I'm trying to solve the following exercise from Silverman:
Let $E:y^2=x^3+x$. Show that $\# E(\Bbb F_p)$ is divisible by $4$ for prime $p\ge 3$.
My approach is as follows: if $\phi$ denotes the $p$-power Frobenius morphism, then $P\in E(\Bbb F_p)$ if and only if $\phi(P)=P$. Therefore $\# E(\Bbb F_p)=\#\ker(1-\phi)$, which because $1-\phi$ is separable is equal to $\deg(1-\phi)$. I have therefore reduced my problem to showing $\deg(1-\phi)$ is divisible by $4$.
To do this, I would like to use the more general fact that if $\psi$ is an endomorphism of $E$, and we let $\psi_{\ell}$ denote the induced endomorphism on the Tate module $T_{\ell}(E)$ for any prime $\ell$ not equal to the characteristic, then we have
$$\deg(\psi)=\det(\psi_{\ell}).$$
I would like to apply this to our case, when $\psi=1-\phi$ and $\ell=2$. So then I just need to calculate $\det((1-\phi)_2)$. I don't know a good way to do this though, as I'm not sure how to get an explicit $\Bbb Z_{\ell}$-basis of $T_{\ell}(E)\cong\Bbb Z_{\ell}\times\Bbb Z_{\ell}$ (the proof of the latter isomorphism was very non-constructive). I don't even know how to give any explicit points of $E(\Bbb F_p)$ besides $O$, as it relies on $x^3+x$ being a quadratic residue modulo $p$.
Does anybody have any suggestions on how to proceed? Does this approach even seem viable, or is there an easier way to do this?
All you have to do is find a subgroup of order $4$ in $E(\Bbb F_p)$. This is easy if $p\equiv1\pmod4$. In this case $(0,0)$, $(i,0)$ and $(-i,0)$ are elements of order $2$, where $i^2\equiv-1\pmod p$.
When $p\equiv 3\pmod 4$ you only have one element of order $2$, viz., $(0,0)$. I suppose in this case you need to find $(x,y)\in E(\Bbb F_p)$ with $[2](x,y)=(0,0)$. Too much like hard work for me at the moment $\smile$.