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\title{2015 Elliptic Curves Solutions.}
\author{Minhyong Kim}
\begin{document}
\begin{flushleft}

1.

(a) State and prove Hensel's lemma.

\ms

\ms

Let~$K$ be a field, complete with respect to a non-Archimedean
valuation~$|\ \, |$, with valuation ring~$R = \{ x\in K : |x| \leqslant 1\}$.
If $f(x) \in R[x]$ and $a_0\in R$ satisfies
$| f(a_0) | < | f'(a_0) |^2$, then there exists a unique $a\in R$ such that
$f(a) = 0$ and $| a - a_0 | \leqslant | f(a_0) |/ | f'(a_0) |$.

\ms

\begin{proof}

[{\it From lectures}] Define $f_j(x)$ by:
\par $f(x + y) = f_0(x) + f_1(x) y + f_2(x) y^2 + \ldots,$
\par\noindent 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 $| f'(a_1) - f'(a_0) | = | f'(a_0+b_0) - f'(a_0) |
= | (\hbox{poly in }a_0) b_0 + (\hbox{poly in }a_0) b_0^2 + \ldots |$
\par \ \ \ \ \ \ \ \ \ \ \ $\leqslant | b_0 | < | f'(a_0) |$
\ \ (by $(*)$),
\par\noindent  so that $| f'(a_1) | = | f'(a_0) |$.
\par 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 $f_0(a_0) + f_1(a_0)b_0 = 0$]
\par\ \ \ $\leqslant \hbox{max}_{j\geqslant 2} |f_j(a_0)| |b_0|^j
\leqslant | b_0 |^2 = \frac{ |f(a_0 )|^2}{|f'(a_0 )|^2}
= \rho | f(a_0) | < | f(a_0) |$,
where $\rho = \frac{ |f(a_0 )|}{|f'(a_0 )|^2} < 1$.
\par Summarising: $ |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 For all~$n$, given $a_n\in R$, define
$b_n = -f(a_n)/f'(a_n)$ and $a_{n+1} = a_n + b_n = a_n -f(a_n)/f'(a_n)$.
\par\noindent Assume, as induction hypothesis, that:
\par $|f'(a_n)| = \ldots = |f'(a_1)| = |f'(a_0)|$
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 the case~$n=0$ above]
\par \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
$= \frac{ | f(a_n) |^2 }{ | f'(a_n) |^2 }
= \frac{ | f(a_n) |^2 }{ | f'(a_0) |^2 }$\ \ [by~(1), the induction hypothesis]
\par \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
$\leqslant \frac{ | f(a_0) | }{ | f'(a_0) |^2 } |f(a_n)|$
\ \ [since $|f(a_n)| \leqslant |f(a_0)|$ by~(1), the induction hypothesis]
\par \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
$= \rho | f(a_n) | \leqslant \rho^{n+1} | f(a_0) |$
\ \ [by~(1), the induction hypothesis].
\par\noindent By induction, $\forall n$, $|f'(a_n)| = |f'(a_0)|$ 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 the theorem 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\noindent $| a - a_0 | = | \sum b_n | \leqslant
\max | b_n | = \max \frac{| f(a_n) |}{|f'(a_n)|}
= \max \frac{| f(a_n) |}{|f'(a_0)|}
= \frac{| f(a_0) |}{|f'(a_0)|}$ [by~(2)], as required.
\par\noindent For uniqueness, imagine that $\hat a \not= a$ also satisfied
$f(\hat a) = 0$ and $| \hat a - a_0 | \leqslant | f(a_0) |/ | f'(a_0) |$.
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.
\end{proof}
\hfill {\bf [10~marks, bookwork]}
\ms

\ms

(b)

Which of the following numbers has a cube root in $\Q_3$? Be sure to 
justify your answer.
$$(i) \ \ \ \ 35 \ \ \ \ \ \  (ii) \ \ \ \ 132 \ \ \ \ \ \  
(iii) \ \ \ \ 10/459.$$

\ms

\ms
(i) 35 is a unit in $\Z_3$ and $35 \equiv 8 \mod 27$. Thus, for $f(x)=x^3-35$, 
we have $$|f(2)|_3\leq 1/27< 1/9=|f'(2)|_3^2.$$ Therefore, by Hensel's 
Lemma, $f(x)$ has a root in $\Z_3$, which is a cube root of 35.
\ms

(ii) $132=3\times 44$, so that $v_3(132)=3^{-1}$. If we had $\a^3=132$, 
then we would have
$|\a|_3=(1/3)^{1/3}$, which is impossible. So 132 has no cube root in $\Q_3$.
\ms

