) /\hat\phi(\D(\Q))$.
Here $\hat q : \C(\Q) /\hat\phi(\D(\Q)) \longrightarrow \qmods$.
Note $\qhat : (0,0) \mapsto b = -p_1 p_2$ and $r | b$,
so that $\{ 1, -p_1 p_2 \} \leqslant 
\{ \pm 1, \pm p_1, \pm p_2 \pm p_1 p_2 \}$. There are three nontrivial
cosets; take $-1, p_1, -p_1$ as representatives.
Here ${\widehat W}_r :$ $r\ell^4 + a \ell^2 m^2 + (b/r) m^4 = n^2$,
some $\ell, m, n \in \Z,
\hbox{ not all $0$, with gcd}(\ell,m) = 1$.
\par\noindent Lemma $(**)$.
${\widehat W}_r : r \ell^4 - (p_1 p_2/r) m^4 = n^2$,
$i \in \{ 1,2 \}$; if $p_i \nmid r$ then $p_i \nmid \ell$. 
Proof: $p_i | \ell$ gives $p_i | n$, so:
$p_i^2 | \ell^4$, $p_i^2 | n^2$,
so that $p_i^2 | (p_1 p_2/r) m^4$ and $p_i | m$, contradiction.
\par
${\widehat W}_{-1} : -\ell^4 + p_1 p_2 m^4 = n^2$.
$-\ell^4 \equiv n^2$ (mod~$p_1$). By $(**)$: $p_1 \nmid \ell$.
$-1 \equiv (n/\ell^2)^2$
so $\bigl( \frac{-1}{p_1} \bigr) = 1$, impossible since 
$p_1 \equiv 3$ (mod~$4$). 
Hence $-1 \not\in \im \qhat$.
\par
${\widehat W}_{p_1} : p_1 \ell^4 - p_2 m^4 = n^2$.
$p_1 \ell^4 \equiv n^2$ (mod~$p_2$). By $(**)$: $p_2 \nmid \ell$.
$p_1 \equiv (n/\ell^2)^2$
so $\bigl( \frac{p_1}{p_2} \bigr) = 1$, impossible since
given $\bigl( \frac{p_1}{p_2} \bigr) = -1$.
Hence $p_1 \not\in \im \qhat$.
\par
${\widehat W}_{-p_1} : -p_1 \ell^4 + p_2 m^4 = n^2$.
$-p_1 \ell^4 \equiv n^2$ (mod~$p_2$). By $(**)$: $p_2 \nmid \ell$.
$-p_1 \equiv (n/\ell^2)^2$
so $\bigl( \frac{-p_1}{p_2} \bigr) = 1$, impossible since
$\bigl( \frac{-p_1}{p_2} \bigr) = 
\bigl( \frac{-1}{p_2} \bigr) \bigl( \frac{p_1}{p_2} \bigr)
= 1\cdot(-1) = -1$.
Hence $-p_1 \not\in \im \qhat$.
\par
So: $\im \qhat = \{ 1, - p_1 p_2 \}$ and
$\C(\Q) /\hat\phi(\D(\Q)) = \{ \oo, (0,0) \} = \langle (0,0) \rangle
\cong C_2$.
\par\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 $\hat\phi\bigl( \D(\Q) /\phi(\C(\Q)) \bigr)$
$= \hat\phi \bigl( \langle (0,0) \rangle \bigr)$
$= \{ \oo \}$, 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
$( \pm \sqrt{p_1 p_2}, 0 ) \not\in \C(\Q)$. Hence the rank is~$0$.
[{\bf 10 marks}; Unseen.]

\medskip
\hrule

\end{document}
