v - 4)
(v^3 - 4 v^2 + 7 v + 4)
\neq 0,$$ let $w$ be a square root of $v$ in ${{K\llap{$\overline{\phantom{\rmK}}$}}}$, and define numbers $\rho_1,\ldots,\rho_6$ by setting $$\begin{aligned}
\rho_1 & = \frac{(-2+w)(1+w)}{2w^2}    
       & \qquad \qquad \rho_2 & = \frac{(-2-w)(1-w)}{2w^2}\\
\rho_3 & = \frac{-2(2+w)}{(-2+w)(1+w)} 
       & \qquad \qquad \rho_4 & = \frac{(-2-w)(1-w)}{(-w)(-1-w)} \\
\rho_5 & = \frac{(-2+w)(1+w)}{w(-1+w)} 
       & \qquad \qquad \rho_6 & = \frac{-2(2-w)}{(-2-w)(1-w)}.\end{aligned}$$ The $\rho_i$ are distinct from one another, so that if we let $f = \prod(x-\rho_i)$ then the curve $D$ over $K$ defined by $y^2 = f$ has genus $2$. Set $$\begin{aligned}
g_1 & = (x-\rho_1)(x-\rho_5)    
    & \qquad \qquad g_1' & = (x-\rho_1)(x-\rho_3)\\
g_2 & = (x-\rho_2)(x-\rho_4)    
    & \qquad \qquad g_2' & = (x-\rho_2)(x-\rho_6)\\
g_3 & = (x-\rho_3)(x-\rho_6)    
    & \qquad \qquad g_3' & = (x-\rho_4)(x-\rho_5).\end{aligned}$$ The Richelot duals $C$ and $C'$ of $D$ with respect to the factorizations $f = g_1 g_2 g_3$ and $f = g_1' g_2' g_3'$ exist, and their Jacobians become isomorphic to one another over $K(\sqrt{v(v-4)}\,)$. The curves $C$ and $C'$ are geometrically non-isomorphic unless one of the following conditions holds[:]{}

-   ${\operatorname{char}}K = 3$ and $v^{10} - v^8 + v^7 - v^6 - v^5 + v + 1 = 0$[;]{}

-   