(iii) We have $10/459=10\cdot 17^{-1}\cdot 3^{-3}.$ Therefore, $10/459$ 
has a cube root if and only if $10 \cdot 17^{-1}$ has a cube root. 
Now, $4^3=64=10 \mod 27$, so 10 has a cube root in $\Z_3$ by considering 
the polynomial $x^3-10$ and using Hensel's Lemma as in (i). Similarly, 
$17=-10=(-4)^3 \mod 27$ so, 17 has a cube root in $\Z_3$. Therefore, 
$10/17$ has a cube root in $\Q_3$.
\hfill {\bf [5~marks, unseen]}
\ms



\ms

\ms

\ms


(c)


Prove that $2y^2=x^4-17$ has a solution in $\R$ and in  $\Q_p$ for 
all primes $p$.
\ms

[{\em You may use any of the theorems stated during lectures.}]
\bigskip

$(x^4-17)/2$ is positive for $x$ large, so the equation clearly has real 
solutions.
For any algebraically closed field of characteristic different from 
2 and 17, $x^4-17$ has 4 distinct roots, so $2y^2=x^4-17$ is non-singular 
over $\F_p$ for all $p\neq 2,17$. If we denote by $N_p$ the number 
of solutions in $\F_p$, then
Hasse's bound gives 
$$N_p\geq p-1-2\sqrt{p}$$
for any $p\neq 2, 17$. Therefore, for any such $p\geq 7$, there is 
an $\F_p$-solution $(a,b)$.
If we put $f(x,y)=2y^2-x^4+17$, then either $df/dx(a,b)\neq 0$ or 
$df/dy(a,b)\neq 0$. In the first case, if we put
$g(x)=f(x,b)$, then $g'(a)\neq 0$. Therefore, by Hensel's lemma, we can 
lift $(a,b)$ to a solution to $f(x,y)=0$ in $\Q_p$. A similar argument 
using $h(y)=f(a,y)$ gives a $\Q_p$ solution in the second case.
So we need only examine for solutions in $\Q_{17}$, $\Q_{5}$, $\Q_3$, 
and $\Q_2$.
\ms

$\Q_{17}$. If we put $g(y)=y^2-(1/2)(1-17) \in \Z_{17}[x]$, then 
$g(3)\equiv 0 \mod 17$ and $g'(3)=6 \in \Z_{17}^*$. So by Hensel's 
lemma, $g(y)$ has a root $b\in \Z_{17}$. Then $(1,b)$ is  a zero 
of $2y^2-x^4+17$.
\ms

$\Q_5$. Now put $g(y)=2y^2+17$. Then $g(2)\equiv 0\mod 5$ and 
$g'(2)=8\in \Z_5^*$. So $g(y)$ has a zero $b\in \Z_5$ and $(0,b)$ 
is a zero of $2y^2-x^4+17$.
\ms

$\Q_3$.  Put $g(y)= 2y^2-(1-17)=2y^2+16$. Then $g(1)\equiv 0 \mod 3$ and
$g'(1)=2\in \Z_3^*$. So $g(y)$ has a root  $b$ in $\Z_3$ by Hensel's 
Lemma, and $(1,b)$ is a solution of $f(x,y)=0$.
\ms


$\Q_2$. It suffices to show that 17 has a fourth root in $\Z_2$, so 
consider $g(x)=x^2-17$. We have $g(1)=1-17\equiv 0 \mod 16$, so that 
$|g(1)|_2=2^{-4}$, while $g'(1)=2$, so $|g'(1)|_2=2^{-1}$. Hence,
$|g(1)|_2< |g'(1)|_2^2$, and $g(x)$ has a root in $\a \in \Z_2$ such that
$|\a-1|_2\leq |g(1)|_2/|g'(1)|_2=2^{-3}.$
Now we can consider $h(x)=x^2-\a$. Then $|h(1)|_2\leq 2^{-3}$, while
$|h'(1)|_2=2^{-1}$, so that $h(x)$ has a root $\b\in \Z_2$. Then $\b $ 
is a 4-th root of 17.


\hfill {\bf [10~marks, seen]}
\ms

\ms


\ms



2.


