Why does 4 | E(K) for Montgomery curves? Given a Montgomery curve over some finite field $K$ in the form 
$E/K: by^2 = x^3 + ax^2 + x$ and using $E(K)$ for the $K$-rational points. I've just read that the number of points in $E(K)$ is always divisible by 4.
This is from crypto and usually it is assumed that $\operatorname{char}(K) \neq 2$ and $\neq 3$. In this specific case we also have $K = \mathbb{F}_{p^2}$ for some prime $p$.
Can somebody tell me why this is the case or if I have overlooked something?
 A: Quoting from: Montgomery: Speeding the Pollard and Elliptic Curve Methods of Factorization, Mathematics of Computation, Volume 48. Number 177, January 1987, pages 243–264:

[p. 260:]
  $$\tag{10.3.1.1} By^2 = x^3 + Ax^2 + x$$
  [p. 262:]
Let $p$ be a prime which does not divide $B(A + 2)(A - 2)$.
  Suyama [31] observes that the order of the group associated with
  $(10.3.1.1)$ modulo $p$ will always be divisible by $4$.
  If $B(A + 2)$ is a quadratic residue, then the point $(1, \sqrt{(A + 2)/B})$
  has order $4$.
  If $B(A - 2)$ is a quadratic residue, then the point $(-1, \sqrt{(A - 2)/B})$
  has order $4$.
  If $(A + 2)(A - 2)$ is a quadratic residue,
  then the cubic has three linear factors,
  and again there is a subgroup of order $4$.
[p. 264:]

  
*Hiromi Suyama, "Informal preliminary report (8)," 25 Oct. 1985
  

For easier understanding, note that:


*

*Doubling the mentioned points of order $4$
gives the point $(0,0)$ of order $2$.

*The discriminant of the cubic's quadratic factor is $A^2-4$.

*If the cubic has three linear factors,
then the three roots correspond to points of order $2$ which generate a
subgroup isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$.


Following the quoted snippet, further conditions are mentioned,
given by Suyama, ensuring the existence of points of order $3$,
hence of group order divisible by $12$.
