I'm trying to solve the following problem without resorting to a direct calculation:
Let $E : Y^2 = X^3 + AX$, where $A \in \mathbb{Z}$ and $A \ne 0$. Let $F(X, Y )$ be the formal group associated to $E$ and let $F(X, Y ) = \sum F_n(X, Y )$, where each $F_n(X, Y )$ is homogeneous of degree $n$. Show that $F_n(X, Y ) = 0$ unless $n \equiv 1 \mod 4$.
I believe that there is some order 4 group property which should make this come out quite nicely, but I can't really see it. Can anyone help me with this?
My copy of Silverman is in my office, so I cannot recall/check all the key bits. I just have this hunch that this is related to the existence of an order four automorphism of elliptic curves of this type. This is too long to fit into a comment, so an answer it is.
To move the identity element to the origin, $(z,w)=(0,0)$ we do the usual change of coordinates $z=-x/y$, $w=-1/y$. The equation of this curve then becomes $$w=z^3+Azw^2\quad(*)$$ making the automorphism $\sigma:z\mapsto iz, w\mapsto -iw$ stand out. The formal group law $F(z_1,z_2)$ basically works with that $z$-coordinate. Given that $\sigma$ is an automorphism of the elliptic curve, the formal group law must satisfy $$ F(\sigma(z_1),\sigma(z_2))=\sigma(F(z_1,z_2)). $$ In other words, we must have $$ F(iz_1,iz_2)=iF(z_1,z_2)\qquad (**) $$ for all $z_1,z_2\in\Bbb{C}$.
Your claim follows immediately from $(**)$, because a non-zero homogeneous term $F_n(z_1,z_2)$ of degree $n$ must satisfy both $F_n(iz_1,iz_2)=iF_n(z_1,z_2)$ and $F_n(iz_1,iz_2)=i^nF(z_1,z_2)$. Therefore $i^n=i$ whenever $F_n\neq0$.
The automorphism $\sigma$ also forces the power series solution $w=w(z)\in\Bbb{C}[[z]]$ of $(*)$ to only have terms of degrees $\equiv 3\pmod 4$. Being rusty, I first thought that we need to use that somehow, but those calculations scare me.