(a)

For the elliptic curve  $\cE$ with equation
$y^2=x^3+1$
compute the groups
\ms

(i) $\cE(\F_5)$

(ii) $\cE(\F_7)$
\ms

For case (i) describe directly  the group law  on the points of the curve.
\bigskip

(i) As we go through the values $0,1,2,3,4$ for $x$, we get $1,2,4,3,0$ 
for $x^3+1$. These correspond to the points
$$P_1=(0,1), P_2=(0,4), P_3=(2, 2), P_4=(2,3), P_5=(4,0),$$
which, together with the origin $O$ make up the group of point. This group 
is clearly isomorphic to $C_6$. Note that $P_5$ is the only point of 
order 2. We compute $2P_3$ as follows. The slope of the tangent is
$y'=3x^2/2y=2/4=3.$ Thus, the equation of the tangent line is
$y=3(x-2)+2=3x-4=3x+1.$ We find the points of intersection using the roots of
$$(3x+1)^2-x^3-1=4x^2+x+1+4x^3-1=x(4x^2+4x+1)=x(2x+1)^2.$$
Thus, $2P_3=(0,1)=P_1$. We check easily that $P_1$ and $P_2$ have order 3. So
$4P_3=2P_1=P_2$. Since $P_3$ must have order 6, we have
$3P_3=P_5$. Hence, we must have $5P_3=P_4$, the other point of order 6.
Finally, $6P_3=O$. To summarize, the group of points is given by
$$O, \ \ \ P_3, \ \ \ 2P_3=P_1, \ \ \ 3P_3=P_5, 
\ \ \ 4P_3=P_2, \ \ \ 5P_3=P_4,$$
which also contains the information for the group table.

\hfill {\bf [5~marks, seen]}
\ms

\ms

\ms

(ii) As we go through the values $0,1,2,3,4,5,6$ for $x$, we get for $x^3+1$,
$1,2,2, 0, 2, 0, 0$ giving us the points
$$(0,\pm 1), (1,\pm 3), (2, \pm 3), (3,0), (4,\pm 3), (5,0), (6,0).$$
Together with the origin, we see that the group has order 12. By examining 
the points of order two, we see that there is a subgroup isomorphic 
to $C_2\times C_2$. Hence, the group is isomorphic to
$C_2\times C_2\times C_3$.

\hfill {\bf [5~marks, seen]}
\ms

\ms


(b) Compute the torsion subgroups of the rational points of the elliptic 
curves over $\Q$ given by the equations
$$D_1: \ \ \  y^2=x^3+x^2+2x+1;  \ \ \ \ D_2: y^2=x^3+64.$$
\ms


[{\em You may use any of the theorems stated during lectures.}]
\bigskip

$D_1$.

We have the point $(0,1)$. The slope of the tangent line there is
$y'=2/2=1$, so that the tangent line is $y=x+1$. To find the points 
of intersection, we solve
$$x^3+x^2+2x+1=(x+1)^2,$$
which has a triple root $x=0$. Therefore, $(0,1)$ is a point of order 3.

Now we consider the points in $\F_3$. Over $\F_3$, we have
$$(x^3+x^2+2x+1)'=2x+2,$$
which has only $x=2$ as zero. But $2^3+2^2+2\times 2+1=2\neq 0$, so 
the curve has good reduction.  In fact, using the same computation, 
we see that the only points are
$$(0, \pm 1), O.$$
But we have the injection
$$D_1(\Q)_{tor}\rInto D_1(\F_3).$$
Therefore, we must have $D_1(\Q)_{tor}=C_3$, generated by $(0,1)$.

\hfill {\bf [5~marks, unseen]}
\ms

\ms

$D_2$.

We have the point $(0,8)$, which is again easily seen to have order 3. 
There is also the point $(-4,0)$ of order 2. Thus, $D_2(\Q)_{tor}$ 
has order at least 6.

Now we compute the points over $\F_5$, where the curve takes the form
$y^2=x^3-1,$ which is non-singular. As we run through, $x=0,1,2,3,4$, we get
$x^3-1=4,0,2, 1, 3.$ Thus, we get the points
$$(0,\pm 2), (1,0), (3, \pm 1), O.$$
That is, $D_2(\F_5)$ has order 6. Hence, since
$$D_2(\Q)_{tor}\rInto D_2(\F_5),$$
we see that $D_2(\Q)_{tor}=C_6$, generated by $(-4,0)+(0,8)$.

\hfill {\bf [5~marks, unseen]}
\ms

\ms




(c) Let $\cE$ be the elliptic curve over $\Q$ given by the equation 
$$y^2=x^3+8.$$
 Prove or disprove: $\cE$ has infinitely many rational points.
 
