I was working my way through the Exercises in the Book Elliptic Curves: Number Theory and Cryptography by L. Washington. Exercise 4.12 tells us to:
Let $p$ be a prime and let $E$ be a supersingular elliptic curve over the finite field $\mathbf{F}_p$. Let $\phi_p$ be the Frobenius endomorphism. Show that some power of $\phi_p$ is an integer. (Note: This is easy when $p \geq 5$. The cases $p = 2,3$ can be done by a case-bye-case calculation.)
My solution is:
For $p \geq 5$ we have if $E$ is supersingular then $a = 0$ and since $\phi_p^2 -a\phi_p + p = 0$ we have $\phi_p^2 = -p$
For $p = 2$ since $ a\equiv 0$ mod 2, with Hasse we have that $a = \{-2,0,2\}$ and since $\phi_p$ is the identity on $\mathbf{F}_p$ we get that $\phi_p^2 = \{0,-2,-4\}$.
Analog for $p=3$
My first question: Is this argumentation correct?
Second question: Since if $p$ is a prime number and $\phi_p = (x^p, y^p)$ with little Fermat we have $(x^p, y^p) = (x,y)$. Does that mean we can find for any elliptic curve an integer, such that $\phi_p^2$ is an integer or did i forget something?
Third question: Since $\phi_p$ is the identity on $\mathbf{F}_p$ and the question asked to show that the power of $\phi_p$ is an integer, could i just say $\phi_p^1 = 1$ and be done?
Link to similar question: Show that $\phi_p^2 = -p$