Descent by 2-isogeny Im practicing with a exercise about 2-isogenies, but struggling a bit. Im doing the following exercise:
Given two elliptic curves over $\mathbb Q$.
\begin{equation}E: y^2 = x(x^2-5) \quad E':y^2 = x(x^2+20)\end{equation}
These are related by a 2-isogeny: $\phi((x,y)) = (x-\frac 5x, y + \frac{5y}{x^2})$ if $x\neq 0$, $\phi((0,0)) = O$.
(a) Show that the group $E'(\mathbb Q)/\phi(E(\mathbb Q))$ has order 2.
(b) Compute $E(\mathbb Q)/\hat\phi(E'(\mathbb Q))$, and hence calculate the rank of $E(\mathbb Q)$.
This is what I have found so far:
Define a function (which is a homomorphism of groups) $q:E'(\mathbb Q) \to \mathbb Q^*/\mathbb Q ^{*^2}$
$q((u,v)) = [u]$, if $u\neq 0$, $q((0,0)) = [20], q(O) = [1]$.
Then the sequence 
\begin{equation} E(\mathbb Q)\to^\phi E'(\mathbb Q)\to^q \mathbb Q^*/\mathbb Q ^{*^2}\end{equation} is exact.
$\textbf{Lemma.}$ Let $r$ be a squarefree integer. $[r]\in\mathbb Q^*/\mathbb Q ^{*^2}$ is in the image of $q$ if and only if
\begin{equation}r^2l^4+20m^4=rn^2\end{equation}
has a integer solution. This can only happen if $r|20$.
This is what i have done so far:
The square free integers dividing 20 are $r = \pm1, \pm2,\pm5,\pm10,\pm20$.
One easily verifies that $r = 1$ gives rise to the solution $(1,0,1)$ and $r = 20$ to $(0,1,1)$.
Easily reasoning gives that negative $r$ cannot work, as it gives a positive right hand side and negative left hand side.
So we are left with $r = 1,2,5,10,20$. Do I need to prove for $r = 2,5,10$ that solutions doesn't exist? Or are there easier methods for showing only 1 and 20 satisfy this condition?
For (b), the squarefree integers dividing 5 are $r = \pm1,\pm5$. Again, $r=1,5$ give rise to solutions. For $r = -1$, we have
[begin{equation} -l^4+5m^4=n^2\end{equation}
which has solution $(1,1,2)$. Hence $\text{im}(q) = \{\pm1,\pm5\}$
and $E(\mathbb Q)/\hat\phi(E'(\mathbb Q))$ has order 4. Finding generators gives that it is generated by $(0,0)$ and $(-1,2)$.
Is this correct? please correct me if im wrong.
And from this point, how to calculate the rank of $E(\mathbb Q)$.
Thanks a lot in advance :)
 A: I will take for granted that
$$
E (\mathbb Q)\overset\phi\longrightarrow 
E'(\mathbb Q)\overset   q\longrightarrow 
\mathbb Q^\times/(\mathbb Q ^\times)^2
$$
is an exact sequence.
I will also take the lemma for granted.
(a)
Then the image in 
$\mathbb Q^\times/(\mathbb Q ^\times)^2
$
of the positive divisors $1,2,4,5,10,20$ of $20$ has representatives
$1,2,5,10$.
We already know that $1$ and $20\equiv 5$ (modulo rational squares) are in the image.
Because of the group structure it is enough to show that $2$ is not in the image, by the lemma we consider the equation in integers
$$
4l^4+20m^4=2n^2
$$
and show it has no solutions. If $l\ne 0$ modulo $5$, then also $n\ne 0$ modulo $5$, and it follows that $2$ is a quadratic residue modulo $5$, contradiction. So $l=5l_1$, so $n=5n_1$ and we get 
$$
4\cdot 5^3l_1^4+4m^4=2\cdot 5n_1^2\ .
$$ 
So $m=5m_1$, and we can further write
$$
4\cdot 5^3l_1^4+4\cdot 5^4m_1^4=2\cdot 5n_1^2\ .
$$ 
So $n_1=5n_2$, and we can further write
$$
4\cdot 5^3l_1^4+4\cdot 5^4m_1^4=2\cdot 5^3n_2^2\ .
$$
Simplifying with $5^3$ delivers a smaller solution.
We could have started with a minimal solution, or here we can invoke an infinite descent, thus getting a contradiction.
We have so far:
$$
\begin{aligned}
E'(\Bbb Q)/\phi E(\Bbb Q)
&\overset\cong{\underset q\longrightarrow}
\operatorname{Image} (q) = \langle 5\rangle\subseteq 
\mathbb Q^\times/(\mathbb Q ^\times)^2
\\\\
&\text{has one generator and two elements.}
\end{aligned}
$$
(b)
It is ok, we have on $E$ the $\Bbb Q$-rational points $(0,0)$, a torsion point, mapping to $5$, and $(-1,2)$, mapping to $-1$, so if the Lemma also aplies for the dual map we have:
$$
\begin{aligned}
E(\Bbb Q)/\hat\phi E'(\Bbb Q)
&\overset\cong{\underset {q'}\longrightarrow}
\operatorname{Image} (q') = \langle -1,5\rangle\subseteq 
\mathbb Q^\times/(\mathbb Q ^\times)^2
\\\\
&\text{has two generators and four elements.}
\end{aligned}
$$
Consider now the composition of the two isogenies, $[2]=[2]_E=\hat\phi\circ\phi$, which leads
to the following exact diagram:
$\require{AMScd}$
\begin{CD}
    @.    0        @.       0       @.            0   \\
@.      @VVV              @VVV                  @VVV  \\
0 @>>> \hat\phi\phi E(\Bbb Q) 
                @>>> \hat \phi E'(\Bbb Q) @>>> \boxed{0}  @>>> 0\\
@.       @|             @VVV          @VVV  \\
0 @>>> [2] E(\Bbb Q) @>>> E (\Bbb Q) @>>>         (?) @>>> 0\\
@.      @VVV              @VVV                    @VV{\cong}V  \\
0 @>>>    0   @>>> \langle 5,-1\rangle @= \langle 5,-1 \rangle @>>> 0\\
@.      @VVV              @VVV            @VVV  \\
    @.    0        @.       0       @.      0   
\end{CD}
(The left upper square is made of inclusions.)
The entry $\boxed{0}$ in the first horizontal short exact row is indeed vanishing, because we know that $(0,0)$ generates $E'(\Bbb Q)$ modulo the image of $\phi$ (because it is mapped by $\phi$ to a non-trivial element), but we pass to the image via $\hat \phi$, and this torsion point is mapped to the zero element $\infty\in E(\Bbb Q)$.
It follows that the $(?)$ entry is also a vector space of dimension two over $\Bbb F_2=\Bbb Z/2$. One part of it comes from the $2$-torsion element $(0,0)$ of $E(\Bbb Q)$, and this is the only $2$-torsion part. So the rest comes from the generator of $E(\Bbb Q)$ modulo torsion, the preimage of $-1$ is such a generator, so $(-1,2)\in E(\Bbb Q)$. 
