Curve with given Frobenius polynomial Does there exist a prime $p$ and a smooth genus 2 curve $C / \mathbf{F}_p$ such that the characteristic polynomial of Frobenius on the Tate module of $J(C)$ is given by $(T^2 - p)^2$?
More generally, for a curve of arbitrary genus, it possible that both $\sqrt{p}$ and $-\sqrt{p}$ can occur as eigenvalues of the Frobenius?
(This is essentially an idle question, which came up when I was giving an expository talk on the Weil conjectures; I wanted to state the functional equation in the form "we can write the char poly of Frobenius as $\prod_{i=1}^{2g} (T - \lambda_i)$ where $\lambda_{2g+1-i} = p / \lambda_i$", and later I realised that this wouldn't work if both $\sqrt{p}$ and $-\sqrt{p}$ showed up as eigenvalues.)
EDIT. As David Speyer points out, I made a mistake in formulating my original question; the roots of $(T^2 - p)^2$ can still be arranged in this way. Perhaps a better question is

Does there exist a curve over $\mathbf{F}_{p^2}$, of any genus, for which both $p$ and $-p$ occur to odd multiplicity as Frobenius eigenvalues?

 A: In its action on the Tate module, $\operatorname{Frob}_q$ is an element of $GSP_{2g}(\mathbb Q_\ell)$ whose action on the symplectic form is multiplication by $q$. This is due to the Weil pairing, or Poincare duality for etale cohomology.
Every such matrix has eigenvalues $\lambda_1, \dots, \lambda_{2g}$ whose with $\lambda_i \lambda_{2g-i} = q$. This is because the maximal torus in $GSP_{2g}$ has eigenvalues of that form, and every semisimple element is conjugate to an element of the maximal torus (so take the semisimplification of Frobenius, or prove that it is semisimple).
A: After thinking about this a bit more, I realised that no such example exists. Let $q = p^{2}$ (or $p^{2f}$ for any integer $f$), let $C$ be a curve over $\mathbf{F}_q$, and let $J = J(C)$. The Frobenius must be compatible with the Weil pairing $T_\ell J \times T_\ell J \to \mathbf{Z}_\ell(1)$, and Frobenius acts as multiplication by $q$ on the latter.
Since the Weil pairing is a perfect skew-symmetric pairing, this implies that the determinant of the Frobenius on $T_\ell J$ is $q^g$, where $g$ is the genus of $C$. (This follows from the same argument as is used classically to show that a symplectic matrix must have determinant $+1$.) 
However, all the eigenvalues of Frobenius on $T_\ell J$ are either non-real, or are equal to $\pm \sqrt{q}$. The non-real eigenvalues occur in conjugate pairs, each of which has product $q$. So if $-\sqrt{q}$ occurs to odd multiplicity, the determinant would have to be $< 0$, a contradiction.
A: For genus $1$ any supersingular elliptic curve will have trace of Frobenius $0$, and so have minimal polynomial of Frobenius $T^2-p$. If you permit an abelian surface, not just a Jacobian, then the product of this elliptic curve with itself will have the desired minimal polynomial of Frobenius. 
For genus $2$ hyperelliptic curves a supersingular hyperelliptic curve is one whose Jacobian is isogenous to a product of $2$ supersingular elliptic curves, and they are known to exist. See Choie, Jeong and Lee "Supersingular Hyperelliptic Curves of Genus 2 over Finite Fields".
