If $M =(u,v)$ is a point on the Montgomery curve, then the $u$-coordinate of $2M$ is $(u^2-1)^2/(4v^2)$ is necessarily square. It follows that if $(x,y)$ is a point on $E_{a,d}$, and $a-d$ is square, then $(1+y)/(1-y)$ is also square.
\todo{Nega montgomery}
Likewhise, when $d-a$ is square in \F, $E_{a,d}$ is isomorphic to the Montgomery curve
$$v^2= u\cdot\left(u^2-2\cdot\frac{a+d}{a-d}\cdot u +1\right)$$