Square-class index formula for two-isogeny descent ($2^r=\#\alpha (E)\#{\alpha }^{\prime }(E^{\prime })/4$)

For $E:y^2=x(x^2+ax+b)$ and $E^{\prime }:y^2=x(x^2-2ax+a^2-4b)$ over $\mathbb{Q}$, define the square-class homomorphisms by $\alpha (x,y)=[x]$, $\alpha (O)=1$, $\alpha ((0,0))=[b]$, and similarly ${\alpha }^{\prime }$ with $b^{\prime }=a^2-4b$. Their kernels are the images of the dual degree-two isogenies of elliptic curves. If $r$ is the rank of an abelian group of $E(\mathbb{Q})$, then $2^r=\#\alpha (E)\#{\alpha }^{\prime }(E^{\prime })/4$. To check the factor four, put $\delta =[\ker \hat{\varphi }:\ker \hat{\varphi }\cap \varphi E]$. The index of $2E$ in $E$ is $\#\alpha (E)\#{\alpha }^{\prime }(E^{\prime })/\delta$. If $b^{\prime }$ is a square, $\delta =1$ and $\#E(\mathbb{Q})[2]=4$; otherwise $\delta =2$ and $\#E(\mathbb{Q})[2]=2$. In either case $\delta \#E(\mathbb{Q})[2]=4$.