Calculate the number of points of an elliptic curve in medium Weierstrass form over finite field
Let $\phi^k(x,y)= (x^{3^k},y^{3^k})$ then $\#E(\mathbb{F}_{3^k}) =\deg_s(\phi^k-1)$. Is the endomorphism $\phi^k-1$ separable ? Yes because inserapable endomorphisms are of the form $\rho \circ \phi$. Then $$\deg_s(\phi^k-1) = \deg(\phi^k-1)=((\phi^*)^k-1)(\phi^k-1)\\= (\phi^*\phi)^k+1-(\phi^*)^k-\phi^k = 3^k+1-\alpha^k-(\alpha^*)^k$$ where $\phi^*$ is the dual isogeny such that $\phi^* \phi = \deg(\phi) = 3$ and $\phi+\phi^* = t = 3+1-\#E(\mathbb{F}_{3})$ and $\alpha$ is the root of the minimal polynomial $X^2-t X + 3 = 0$ of the Frobenius
magma code
F := FiniteField(3); A<x,y> := AffineSpace(F,2);
C := Curve(A,y^2-x^3-x^2-x-1);
t :=3+1- #Points(ProjectiveClosure(C));
P<z> := PolynomialRing(Integers()); K<a> := NumberField(z^2-t*z+3); aa := Norm(a)/a;
for k in [2..10] do
Ck := BaseChange(C,FiniteField(3^k));
Ek := #Points(ProjectiveClosure(Ck));
[Ek,3^k+1-a^k-aa^k];
end for;
To obtain the minimal polynomial of endomorphisms :
Write that $E(\overline{\mathbb{F}_3}) $ is a subgroup of $\mathbb{Q}/\mathbb{Z}\times \mathbb{Q}/\mathbb{Z}$ so any group homomorphism acts as a matrix $A=\pmatrix{a & b \\c & d} \in M_2(\widehat{\mathbb{Z}})$ (matrix of profinite integers). Then the dual homomorphism is $A^*=\pmatrix{d & -b \\-c & a}$ so that $A^* A = \pmatrix{ad-bc& 0 \\ 0 & ad-bc}$ and $A + A^* = \pmatrix{a+d & 0 \\0 & a+d}$, so they both act as direct multiplication by an element in $\widehat{\mathbb{Z}}$. If $A$ is an endomorphism (defined by polynomial equations) then so are $A^*,A + A^*,A^*A$ so the latter must act as multiplication by elements in $\mathbb{Z}$.