\documentclass{article} \usepackage{kotex} \setlength{\textheight}{22cm} \headsep=15pt %%change hight page \setlength{\textwidth}{15.5cm} %%change width page \setlength{\oddsidemargin}{0.5cm} \setlength{\topmargin}{-.5cm} \setlength{\evensidemargin}{\oddsidemargin} \def\sms{\smallskip} \def\supp{\mbox{supp}} \usepackage{mathtools} \def\g{\gamma} \usepackage{tikz} \usetikzlibrary{calc} \usetikzlibrary{matrix} \usetikzlibrary{positioning} \def\fX{\mathfrak{X}} \def\Map{\mbox{Map}} \def\fY{\mathfrak{Y}} \def\Spl{\mbox{Spl}} \def\fP{\mathfrak{P}} \def\max{\mbox{max}} %\usepackage{mypack} \usepackage{graphicx} \usepackage{amsmath, amsthm, amssymb} \def\Kbar{\bar{K}} \def\extet{\Ext^1_{\Un(\bX)}} \def\ext2et{\Ext^2_{\Un(\bX)}} \def\homqp{\Hom_{\Q_p}} \def\homet{\Hom_{\Un(\bX) }} \def\I{\A^{\times}} \def\cG{\mathfrak{G}} \def\cO{\mathfrak{O}} \def\ms{\medskip} \def\S{\Sigma} \def\ZS{\Z\langle S \rangle } \def\ZSB{\Z\langle S; B \rangle } \def\ZSC{\Z\langle S; C \rangle } \def\ZSD{\Z\langle S; D\rangle } \def\Fx{F^{\times}} \def\bcE{\overline{\cE}} \def\bcM{\overline{\cM}_{1,1}} \def\calG{{\cal G}} \def\cMM{\cM_{1,1}} \def\cE{{\cal E}} \def\cC{{\cal C}} \def\cD{{\cal D}} \def\F{{\mathbb F}} \def\Z{{\mathbb Z}} \def\R{{\mathbb R}} \def\Q{{\mathbb Q}} \def\a{{\alpha}} \def\b{{\beta}} \def\rTo{{\mapsto}} \def\rInto{{\hookrightarrow}} %\usepackage{mypack, color, sgame, sgamevar, tikz} \usepackage{color, sgame, sgamevar, tikz} \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]} \end{flushleft} \end{document}