= Solution
{c}
Use the <two-isogeny formula> formula with $a=\alpha$ and $b=-\alpha$. Its companion is
$$
D:v^2=u\bigl(u^2-2\alpha u+\alpha(\alpha+4)\bigr).
$$
Both curves are nonsingular because $\alpha\neq0,-4$. With $b'=\alpha(\alpha+4)$, choose
$$
\phi(x,y)=\left(x+\alpha-\frac{\alpha}{x},
y\left(1+\frac{\alpha}{x^2}\right)\right),
$$
$$
\widehat\phi(u,v)=
\left(\frac{u-2\alpha+b'/u}{4},
\frac{v(1-b'/u^2)}8\right).
$$
Each map extends across its exceptional affine points, and $\widehat\phi\phi=[2]$. On $D$, the first coordinate of the dual can also be written $v^2/(4u^2)$.
Let $P=(1,1)$ and solve $\widehat\phi(u,v)=P$. The first coordinate gives $v^2=4u^2$. Since $u\neq0$, the equation of $D$ then gives
$$
u^2-(2\alpha+4)u+\alpha(\alpha+4)=0
\quad\Longleftrightarrow\quad
(u-\alpha)(u-\alpha-4)=0.
$$
The second coordinate selects the signs. The two rational preimages are exactly
$$
Q_1=(\alpha,-2\alpha),\qquad
Q_2=(\alpha+4,2(\alpha+4)).
$$
Substituting into the dual formula checks that both map to $(1,1)$. Their coordinates are nonexceptional under the assumptions, and the degree-two map has no further preimages.
The <two-torsion square-class homomorphism> $\alpha_D$ has kernel $\phi(C(\mathbb Q))$, and on these two points it is their nonzero $u$-coordinate modulo squares. Thus $Q_1$ is in the image of $\phi$ exactly when $\alpha$ is a rational square, and $Q_2$ is in that image exactly when $\alpha+4$ is a rational square.
If $P=2R$, then $\phi(R)$ is one of the two preimages, because $\widehat\phi\phi(R)=2R=P$. Conversely, if $Q_j=\phi(R)$, applying the dual shows $P=2R$. Therefore
$$
P\in2C(\mathbb Q)
\Longleftrightarrow
\alpha\in\mathbb Q^2\ \text{or}\ \alpha+4\in\mathbb Q^2.
$$
Finally, if an <integer> equals $(a/b)^2$ in lowest terms, then $b^2\mid a^2$ forces $b=1$. Thus rational squareness of these <integer> parameters is exactly <integer> squareness, and
$$
\boxed{(1,1)\in2C(\mathbb Q)
\iff\alpha\text{ or }\alpha+4\text{ is an integer square}.}
$$
This is the <two-isogeny halving criterion for the alpha family>.
\Image[/past-exam-of-the-analytic-number-theory-and-elliptic-curves-course-of-the-university-of-oxford/2009-halving-isogenies.png]
{title=Halving the point is equivalent to lifting either of its two dual-isogeny preimages}
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