# Show that ${\phi^2}_p =-p$

Let $p$ be a prime number, let $E$ be a supersingular elliptic curve over $\mathbb F_p$, and let $\phi_p\in End(E)$ be the Frobenius endomorphism. Show that $\phi^2_p =-p$

For $p\ge5$ the exercise is OK, because in general $a:=p+1-\#E(\mathbb F_p)$ satisfies $\phi_p^2-a\phi_p+p=0$ and if $p\ge5$ then $a=0$

But for $p=2,3$ there must be some cases where, $a\neq0$ and for equality to hold, i.e. $a\phi_p=0$, I guess $\phi_p$ must be some trivial endomorphism, or is there another possibility ?

Let $E/\mathbb{F}_2:y^2+y=x^3+x$ be an elliptic curve. It is easily checked that it is supersingular, and that $\#E(\mathbb{F}_2)=5.$ That means that $a=p+1-\#E(\mathbb{F}_2)=-2$.
Suppose that $\phi_2^2=-2$.
Then from $\phi_2^2-a\phi_2+2=0$ we get that $a\phi_2=0$. In particular, $[a]\phi_2(Q)=\mathcal{O}$ for all $Q\in E(\mathbb{F}_2)$. But for all $Q\in E(\mathbb{F}_2)$ we know that $\phi_2(Q)=Q$, and therefore we have that $[a]Q=\mathcal{O}$ for all $Q$. Since $a=-2$, we must have that $[2]Q=\mathcal{O}$ for all $Q\in E(\mathbb{F}_2)$. But $E(\mathbb{F}_2)$ is a group of order 5, so there is only a single element such that its order divides 2. We have arrived at a contradiction.
Therefore $\phi_2^2\neq -2$.
For $p=3$ consider $C/\mathbb{F}_3:y^2+2y=x^3+2x$. This is a supersingular elliptic curve with $\#C(\mathbb{F}_3)=7$ and $a=-3$. By the same argument as above, $\phi_3^2\ne -3$.
Your statement is false for $p=2$ and $p=3$.