We do not claim that our construction will produce all such curves.

First we recall Bending’s construction. Let $K$ be a field of characteristic not $2$ and let $A$, $P$, and $Q$ be elements of $K$ with $P$ nonzero. Define $$\begin{aligned}
B &= (APQ - Q^2 + 4P^2 + 1)/P^2\\
C &= 4(AP - Q)/P\\
R &= 4P\end{aligned}$$ and let $\alpha_1$, $\alpha_2$, $\alpha_3$ be the roots of $T^3 + AT^2 + BT + C$ in a separable closure ${{K\llap{$\overline{\phantom{\rmK}}$}}}$ of $K$. For $i = 1,2,3$ let $$G_i = X^2 - \alpha_i X + P\alpha_i^2 + Q\alpha_i + R,$$ and suppose that the product $G_1 G_2 G_3 \in K[X]$ has nonzero discriminant. Let $D$ be a nonzero element of $K$.

\[T-Bending\] The Jacobian of the genus-$2$ curve $D Y^2 = G_1 G_2 G_3$ has a $K$-rational endomorphism that is fixed by the Rosati involution and whose square is $2$. Furthermore, if $\#K>5$ then every curve over $K$ whose Jacobian has such an endomorphism is isomorphic to a curve that arises in this way from some choice of $A$, $P$, $Q$, and $D$ in $K$.

See [@Bending Theorem 4.1]. Bending assumes that the base field $K$ has characteristic $0$, but his proof 