 \ms
 [{\em You may use any of the theorems stated during lectures.}]

\bigskip

We examine the point $(1,3)$. Compute $2(1,3)$. The tangent line at 
$(1,3)$ has slope $(3x^2/2y)|_{(1,3)}=1/2$. Thus, the equation of the 
tangent line is
$y=(1/2)x+(5/2)$. To find the point of intersection, we substitute
$$(1/4)(x+5)^2=x^3+8.$$
Thus, the $x$-coordinate $\a$ of the intersection point satisfies 
$\a+1+1=1/4$, and hence $\a=-7/4$. Thus, by the Nagell-Lutz theorem, 
$2(1,3)$ is not a torsion point. Hence $(1,3)$ is not a torsion point. 
Therefore, the curve has infinitely many rational points.
\hfill {\bf [5~marks, unseen]}
\bigskip

\bigskip

3.



(a)

Find the structure of the Mordell-Weil group of $${\cal C}: \ \ y^2=x^3-10x.$$

\ms
\ms

First we compute the points over $\F_3$. For $x=0,1,2$, we get
$x^3-10x=x^3-x=0,0,0.$ Thus, all the $\F_3$ points are 2-torsion points, and
$$\cC(\F_3)=C_2\times C_2.$$
Now, $\cC(\Q)_{tor}\rInto C_2\times C_2$ via reduction modulo 3. 
Thus, $\cC(\Q)_{tor}$ must also consist entirely of 2-torsion points. 
However, since $x^3-10x=x(x^2-10)$ and $10$ does not have a rational 
square root, we see that
$$\cC(\Q)_{tor}=\langle (0,0)\rangle\simeq C_2.$$ 

\hfill {\bf [5~marks, unseen, but similar seen]}
\ms

We go on to compute the rank. We have dual isogenies
$$\phi: \cC\rTo \cD$$
$$\psi: \cD\rTo \cC$$
where $$\cD: V^2=U(U^2+40),$$
and an exact sequence
$$\cD(\Q)/\phi\rTo^{\psi}\cC(\Q)/2\cC(\Q)\rTo \cC(\Q)/\psi\rTo 0.$$

$\cD(\Q)/\phi$. This is isomorphic to the subgroup 
$Im(q')\subset \Q^*/(\Q^*)^2$ represented by 1, 40, and  square-free 
$r|40=2^3\cdot 5$ for which there are
$l,m,n$ not all zero $(l,m)=1$ such that
$$rl^4+(40/r)m^4=n^2.$$
We have to consider $r\in \{\pm 1, \pm 2, \pm 5, \pm 10.\}$.
\ms

-1,-2,-5,-10: In these cases, there are clearly no solutions by positivitiy.
\ms

$r=5$:

$$5l^4+8m^4=n^2.$$
Reducing modulo 5, we get
$$3m^4=n^2.$$
Since 3 has no square root in $\F_5$, we must have both $m$ and $n$ zero.
Thus, $5|m$ and $5|n$. But then, $5^2|5l^4$, so that $5|l$. Therefore, 
there are no solutions with $(l,m)=1$ and $5\notin Im(q')$. 
Since $10=40/2^2\in Im(q')$, we see that $2\notin Im(q')$. Therefore,
$Im(q')=\{1,10\}\simeq C_2$.
\ms

\ms

Now we consider
$\cC(\Q)/\psi.$ This is isomorphic to   the subgroup 
$Im(q)\subset \Q^*/(\Q^*)^2$ represented by 1, -10, and  
square-free $r|10=2\cdot 5$ for which there are
$l,m,n$ not all zero $(l,m)=1$ such that
$$rl^4-(10/r)m^4=n^2.$$
We have to consider $r\in \{\pm 1, \pm 2, \pm 5, \pm 10.\}$.

\ms

$r=5$:

$$5l^4-2m^4=n^2.$$
Reducing modulo 5, we get
$$3m^4=n^2.$$
Since 3 has no square root in $\F_5$, we see that $5|m$ and $5|n$. 
As above, we see that $5|l$ so that there is no solution with $(l,m)=1$.
Once again, since $-10\in Im(q)$, we see that $-2\notin Im(q)$. 

$r=-5$:
$$-5l^4+2m^4=n^2.$$
Reducing modulo 5, we get
$$2m^4=n^2.$$
Since 2 has no square root in $\F_5$, we see that $5|m$ and $5|n$. 
As above, we see that $5|l$ so that there is no solution with $(l,m)=1$.
This also implies that $2$ is not in the image.

$r=-1$.

$$-l^4+10m^4=n^2.$$

In this case, we see the solution $(1,1,3)$. Therefore, we see that
$$\cC(\Q)/\psi \simeq \{\pm 1, \pm 10\}\simeq  C_2\times C_2.$$

\ms

\ms

Now, $\cD(\Q)/\phi$ is represented by
$O, (0,0)$, which then goes to $O\in \cC(\Q)/2$ under the map 
$\psi: \cD(\Q)/\phi \rTo \cC(\Q)/2.$ Therefore,
$$\cC(\Q)/2\simeq \cC(\Q)/\psi\simeq C_2\times C_2.$$

Hence, $$\cC(\Q)\simeq C_2\times \Z.$$
A point of infinite order is $(-1,3)$.


