(r s^2 - r s t - r s - r t - 8 s t)\end{aligned}$$ and suppose that the polynomial $T^3 - c_2 T^2 + c_1 T - c_0$ has three distinct roots $\beta_1$, $\beta_2$, $\beta_3$ in ${{K\llap{$\overline{\phantom{\rmK}}$}}}$. For $i = 1,2,3$ let $$g_i = x^2 - 
        2\beta_i x + 
        (1-s)\beta_i^2 - 4 s (s - 1)^2 t (s - t - 1),$$ and suppose that the discriminant of the product $f = g_1 g_2 g_3$ is nonzero. Let ${{\mathcal C}}(r,s,t)$ be the curve over $K$ defined by $y^2 = f$.

\[T-RM\] The Richelot dual of ${{\mathcal C}}(r,s,t)$ associated to the factorization $f = g_1 g_2 g_3$ is isomorphic over $K(\sqrt{st}\,)$ to ${{\mathcal C}}(r,s,t)$. The endomorphism of ${\operatorname{Jac}}{{\mathcal C}}(r,s,t)$ [(]{}over $K(\sqrt{st}\,)$[)]{} obtained by composing the Richelot isogeny with the natural isomorphism from the dual curve to ${{\mathcal C}}(r,s,t)$ is fixed by the Rosati involution, and its square is the multiplication-by-$2$ endomorphism.

This theorem can be proven by direct calculation, but here we will prove it by relating the curve ${{\mathcal C}}(r,s,t)$ back to Bending’s construction. Let $Q$ be a square root of $st$ in ${{K\llap{$\overline{\phantom{\rmK}}$}}}$, let $P = (1 - s)/4$, let $A = (r + 6 s t 