I guess that the answer should be well known, but I fail to find a reference for it. For any prime $p$, is there a supersingular elliptic curve actually defined over $\mathbb F_p$ ?
In Katz and Mazur's book in the proof of Theorem 2.9.4, the authors show that any supersingular elliptic curve over $\overline{\mathbb F_p}$ can be defined over $\mathbb F_{p^2}$ if $p$ is odd, and over $\mathbb F_{16}$ if $p=2$. But is there at least one of them which actually stems from $\mathbb F_p$ ?
Edit: In Li and Oort's book "Moduli of supersingular abelian varieties", at the top of page 9, the authors write "For every $p$ there exists a supersingular elliptic curve over $\mathbb F_p$."
Therefore it seems that the answer is affirmative, however they do not offer a justification as to why it is true.
Best Answer
Yes, there is always at least one supersingular elliptic curve over $\mathbf{F}_p$.
Assume that $p$ is odd. (For $p=2$ take $j=0=1728$). Let $q \equiv -1 \bmod 4$ be a prime such that $-q$ is not a quadratic residue modulo $p$ (they exist). Let $F = \mathbf{Q}(\sqrt{-q})$. The class number $h$ of $F$ is odd, and if $H$ is the Hilbert class field of $F$, then $\mathrm{Gal}(H/\mathbf{Q})$ is dihedral, and Frobenius at $p$ is a reflection (since it is non-trivial in the quotient $\mathrm{Gal}(F/\mathbf{Q})$). In particular, if $H^{+} \subset H$ is any subfield of degree $h$ over $\mathbf{Q}$ (all such fields are isomorphic), then there is a (unique as it happens) prime above $p$ in $H^{+}$ of degree one.
Now by the theory of complex multiplication, there exists an elliptic curve $E$ with CM by the ring of integers of $F$ which is defined over a subfield $H^{+}$ of degree $h$. For example, one can take it to have $j$-invariant
$$j = j \left( \frac{1 + \sqrt{-q}}{2} \right) \in \mathbf{R},$$
(Since $F(j) = H$ and $j$ is real it follows that $H^{+} = \mathbf{Q}(j)$ has index two in $H$).
By assumption, there is a prime above $p$ in $H^{+}$ of degree one, and reducing $E$ modulo this prime we get an elliptic curve $E/\mathbf{F}_p$. Because $p$ is inert in $F$ this will be supersingular.
An example. If $p=3$, then you could take $q=7, 19, 31, \ldots$. If $q=7$, then
$$j = -3375 \equiv 0 \bmod 3,$$
If $q = 31$ on the other hand (to give an example with $h > 1$) then $j$ is a root of
$$1566028350940383 - 58682638134x + 39491307x^2 + x^3 = 0.$$
There is a unique solution in $\mathbf{Q}_3$ given by
$$j = 3^3 + 3^5 + 2 \cdot 3^6 + 2 \cdot 3^8 + 3^9 + 3^{10} + \ldots $$
and (of course again) $j = 0 \bmod 3$.