Let $E$ be an elliptic curve over $\mathbb{F}_p$, then the $p$th-power Frobenius morphism, denoted $\pi: (x,y) \mapsto (x^p, y^p)$, is an endomorphism on $E$.
There is an argument showing that if $E$ is supersingular, then $\text{End}_{\mathbb{F}_p}E$ is an order in the imagimary quadratic field $\mathbb{Q}(\sqrt{-p})$, containing $\mathbb{Z}(\sqrt{-p})$. I don't know why $\sqrt{-p}$, regarded as an endomorphism on $E$, equals $\pi$ which means $\pi^2+[p]=0$.
As I know of, if $E$ is supersingular then $[p]=\hat{\pi}\circ\pi$ is purely inseparable and we must have $\pi=\hat{\pi}$, i.e., $\pi^2=[p]$.