\hfill {\bf [10~marks, unseen, but similar seen]}




\bigskip
	
(b)

Let $p$ be a prime
such that $p\equiv 5 \mod 8$. Show that
$$\cE_p: \ \ y^2=x^3+px$$
 has rank at most 1.

\bigskip


\bigskip


In this case, the there are isogenies
$$\phi: E_p\rTo E_p'$$
and
$$\psi: E_p'\rTo E_p,$$
where
$$E_p': V^2=U^3-4pU.$$

We first consider $E_p$. 
$E_p(\Q)/\psi$ is  isomorphic to   the subgroup 
$Im(q)\subset \Q^*/(\Q^*)^2$ represented by 1, $p$, and  square-free $r|p$ 
for which there are
$l,m,n$ not all zero $(l,m)=1$ such that
$$rl^4+(p/r)m^4=n^2.$$
We need to consider $r\in \{\pm 1, \pm p\}.$
Clearly, no negative $r$ will work. So
$$E_p(\Q)/\psi\simeq C_2.$$

\ms
$E'_p(\Q)/\phi$ is  isomorphic to   the subgroup 
$Im(q')\subset \Q^*/(\Q^*)^2$ represented by 1, $-4p \equiv -p$, and  square-free $r|4p$ for which there are
$l,m,n$ not all zero $(l,m)=1$ such that
$$rl^4-(4p/r)m^4=n^2.$$
We need to consider $r\in \{\pm 1, \pm 2, \pm p, \pm 2p\}.$ 
\ms



$r=2$.

$$2l^4-2pm^4=n^2.$$
Reducing modulo $p$, we get that $2l^4=n^2$. But since $p\equiv 5 \mod 8$, 
2 has no square root in $\F_p$. Hence, $p|l$ and $p|n$. Therefore, $p|m$, 
and there is no solution such that $(l,m)=1$.
Thus, $2\notin Im(q')$. 
\ms

$r=-2$.

$$-2l^4+2pm^4=n^2.$$
Reducing modulo $p$, we get that $-2l^4=n^2$. But since $p\equiv 5 \mod 8$, 
we also have $p\equiv 1 \mod 4$. So $-1$ has a square root in $\F_p$. Hence,
$-2$ has no square root in $\F_p$. Hence, $p|l$ and $p|n$. Therefore, 
$p|m$, and there is no solution such that $(l,m)=1$.
Thus, $-2\notin Im(q')$. 

From this, we also conclude that $\pm 2p\notin Im(q')$.

 Now we consider $r=p$. Then we have the equation
$$pl^4-4m^4=n^2.$$
If there is a non-trivial solution, then $p\in Im(q')$, so $-1\in Im(q')$. 
Thus, $$E'_p(\Q)/\phi\simeq \{\pm 1, \pm p\}\simeq C_2\times C_2.$$
Since the class of $(0,0)$ goes to zero in $E_p(\Q)/2$, we get that
$\dim_{\F_2}E_p(\Q)/2=2.$ Therefore, in this case,
the rank of $E_p(\Q)$ is 1. 
\ms

If there is no non-trivial solution, then $p\notin Im (q')$ 
so $-1\notin Im (q')$.
Hence, 
$$E'_p(\Q)/\phi\simeq \{ 1, -p\}\simeq C_2$$ and $E'_p(\Q)/\phi$ 
goes to zero in $E_p(\Q)/2.$ Therefore,
$ E_p(\Q)/2\simeq E_p(\Q)/\psi\simeq C_2$
and $E_p(\Q)$ has rank zero.
In any case, the rank is at most 1.

\hfill {\bf [7~marks, unseen]}

(c)


What is the rank of $\cE_p$ for $p=5$, $p=13$,and $p=29$?





\bigskip




\bigskip
For $p=5$, the determining equation becomes
$$5l^4-4m^4=n^2$$
so that we get the solution $(l,m, n)=(1,1,1)$. Therefore, 
$E_5(\Q)$ has rank 1.

\ms

For $p=13$, the equation is
$$13l^4-4m^4=n^2$$
and we get the solution $(l,m,n)=(1,1,3)$. Therefore, $E_{13}(\Q)$ has rank 1.

\ms

For $p=29$, we have
$$29l^4-4m^4=n^2$$
with the solution $(1,1, 5)$. Therefore, $E_{29}(\Q)$ has rank 1.

\hfill {\bf [3~marks, unseen]}

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