Montgomery form elliptic curve with given j-invariant

Question

For a given $j\in K$, I would like to find a Montgomery form elliptic curve $M_{A,B}: By^2=x^3+Ax^2+x$ such that j-invariant of $M_{A,B}$ is j (if it is possible). In the form $y^2=x^3+Ax^2+x$ is also OK.

Answer 1

Let $K = \mathbb{F}_q$. Then the $j$-invariant of a montgomery curve is given by $$j = \frac{256(A^2-3)^3}{A^2 - 4}$$ and therefore the isomorphism class of $M_{A,B}$ over $\mathbb{F}_q^{\text{alg}}$ is completely determined by the number $A^2$.

Remark: $q$ is a power of some odd prime, i.e. $256 \neq 0$.

One can show that not every element arises as the right hand side such that you cannot expect to find a montgomery curve for arbitrary $j$.